Solutions to Axler’s Linear Algebra Done Right

By 1.1, every complex number is by definition an ordered pair of real numbers written in the form \(a + bi\), and addition on \(\mathbf{C}\) is defined by

\begin{equation*} (a + bi) + (c + di) = (a + c) + (b + d)i, \end{equation*}

where \(a, b, c, d \in \mathbf{R}\).

So suppose \(\alpha, \beta \in \mathbf{C}\), and write

\begin{equation*} \alpha = a + bi, \qquad \beta = c + di \end{equation*}

with \(a, b, c, d \in \mathbf{R}\). Then the definition of complex addition gives

\begin{equation*} \alpha + \beta = (a + bi) + (c + di) = (a + c) + (b + d)i \end{equation*}

and, applying the same definition with the roles of the two numbers interchanged,

\begin{equation*} \beta + \alpha = (c + di) + (a + bi) = (c + a) + (d + b)i. \end{equation*}

Now \(a, b, c, d\) are real numbers, and addition of real numbers is commutative. Hence \(a + c = c + a\) and \(b + d = d + b\). Therefore the two complex numbers displayed above have the same real part and the same imaginary part, so by the definition of equality of complex numbers as ordered pairs of reals (1.1 together with 1.8, since two lists are equal exactly when they have the same entries in the same order),

\begin{equation*} \alpha + \beta = (a + c) + (b + d)i = (c + a) + (d + b)i = \beta + \alpha. \end{equation*}

Since \(\alpha\) and \(\beta\) were arbitrary elements of \(\mathbf{C}\), this proves the commutativity of addition asserted in 1.3.

Suppose \(\alpha, \beta, \lambda \in \mathbf{C}\), and write

\begin{equation*} \alpha = a + bi, \qquad \beta = c + di, \qquad \lambda = e + fi \end{equation*}

with \(a, b, c, d, e, f \in \mathbf{R}\).

Applying the definition of complex addition from 1.1 twice, first to \(\alpha\) and \(\beta\) and then to the result and \(\lambda\), we get

\begin{equation*} \begin{aligned} (\alpha + \beta) + \lambda &= \bigl((a + bi) + (c + di)\bigr) + (e + fi) \\ &= \bigl((a + c) + (b + d)i\bigr) + (e + fi) \\ &= \bigl((a + c) + e\bigr) + \bigl((b + d) + f\bigr)i. \end{aligned} \end{equation*}

Applying the same definition in the other grouping, first to \(\beta\) and \(\lambda\) and then to \(\alpha\) and the result, we get

\begin{equation*} \begin{aligned} \alpha + (\beta + \lambda) &= (a + bi) + \bigl((c + di) + (e + fi)\bigr) \\ &= (a + bi) + \bigl((c + e) + (d + f)i\bigr) \\ &= \bigl(a + (c + e)\bigr) + \bigl(b + (d + f)\bigr)i. \end{aligned} \end{equation*}

The numbers \(a, c, e\) are real and addition of real numbers is associative, so \((a + c) + e = a + (c + e)\). Likewise \(b, d, f\) are real, so \((b + d) + f = b + (d + f)\). Thus the two complex numbers computed above have the same real part and the same imaginary part, and hence are equal:

\begin{equation*} (\alpha + \beta) + \lambda = \alpha + (\beta + \lambda). \end{equation*}

Since \(\alpha, \beta, \lambda\) were arbitrary elements of \(\mathbf{C}\), this proves the associativity of addition asserted in 1.3.

Recall from 1.1 that multiplication on \(\mathbf{C}\) is defined by

\begin{equation*} (a + bi)(c + di) = (ac - bd) + (ad + bc)i, \end{equation*}

where \(a, b, c, d \in \mathbf{R}\).

Suppose \(\alpha, \beta, \lambda \in \mathbf{C}\), and write

\begin{equation*} \alpha = a + bi, \qquad \beta = c + di, \qquad \lambda = e + fi \end{equation*}

with \(a, b, c, d, e, f \in \mathbf{R}\).

First compute \((\alpha\beta)\lambda\). The definition of multiplication gives \(\alpha\beta = (ac - bd) + (ad + bc)i\), and applying the definition again with the real number \(ac - bd\) in place of \(a\) and \(ad + bc\) in place of \(b\) yields

\begin{equation*} \begin{aligned} (\alpha\beta)\lambda &= \bigl((ac - bd) + (ad + bc)i\bigr)(e + fi) \\ &= \bigl((ac - bd)e - (ad + bc)f\bigr) + \bigl((ac - bd)f + (ad + bc)e\bigr)i \\ &= (ace - bde - adf - bcf) + (acf - bdf + ade + bce)i, \end{aligned} \end{equation*}

where the last step uses only the distributive and associative laws for real numbers.

Now compute \(\alpha(\beta\lambda)\). The definition of multiplication gives \(\beta\lambda = (ce - df) + (cf + de)i\), so

\begin{equation*} \begin{aligned} \alpha(\beta\lambda) &= (a + bi)\bigl((ce - df) + (cf + de)i\bigr) \\ &= \bigl(a(ce - df) - b(cf + de)\bigr) + \bigl(a(cf + de) + b(ce - df)\bigr)i \\ &= (ace - adf - bcf - bde) + (acf + ade + bce - bdf)i. \end{aligned} \end{equation*}

Comparing the two results, the real parts are

\begin{equation*} ace - bde - adf - bcf \qquad\text{and}\qquad ace - adf - bcf - bde, \end{equation*}

which are equal because addition of real numbers is commutative and associative, so the same four real summands may be rearranged freely. The imaginary parts are

\begin{equation*} acf - bdf + ade + bce \qquad\text{and}\qquad acf + ade + bce - bdf, \end{equation*}

which are equal for the same reason.

Hence \((\alpha\beta)\lambda\) and \(\alpha(\beta\lambda)\) have the same real part and the same imaginary part, so they are equal. Since \(\alpha, \beta, \lambda\) were arbitrary elements of \(\mathbf{C}\), this proves the associativity of multiplication asserted in 1.3.

Suppose \(\lambda, \alpha, \beta \in \mathbf{C}\), and write

\begin{equation*} \lambda = e + fi, \qquad \alpha = a + bi, \qquad \beta = c + di \end{equation*}

with \(a, b, c, d, e, f \in \mathbf{R}\).

By the definition of complex addition in 1.1 we have \(\alpha + \beta = (a + c) + (b + d)i\), so the definition of complex multiplication in 1.1 gives

\begin{equation*} \begin{aligned} \lambda(\alpha + \beta) &= (e + fi)\bigl((a + c) + (b + d)i\bigr) \\ &= \bigl(e(a + c) - f(b + d)\bigr) + \bigl(e(b + d) + f(a + c)\bigr)i \\ &= (ea + ec - fb - fd) + (eb + ed + fa + fc)i, \end{aligned} \end{equation*}

where the last equality uses the distributive law for real numbers.

On the other hand, the definition of complex multiplication gives

\begin{equation*} \lambda\alpha = (ea - fb) + (eb + fa)i \qquad\text{and}\qquad \lambda\beta = (ec - fd) + (ed + fc)i, \end{equation*}

so by the definition of complex addition,

\begin{equation*} \lambda\alpha + \lambda\beta = \bigl((ea - fb) + (ec - fd)\bigr) + \bigl((eb + fa) + (ed + fc)\bigr)i. \end{equation*}

The real part \((ea - fb) + (ec - fd)\) equals \(ea + ec - fb - fd\), and the imaginary part \((eb + fa) + (ed + fc)\) equals \(eb + ed + fa + fc\); in each case the two expressions are sums of the same four real numbers, hence equal by the commutativity and associativity of addition in \(\mathbf{R}\).

Therefore \(\lambda(\alpha + \beta)\) and \(\lambda\alpha + \lambda\beta\) have the same real part and the same imaginary part, so

\begin{equation*} \lambda(\alpha + \beta) = \lambda\alpha + \lambda\beta. \end{equation*}

Since \(\lambda, \alpha, \beta\) were arbitrary elements of \(\mathbf{C}\), this proves the distributive property asserted in 1.3.

Throughout, \(0\) denotes the complex number \(0 + 0i\); this is consistent with the identification in 1.1 of \(a + 0i\) with the real number \(a\). The proof below works directly from the definitions in 1.1 and assumes no part of 1.3, since 1.3 is what this exercise and its companions establish. (The optional remark at the end does invoke the identity clause \(\lambda + 0 = \lambda\) of 1.3, but that clause is not among the properties left to these exercises, and it is immediate from 1.1.)

Suppose \(\alpha \in \mathbf{C}\) and write \(\alpha = a + bi\) with \(a, b \in \mathbf{R}\).

Existence. Let \(\beta = (-a) + (-b)i\), which is an element of \(\mathbf{C}\) because \(-a\) and \(-b\) are real numbers. Then the definition of complex addition in 1.1 gives

\begin{equation*} \alpha + \beta = (a + bi) + \bigl((-a) + (-b)i\bigr) = \bigl(a + (-a)\bigr) + \bigl(b + (-b)\bigr)i = 0 + 0i = 0. \end{equation*}

Uniqueness. Suppose \(\beta \in \mathbf{C}\) satisfies \(\alpha + \beta = 0\), and write \(\beta = c + di\) with \(c, d \in \mathbf{R}\). Then

\begin{equation*} 0 + 0i = \alpha + \beta = (a + c) + (b + d)i. \end{equation*}

Two complex numbers are equal precisely when they are equal as ordered pairs of real numbers, that is, when their real parts agree and their imaginary parts agree (1.1). Hence

\begin{equation*} a + c = 0 \qquad\text{and}\qquad b + d = 0. \end{equation*}

Since \(a, b, c, d\) are real numbers and each real number has a unique additive inverse in \(\mathbf{R}\), these two equations force \(c = -a\) and \(d = -b\). Thus \(\beta = (-a) + (-b)i\), which is exactly the element exhibited in the existence step.

Therefore for every \(\alpha \in \mathbf{C}\) there is one and only one \(\beta \in \mathbf{C}\) with \(\alpha + \beta = 0\), namely \(\beta = (-a) + (-b)i\). This proves the additive inverse property asserted in 1.3 and justifies the notation \(-\alpha\) introduced in 1.5.

Remark. Uniqueness also follows abstractly from the properties already established, without coordinates: if \(\alpha + \beta = 0\) and \(\alpha + \beta^{\prime} = 0\), then using Exercise 1A.1 (commutativity), Exercise 1A.2 (associativity), and the additive identity property from 1.3,

\begin{equation*} \beta = \beta + 0 = \beta + (\alpha + \beta^{\prime}) = (\beta + \alpha) + \beta^{\prime} = (\alpha + \beta) + \beta^{\prime} = 0 + \beta^{\prime} = \beta^{\prime}. \end{equation*}

Suppose \(\alpha \in \mathbf{C}\) with \(\alpha \neq 0\), and write \(\alpha = a + bi\) with \(a, b \in \mathbf{R}\). Here \(1\) denotes the complex number \(1 + 0i\) and \(0\) denotes \(0 + 0i\).

First note that \(a^2 + b^2 > 0\). Indeed, since \(\alpha \neq 0\), the ordered pair \((a, b)\) is not \((0, 0)\), so at least one of \(a, b\) is nonzero; squares of real numbers are nonnegative, and the square of a nonzero real number is positive, so \(a^2 + b^2 > 0\). In particular we may divide real numbers by \(a^2 + b^2\).

Existence. Let

\begin{equation*} \beta = \frac{a}{a^2 + b^2} + \frac{-b}{a^2 + b^2}\, i, \end{equation*}

which lies in \(\mathbf{C}\) because both coefficients are real numbers. By the definition of complex multiplication in 1.1,

\begin{equation*} \begin{aligned} \alpha\beta &= \left(a \cdot \frac{a}{a^2 + b^2} - b \cdot \frac{-b}{a^2 + b^2}\right) + \left(a \cdot \frac{-b}{a^2 + b^2} + b \cdot \frac{a}{a^2 + b^2}\right) i \\ &= \frac{a^2 + b^2}{a^2 + b^2} + \frac{-ab + ab}{a^2 + b^2}\, i \\ &= 1 + 0i = 1. \end{aligned} \end{equation*}

Uniqueness. Suppose \(\beta \in \mathbf{C}\) satisfies \(\alpha\beta = 1\), and write \(\beta = c + di\) with \(c, d \in \mathbf{R}\). By the definition of complex multiplication,

\begin{equation*} 1 + 0i = \alpha\beta = (ac - bd) + (ad + bc)i, \end{equation*}

and equality of complex numbers means equality of real and imaginary parts, so

\begin{equation*} ac - bd = 1 \qquad\text{and}\qquad ad + bc = 0. \end{equation*}

Multiply the first equation by \(a\) and the second by \(b\) and add:

\begin{equation*} a^2 c - abd + abd + b^2 c = a, \qquad\text{that is,}\qquad (a^2 + b^2)c = a. \end{equation*}

Multiply the first equation by \(-b\) and the second by \(a\) and add:

\begin{equation*} -abc + b^2 d + a^2 d + abc = -b, \qquad\text{that is,}\qquad (a^2 + b^2)d = -b. \end{equation*}

Since \(a^2 + b^2 \neq 0\), these determine \(c\) and \(d\) uniquely:

\begin{equation*} c = \frac{a}{a^2 + b^2}, \qquad d = \frac{-b}{a^2 + b^2}. \end{equation*}

Thus \(\beta\) must be the complex number exhibited in the existence step, and no other element of \(\mathbf{C}\) can satisfy \(\alpha\beta = 1\).

Therefore every nonzero \(\alpha \in \mathbf{C}\) has exactly one multiplicative inverse, proving the multiplicative inverse property asserted in 1.3 and justifying the notation \(1/\alpha\) introduced in 1.5.

Remark. Uniqueness again also follows abstractly: if \(\alpha\beta = 1\) and \(\alpha\beta^{\prime} = 1\), then using the commutativity of complex multiplication proved in 1.4, the associativity proved in Exercise 1A.3, and the multiplicative identity property from 1.3,

\begin{equation*} \beta = \beta \cdot 1 = \beta(\alpha\beta^{\prime}) = (\beta\alpha)\beta^{\prime} = (\alpha\beta)\beta^{\prime} = 1 \cdot \beta^{\prime} = \beta^{\prime}. \end{equation*}

Let

\begin{equation*} \omega = \frac{-1 + \sqrt{3}\, i}{2}. \end{equation*}

First we identify \(\omega\) in the standard form \(a + bi\) with \(a, b \in \mathbf{R}\). By the definition of division in 1.5, \(\omega = (-1 + \sqrt{3}\, i)\bigl(1/2\bigr)\), where \(1/2\) is the multiplicative inverse of the complex number \(2 = 2 + 0i\); that inverse is the real number \(\tfrac12\), viewed as \(\tfrac12 + 0i\), since \((2 + 0i)(\tfrac12 + 0i) = 1 + 0i\). Hence by the definition of complex multiplication in 1.1,

\begin{equation*} \omega = (-1 + \sqrt{3}\, i)\left(\tfrac12 + 0i\right) = \left(-1 \cdot \tfrac12 - \sqrt{3} \cdot 0\right) + \left(-1 \cdot 0 + \sqrt{3} \cdot \tfrac12\right) i = -\frac{1}{2} + \frac{\sqrt{3}}{2}\, i. \end{equation*}

By the definition of powers given in Section 1A, \(\omega^3 = \omega\omega\omega\); the associativity proved in Exercise 1A.3 shows that the grouping is irrelevant, so we may compute \(\omega^3 = (\omega^2)\omega\).

Compute \(\omega^2\) using the definition of complex multiplication, with \(a = -\tfrac12\) and \(b = \tfrac{\sqrt{3}}{2}\):

\begin{equation*} \begin{aligned} \omega^2 &= \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}\, i\right)\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}\, i\right) \\ &= \left(\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right) - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}\right) + \left(\left(-\frac{1}{2}\right)\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}\left(-\frac{1}{2}\right)\right) i \\ &= \left(\frac{1}{4} - \frac{3}{4}\right) + \left(-\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4}\right) i \\ &= -\frac{1}{2} - \frac{\sqrt{3}}{2}\, i, \end{aligned} \end{equation*}

where we used \(\sqrt{3} \cdot \sqrt{3} = 3\).

Now multiply by \(\omega\) once more:

\begin{equation*} \begin{aligned} \omega^3 &= \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}\, i\right)\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}\, i\right) \\ &= \left(\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right) - \left(-\frac{\sqrt{3}}{2}\right)\frac{\sqrt{3}}{2}\right) + \left(\left(-\frac{1}{2}\right)\frac{\sqrt{3}}{2} + \left(-\frac{\sqrt{3}}{2}\right)\left(-\frac{1}{2}\right)\right) i \\ &= \left(\frac{1}{4} + \frac{3}{4}\right) + \left(-\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4}\right) i \\ &= 1 + 0i = 1. \end{aligned} \end{equation*}

Thus \(\omega^3 = 1\), so \(\omega = \dfrac{-1 + \sqrt{3}\, i}{2}\) is a cube root of \(1\).

Remark. Note that \(\omega \neq 1\), since \(\omega\) has imaginary part \(\tfrac{\sqrt{3}}{2} \neq 0\) while \(1 = 1 + 0i\) has imaginary part \(0\). So \(\omega\) is a cube root of \(1\) different from the obvious one. The computation above also shows that \(\omega^2 = \dfrac{-1 - \sqrt{3}\, i}{2}\), and since \((\omega^2)^3 = \omega^6 = (\omega^3)^2 = 1\) by the power rules recorded in Section 1A, that number is a cube root of \(1\) as well. It is distinct from both \(1\) and \(\omega\), because its imaginary part is \(-\tfrac{\sqrt{3}}{2}\), which differs from \(0\) and from \(\tfrac{\sqrt{3}}{2}\). Hence \(1\), \(\omega\), \(\omega^2\) are three distinct cube roots of \(1\). Equivalently, one can see all of this at once from the factorization

\begin{equation*} z^3 - 1 = (z - 1)(z^2 + z + 1), \end{equation*}

because \(\omega\) satisfies \(\omega^2 + \omega + 1 = \left(-\tfrac12 - \tfrac{\sqrt{3}}{2}i\right) + \left(-\tfrac12 + \tfrac{\sqrt{3}}{2}i\right) + 1 = 0\).

We look for \(z \in \mathbf{C}\) with \(z^2 = i\). Write \(z = a + bi\) with \(a, b \in \mathbf{R}\). Then, using the definition 1.1 of complex multiplication,

\begin{equation*} z^2 = (a + bi)(a + bi) = (a^2 - b^2) + (2ab)i . \end{equation*}

Two complex numbers are equal exactly when their real parts agree and their imaginary parts agree, so \(z^2 = i = 0 + 1i\) forces

\begin{equation*} a^2 - b^2 = 0 \quad \text{and} \quad 2ab = 1 . \end{equation*}

The first equation gives \(b = a\) or \(b = -a\). If \(b = -a\), then \(2ab = -2a^2 \leq 0\), which contradicts \(2ab = 1\). Hence \(b = a\), and then \(2a^2 = 1\), so \(a = \tfrac{1}{\sqrt{2}}\) or \(a = -\tfrac{1}{\sqrt{2}}\).

Thus the only candidates are

\begin{equation*} z = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\, i = \frac{\sqrt{2}}{2}(1 + i) \qquad \text{and} \qquad z = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\, i = -\frac{\sqrt{2}}{2}(1 + i) . \end{equation*}

Both really are square roots of \(i\). Indeed,

\begin{equation*} \left( \frac{1 + i}{\sqrt{2}} \right)^{2} = \frac{(1 + i)^2}{2} = \frac{1 + 2i + i^2}{2} = \frac{1 + 2i - 1}{2} = \frac{2i}{2} = i , \end{equation*}

and for the second candidate, applying the definition 1.1 of complex multiplication directly,

\begin{equation*} \left( -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\, i \right)^{2} = \left( \tfrac{1}{2} - \tfrac{1}{2} \right) + 2 \left( -\tfrac{1}{\sqrt{2}} \right)\left( -\tfrac{1}{\sqrt{2}} \right) i = 0 + i = i . \end{equation*}

Finally the two numbers are distinct: their difference is \(\sqrt{2}(1 + i) \neq 0\). So

\begin{equation*} \frac{\sqrt{2}}{2}(1 + i) \qquad \text{and} \qquad -\frac{\sqrt{2}}{2}(1 + i) \end{equation*}

are two distinct square roots of \(i\). The computation above also shows that these are the only square roots of \(i\), which is consistent with the fact that a nonzero complex number has exactly two square roots.

Write \(x = (x_1, x_2, x_3, x_4)\) with each \(x_k \in \mathbf{R}\). By the definition 1.18 of scalar multiplication we have \(2x = (2x_1, 2x_2, 2x_3, 2x_4)\), and then by the definition 1.13 of addition in \(\mathbf{F}^n\),

\begin{equation*} (4, -3, 1, 7) + 2x = (4 + 2x_1,\; -3 + 2x_2,\; 1 + 2x_3,\; 7 + 2x_4). \end{equation*}

Two lists are equal exactly when they agree in each coordinate, so the required equation is equivalent to the four real equations

\begin{equation*} 4 + 2x_1 = 5, \qquad -3 + 2x_2 = 9, \qquad 1 + 2x_3 = -6, \qquad 7 + 2x_4 = 8 , \end{equation*}

whose unique solutions are

\begin{equation*} x_1 = \tfrac{1}{2}, \qquad x_2 = 6, \qquad x_3 = -\tfrac{7}{2}, \qquad x_4 = \tfrac{1}{2} . \end{equation*}

Hence

\begin{equation*} x = \left( \tfrac{1}{2},\, 6,\, -\tfrac{7}{2},\, \tfrac{1}{2} \right). \end{equation*}

Check: \(2x = (1, 12, -7, 1)\), and

\begin{equation*} (4, -3, 1, 7) + (1, 12, -7, 1) = (5, 9, -6, 8), \end{equation*}

as required. Since each coordinate equation had a unique solution, this \(x\) is the only element of \(\mathbf{R}^4\) with the stated property.

Suppose, for the sake of contradiction, that such a \(\lambda \in \mathbf{C}\) exists. By the definition 1.18 of scalar multiplication,

\begin{equation*} \lambda(2 - 3i,\; 5 + 4i,\; -6 + 7i) = \bigl( \lambda(2 - 3i),\; \lambda(5 + 4i),\; \lambda(-6 + 7i) \bigr), \end{equation*}

and equality of lists means equality in each coordinate. So \(\lambda\) would have to satisfy all three of

\begin{equation*} \lambda(2 - 3i) = 12 - 5i, \qquad \lambda(5 + 4i) = 7 + 22i, \qquad \lambda(-6 + 7i) = -32 - 9i . \end{equation*}

The first equation already pins down \(\lambda\). Since \(2 - 3i \neq 0\), it has a multiplicative inverse in \(\mathbf{C}\) (see 1.3 and 1.5), and multiplying the first equation by \(\dfrac{1}{2 - 3i}\) gives

\begin{equation*} \lambda = \frac{12 - 5i}{2 - 3i} = \frac{(12 - 5i)(2 + 3i)}{(2 - 3i)(2 + 3i)} = \frac{24 + 36i - 10i - 15i^2}{4 + 9} = \frac{39 + 26i}{13} = 3 + 2i . \end{equation*}

Thus \(\lambda = 3 + 2i\) is the only possible value.

Now test this value in the third coordinate:

\begin{equation*} (3 + 2i)(-6 + 7i) = -18 + 21i - 12i + 14i^2 = -18 + 9i - 14 = -32 + 9i . \end{equation*}

But the third coordinate on the right side of the desired equation is \(-32 - 9i\), and \(-32 + 9i \neq -32 - 9i\) because their imaginary parts \(9\) and \(-9\) differ. This contradicts the third equation above.

Hence no \(\lambda \in \mathbf{C}\) satisfies all three coordinate equations, so no such \(\lambda\) exists.

Remark: it is only the third coordinate that fails. The first two are consistent with \(\lambda = 3 + 2i\), since

\begin{equation*} (3 + 2i)(5 + 4i) = 15 + 12i + 10i + 8i^2 = 15 + 22i - 8 = 7 + 22i . \end{equation*}

Geometrically, the point is that the list \((12 - 5i,\, 7 + 22i,\, -32 - 9i)\) is not a scalar multiple of \((2 - 3i,\, 5 + 4i,\, -6 + 7i)\): a single scalar must act simultaneously on every coordinate, and no one scalar works here.

Recall that \(\mathbf{F}\) denotes \(\mathbf{R}\) or \(\mathbf{C}\) (see 1.6), and that in either case addition in \(\mathbf{F}\) is associative: for \(\mathbf{C}\) this is the associativity statement in 1.3, and \(\mathbf{R}\) inherits it as a subset of \(\mathbf{C}\) (or one may take it as a familiar property of the real numbers).

Let \(x, y, z \in \mathbf{F}^n\). By the definition 1.11 of \(\mathbf{F}^n\) we may write

\begin{equation*} x = (x_1, \ldots, x_n), \qquad y = (y_1, \ldots, y_n), \qquad z = (z_1, \ldots, z_n), \end{equation*}

where \(x_k, y_k, z_k \in \mathbf{F}\) for each \(k \in \{1, \ldots, n\}\). Using the definition 1.13 of addition in \(\mathbf{F}^n\) twice, then associativity of addition in \(\mathbf{F}\) in each of the \(n\) coordinates, then the definition 1.13 twice more, we compute

\begin{equation*} \begin{aligned} (x + y) + z &= \bigl( (x_1, \ldots, x_n) + (y_1, \ldots, y_n) \bigr) + (z_1, \ldots, z_n) \\ &= (x_1 + y_1, \ldots, x_n + y_n) + (z_1, \ldots, z_n) \\ &= \bigl( (x_1 + y_1) + z_1, \; \ldots, \; (x_n + y_n) + z_n \bigr) \\ &= \bigl( x_1 + (y_1 + z_1), \; \ldots, \; x_n + (y_n + z_n) \bigr) \\ &= (x_1, \ldots, x_n) + (y_1 + z_1, \ldots, y_n + z_n) \\ &= (x_1, \ldots, x_n) + \bigl( (y_1, \ldots, y_n) + (z_1, \ldots, z_n) \bigr) \\ &= x + (y + z). \end{aligned} \end{equation*}

The fourth equality is the only nonformal step: it holds because \((a + b) + c = a + (b + c)\) for all \(a, b, c \in \mathbf{F}\), applied with \(a = x_k\), \(b = y_k\), \(c = z_k\) for each \(k = 1, \ldots, n\). All other equalities are simply the definition of addition in \(\mathbf{F}^n\).

Since \(x, y, z \in \mathbf{F}^n\) were arbitrary, addition in \(\mathbf{F}^n\) is associative.

Note first that the two sides of the claimed identity are built from different operations: on the left, \(ab\) is a product of two elements of \(\mathbf{F}\) and then that scalar multiplies the vector \(x\); on the right, \(bx\) is a scalar times a vector and then \(a\) multiplies the resulting vector. So the statement is an associativity statement linking multiplication in \(\mathbf{F}\) with scalar multiplication in \(\mathbf{F}^n\).

Multiplication in \(\mathbf{F}\) is associative: for \(\mathbf{F} = \mathbf{C}\) this is the associativity statement in 1.3, and for \(\mathbf{F} = \mathbf{R}\) it is the familiar property of real multiplication (also a special case of 1.3, since \(\mathbf{R} \subseteq \mathbf{C}\)).

Let \(a, b \in \mathbf{F}\) and let \(x = (x_1, \ldots, x_n) \in \mathbf{F}^n\), where each \(x_k \in \mathbf{F}\) (definition 1.11). Applying the definition 1.18 of scalar multiplication in \(\mathbf{F}^n\) repeatedly,

\begin{equation*} \begin{aligned} (ab)x &= (ab)(x_1, \ldots, x_n) \\ &= \bigl( (ab)x_1, \; \ldots, \; (ab)x_n \bigr) \\ &= \bigl( a(bx_1), \; \ldots, \; a(bx_n) \bigr) \\ &= a(bx_1, \ldots, bx_n) \\ &= a \bigl( b(x_1, \ldots, x_n) \bigr) \\ &= a(bx). \end{aligned} \end{equation*}

The third equality uses associativity of multiplication in \(\mathbf{F}\), applied in each coordinate with the elements \(a\), \(b\), \(x_k\) of \(\mathbf{F}\) for \(k = 1, \ldots, n\). The second and fourth equalities are the definition 1.18 (in the fourth, read from right to left, with scalar \(a\)), and the fifth is again 1.18 with scalar \(b\).

Since \(a, b \in \mathbf{F}\) and \(x \in \mathbf{F}^n\) were arbitrary, we conclude that \((ab)x = a(bx)\) always holds. This is the associativity axiom for scalar multiplication that will appear again in the definition of a vector space in Section 1B.

Here \(1\) denotes the multiplicative identity of \(\mathbf{F}\), and \(1x\) means the scalar \(1\) multiplying the vector \(x \in \mathbf{F}^n\) in the sense of definition 1.18.

The key fact about \(\mathbf{F}\) is the multiplicative identity property from 1.3: \(\lambda 1 = \lambda\), and hence by commutativity of multiplication \(1\lambda = \lambda\), for every \(\lambda \in \mathbf{F}\). (For \(\mathbf{F} = \mathbf{C}\) this is exactly the identities statement in 1.3; for \(\mathbf{F} = \mathbf{R}\) it is the familiar property of \(1\), and also a special case since \(\mathbf{R} \subseteq \mathbf{C}\).)

Let \(x \in \mathbf{F}^n\) and write \(x = (x_1, \ldots, x_n)\) with each \(x_k \in \mathbf{F}\), as in definition 1.11. Then by the definition 1.18 of scalar multiplication,

\begin{equation*} 1x = 1(x_1, \ldots, x_n) = (1x_1, \ldots, 1x_n) = (x_1, \ldots, x_n) = x , \end{equation*}

where the third equality holds because \(1x_k = x_k\) in \(\mathbf{F}\) for each \(k \in \{1, \ldots, n\}\).

Since \(x \in \mathbf{F}^n\) was arbitrary, \(1x = x\) for all \(x \in \mathbf{F}^n\). Note that this property is genuinely a separate assumption in the abstract setting: when the definition of a vector space is given in Section 1B, the equation \(1v = v\) must be listed as an axiom, because it does not follow from the other axioms. To see this, take \(V = \mathbf{R}\) with its usual addition but with scalar multiplication redefined by \(\lambda \cdot v = 0\) for all \(\lambda \in \mathbf{R}\) and all \(v \in V\). Every other vector space axiom holds for this \(V\) (both sides of the associativity and distributivity axioms are \(0\)), yet \(1 \cdot 1 = 0 \neq 1\).

This is the distributive property for scalar multiplication over vector addition in \(\mathbf{F}^n\); it will be proved by reducing it, coordinate by coordinate, to the distributive property in \(\mathbf{F}\).

Multiplication distributes over addition in \(\mathbf{F}\): for \(\mathbf{F} = \mathbf{C}\) this is the distributive property in 1.3, namely \(\lambda(\alpha + \beta) = \lambda\alpha + \lambda\beta\) for all \(\lambda, \alpha, \beta \in \mathbf{C}\); for \(\mathbf{F} = \mathbf{R}\) it is the same statement restricted to real numbers.

Let \(\lambda \in \mathbf{F}\) and let \(x, y \in \mathbf{F}^n\), written as \(x = (x_1, \ldots, x_n)\) and \(y = (y_1, \ldots, y_n)\) with all \(x_k, y_k \in \mathbf{F}\) (definition 1.11). Then

\begin{equation*} \begin{aligned} \lambda(x + y) &= \lambda \bigl( (x_1, \ldots, x_n) + (y_1, \ldots, y_n) \bigr) \\ &= \lambda(x_1 + y_1, \; \ldots, \; x_n + y_n) \\ &= \bigl( \lambda(x_1 + y_1), \; \ldots, \; \lambda(x_n + y_n) \bigr) \\ &= (\lambda x_1 + \lambda y_1, \; \ldots, \; \lambda x_n + \lambda y_n) \\ &= (\lambda x_1, \ldots, \lambda x_n) + (\lambda y_1, \ldots, \lambda y_n) \\ &= \lambda(x_1, \ldots, x_n) + \lambda(y_1, \ldots, y_n) \\ &= \lambda x + \lambda y . \end{aligned} \end{equation*}

Here the second equality is the definition 1.13 of addition in \(\mathbf{F}^n\), the third is the definition 1.18 of scalar multiplication, the fourth applies the distributive property in \(\mathbf{F}\) separately in each of the \(n\) coordinates, the fifth is again 1.13 (read from right to left), and the sixth is again 1.18.

Since \(\lambda \in \mathbf{F}\) and \(x, y \in \mathbf{F}^n\) were arbitrary, the identity \(\lambda(x + y) = \lambda x + \lambda y\) holds throughout \(\mathbf{F}^n\).

Fix \(a, b \in \mathbf{F}\) and \(x \in \mathbf{F}^n\), and write \(x = (x_1, \dots, x_n)\) with each \(x_j \in \mathbf{F}\).

The definition of scalar multiplication in \(\mathbf{F}^n\) (1.18) says that a scalar multiplies a list coordinate by coordinate. Applying it with the scalar \(a+b\) gives

\begin{equation*} (a+b)x = \bigl((a+b)x_1, \dots, (a+b)x_n\bigr). \end{equation*}

Now compute the right side of the desired identity. Two more applications of 1.18 give \(ax = (ax_1, \dots, ax_n)\) and \(bx = (bx_1, \dots, bx_n)\), and then the definition of addition in \(\mathbf{F}^n\) (1.13), which adds corresponding coordinates, gives

\begin{equation*} ax + bx = (ax_1, \dots, ax_n) + (bx_1, \dots, bx_n) = (ax_1 + bx_1, \dots, ax_n + bx_n). \end{equation*}

Thus both \((a+b)x\) and \(ax + bx\) are lists of length \(n\), and it suffices to show that their \(j\)-th coordinates agree for each \(j \in \{1, \dots, n\}\). The \(j\)-th coordinate of \((a+b)x\) is \((a+b)x_j\) and the \(j\)-th coordinate of \(ax + bx\) is \(ax_j + bx_j\). These are equal because \(a\), \(b\), and \(x_j\) all lie in \(\mathbf{F}\), and multiplication in \(\mathbf{F}\) distributes over addition in \(\mathbf{F}\). Recall that by 1.6 the symbol \(\mathbf{F}\) stands for either \(\mathbf{R}\) or \(\mathbf{C}\), so there are two cases and both are already available to us. For \(\mathbf{F} = \mathbf{R}\) the equality \((a+b)x_j = ax_j + bx_j\) is the familiar distributive property of real arithmetic. For \(\mathbf{F} = \mathbf{C}\), the distributive property recorded in 1.3 is stated in the form \(\lambda(\alpha + \beta) = \lambda\alpha + \lambda\beta\), with the sum inside the second factor, which is not literally the form we need; but 1.3 also records that complex multiplication is commutative, and combining the two gives

\begin{equation*} (a+b)x_j = x_j(a+b) = x_j a + x_j b = a x_j + b x_j, \end{equation*}

which is exactly the required identity. (Nothing here uses any special feature of \(\mathbf{R}\) or \(\mathbf{C}\) beyond commutativity and distributivity of multiplication, so the same argument works verbatim over any field, as the digression on fields at the end of Section 1A points out.)

Since the coordinates agree in every slot, the two lists are equal:

\begin{equation*} (a+b)x = ax + bx. \end{equation*}

As \(a, b \in \mathbf{F}\) and \(x \in \mathbf{F}^n\) were arbitrary, the identity holds in general.

This is one of the two distributive properties required in the definition of a vector space (1.20). Together with Exercises 11 through 14 of this section (associativity of addition, associativity of scalar multiplication, the multiplicative identity, and the other distributive property) and the commutativity of addition (1.14) along with the additive identity \(0 \in \mathbf{F}^n\) (defined in 1.15, with \(x + 0 = x\) noted in 1.16) and additive inverses (1.17), it confirms that \(\mathbf{F}^n\), equipped with the operations of 1.13 and 1.18, is a vector space over \(\mathbf{F}\).

Let \(v \in V\). By 1.28, the notation \(-v\) denotes the additive inverse of \(v\), and the notation \(-(-v)\) denotes the additive inverse of \(-v\); these notations are unambiguous because every element of a vector space has a unique additive inverse (1.27).

By the definition of \(-v\) we have

\begin{equation*} v + (-v) = 0 . \end{equation*}

Using commutativity of addition, this says

\begin{equation*} (-v) + v = 0 , \end{equation*}

which is exactly the statement that \(v\) is an additive inverse of the vector \(-v\). But \(-(-v)\) is also an additive inverse of \(-v\). Since additive inverses are unique (1.27), these two vectors are equal:

\begin{equation*} -(-v) = v , \end{equation*}

as desired.

(Alternatively, one may argue directly:

\begin{equation*} -(-v) = -(-v) + 0 = -(-v) + \bigl((-v) + v\bigr) = \bigl(-(-v) + (-v)\bigr) + v = 0 + v = v , \end{equation*}

where the first equality is the additive identity condition, the second uses \((-v) + v = v + (-v) = 0\) (commutativity and the definition of \(-v\)), the third is associativity, the fourth uses \(-(-v) + (-v) = (-v) + \bigl(-(-v)\bigr) = 0\) (commutativity and the definition of \(-(-v)\)), and the last uses commutativity together with the additive identity condition. This is just the proof of 1.27 written out for this special case.)

Recall that \(\mathbb{F}\) denotes \(\mathbb{R}\) or \(\mathbb{C}\) (1.1), so every nonzero element of \(\mathbb{F}\) has a multiplicative inverse in \(\mathbb{F}\).

Suppose \(a \in \mathbb{F}\), \(v \in V\), and \(av = 0\). If \(a = 0\), then the conclusion “\(a = 0\) or \(v = 0\)” holds and there is nothing more to prove. So assume \(a \neq 0\). Then \(\frac{1}{a} \in \mathbb{F}\), and we compute

\begin{equation*} v = 1v = \left( \frac{1}{a} \, a \right) v = \frac{1}{a} (av) = \frac{1}{a} \, 0 = 0 , \end{equation*}

where the first equality holds by the multiplicative identity property in the definition of a vector space (1.20), the second holds because \(\frac{1}{a} a = 1\) in \(\mathbb{F}\), the third holds by the associativity property \((bc)v = b(cv)\) in 1.20, the fourth holds because \(av = 0\) by hypothesis, and the last equality holds by 1.31 (a number times the vector \(0\) equals \(0\)).

Thus if \(a \neq 0\) then \(v = 0\). Hence in every case \(a = 0\) or \(v = 0\), as desired.

Here \(3\) denotes the scalar \(3 \in \mathbb{F}\); since \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), we have \(3 \neq 0\) and hence \(\frac{1}{3} \in \mathbb{F}\).

Existence. Put

\begin{equation*} x = \tfrac{1}{3}(w - v) = \tfrac{1}{3}\bigl(w + (-v)\bigr) \in V . \end{equation*}

Then, using associativity of scalar multiplication and the multiplicative identity property from 1.20,

\begin{equation*} 3x = 3\left( \tfrac{1}{3}(w - v) \right) = \left( 3 \cdot \tfrac{1}{3} \right)(w - v) = 1(w - v) = w + (-v) . \end{equation*}

Therefore, using commutativity and associativity of addition,

\begin{equation*} v + 3x = v + \bigl(w + (-v)\bigr) = w + \bigl(v + (-v)\bigr) = w + 0 = w , \end{equation*}

where the last equality uses the additive identity property. So this \(x\) satisfies the equation.

Uniqueness. Suppose \(x, x^{\prime} \in V\) both satisfy \(v + 3x = w\) and \(v + 3x^{\prime} = w\). Then \(v + 3x = v + 3x^{\prime}\). Add \(-v\) to both sides on the left, using commutativity, the additive identity condition, and associativity:

\begin{equation*} 3x = 0 + 3x = \bigl((-v) + v\bigr) + 3x = (-v) + (v + 3x) = (-v) + (v + 3x^{\prime}) = \bigl((-v) + v\bigr) + 3x^{\prime} = 3x^{\prime} . \end{equation*}

Multiplying both sides by the scalar \(\frac{1}{3}\) and using associativity of scalar multiplication together with the multiplicative identity property gives

\begin{equation*} x = 1x = \left( \tfrac{1}{3} \cdot 3 \right) x = \tfrac{1}{3}(3x) = \tfrac{1}{3}(3x^{\prime}) = \left( \tfrac{1}{3} \cdot 3 \right) x^{\prime} = 1x^{\prime} = x^{\prime} . \end{equation*}

Hence there is exactly one \(x \in V\) with \(v + 3x = w\), namely \(x = \frac{1}{3}(w - v)\).

Note that the hypothesis that \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\) was used only through the fact that the scalar \(3\) is invertible in \(\mathbb{F}\); over a field of characteristic \(3\) the statement would fail.

The requirement that fails is the additive identity condition: there is no element \(0 \in \emptyset\), because \(\emptyset\) has no elements at all.

To see that this is the only failure, go through the definition 1.20 with \(V = \emptyset\).

First, the empty set does carry an addition and a scalar multiplication in the sense of 1.19. An addition on \(V\) is a function from \(V \times V\) to \(V\); here \(\emptyset \times \emptyset = \emptyset\), and the empty function \(\emptyset \to \emptyset\) is a (unique, legitimate) such function. Likewise a scalar multiplication is a function from \(\mathbb{F} \times V\) to \(V\); here \(\mathbb{F} \times \emptyset = \emptyset\), and again the empty function serves. So there is nothing wrong with the operations.

Now examine each listed property.

• Commutativity: “\(u + v = v + u\) for all \(u, v \in V\)” is a statement universally quantified over elements of \(V\). Since \(V = \emptyset\) has no elements, the statement holds vacuously.
• Associativity: “\((u + v) + w = u + (v + w)\) for all \(u, v, w \in V\)” is likewise vacuous, and “\((ab)v = a(bv)\) for all \(v \in V\) and all \(a, b \in \mathbb{F}\)” is vacuous in \(v\).
• Additive inverse: “for every \(v \in V\), there exists \(w \in V\) such that \(v + w = 0\)” begins with a universal quantifier over \(V\), so it too is vacuously satisfied; there is no \(v\) for which a \(w\) must be produced.
• Multiplicative identity: “\(1v = v\) for all \(v \in V\)” is vacuous.
• Distributive properties: both “\(a(u + v) = au + av\)” and “\((a + b)v = av + bv\)” are universally quantified over elements of \(V\), hence vacuous.

The additive identity condition is the only one of the six that asserts the existence of an element of \(V\) rather than a property of all elements of \(V\). An existence statement cannot hold in the empty set. Hence \(\emptyset\) satisfies every requirement in 1.20 except the additive identity condition, and that single failure is what disqualifies it from being a vector space.

Let \(V\) be a set equipped with an addition and a scalar multiplication (1.19), and consider the five conditions of 1.20 other than the additive inverse condition, namely commutativity, associativity, additive identity, multiplicative identity, and the distributive properties. Call these the common conditions. We must show:

\(V\) satisfies the common conditions together with the additive inverse condition if and only if \(V\) satisfies the common conditions together with the condition \(0v = 0\) for all \(v \in V\).

First note that under the common conditions alone the phrase “the additive identity of \(V\)” is unambiguous: the proof of 1.26 shows that if \(0\) and \(0^{\prime}\) are both additive identities then \(0^{\prime} = 0^{\prime} + 0 = 0 + 0^{\prime} = 0\), and this argument uses only the additive identity condition and commutativity. So the new condition \(0v = 0\) has a well-defined meaning in either setting.

Suppose first that \(V\) satisfies the common conditions and the additive inverse condition, i.e. \(V\) is a vector space. Then \(0v = 0\) for every \(v \in V\) by 1.30. Explicitly, the distributive property gives

\begin{equation*} 0v = (0 + 0)v = 0v + 0v , \end{equation*}

and adding an additive inverse \(w\) of \(0v\) to both sides gives

\begin{equation*} 0 = 0v + w = (0v + 0v) + w = 0v + (0v + w) = 0v + 0 = 0v , \end{equation*}

using associativity and the additive identity condition. Hence the new condition holds.

Conversely, suppose \(V\) satisfies the common conditions and satisfies \(0v = 0\) for all \(v \in V\). Let \(v \in V\) and set \(w = (-1)v\), which lies in \(V\) because scalar multiplication maps \(\mathbb{F} \times V\) into \(V\). Then

\begin{equation*} v + w = 1v + (-1)v = \bigl(1 + (-1)\bigr) v = 0v = 0 , \end{equation*}

where the first equality uses the multiplicative identity condition, the second uses the distributive property \((a + b)v = av + bv\), the third uses \(1 + (-1) = 0\) in \(\mathbb{F}\), and the fourth is the assumed new condition. Thus \(v\) has an additive inverse in \(V\), so the additive inverse condition holds.

Since each version of the definition implies the other, the collection of objects satisfying the definition is unchanged when the additive inverse condition is replaced by the condition \(0v = 0\) for all \(v \in V\). In other words, the replacement is legitimate.

No. Write \(W = \mathbb{R} \cup \{\infty, -\infty\}\). The set \(W\) with these operations is not a vector space over \(\mathbb{R}\), because addition on \(W\) is not associative.

Indeed, take \(u = -\infty\), \(v = \infty\), and \(w = 1\). Then

\begin{equation*} (u + v) + w = \bigl((-\infty) + \infty\bigr) + 1 = 0 + 1 = 1 , \end{equation*}

while

\begin{equation*} u + (v + w) = (-\infty) + (\infty + 1) = (-\infty) + \infty = 0 , \end{equation*}

using \(\infty + 1 = \infty\) from the definition. Since \(1 \neq 0\), the associativity requirement of 1.20 fails, so \(W\) is not a vector space over \(\mathbb{R}\).

A distributive property fails as well, which gives a second independent reason. Take \(a = 2\), \(b = -1\), and \(v = \infty\). Then

\begin{equation*} (a + b)v = 1 \cdot \infty = \infty , \end{equation*}

whereas

\begin{equation*} av + bv = 2\infty + (-1)\infty = \infty + (-\infty) = 0 . \end{equation*}

Since \(\infty \neq 0\), the requirement \((a + b)v = av + bv\) fails.

It is worth noting that these are the only failures: all the remaining requirements of 1.20 do hold for \(W\).

• Commutativity of addition holds because every clause in the definition of addition is stated symmetrically in its two arguments, and addition of real numbers is commutative.
• The additive identity is the real number \(0\): for \(t \in \mathbb{R}\) we have \(t + 0 = t\) as usual, and \(\infty + 0 = \infty\) and \((-\infty) + 0 = -\infty\) by the displayed rules with the real number \(t = 0\).
• Every element of \(W\) has an additive inverse: for \(t \in \mathbb{R}\) take \(-t\); for \(\infty\) take \(-\infty\), since \(\infty + (-\infty) = 0\); for \(-\infty\) take \(\infty\), since \((-\infty) + \infty = 0\).
• Associativity of scalar multiplication, \((ab)v = a(bv)\), holds. For \(v \in \mathbb{R}\) this is associativity of multiplication of real numbers. For \(v = \pm\infty\), both sides equal \(0\) whenever \(a = 0\) or \(b = 0\) (note \(0 \cdot (\pm\infty) = 0\), and \(a \cdot 0 = 0\) for every \(a \in \mathbb{R}\) since \(0 \in \mathbb{R}\)); and when \(a, b \neq 0\) both sides equal \(\infty\) or \(-\infty\) according to the sign of \(ab\), since scalar multiplication by a nonzero \(t\) preserves \(\pm\infty\) when \(t > 0\) and swaps them when \(t < 0\).
• The multiplicative identity condition \(1v = v\) holds: it is clear for \(v \in \mathbb{R}\), and \(1 \cdot \infty = \infty\), \(1 \cdot (-\infty) = -\infty\) because \(1 > 0\).
• The other distributive property, \(a(u + v) = au + av\), does hold. If \(a = 0\) both sides equal \(0\). If \(a > 0\), then multiplication by \(a\) fixes \(\infty\) and \(-\infty\) and acts as usual on \(\mathbb{R}\); checking the cases (both of \(u, v\) real; one real and one infinite; both infinite of the same sign; both infinite of opposite signs, where each side equals \(0\)) gives equality in every case. If \(a < 0\), multiplication by \(a\) interchanges \(\infty\) and \(-\infty\) and negates the sign of real numbers, and the same case check again gives equality, the opposite-signs case once more giving \(0\) on both sides.

So \(W\) satisfies commutativity, the existence of an additive identity, the existence of additive inverses, associativity of scalar multiplication, the multiplicative identity condition, and the distributive property \(a(u + v) = au + av\), but it fails associativity of addition and the distributive property \((a + b)v = av + bv\). One failure is enough: \(W\) is not a vector space over \(\mathbb{R}\).

Throughout, \(V\) is a vector space over \(\mathbb{F}\) (1.29), and we use repeatedly that two functions \(f, g \colon S \to V\) are equal if and only if \(f(x) = g(x)\) for every \(x \in S\).

Definition of the operations. Imitating 1.24, define addition and scalar multiplication on \(V^S\) pointwise: for \(f, g \in V^S\) let \(f + g \in V^S\) be the function given by

\begin{equation*} (f + g)(x) = f(x) + g(x) \quad \text{for all } x \in S , \end{equation*}

and for \(\lambda \in \mathbb{F}\) and \(f \in V^S\) let \(\lambda f \in V^S\) be the function given by

\begin{equation*} (\lambda f)(x) = \lambda \bigl( f(x) \bigr) \quad \text{for all } x \in S . \end{equation*}

The right sides make sense because \(f(x), g(x) \in V\) and \(V\) has an addition and a scalar multiplication; and the results are again functions from \(S\) to \(V\), so these are indeed an addition and a scalar multiplication on \(V^S\) in the sense of 1.19.

We now verify each requirement of 1.20; in every case the verification is the corresponding property of \(V\) applied at each point \(x \in S\).

Commutativity. Let \(f, g \in V^S\). For every \(x \in S\),

\begin{equation*} (f + g)(x) = f(x) + g(x) = g(x) + f(x) = (g + f)(x) , \end{equation*}

the middle equality by commutativity in \(V\). Hence \(f + g = g + f\).

Associativity. Let \(f, g, h \in V^S\). For every \(x \in S\),

\begin{equation*} \bigl((f + g) + h\bigr)(x) = \bigl(f(x) + g(x)\bigr) + h(x) = f(x) + \bigl(g(x) + h(x)\bigr) = \bigl(f + (g + h)\bigr)(x) , \end{equation*}

using associativity of addition in \(V\). Hence \((f + g) + h = f + (g + h)\). Similarly, for \(a, b \in \mathbb{F}\) and \(f \in V^S\), for every \(x \in S\),

\begin{equation*} \bigl((ab)f\bigr)(x) = (ab)\bigl(f(x)\bigr) = a\bigl(b f(x)\bigr) = a\bigl((bf)(x)\bigr) = \bigl(a(bf)\bigr)(x) , \end{equation*}

using associativity of scalar multiplication in \(V\). Hence \((ab)f = a(bf)\).

Additive identity. Let \(\mathbf{0} \in V^S\) be the function defined by \(\mathbf{0}(x) = 0\) for all \(x \in S\), where \(0\) is the additive identity of \(V\). For \(f \in V^S\) and every \(x \in S\),

\begin{equation*} (f + \mathbf{0})(x) = f(x) + \mathbf{0}(x) = f(x) + 0 = f(x) , \end{equation*}

so \(f + \mathbf{0} = f\). Thus \(\mathbf{0}\) is an additive identity for \(V^S\).

Additive inverse. Let \(f \in V^S\) and define \(-f \in V^S\) by \((-f)(x) = -\bigl(f(x)\bigr)\) for all \(x \in S\), where \(-\bigl(f(x)\bigr)\) is the additive inverse of \(f(x)\) in \(V\) (which exists and is unique by 1.27). Then for every \(x \in S\),

\begin{equation*} \bigl(f + (-f)\bigr)(x) = f(x) + \bigl(-f(x)\bigr) = 0 = \mathbf{0}(x) , \end{equation*}

so \(f + (-f) = \mathbf{0}\). Thus every element of \(V^S\) has an additive inverse.

Multiplicative identity. For \(f \in V^S\) and every \(x \in S\),

\begin{equation*} (1f)(x) = 1 \bigl( f(x) \bigr) = f(x) , \end{equation*}

by the multiplicative identity property in \(V\). Hence \(1f = f\).

Distributive properties. Let \(a, b \in \mathbb{F}\) and \(f, g \in V^S\). For every \(x \in S\),

\begin{equation*} \bigl(a(f + g)\bigr)(x) = a\bigl(f(x) + g(x)\bigr) = a f(x) + a g(x) = (af)(x) + (ag)(x) = (af + ag)(x) , \end{equation*}

so \(a(f + g) = af + ag\); and

\begin{equation*} \bigl((a + b)f\bigr)(x) = (a + b)\bigl(f(x)\bigr) = a f(x) + b f(x) = (af)(x) + (bf)(x) = (af + bf)(x) , \end{equation*}

so \((a + b)f = af + bf\). Both computations use the distributive properties in \(V\).

All requirements of 1.20 are satisfied, so \(V^S\) with these operations is a vector space over \(\mathbb{F}\). Its additive identity is the function \(\mathbf{0}\) that sends every element of \(S\) to \(0 \in V\), and the additive inverse of \(f\) is the function \(x \mapsto -f(x)\).

Two remarks. First, taking \(V = \mathbb{F}\) recovers exactly the example \(\mathbb{F}^S\) of 1.24 and 1.25. Second, the hypothesis that \(S\) is nonempty was never used: if \(S = \emptyset\) then \(V^S\) consists of the single (empty) function, and the argument above still applies, exhibiting \(V^S\) as the zero vector space \(\{ \mathbf{0} \}\). Nonemptiness of \(S\) merely matches the convention used in 1.25.

Throughout, \(V\) is a vector space over \(\mathbf{R}\), so all eight properties in the definition of a vector space (1.20) are available for \(V\) with real scalars. Since an element of \(V_{\mathbf{C}}\) is by definition the ordered pair \((u,v)\) merely written as \(u+iv\), we have the equality criterion

\begin{equation*} u_1 + i v_1 = u_2 + i v_2 \quad\Longleftrightarrow\quad u_1 = u_2 \ \text{ and } \ v_1 = v_2 . \end{equation*}

Both operations defined above produce an expression of the form (element of \(V\)) \(+\, i\,\)(element of \(V\)), so addition is a function from \(V_{\mathbf{C}} \times V_{\mathbf{C}}\) to \(V_{\mathbf{C}}\) and complex scalar multiplication is a function from \(\mathbf{C} \times V_{\mathbf{C}}\) to \(V_{\mathbf{C}}\), as required. Note also that every complex number has a unique representation \(a+bi\) with \(a,b \in \mathbf{R}\), so the formula for scalar multiplication is unambiguous.

Preliminary remark on the notation \(au - bv\). By 1.28, \(au - bv\) means \(au + \bigl(-(bv)\bigr)\). We first check that \(-(bv) = (-b)v\) for all \(b \in \mathbf{R}\) and \(v \in V\). Indeed,

\begin{equation*} bv + (-b)v = \bigl(b + (-b)\bigr)v = 0v = 0, \end{equation*}

where the first equality is the distributive property in \(V\) and the last is 1.30. Hence \((-b)v\) is an additive inverse of \(bv\), and additive inverses are unique (1.27), so \(-(bv) = (-b)v\). Consequently every expression below of the form \(\alpha u + \beta v\) with \(\alpha,\beta \in \mathbf{R}\) may be manipulated using only the vector space properties of \(V\) over \(\mathbf{R}\), with subtraction of real coefficients handled as addition of negatives. We use this silently from now on.

We now verify the eight properties of 1.20 for \(V_{\mathbf{C}}\) over \(\mathbf{F} = \mathbf{C}\). Fix arbitrary \(u, v, u_1, v_1, u_2, v_2, u_3, v_3 \in V\) and arbitrary \(\lambda = a + bi\), \(\mu = c + di\) in \(\mathbf{C}\), where \(a,b,c,d \in \mathbf{R}\).

Commutativity. Using commutativity of addition in \(V\),

\begin{equation*} (u_1 + iv_1) + (u_2 + iv_2) = (u_1 + u_2) + i(v_1 + v_2) = (u_2 + u_1) + i(v_2 + v_1) = (u_2 + iv_2) + (u_1 + iv_1). \end{equation*}

Associativity of addition. Using associativity of addition in \(V\),

\begin{equation*} \begin{aligned} \bigl[(u_1 + iv_1) + (u_2 + iv_2)\bigr] + (u_3 + iv_3) &= \bigl[(u_1 + u_2) + i(v_1 + v_2)\bigr] + (u_3 + iv_3)\\ &= \bigl((u_1 + u_2) + u_3\bigr) + i\bigl((v_1 + v_2) + v_3\bigr)\\ &= \bigl(u_1 + (u_2 + u_3)\bigr) + i\bigl(v_1 + (v_2 + v_3)\bigr)\\ &= (u_1 + iv_1) + \bigl[(u_2 + u_3) + i(v_2 + v_3)\bigr]\\ &= (u_1 + iv_1) + \bigl[(u_2 + iv_2) + (u_3 + iv_3)\bigr]. \end{aligned} \end{equation*}

Associativity of scalar multiplication. We must show \((\lambda\mu)(u+iv) = \lambda\bigl(\mu(u+iv)\bigr)\). In \(\mathbf{C}\) we have \(\lambda\mu = (ac - bd) + (ad + bc)i\), so by definition

\begin{equation*} (\lambda\mu)(u + iv) = \bigl((ac-bd)u - (ad+bc)v\bigr) + i\bigl((ac-bd)v + (ad+bc)u\bigr). \end{equation*}

On the other hand \(\mu(u+iv) = (cu - dv) + i(cv + du)\), and therefore

\begin{equation*} \begin{aligned} \lambda\bigl(\mu(u+iv)\bigr) &= \Bigl(a(cu - dv) - b(cv + du)\Bigr) + i\Bigl(a(cv + du) + b(cu - dv)\Bigr)\\ &= \Bigl((ac)u - (ad)v - (bc)v - (bd)u\Bigr) + i\Bigl((ac)v + (ad)u + (bc)u - (bd)v\Bigr)\\ &= \bigl((ac - bd)u - (ad + bc)v\bigr) + i\bigl((ac - bd)v + (ad + bc)u\bigr), \end{aligned} \end{equation*}

where the second equality uses the distributive property \(a(x+y) = ax + ay\) and associativity of scalar multiplication \(a(cu) = (ac)u\) in \(V\), and the third uses the distributive property \((\alpha + \beta)x = \alpha x + \beta x\) in \(V\) together with commutativity and associativity of addition in \(V\). The two displayed expressions agree, as desired.

Additive identity. Let \(0\) denote the additive identity of \(V\) and set \(\mathbf{0} = 0 + i0 \in V_{\mathbf{C}}\). Then

\begin{equation*} (u + iv) + (0 + i0) = (u + 0) + i(v + 0) = u + iv, \end{equation*}

so \(\mathbf{0}\) is an additive identity for \(V_{\mathbf{C}}\) (and it is the only one, by 1.26).

Additive inverse. Given \(u + iv \in V_{\mathbf{C}}\), put \(w = (-u) + i(-v)\), which lies in \(V_{\mathbf{C}}\) because \(-u, -v \in V\). Then

\begin{equation*} (u + iv) + w = \bigl(u + (-u)\bigr) + i\bigl(v + (-v)\bigr) = 0 + i0 = \mathbf{0}. \end{equation*}

Thus every element of \(V_{\mathbf{C}}\) has an additive inverse.

Multiplicative identity. The complex number \(1\) is \(1 + 0i\), so

\begin{equation*} 1(u + iv) = (1u - 0v) + i(1v + 0u) = (u - 0) + i(v + 0) = u + iv, \end{equation*}

using \(1u = u\) in \(V\) and \(0v = 0u = 0\) by 1.30. Here \(u - 0v\) means \(u + \bigl(-(0v)\bigr)\), and by the preliminary remark \(-(0v) = (-0)v = 0v = 0\), so \(u - 0v = u + 0 = u\); similarly \(v + 0u = v + 0 = v\).

First distributive property. We show \(\lambda(x + y) = \lambda x + \lambda y\). With \(x = u_1 + iv_1\) and \(y = u_2 + iv_2\),

\begin{equation*} \begin{aligned} \lambda(x + y) &= (a + bi)\bigl((u_1 + u_2) + i(v_1 + v_2)\bigr)\\ &= \Bigl(a(u_1 + u_2) - b(v_1 + v_2)\Bigr) + i\Bigl(a(v_1 + v_2) + b(u_1 + u_2)\Bigr)\\ &= \bigl(au_1 + au_2 - bv_1 - bv_2\bigr) + i\bigl(av_1 + av_2 + bu_1 + bu_2\bigr)\\ &= \bigl((au_1 - bv_1) + (au_2 - bv_2)\bigr) + i\bigl((av_1 + bu_1) + (av_2 + bu_2)\bigr)\\ &= \bigl((au_1 - bv_1) + i(av_1 + bu_1)\bigr) + \bigl((au_2 - bv_2) + i(av_2 + bu_2)\bigr)\\ &= \lambda x + \lambda y, \end{aligned} \end{equation*}

where the third equality uses the distributive property \(\alpha(x+y) = \alpha x + \alpha y\) in \(V\) (applied with the real scalars \(a\) and \(-b\)), and the fourth uses commutativity and associativity of addition in \(V\) to regroup the four summands.

Second distributive property. We show \((\lambda + \mu)x = \lambda x + \mu x\). Since \(\lambda + \mu = (a + c) + (b + d)i\), with \(x = u + iv\) we get

\begin{equation*} \begin{aligned} (\lambda + \mu)x &= \bigl((a+c)u - (b+d)v\bigr) + i\bigl((a+c)v + (b+d)u\bigr)\\ &= \bigl(au + cu - bv - dv\bigr) + i\bigl(av + cv + bu + du\bigr)\\ &= \bigl((au - bv) + (cu - dv)\bigr) + i\bigl((av + bu) + (cv + du)\bigr)\\ &= \bigl((au - bv) + i(av + bu)\bigr) + \bigl((cu - dv) + i(cv + du)\bigr)\\ &= \lambda x + \mu x, \end{aligned} \end{equation*}

where the second equality uses the distributive property \((\alpha + \beta)x = \alpha x + \beta x\) in \(V\) (with the real scalars \(a, c\) and \(-b, -d\), noting \(-(b+d) = (-b) + (-d)\)), and the third uses commutativity and associativity of addition in \(V\).

All eight properties of 1.20 hold, so \(V_{\mathbf{C}}\) is a vector space over \(\mathbf{C}\).

Remark on the identification. Identify \(u \in V\) with \(u + i0 \in V_{\mathbf{C}}\). This identification respects the operations inherited from \(V\): for \(u_1, u_2 \in V\) we have \((u_1 + i0) + (u_2 + i0) = (u_1 + u_2) + i0\), and for a real scalar \(a\), viewed as the complex number \(a + 0i\), applying the definition with \(b = 0\) and \(v = 0\) gives

\begin{equation*} (a + 0i)(u + i0) = \bigl(au - 0 \cdot 0\bigr) + i\bigl(a0 + 0u\bigr) = au + i0, \end{equation*}

since \(0 \cdot 0 = 0\) and \(0u = 0\) by 1.30 while \(a0 = 0\) by 1.31 (and \(-0 = 0\), as noted above). So the copy of \(V\) inside \(V_{\mathbf{C}}\) carries exactly its original real vector space structure. Taking \(V = \mathbf{R}^n\) recovers \(\mathbf{C}^n\): the pair \((x, y)\) with \(x, y \in \mathbf{R}^n\) corresponds to \(x + iy = (x_1 + iy_1, \dots, x_n + iy_n) \in \mathbf{C}^n\), and the addition and scalar multiplication defined above become the usual coordinatewise operations on \(\mathbf{C}^n\).

Throughout we use 1.34: a subset \(U\) of a vector space \(V\) is a subspace if and only if \(0 \in U\), \(U\) is closed under addition, and \(U\) is closed under scalar multiplication.

(a) This is a subspace. Write \(U = \{(x_1, x_2, x_3) \in \mathbf{F}^3 : x_1 + 2x_2 + 3x_3 = 0\}\).

Additive identity: \(0 + 2 \cdot 0 + 3 \cdot 0 = 0\), so \((0,0,0) \in U\).

Closed under addition: suppose \((x_1, x_2, x_3)\) and \((y_1, y_2, y_3)\) belong to \(U\). Then

\begin{equation*} (x_1 + y_1) + 2(x_2 + y_2) + 3(x_3 + y_3) = (x_1 + 2x_2 + 3x_3) + (y_1 + 2y_2 + 3y_3) = 0 + 0 = 0, \end{equation*}

so \((x_1, x_2, x_3) + (y_1, y_2, y_3) = (x_1 + y_1, x_2 + y_2, x_3 + y_3) \in U\).

Closed under scalar multiplication: if \(\lambda \in \mathbf{F}\) and \((x_1, x_2, x_3) \in U\), then

\begin{equation*} \lambda x_1 + 2\lambda x_2 + 3 \lambda x_3 = \lambda(x_1 + 2x_2 + 3x_3) = \lambda \cdot 0 = 0, \end{equation*}

so \(\lambda(x_1, x_2, x_3) \in U\). Hence \(U\) is a subspace of \(\mathbf{F}^3\).

(b) This is not a subspace, because it does not contain the additive identity: \(0 + 2 \cdot 0 + 3 \cdot 0 = 0 \neq 4\), so \((0,0,0)\) is not in the set. By 1.34 the set is not a subspace of \(\mathbf{F}^3\). (It also fails the other two conditions: \((4,0,0)\) lies in the set, but \((4,0,0) + (4,0,0) = (8,0,0)\) does not, since \(8 \neq 4\); and \(2(4,0,0) = (8,0,0)\) does not either.)

(c) This is not a subspace. The set contains \((1,1,0)\) and \((0,0,1)\), since \(1 \cdot 1 \cdot 0 = 0\) and \(0 \cdot 0 \cdot 1 = 0\). But

\begin{equation*} (1,1,0) + (0,0,1) = (1,1,1) \end{equation*}

and \(1 \cdot 1 \cdot 1 = 1 \neq 0\), so the sum is not in the set. Thus the set is not closed under addition and hence is not a subspace of \(\mathbf{F}^3\). (Note that \(0\) does lie in the set and the set is closed under scalar multiplication, since \((\lambda x_1)(\lambda x_2)(\lambda x_3) = \lambda^3 x_1 x_2 x_3\); only closure under addition fails.)

(d) This is a subspace. Write \(U = \{(x_1, x_2, x_3) \in \mathbf{F}^3 : x_1 = 5x_3\}\).

Additive identity: \(0 = 5 \cdot 0\), so \((0,0,0) \in U\).

Closed under addition: if \((x_1, x_2, x_3), (y_1, y_2, y_3) \in U\), then \(x_1 = 5x_3\) and \(y_1 = 5y_3\), so

\begin{equation*} x_1 + y_1 = 5x_3 + 5y_3 = 5(x_3 + y_3), \end{equation*}

which says exactly that \((x_1 + y_1, x_2 + y_2, x_3 + y_3) \in U\).

Closed under scalar multiplication: if \(\lambda \in \mathbf{F}\) and \((x_1, x_2, x_3) \in U\), then \(\lambda x_1 = \lambda(5x_3) = 5(\lambda x_3)\), so \(\lambda(x_1, x_2, x_3) \in U\).

Hence \(U\) is a subspace of \(\mathbf{F}^3\).

Example 1.35 makes five assertions. We verify each in turn, using the criterion 1.34 throughout.

(a) If \(b \in \mathbf{F}\), then \(U = \{(x_1, x_2, x_3, x_4) \in \mathbf{F}^4 : x_3 = 5x_4 + b\}\) is a subspace of \(\mathbf{F}^4\) if and only if \(b = 0\).

Suppose first that \(b \neq 0\). Then \((0,0,0,0) \notin U\), because \(0 = 5 \cdot 0 + b\) would force \(b = 0\). By 1.34, \(U\) is not a subspace of \(\mathbf{F}^4\).

Now suppose \(b = 0\), so \(U = \{(x_1, x_2, x_3, x_4) \in \mathbf{F}^4 : x_3 = 5x_4\}\). Since \(0 = 5 \cdot 0\), we have \((0,0,0,0) \in U\). If \((x_1, x_2, x_3, x_4)\) and \((y_1, y_2, y_3, y_4)\) lie in \(U\), then \(x_3 = 5x_4\) and \(y_3 = 5y_4\), hence \(x_3 + y_3 = 5(x_4 + y_4)\), so the sum lies in \(U\). If \(\lambda \in \mathbf{F}\) and \((x_1, x_2, x_3, x_4) \in U\), then \(\lambda x_3 = \lambda(5x_4) = 5(\lambda x_4)\), so \(\lambda(x_1, x_2, x_3, x_4) \in U\). Thus \(U\) is a subspace of \(\mathbf{F}^4\).

(b) The set \(C[0,1]\) of continuous real-valued functions on \([0,1]\) is a subspace of \(\mathbf{R}^{[0,1]}\).

The zero function is constant, hence continuous, so \(0 \in C[0,1]\). If \(f, g\) are continuous on \([0,1]\), then \(f + g\) is continuous on \([0,1]\) (the sum of two continuous functions is continuous), so \(C[0,1]\) is closed under addition. If \(\lambda \in \mathbf{R}\) and \(f\) is continuous, then \(\lambda f\) is continuous, so \(C[0,1]\) is closed under scalar multiplication. Hence \(C[0,1]\) is a subspace of \(\mathbf{R}^{[0,1]}\).

(c) The set \(D\) of differentiable real-valued functions on \(\mathbf{R}\) is a subspace of \(\mathbf{R}^{\mathbf{R}}\).

The zero function is differentiable with derivative identically \(0\), so \(0 \in D\). If \(f, g \in D\), then \(f + g\) is differentiable with \((f+g)^{\prime} = f^{\prime} + g^{\prime}\), so \(f + g \in D\). If \(\lambda \in \mathbf{R}\) and \(f \in D\), then \(\lambda f\) is differentiable with \((\lambda f)^{\prime} = \lambda f^{\prime}\), so \(\lambda f \in D\). Hence \(D\) is a subspace of \(\mathbf{R}^{\mathbf{R}}\).

(d) The set \(U = \{f \in \mathbf{R}^{(0,3)} : f \text{ is differentiable and } f^{\prime}(2) = b\}\) is a subspace of \(\mathbf{R}^{(0,3)}\) if and only if \(b = 0\).

If \(b \neq 0\), then the zero function, whose derivative at \(2\) equals \(0\), does not lie in \(U\); by 1.34, \(U\) is not a subspace.

If \(b = 0\): the zero function is differentiable and its derivative at \(2\) is \(0\), so \(0 \in U\). If \(f, g \in U\), then \(f + g\) is differentiable and \((f+g)^{\prime}(2) = f^{\prime}(2) + g^{\prime}(2) = 0 + 0 = 0\), so \(f + g \in U\). If \(\lambda \in \mathbf{R}\) and \(f \in U\), then \(\lambda f\) is differentiable and \((\lambda f)^{\prime}(2) = \lambda f^{\prime}(2) = \lambda \cdot 0 = 0\), so \(\lambda f \in U\). Hence \(U\) is a subspace of \(\mathbf{R}^{(0,3)}\).

(e) The set \(U\) of all sequences of complex numbers with limit \(0\) is a subspace of \(\mathbf{C}^{\infty}\).

The sequence \((0, 0, \dots)\) has limit \(0\), so \(0 \in U\). If \((x_1, x_2, \dots)\) and \((y_1, y_2, \dots)\) have limit \(0\), then by the limit law for sums \(\lim_{n \to \infty} (x_n + y_n) = \lim_{n \to \infty} x_n + \lim_{n \to \infty} y_n = 0\), so the sum lies in \(U\). If \(\lambda \in \mathbf{C}\) and \((x_1, x_2, \dots) \in U\), then \(\lim_{n \to \infty} \lambda x_n = \lambda \lim_{n \to \infty} x_n = 0\), so \(\lambda(x_1, x_2, \dots) \in U\). Hence \(U\) is a subspace of \(\mathbf{C}^{\infty}\).

Let

\begin{equation*} U = \{f \in \mathbf{R}^{(-4,4)} : f \text{ is differentiable on } (-4,4) \text{ and } f^{\prime}(-1) = 3f(2)\}. \end{equation*}

Note that \(-1\) and \(2\) both lie in \((-4,4)\), so the condition makes sense. We verify the three conditions of 1.34.

Additive identity. The zero function \(0\) on \((-4,4)\) is differentiable, with \(0^{\prime}(x) = 0\) for all \(x\). Hence \(0^{\prime}(-1) = 0 = 3 \cdot 0 = 3 \cdot 0(2)\), so \(0 \in U\).

Closed under addition. Suppose \(f, g \in U\). The sum of two differentiable functions is differentiable, and \((f+g)^{\prime} = f^{\prime} + g^{\prime}\). Therefore

\begin{equation*} (f+g)^{\prime}(-1) = f^{\prime}(-1) + g^{\prime}(-1) = 3f(2) + 3g(2) = 3\bigl(f(2) + g(2)\bigr) = 3(f+g)(2), \end{equation*}

where the second equality uses \(f, g \in U\) and the last uses the definition of the sum of functions. Hence \(f + g \in U\).

Closed under scalar multiplication. Suppose \(\lambda \in \mathbf{R}\) and \(f \in U\). Then \(\lambda f\) is differentiable with \((\lambda f)^{\prime} = \lambda f^{\prime}\), so

\begin{equation*} (\lambda f)^{\prime}(-1) = \lambda f^{\prime}(-1) = \lambda \cdot 3 f(2) = 3\bigl(\lambda f(2)\bigr) = 3 (\lambda f)(2). \end{equation*}

Hence \(\lambda f \in U\).

By 1.34, \(U\) is a subspace of \(\mathbf{R}^{(-4,4)}\).

Let

\begin{equation*} U = \Bigl\{f \in \mathbf{R}^{[0,1]} : f \text{ is continuous on } [0,1] \text{ and } \int_0^1 f = b\Bigr\}. \end{equation*}

Every continuous function on the closed bounded interval \([0,1]\) is Riemann integrable there, so the defining condition makes sense for every continuous \(f\).

First suppose \(b \neq 0\). The zero function \(0\) on \([0,1]\) satisfies \(\int_0^1 0 = 0 \neq b\), so \(0 \notin U\). By 1.34, \(U\) is not a subspace of \(\mathbf{R}^{[0,1]}\).

Now suppose \(b = 0\), so that \(U = \{f \in \mathbf{R}^{[0,1]} : f \text{ continuous},\ \int_0^1 f = 0\}\). We check the three conditions of 1.34.

Additive identity. The zero function is continuous and \(\int_0^1 0 = 0\), so \(0 \in U\).

Closed under addition. Suppose \(f, g \in U\). Then \(f + g\) is continuous (the sum of continuous functions is continuous), and by additivity of the integral

\begin{equation*} \int_0^1 (f + g) = \int_0^1 f + \int_0^1 g = 0 + 0 = 0. \end{equation*}

Hence \(f + g \in U\).

Closed under scalar multiplication. Suppose \(\lambda \in \mathbf{R}\) and \(f \in U\). Then \(\lambda f\) is continuous, and by homogeneity of the integral

\begin{equation*} \int_0^1 (\lambda f) = \lambda \int_0^1 f = \lambda \cdot 0 = 0. \end{equation*}

Hence \(\lambda f \in U\).

By 1.34, \(U\) is a subspace of \(\mathbf{R}^{[0,1]}\) when \(b = 0\). Combining the two directions, \(U\) is a subspace of \(\mathbf{R}^{[0,1]}\) if and only if \(b = 0\).

No.

Here \(\mathbf{R}^2\) is regarded as the subset \(\{(x_1, x_2) \in \mathbf{C}^2 : x_1, x_2 \in \mathbf{R}\}\) of \(\mathbf{C}^2\), and \(\mathbf{C}^2\) is regarded as a vector space over \(\mathbf{C}\), so that the scalars available are all complex numbers.

This subset does contain \((0,0)\) and is closed under addition, since the sum of two real numbers is real. However, it is not closed under scalar multiplication by complex scalars. Indeed, \((1, 0) \in \mathbf{R}^2\) and \(i \in \mathbf{C}\), but

\begin{equation*} i(1,0) = (i, 0) \notin \mathbf{R}^2, \end{equation*}

because \(i\) is not a real number. Since the third condition of 1.34 fails, \(\mathbf{R}^2\) is not a subspace of the complex vector space \(\mathbf{C}^2\).

It is worth noting where this leaves things: \(\mathbf{R}^2\) is a vector space in its own right and it sits inside \(\mathbf{C}^2\) with the same addition, but it fails to be a subspace of \(\mathbf{C}^2\) precisely because the scalar field of \(\mathbf{C}^2\) is \(\mathbf{C}\) rather than \(\mathbf{R}\). If instead we view \(\mathbf{C}^2\) as a real vector space (that is, restrict the scalars to \(\mathbf{R}\)), then \(\mathbf{R}^2\) is a subspace of it.

(a) Yes.

The key point is that the function \(t \mapsto t^3\) is injective on \(\mathbf{R}\). To see this, suppose \(a, b \in \mathbf{R}\) and \(a^3 = b^3\). Then

\begin{equation*} 0 = a^3 - b^3 = (a - b)(a^2 + ab + b^2). \end{equation*}

Now

\begin{equation*} a^2 + ab + b^2 = \Bigl(a + \tfrac{b}{2}\Bigr)^2 + \tfrac{3}{4}b^2 \geq 0, \end{equation*}

and this quantity equals \(0\) only if \(b = 0\) and \(a + \tfrac{b}{2} = 0\), that is, only if \(a = b = 0\). So either \(a = b = 0\), or \(a^2 + ab + b^2 \neq 0\) and the displayed equation forces \(a - b = 0\). In both cases \(a = b\). The converse implication is trivial, so for real \(a, b\) we have

\begin{equation*} a^3 = b^3 \iff a = b. \end{equation*}

(Note that this argument uses that \(\mathbf{R}\) is an ordered field in which squares are nonnegative; it fails over \(\mathbf{C}\), as part (b) shows.)

Consequently

\begin{equation*} U = \{(a,b,c) \in \mathbf{R}^3 : a^3 = b^3\} = \{(a, a, c) \in \mathbf{R}^3 : a, c \in \mathbf{R}\}, \end{equation*}

which is easily seen to be a subspace: \((0,0,0) \in U\); if \((a,a,c)\) and \((a^{\prime}, a^{\prime}, c^{\prime})\) are in \(U\) then their sum \((a + a^{\prime}, a + a^{\prime}, c + c^{\prime})\) has its first two coordinates equal, so it is in \(U\); and for \(\lambda \in \mathbf{R}\), \(\lambda(a,a,c) = (\lambda a, \lambda a, \lambda c) \in U\). By 1.34, \(U\) is a subspace of \(\mathbf{R}^3\).

(b) No.

Over \(\mathbf{C}\) the cube function is not injective, and that breaks closure under addition. Let \(\omega = e^{2\pi i/3} = -\tfrac{1}{2} + \tfrac{\sqrt{3}}{2} i\), a primitive cube root of \(1\), so that \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\). Put

\begin{equation*} u = (1, 1, 0) \quad\text{and}\quad v = (1, \omega, 0). \end{equation*}

Both lie in the set: for \(u\) we have \(1^3 = 1^3\), and for \(v\) we have \(1^3 = 1 = \omega^3\).

Their sum is \(u + v = (2, 1 + \omega, 0)\). Now \(2^3 = 8\), while, using \(\omega^3 = 1\) and \(\omega + \omega^2 = -1\),

\begin{equation*} (1 + \omega)^3 = 1 + 3\omega + 3\omega^2 + \omega^3 = 1 + 3(\omega + \omega^2) + 1 = 2 + 3(-1) = -1. \end{equation*}

Since \(8 \neq -1\), the vector \(u + v\) does not belong to the set. Thus the set is not closed under addition, and by 1.34 it is not a subspace of \(\mathbf{C}^3\).

The statement is false. Here is a counterexample.

Let

\begin{equation*} U = \mathbf{Z}^2 = \{(m, n) \in \mathbf{R}^2 : m, n \text{ are integers}\}. \end{equation*}

\(U\) is nonempty: for instance \((0,0) \in U\).

\(U\) is closed under addition: if \((m, n)\) and \((m^{\prime}, n^{\prime})\) are in \(U\), then \(m + m^{\prime}\) and \(n + n^{\prime}\) are integers, so \((m,n) + (m^{\prime},n^{\prime}) = (m + m^{\prime}, n + n^{\prime}) \in U\).

\(U\) is closed under taking additive inverses: if \((m,n) \in U\), then \(-m\) and \(-n\) are integers, so \(-(m,n) = (-m,-n) \in U\).

But \(U\) is not a subspace of \(\mathbf{R}^2\), because it is not closed under scalar multiplication: \((1,0) \in U\) and \(\tfrac{1}{2} \in \mathbf{R}\), yet

\begin{equation*} \tfrac{1}{2}(1,0) = \bigl(\tfrac{1}{2}, 0\bigr) \notin U, \end{equation*}

since \(\tfrac{1}{2}\) is not an integer. By 1.34, \(U\) is not a subspace of \(\mathbf{R}^2\).

Remark on why the hypotheses are not enough. Closure under addition and additive inverses (together with nonemptiness) makes \(U\) a subgroup of the additive group of \(\mathbf{R}^2\); in particular \(0 \in U\), since picking any \(u \in U\) gives \(0 = u + (-u) \in U\). Iterating addition and inverses yields \(nu \in U\) for every integer \(n\), so \(U\) is automatically closed under multiplication by integer scalars. What the hypotheses cannot deliver is closure under multiplication by non-integer real scalars, and that is exactly the gap the counterexample exploits. The set \(\mathbf{Q}^2 = \{(x,y) \in \mathbf{R}^2 : x, y \in \mathbf{Q}\}\) is another counterexample, for the same reason: it is closed under addition and negation, but \(\sqrt{2}\,(1,0) = (\sqrt{2}, 0) \notin \mathbf{Q}^2\).

Note that over some other scalar fields the analogous statement can be true. For example, if \(\mathbf{F} = \mathbf{Z}/2\) then every scalar is \(0\) or \(1\), and a nonempty subset closed under addition automatically contains \(0\) and is closed under scalar multiplication. The failure above is specific to scalar fields, such as \(\mathbf{R}\), that contain scalars not generated from \(1\) by repeated addition and negation.

Let \(U\) be the union of the two coordinate axes:

\begin{equation*} U = \{(x_1, x_2) \in \mathbb{R}^2 : x_1 x_2 = 0\}. \end{equation*}

\(U\) is nonempty, since \((0,0) \in U\).

\(U\) is closed under scalar multiplication. Indeed, suppose \(\lambda \in \mathbb{R}\) and \((x_1, x_2) \in U\), so that \(x_1 x_2 = 0\). Then

\begin{equation*} (\lambda x_1)(\lambda x_2) = \lambda^2 (x_1 x_2) = \lambda^2 \cdot 0 = 0, \end{equation*}

so \(\lambda (x_1, x_2) = (\lambda x_1, \lambda x_2) \in U\).

However \(U\) is not a subspace of \(\mathbb{R}^2\), because it is not closed under addition: we have \((1,0) \in U\) and \((0,1) \in U\) (both products of coordinates are \(0\)), but

\begin{equation*} (1,0) + (0,1) = (1,1) \notin U, \end{equation*}

since \(1 \cdot 1 = 1 \neq 0\). Because closure under addition is one of the three conditions in 1.34, \(U\) is not a subspace of \(\mathbb{R}^2\).

Geometrically, \(U\) is the union of two lines through the origin. Each line is itself a subspace, and scaling never moves a point off the line it lies on; but adding a nonzero point of one axis to a nonzero point of the other produces a point on neither axis. This example is the two-subspace case of Exercise 12: a union of two subspaces, neither contained in the other, is never a subspace.

No. Let \(P \subseteq \mathbb{R}^{\mathbb{R}}\) denote the set of periodic functions from \(\mathbb{R}\) to \(\mathbb{R}\). Two of the three conditions of 1.34 hold, but closure under addition fails.

First the two conditions that do hold. The zero function is periodic (every positive number is a period for it), so \(0 \in P\). And if \(f \in P\) has period \(p\) and \(\lambda \in \mathbb{R}\), then for all \(x \in \mathbb{R}\) we have \((\lambda f)(x + p) = \lambda f(x+p) = \lambda f(x) = (\lambda f)(x)\), so \(\lambda f\) is periodic with the same period \(p\); thus \(P\) is closed under scalar multiplication.

Now we show \(P\) is not closed under addition. Define \(f, g \colon \mathbb{R} \to \mathbb{R}\) by

\begin{equation*} f(x) = \cos x, \qquad g(x) = \cos(\sqrt{2}\, x). \end{equation*}

Then \(f\) is periodic with period \(2\pi\) and \(g\) is periodic with period \(2\pi/\sqrt{2} = \sqrt{2}\,\pi\), so \(f, g \in P\). We claim \(h = f + g\) is not periodic.

Because \(\cos t \leq 1\) for all \(t \in \mathbb{R}\), we have \(h(x) \leq 2\) for all \(x\), and \(h(0) = 1 + 1 = 2\). Hence \(2\) is the maximum value of \(h\), and \(h(x) = 2\) forces \(\cos x = 1\) and \(\cos(\sqrt{2}\,x) = 1\) (if either summand were less than \(1\), the sum would be less than \(2\)). Now \(\cos x = 1\) if and only if \(x = 2\pi m\) for some \(m \in \mathbb{Z}\), and \(\cos(\sqrt{2}\, x) = 1\) if and only if \(\sqrt{2}\,x = 2\pi n\) for some \(n \in \mathbb{Z}\). Suppose both hold. Substituting \(x = 2\pi m\) into the second equation gives \(2\sqrt{2}\,\pi m = 2\pi n\), hence

\begin{equation*} \sqrt{2}\, m = n. \end{equation*}

If \(m \neq 0\), this would give \(\sqrt{2} = n/m \in \mathbb{Q}\), contradicting the irrationality of \(\sqrt{2}\). Thus \(m = 0\), and then \(n = 0\) and \(x = 0\). In other words,

\begin{equation*} \{x \in \mathbb{R} : h(x) = 2\} = \{0\}. \end{equation*}

Suppose now, seeking a contradiction, that \(h\) were periodic, say \(h(x) = h(x + p)\) for all \(x \in \mathbb{R}\), with \(p > 0\). Taking \(x = 0\) gives \(h(p) = h(0) = 2\), so by the displayed equation above \(p = 0\), contradicting \(p > 0\). Hence \(h = f + g\) is not periodic.

So \(f, g \in P\) but \(f + g \notin P\), and therefore \(P\) is not a subspace of \(\mathbb{R}^{\mathbb{R}}\).

Remark. The obstruction is exactly incommensurability of periods. If \(f\) has period \(p\) and \(g\) has period \(q\) with \(p/q\) rational, say \(p/q = a/b\) with \(a, b\) positive integers, then \(bp = aq\) is a common period and \(f + g\) is periodic. The set of functions periodic with a fixed period \(p\) (i.e. satisfying \(f(x) = f(x+p)\) for all \(x\)) is a subspace of \(\mathbb{R}^{\mathbb{R}}\); it is the union of these subspaces over all \(p > 0\) that fails to be one.

We verify the three conditions of 1.34 for the subset \(V_1 \cap V_2\) of \(V\).

Additive identity. Because \(V_1\) is a subspace of \(V\), we have \(0 \in V_1\); because \(V_2\) is a subspace of \(V\), we have \(0 \in V_2\). Hence \(0 \in V_1 \cap V_2\).

Closed under addition. Suppose \(u, w \in V_1 \cap V_2\). Then \(u, w \in V_1\), and \(V_1\) is closed under addition, so \(u + w \in V_1\). Likewise \(u, w \in V_2\) and \(V_2\) is closed under addition, so \(u + w \in V_2\). Hence \(u + w \in V_1 \cap V_2\).

Closed under scalar multiplication. Suppose \(a \in \mathbb{F}\) and \(u \in V_1 \cap V_2\). Then \(u \in V_1\), and \(V_1\) is closed under scalar multiplication, so \(au \in V_1\). Likewise \(u \in V_2\) gives \(au \in V_2\). Hence \(au \in V_1 \cap V_2\).

All three conditions of 1.34 hold, so \(V_1 \cap V_2\) is a subspace of \(V\).

Let \(\mathcal{A}\) be a nonempty collection of subspaces of \(V\), indexed as \(\{V_i\}_{i \in I}\) with \(I \neq \varnothing\), and set

\begin{equation*} W = \bigcap_{i \in I} V_i = \{v \in V : v \in V_i \text{ for every } i \in I\}. \end{equation*}

Note \(W \subseteq V\), since each \(V_i \subseteq V\) and \(I\) is nonempty. We verify the three conditions of 1.34.

Additive identity. Each \(V_i\) is a subspace of \(V\), so \(0 \in V_i\) for every \(i \in I\). Hence \(0 \in W\).

Closed under addition. Suppose \(u, w \in W\). Let \(i \in I\). Then \(u \in V_i\) and \(w \in V_i\), and \(V_i\) is closed under addition, so \(u + w \in V_i\). Since \(i \in I\) was arbitrary, \(u + w\) belongs to every \(V_i\), that is, \(u + w \in W\).

Closed under scalar multiplication. Suppose \(a \in \mathbb{F}\) and \(u \in W\). Let \(i \in I\). Then \(u \in V_i\), and \(V_i\) is closed under scalar multiplication, so \(au \in V_i\). Since \(i \in I\) was arbitrary, \(au \in W\).

By 1.34, \(W\) is a subspace of \(V\). Taking \(I\) to have two elements recovers Exercise 10.

Remark on the degenerate case. If the collection is empty, then the intersection, interpreted as usual within the ambient space \(V\) (the set of \(v \in V\) lying in every member of the empty collection), is all of \(V\), which is again a subspace of \(V\). So the assertion holds for every collection, with that standard convention.

Remark. This result is what makes the notion of the smallest subspace containing a given subset \(S \subseteq V\) well defined: the intersection of all subspaces of \(V\) that contain \(S\) is itself a subspace containing \(S\), and it is contained in every such subspace. That intersection is the span of \(S\); compare 1.40, which identifies \(V_1 + \cdots + V_m\) as the smallest subspace containing \(V_1, \ldots, V_m\). Note the contrast with unions, which by Exercises 12 and 13 are almost never subspaces.

Let \(U\) and \(W\) be subspaces of \(V\). We prove both implications.

First suppose one of the subspaces is contained in the other. Without loss of generality \(W \subseteq U\) (otherwise interchange the names of \(U\) and \(W\)). Then \(U \cup W = U\), which is a subspace of \(V\) by hypothesis. Thus \(U \cup W\) is a subspace of \(V\).

Now suppose \(U \cup W\) is a subspace of \(V\); we must show \(U \subseteq W\) or \(W \subseteq U\). Suppose neither inclusion holds. Then there exists \(u \in U\) with \(u \notin W\), and there exists \(w \in W\) with \(w \notin U\). Since \(u, w \in U \cup W\) and \(U \cup W\) is closed under addition, we have

\begin{equation*} u + w \in U \cup W. \end{equation*}

Thus \(u + w \in U\) or \(u + w \in W\). We rule out both.

Case 1: \(u + w \in U\). Since \(U\) is a subspace, \(-u \in U\) and hence

\begin{equation*} w = (u + w) + (-u) \in U, \end{equation*}

contradicting \(w \notin U\).

Case 2: \(u + w \in W\). Since \(W\) is a subspace, \(-w \in W\) and hence

\begin{equation*} u = (u + w) + (-w) \in W, \end{equation*}

contradicting \(u \notin W\).

Both cases are impossible, so our supposition was false: one of \(U\), \(W\) is contained in the other, completing the proof.

Note that only the closure of \(U \cup W\) under addition was used; the conclusion holds under that weaker hypothesis alone. The set \(\{(x_1,x_2) \in \mathbb{R}^2 : x_1 x_2 = 0\}\) of Exercise 8 is the union of the two coordinate axes of \(\mathbb{R}^2\), neither of which contains the other, and indeed it is not a subspace.

Let \(V_1, V_2, V_3\) be subspaces of \(V\) and set \(W = V_1 \cup V_2 \cup V_3\).

The easy direction. Suppose one of the subspaces contains the other two; without loss of generality suppose \(V_2 \subseteq V_1\) and \(V_3 \subseteq V_1\). Then \(W = V_1\), which is a subspace of \(V\).

The other direction. Suppose \(W = V_1 \cup V_2 \cup V_3\) is a subspace of \(V\). Assume, seeking a contradiction, that no one of \(V_1, V_2, V_3\) contains the other two.

Step 1: no \(V_j\) is contained in the union of the other two. Suppose it were, say (after relabelling) \(V_3 \subseteq V_1 \cup V_2\). Then

\begin{equation*} W = V_1 \cup V_2 \cup V_3 = V_1 \cup V_2, \end{equation*}

so \(V_1 \cup V_2\) is a subspace of \(V\). By Exercise 12, one of \(V_1, V_2\) contains the other; without loss of generality \(V_2 \subseteq V_1\). Then \(W = V_1\). But \(V_3 \subseteq W = V_1\) and \(V_2 \subseteq V_1\), so \(V_1\) contains the other two subspaces, contrary to our assumption. The same argument applies with the roles of the indices permuted. Hence for each \(j\) the subspace \(V_j\) is not contained in the union of the other two.

Step 2: choose suitable vectors. By Step 1 we may choose

\begin{equation*} u \in V_1 \setminus (V_2 \cup V_3) \qquad \text{and} \qquad w \in V_2 \setminus (V_1 \cup V_3). \end{equation*}

Step 3: the line \(u + \mathbb{F}w\) lands in \(V_3\). Let \(\lambda \in \mathbb{F}\) with \(\lambda \neq 0\). Since \(u, w \in W\) and \(W\) is a subspace, \(u + \lambda w \in W = V_1 \cup V_2 \cup V_3\). We rule out the first two possibilities.

If \(u + \lambda w \in V_1\), then, since \(u \in V_1\) and \(V_1\) is a subspace, \(\lambda w = (u + \lambda w) - u \in V_1\); multiplying by \(\lambda^{-1}\) (available because \(\lambda \neq 0\)) gives \(w \in V_1\), contradicting the choice of \(w\).

If \(u + \lambda w \in V_2\), then, since \(w \in V_2\) and \(V_2\) is a subspace, \(\lambda w \in V_2\) and so \(u = (u + \lambda w) - \lambda w \in V_2\), contradicting the choice of \(u\).

Therefore

\begin{equation*} u + \lambda w \in V_3 \qquad \text{for every nonzero } \lambda \in \mathbb{F}. \end{equation*}

Step 4: the contradiction. Here we use that \(\mathbb{F}\) (which is \(\mathbb{R}\) or \(\mathbb{C}\)) has at least three elements: the scalars \(1\) and \(2 = 1 + 1\) are both nonzero and are distinct from each other. By Step 3,

\begin{equation*} u + 1 \cdot w \in V_3 \qquad \text{and} \qquad u + 2w \in V_3. \end{equation*}

Since \(V_3\) is a subspace, it is closed under subtraction, so

\begin{equation*} w = (u + 2w) - (u + w) \in V_3, \end{equation*}

contradicting \(w \notin V_3\).

This contradiction shows that our assumption was false. Hence one of \(V_1, V_2, V_3\) contains the other two, completing the proof.

Remark on the two-element field. The hypothesis that \(\mathbb{F}\) has more than two elements is genuinely needed, and Step 4 is exactly where it enters: over a field with only two elements the only nonzero scalar is \(1\), so Step 3 produces just one vector and yields no contradiction. Concretely, let \(\mathbb{F}_2\) denote the field with two elements and take \(V = \mathbb{F}_2^2\) with

\begin{equation*} V_1 = \{(0,0),(1,0)\}, \qquad V_2 = \{(0,0),(0,1)\}, \qquad V_3 = \{(0,0),(1,1)\}. \end{equation*}

Each \(V_j\) is a subspace of \(V\), no one of them contains another, yet \(V_1 \cup V_2 \cup V_3\) contains all four vectors of \(\mathbb{F}_2^2\) and so equals \(V\), a subspace.

Remark. The same argument generalizes: if \(\mathbb{F}\) has more than \(m\) elements (in particular if \(\mathbb{F}\) is infinite, as \(\mathbb{R}\) and \(\mathbb{C}\) are), then a vector space over \(\mathbb{F}\) is never the union of \(m\) proper subspaces.

The answer is

\begin{equation*} U + W = \{(x, y, 2x) \in \mathbb{F}^3 : x, y \in \mathbb{F}\}, \end{equation*}

which in words is: \(U + W\) is the set of elements of \(\mathbb{F}^3\) whose third coordinate is twice the first coordinate.

By 1.36, \(U + W = \{u + w : u \in U,\ w \in W\}\). We prove the displayed equality by proving both inclusions.

First inclusion. A typical element of \(U\) is \((a, -a, 2a)\) with \(a \in \mathbb{F}\), and a typical element of \(W\) is \((b, b, 2b)\) with \(b \in \mathbb{F}\). Their sum is

\begin{equation*} (a, -a, 2a) + (b, b, 2b) = (a + b,\ b - a,\ 2a + 2b) = \bigl(a + b,\ b - a,\ 2(a+b)\bigr). \end{equation*}

Setting \(x = a + b\) and \(y = b - a\), this is \((x, y, 2x)\). Hence every element of \(U + W\) has the form \((x, y, 2x)\), that is,

\begin{equation*} U + W \subseteq \{(x, y, 2x) \in \mathbb{F}^3 : x, y \in \mathbb{F}\}. \end{equation*}

Second inclusion. Suppose \(x, y \in \mathbb{F}\). Because \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), the scalar \(2\) is invertible in \(\mathbb{F}\), so we may set

\begin{equation*} a = \frac{x - y}{2}, \qquad b = \frac{x + y}{2}. \end{equation*}

Then \(a + b = x\) and \(b - a = y\), so by the computation above

\begin{equation*} (a, -a, 2a) + (b, b, 2b) = \bigl(a+b,\ b-a,\ 2(a+b)\bigr) = (x, y, 2x), \end{equation*}

with \((a, -a, 2a) \in U\) and \((b, b, 2b) \in W\). Hence \((x, y, 2x) \in U + W\), giving

\begin{equation*} \{(x, y, 2x) \in \mathbb{F}^3 : x, y \in \mathbb{F}\} \subseteq U + W. \end{equation*}

The two inclusions together prove the claimed equality.

Remark. This sum is in fact a direct sum. If \(v \in U \cap W\), then \(v = (a, -a, 2a) = (b, b, 2b)\) for some \(a, b \in \mathbb{F}\); comparing first coordinates gives \(a = b\), and comparing second coordinates gives \(-a = b\), so \(2a = 0\) and hence \(a = 0\). Thus \(U \cap W = \{0\}\), and \(U + W\) is a direct sum by 1.46. Geometrically (taking \(\mathbb{F} = \mathbb{R}\)), \(U\) and \(W\) are two distinct lines through the origin in \(\mathbb{R}^3\), and \(U \oplus W\) is the plane through the origin that contains them both, namely the plane with equation \(x_3 = 2x_1\).

We claim \(U + U = U\).

By the definition of the sum of subspaces (1.36),

\begin{equation*} U + U = \{u_1 + u_2 : u_1 \in U,\ u_2 \in U\}. \end{equation*}

First we show \(U + U \subseteq U\). If \(u_1, u_2 \in U\), then \(u_1 + u_2 \in U\) because \(U\), being a subspace of \(V\), is closed under addition. Hence every element of \(U + U\) lies in \(U\).

Next we show \(U \subseteq U + U\). Let \(u \in U\). Because \(U\) is a subspace, \(0 \in U\), and therefore

\begin{equation*} u = u + 0 \end{equation*}

exhibits \(u\) as a sum of an element of \(U\) and an element of \(U\). Thus \(u \in U + U\).

Combining the two inclusions gives \(U + U = U\).

(Alternatively, this follows at once from 1.40: \(U + U\) is the smallest subspace of \(V\) containing \(U\) and \(U\), and the smallest subspace of \(V\) containing \(U\) is \(U\) itself, since \(U\) is already a subspace.)

Yes, addition of subspaces is commutative.

Let \(U\) and \(W\) be subspaces of \(V\). By 1.36,

\begin{equation*} U + W = \{u + w : u \in U,\ w \in W\}, \qquad W + U = \{w + u : w \in W,\ u \in U\}. \end{equation*}

Suppose \(v \in U + W\). Then \(v = u + w\) for some \(u \in U\) and some \(w \in W\). Addition in the vector space \(V\) is commutative (this is one of the axioms in the definition 1.20 of a vector space), so

\begin{equation*} v = u + w = w + u, \end{equation*}

and the right side exhibits \(v\) as the sum of an element of \(W\) and an element of \(U\). Hence \(v \in W + U\), and therefore \(U + W \subseteq W + U\).

Interchanging the roles of \(U\) and \(W\) in the argument above gives \(W + U \subseteq U + W\).

Thus \(U + W = W + U\), as desired.

The same reasoning applies to sums of more than two subspaces: because addition in \(V\) is commutative and associative, \(V_1 + \dots + V_m\) is unchanged by permuting \(V_1, \dots, V_m\).

Yes, addition of subspaces is associative. We show that both sides equal

\begin{equation*} S = \{v_1 + v_2 + v_3 : v_1 \in V_1,\ v_2 \in V_2,\ v_3 \in V_3\}, \end{equation*}

which is the set \(V_1 + V_2 + V_3\) of 1.36. (Note that \(S\) is well defined without parentheses, because addition in \(V\) is associative by the vector space axioms in 1.20.)

First we show \((V_1 + V_2) + V_3 = S\).

Suppose \(v \in (V_1 + V_2) + V_3\). By 1.36 applied to the two subspaces \(V_1 + V_2\) and \(V_3\) (recall that \(V_1 + V_2\) is a subspace of \(V\), by 1.40), we can write \(v = w + v_3\) with \(w \in V_1 + V_2\) and \(v_3 \in V_3\). Applying 1.36 again to \(w\), we can write \(w = v_1 + v_2\) with \(v_1 \in V_1\) and \(v_2 \in V_2\). Hence

\begin{equation*} v = (v_1 + v_2) + v_3 = v_1 + v_2 + v_3 \in S. \end{equation*}

Thus \((V_1 + V_2) + V_3 \subseteq S\).

Conversely, suppose \(v = v_1 + v_2 + v_3\) with \(v_k \in V_k\) for each \(k\). Then \(v_1 + v_2 \in V_1 + V_2\), and so

\begin{equation*} v = (v_1 + v_2) + v_3 \in (V_1 + V_2) + V_3. \end{equation*}

Thus \(S \subseteq (V_1 + V_2) + V_3\), and therefore \((V_1 + V_2) + V_3 = S\).

The proof that \(V_1 + (V_2 + V_3) = S\) is the same, with the parentheses moved: an element of \(V_1 + (V_2 + V_3)\) has the form \(v_1 + w\) with \(v_1 \in V_1\) and \(w = v_2 + v_3\) for some \(v_2 \in V_2\), \(v_3 \in V_3\), giving \(v = v_1 + (v_2 + v_3) = v_1 + v_2 + v_3 \in S\); and conversely any \(v_1 + v_2 + v_3\) with \(v_k \in V_k\) equals \(v_1 + (v_2 + v_3) \in V_1 + (V_2 + V_3)\).

Hence

\begin{equation*} (V_1 + V_2) + V_3 = S = V_1 + (V_2 + V_3), \end{equation*}

so addition of subspaces is associative.

Yes: the subspace \(\{0\}\) is an additive identity, and it is the only one. The only subspace with an additive inverse is \(\{0\}\) itself.

The subspace \(\{0\}\) is an additive identity. Let \(U\) be a subspace of \(V\). If \(u \in U\), then \(u = u + 0\) with \(u \in U\) and \(0 \in \{0\}\), so \(u \in U + \{0\}\). Conversely, every element of \(U + \{0\}\) has the form \(u + 0 = u\) for some \(u \in U\), hence lies in \(U\). Therefore

\begin{equation*} U + \{0\} = U \end{equation*}

for every subspace \(U\) of \(V\). By 1C.16 also \(\{0\} + U = U\), so \(\{0\}\) is a two-sided identity.

It is the only additive identity. Suppose \(E\) is a subspace of \(V\) with \(U + E = U\) for every subspace \(U\) of \(V\). Taking \(U = \{0\}\) gives \(\{0\} + E = \{0\}\). On the other hand, the previous paragraph applied with \(U = E\) gives \(E + \{0\} = E\), and hence \(\{0\} + E = E\) by commutativity (1C.16). Comparing the two computations of \(\{0\} + E\) yields \(E = \{0\}\).

Now consider additive inverses. Suppose \(U\) and \(W\) are subspaces of \(V\) with

\begin{equation*} U + W = \{0\}. \end{equation*}

By 1.40 the sum \(U + W\) contains \(U\); explicitly, if \(u \in U\) then \(u = u + 0 \in U + W\). So \(U \subseteq U + W = \{0\}\), forcing \(U = \{0\}\). The same argument gives \(W = \{0\}\).

Consequently a subspace \(U\) has an additive inverse if and only if \(U = \{0\}\), and in that case its additive inverse is \(\{0\}\) itself (indeed \(\{0\} + \{0\} = \{0\}\)).

Thus addition of subspaces of \(V\) is commutative (1C.16), associative (1C.17), and has an identity, but inverses fail badly: the collection of subspaces of \(V\) under addition is a commutative monoid, not a group, unless \(V = \{0\}\).

The statement is false; in other words, addition of subspaces does not satisfy a cancellation law. Here is a counterexample.

Take \(V = \mathbf{F}^2\) and set

\begin{equation*} V_1 = \{(x, 0) \in \mathbf{F}^2 : x \in \mathbf{F}\}, \qquad V_2 = \{(0, y) \in \mathbf{F}^2 : y \in \mathbf{F}\}, \qquad U = \mathbf{F}^2. \end{equation*}

Each of \(V_1\), \(V_2\), \(U\) is a subspace of \(\mathbf{F}^2\): each contains \((0,0)\) and is clearly closed under addition and scalar multiplication, so 1.34 applies.

Now

\begin{equation*} V_1 + U = U = \mathbf{F}^2 \quad\text{and}\quad V_2 + U = U = \mathbf{F}^2. \end{equation*}

Here each equality \(V_k + U = U\) holds because \(V_k \subseteq U\). Indeed, in that case every element \(v + u\) with \(v \in V_k\) and \(u \in U\) lies in \(U\), since \(U\) is closed under addition; this gives \(V_k + U \subseteq U\). Conversely \(U \subseteq V_k + U\), because each \(u \in U\) can be written as \(u = 0 + u\) with \(0 \in V_k\). (Alternatively: by 1.40 the subspace \(V_k + U\) is the smallest subspace of \(V\) containing both \(V_k\) and \(U\), and when \(V_k \subseteq U\) that smallest subspace is \(U\) itself.)

Hence \(V_1 + U = V_2 + U\). But

\begin{equation*} (1, 0) \in V_1 \quad\text{and}\quad (1, 0) \notin V_2, \end{equation*}

so \(V_1 \neq V_2\). This is a counterexample.

The failure has nothing to do with the particular choice above. The obstruction is exactly the failure of additive inverses found in 1C.18: in a commutative monoid, cancellation would follow from the existence of inverses, and here the only invertible element is \(\{0\}\). Indeed, the same phenomenon occurs whenever \(U\) is large enough to swallow both \(V_1\) and \(V_2\): if \(V_1 \subseteq U\) and \(V_2 \subseteq U\), then \(V_1 + U = U = V_2 + U\) regardless of how different \(V_1\) and \(V_2\) are.

Cancellation can also fail with \(U\) a proper subspace of \(V\). In \(V = \mathbf{F}^3\), take

\begin{equation*} V_1 = \{(x,0,0) : x \in \mathbf{F}\},\quad V_2 = \{(0,y,0) : y \in \mathbf{F}\},\quad U = \{(x,y,0) : x, y \in \mathbf{F}\}. \end{equation*}

Then \(V_1 \subseteq U\) and \(V_2 \subseteq U\), so \(V_1 + U = U = V_2 + U\), while \(V_1 \neq V_2\) and \(U \neq V\).

Take

\begin{equation*} W = \{(x, 0, y, 0) \in \mathbf{F}^4 : x, y \in \mathbf{F}\}. \end{equation*}

First, \(W\) is a subspace of \(\mathbf{F}^4\). It contains \((0,0,0,0)\), it is closed under addition since

\begin{equation*} (x, 0, y, 0) + (x^{\prime}, 0, y^{\prime}, 0) = (x + x^{\prime}, 0, y + y^{\prime}, 0), \end{equation*}

and it is closed under scalar multiplication since \(\lambda(x, 0, y, 0) = (\lambda x, 0, \lambda y, 0)\). Thus 1.34 shows \(W\) is a subspace. (The same three checks show \(U\) is a subspace of \(\mathbf{F}^4\).)

Next we show \(\mathbf{F}^4 = U + W\). Let \((a, b, c, d) \in \mathbf{F}^4\). Then

\begin{equation*} (a, b, c, d) = (b, b, d, d) + (a - b, 0, c - d, 0), \end{equation*}

where \((b, b, d, d) \in U\) and \((a - b, 0, c - d, 0) \in W\). Hence \((a,b,c,d) \in U + W\). Since \(U + W \subseteq \mathbf{F}^4\) trivially, we get \(\mathbf{F}^4 = U + W\).

Finally we show the sum is direct. By 1.46 it suffices to prove \(U \cap W = \{0\}\). Suppose \(v \in U \cap W\). Because \(v \in U\), we have \(v = (x, x, y, y)\) for some \(x, y \in \mathbf{F}\). Because \(v \in W\), the second and fourth coordinates of \(v\) are \(0\); that is, \(x = 0\) and \(y = 0\). Hence \(v = (0,0,0,0)\). Therefore \(U \cap W = \{0\}\), and 1.46 gives that \(U + W\) is a direct sum.

Combining the last two paragraphs,

\begin{equation*} \mathbf{F}^4 = U \oplus W. \end{equation*}

This choice of \(W\) is far from unique. For instance \(W^{\prime} = \{(0, x, 0, y) : x, y \in \mathbf{F}\}\) works by the symmetric argument, as does \(W^{\prime\prime} = \{(0, x, y, 0) : x, y \in \mathbf{F}\}\): for the latter, \((a,b,c,d) = (a, a, d, d) + (0, b - a, c - d, 0)\), and a vector \((x,x,y,y)\) in \(W^{\prime\prime}\) must have \(x = 0\) (first coordinate) and \(y = 0\) (fourth coordinate).

Take

\begin{equation*} W = \{(0, 0, z_1, z_2, z_3) \in \mathbf{F}^5 : z_1, z_2, z_3 \in \mathbf{F}\}. \end{equation*}

First, \(W\) is a subspace of \(\mathbf{F}^5\): it contains \((0,0,0,0,0)\), and

\begin{equation*} (0,0,z_1,z_2,z_3) + (0,0,z_1^{\prime},z_2^{\prime},z_3^{\prime}) = (0,0,z_1 + z_1^{\prime}, z_2 + z_2^{\prime}, z_3 + z_3^{\prime}), \qquad \lambda(0,0,z_1,z_2,z_3) = (0,0,\lambda z_1, \lambda z_2, \lambda z_3), \end{equation*}

so \(W\) is closed under addition and scalar multiplication; now apply 1.34.

(Similarly \(U\) is a subspace: it contains the vector obtained by taking \(x = y = 0\), and writing \(u(x,y) = (x, y, x + y, x - y, 2x)\) we have \(u(x,y) + u(x^{\prime},y^{\prime}) = u(x + x^{\prime}, y + y^{\prime})\) and \(\lambda\, u(x,y) = u(\lambda x, \lambda y)\).)

Next we show \(\mathbf{F}^5 = U + W\). Let \((a_1, a_2, a_3, a_4, a_5) \in \mathbf{F}^5\). Put \(x = a_1\) and \(y = a_2\), so that

\begin{equation*} u = (a_1,\, a_2,\, a_1 + a_2,\, a_1 - a_2,\, 2a_1) \in U. \end{equation*}

Then

\begin{equation*} (a_1, a_2, a_3, a_4, a_5) - u = \bigl(0,\, 0,\, a_3 - a_1 - a_2,\, a_4 - a_1 + a_2,\, a_5 - 2a_1\bigr), \end{equation*}

whose first two coordinates are \(0\), so this difference lies in \(W\). Calling it \(w\), we have \((a_1,a_2,a_3,a_4,a_5) = u + w \in U + W\). Hence \(\mathbf{F}^5 = U + W\).

Finally we show the sum is direct, using 1.46: we verify \(U \cap W = \{0\}\). Suppose \(v \in U \cap W\). Since \(v \in U\), we may write

\begin{equation*} v = (x, y, x + y, x - y, 2x) \end{equation*}

for some \(x, y \in \mathbf{F}\). Since \(v \in W\), its first two coordinates vanish, so \(x = 0\) and \(y = 0\). Then all five coordinates of \(v\) are \(0\), that is, \(v = 0\). Hence \(U \cap W = \{0\}\), and 1.46 shows that \(U + W\) is a direct sum.

Therefore

\begin{equation*} \mathbf{F}^5 = U \oplus W. \end{equation*}

Note that the argument used no division and no property of \(\mathbf{F}\) beyond the field axioms; in particular it is uniform in \(\mathbf{F}\), which in this book denotes \(\mathbf{R}\) or \(\mathbf{C}\) (1.6). The underlying idea is general: \(U\) is the set of vectors whose last three coordinates are determined by the first two, so complementing it with the subspace of vectors whose first two coordinates vanish separates the free parameters from the determined ones.

Take

\begin{equation*} W_1 = \{(0, 0, z, 0, 0) : z \in \mathbb{F}\}, \qquad W_2 = \{(0, 0, 0, z, 0) : z \in \mathbb{F}\}, \qquad W_3 = \{(0, 0, 0, 0, z) : z \in \mathbb{F}\}. \end{equation*}

Each of these is a subspace of \(\mathbb{F}^5\) by 1.34: each contains \(0\) (take \(z = 0\)) and each is closed under addition and scalar multiplication, since adding two such vectors or multiplying one by a scalar only adds or scales the single free coordinate and leaves all other coordinates equal to \(0\). None of them equals \(\{0\}\), since each contains the vector with \(1\) in its free slot and \(0\) elsewhere.

The set \(U\) is also a subspace of \(\mathbb{F}^5\), again by 1.34: taking \(x = y = 0\) shows \(0 \in U\), and

\begin{equation*} (x, y, x + y, x - y, 2x) + (x^{\prime}, y^{\prime}, x^{\prime} + y^{\prime}, x^{\prime} - y^{\prime}, 2x^{\prime}) = (x + x^{\prime}, y + y^{\prime}, (x + x^{\prime}) + (y + y^{\prime}), (x + x^{\prime}) - (y + y^{\prime}), 2(x + x^{\prime})), \end{equation*}

\begin{equation*} \lambda (x, y, x + y, x - y, 2x) = (\lambda x, \lambda y, \lambda x + \lambda y, \lambda x - \lambda y, 2\lambda x), \end{equation*}

so \(U\) is closed under addition and scalar multiplication. (We need this because the results on direct sums quoted below apply to subspaces.)

First we check that the sum is all of \(\mathbb{F}^5\). Let \((a_1, a_2, a_3, a_4, a_5) \in \mathbb{F}^5\). Put \(u = (a_1, a_2, a_1 + a_2, a_1 - a_2, 2a_1)\), which lies in \(U\) (take \(x = a_1\) and \(y = a_2\) in the definition of \(U\)). Then

\begin{equation*} (a_1, a_2, a_3, a_4, a_5) - u = \bigl(0, 0, a_3 - a_1 - a_2, a_4 - a_1 + a_2, a_5 - 2a_1\bigr), \end{equation*}

and this vector is the sum

\begin{equation*} (0, 0, a_3 - a_1 - a_2, 0, 0) + (0, 0, 0, a_4 - a_1 + a_2, 0) + (0, 0, 0, 0, a_5 - 2a_1) \end{equation*}

of a vector in \(W_1\), a vector in \(W_2\), and a vector in \(W_3\). Hence

\begin{equation*} (a_1, a_2, a_3, a_4, a_5) \in U + W_1 + W_2 + W_3, \end{equation*}

so \(\mathbb{F}^5 = U + W_1 + W_2 + W_3\) (the reverse inclusion is automatic, because \(U, W_1, W_2, W_3\) are all subsets of \(\mathbb{F}^5\)).

Next we check that the sum is direct. By 1.45 it suffices to show that the only way to write \(0\) as \(u + w_1 + w_2 + w_3\) with \(u \in U\) and \(w_k \in W_k\) is with all four vectors equal to \(0\). So suppose

\begin{equation*} (0,0,0,0,0) = (x, y, x + y, x - y, 2x) + (0, 0, c, 0, 0) + (0, 0, 0, d, 0) + (0, 0, 0, 0, e) \end{equation*}

for some \(x, y, c, d, e \in \mathbb{F}\). Comparing the first coordinates gives \(x = 0\), and comparing the second coordinates gives \(y = 0\). Therefore the \(U\)-term is \((0,0,0,0,0)\), and now comparing the third, fourth, and fifth coordinates gives \(c = 0\), \(d = 0\), and \(e = 0\). Thus all four summands equal \(0\), and by 1.45 the sum \(U + W_1 + W_2 + W_3\) is a direct sum.

Combining the two paragraphs, \(\mathbb{F}^5 = U \oplus W_1 \oplus W_2 \oplus W_3\).

The statement is false. Following the hint, we produce a counterexample in \(V = \mathbb{F}^2\). Let

\begin{equation*} U = \{(x, 0) : x \in \mathbb{F}\}, \qquad V_1 = \{(0, y) : y \in \mathbb{F}\}, \qquad V_2 = \{(y, y) : y \in \mathbb{F}\}. \end{equation*}

Each of these three sets is a subspace of \(\mathbb{F}^2\): each contains \((0,0)\), and each is visibly closed under addition and under scalar multiplication, since in each case the defining condition is a linear condition on the coordinates.

We claim \(\mathbb{F}^2 = V_1 \oplus U\). For the sum, if \((a, b) \in \mathbb{F}^2\) then

\begin{equation*} (a, b) = (0, b) + (a, 0), \end{equation*}

with \((0,b) \in V_1\) and \((a,0) \in U\); hence \(\mathbb{F}^2 = V_1 + U\). For directness, suppose \((a,b) \in V_1 \cap U\). Membership in \(V_1\) forces \(a = 0\) and membership in \(U\) forces \(b = 0\), so \(V_1 \cap U = \{(0,0)\}\). By 1.46 the sum \(V_1 + U\) is a direct sum. Thus \(\mathbb{F}^2 = V_1 \oplus U\).

We claim also that \(\mathbb{F}^2 = V_2 \oplus U\). For the sum, if \((a, b) \in \mathbb{F}^2\) then

\begin{equation*} (a, b) = (b, b) + (a - b, 0), \end{equation*}

with \((b,b) \in V_2\) and \((a - b, 0) \in U\); hence \(\mathbb{F}^2 = V_2 + U\). For directness, suppose \((a,b) \in V_2 \cap U\). Membership in \(U\) gives \(b = 0\), and membership in \(V_2\) gives \(a = b\); hence \(a = b = 0\) and \(V_2 \cap U = \{(0,0)\}\). By 1.46 the sum \(V_2 + U\) is a direct sum. Thus \(\mathbb{F}^2 = V_2 \oplus U\).

Finally, \(V_1 \neq V_2\), because \((1,1) \in V_2\) but \((1,1) \notin V_1\) (every element of \(V_1\) has first coordinate \(0\)).

So \(V_1\) and \(V_2\) are subspaces of \(V = \mathbb{F}^2\) with \(V = V_1 \oplus U = V_2 \oplus U\) and \(V_1 \neq V_2\), which refutes the conjecture.

Geometrically (thinking of \(\mathbb{F} = \mathbb{R}\)), \(U\) is the horizontal axis and \(V_1, V_2\) are two different lines through the origin, neither of them horizontal; each of them, together with the horizontal axis, spans the plane, and meets the horizontal axis only at the origin. So a subspace has many different complements in general, not a unique one. What is true is a weaker statement: \(V_1\) and \(V_2\) must have the same dimension, since by 2.43 both equal \(\dim V - \dim U\) when \(V\) is finite-dimensional.

Here \(\mathbb{R}^{\mathbb{R}}\) denotes the vector space of all functions from \(\mathbb{R}\) to \(\mathbb{R}\), with addition and scalar multiplication defined pointwise. There are three things to prove: that \(V_{\mathrm{e}}\) and \(V_{\mathrm{o}}\) are subspaces of \(\mathbb{R}^{\mathbb{R}}\), that their sum is all of \(\mathbb{R}^{\mathbb{R}}\), and that the sum is direct.

Step 1: \(V_{\mathrm{e}}\) and \(V_{\mathrm{o}}\) are subspaces. We use 1.34. The zero function \(0\) satisfies \(0(-x) = 0 = 0(x)\) and also \(0(-x) = 0 = -0(x)\) for all \(x \in \mathbb{R}\), so the zero function belongs to both \(V_{\mathrm{e}}\) and \(V_{\mathrm{o}}\). If \(f, g \in V_{\mathrm{e}}\) and \(x \in \mathbb{R}\), then

\begin{equation*} (f + g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f + g)(x), \end{equation*}

so \(f + g \in V_{\mathrm{e}}\); and if \(\lambda \in \mathbb{R}\) then

\begin{equation*} (\lambda f)(-x) = \lambda f(-x) = \lambda f(x) = (\lambda f)(x), \end{equation*}

so \(\lambda f \in V_{\mathrm{e}}\). Hence \(V_{\mathrm{e}}\) is a subspace of \(\mathbb{R}^{\mathbb{R}}\). The same computations with the sign flipped work for \(V_{\mathrm{o}}\): if \(f, g \in V_{\mathrm{o}}\) and \(\lambda \in \mathbb{R}\), then for all \(x \in \mathbb{R}\),

\begin{equation*} \begin{aligned} (f + g)(-x) &= f(-x) + g(-x) = -f(x) - g(x) = -(f + g)(x), \\ (\lambda f)(-x) &= \lambda f(-x) = -\lambda f(x) = -(\lambda f)(x), \end{aligned} \end{equation*}

so \(V_{\mathrm{o}}\) is a subspace of \(\mathbb{R}^{\mathbb{R}}\) as well.

Step 2: \(\mathbb{R}^{\mathbb{R}} = V_{\mathrm{e}} + V_{\mathrm{o}}\). Let \(f \in \mathbb{R}^{\mathbb{R}}\). Define \(g, h \colon \mathbb{R} \to \mathbb{R}\) by

\begin{equation*} g(x) = \frac{f(x) + f(-x)}{2}, \qquad h(x) = \frac{f(x) - f(-x)}{2} \end{equation*}

for \(x \in \mathbb{R}\) (this makes sense because \(2\) is invertible in \(\mathbb{R}\)). Then for every \(x \in \mathbb{R}\),

\begin{equation*} g(-x) = \frac{f(-x) + f(x)}{2} = g(x), \end{equation*}

so \(g \in V_{\mathrm{e}}\), and

\begin{equation*} h(-x) = \frac{f(-x) - f(x)}{2} = -\,\frac{f(x) - f(-x)}{2} = -h(x), \end{equation*}

so \(h \in V_{\mathrm{o}}\). Moreover, for every \(x \in \mathbb{R}\),

\begin{equation*} g(x) + h(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2} = f(x), \end{equation*}

so \(f = g + h \in V_{\mathrm{e}} + V_{\mathrm{o}}\). Since \(f\) was arbitrary, \(\mathbb{R}^{\mathbb{R}} = V_{\mathrm{e}} + V_{\mathrm{o}}\) (the inclusion \(V_{\mathrm{e}} + V_{\mathrm{o}} \subseteq \mathbb{R}^{\mathbb{R}}\) is clear, as both are subspaces of \(\mathbb{R}^{\mathbb{R}}\)).

Step 3: the sum is direct. By 1.46 it suffices to show \(V_{\mathrm{e}} \cap V_{\mathrm{o}} = \{0\}\). Suppose \(f \in V_{\mathrm{e}} \cap V_{\mathrm{o}}\) and let \(x \in \mathbb{R}\). Because \(f\) is even, \(f(-x) = f(x)\); because \(f\) is odd, \(f(-x) = -f(x)\). Hence \(f(x) = -f(x)\), so \(2f(x) = 0\) and therefore \(f(x) = 0\). As \(x \in \mathbb{R}\) was arbitrary, \(f\) is the zero function. Thus \(V_{\mathrm{e}} \cap V_{\mathrm{o}} = \{0\}\), and by 1.46 the sum \(V_{\mathrm{e}} + V_{\mathrm{o}}\) is a direct sum.

Combining the three steps, \(\mathbb{R}^{\mathbb{R}} = V_{\mathrm{e}} \oplus V_{\mathrm{o}}\).

Write \(U = \{(x,y,z) \in \mathbf{F}^3 : x+y+z = 0\}\), and consider the list

\begin{equation*} (1,0,-1),\quad (0,1,-1),\quad (1,-1,0),\quad (1,1,-2). \end{equation*}

These four vectors are pairwise distinct (compare first coordinates to separate the second vector from the others, and second coordinates to separate the remaining three). We claim that their span equals \(U\).

First, \(\operatorname{span}\bigl((1,0,-1),(0,1,-1),(1,-1,0),(1,1,-2)\bigr) \subseteq U\). Indeed, \(U\) is a subspace of \(\mathbf{F}^3\): it contains \((0,0,0)\), and if \((x,y,z),(x^{\prime},y^{\prime},z^{\prime}) \in U\) and \(\lambda \in \mathbf{F}\) then

\begin{equation*} (x + x^{\prime}) + (y + y^{\prime}) + (z + z^{\prime}) = (x+y+z) + (x^{\prime}+y^{\prime}+z^{\prime}) = 0 \end{equation*}

and \(\lambda x + \lambda y + \lambda z = \lambda(x+y+z) = 0\), so \(U\) is closed under addition and scalar multiplication. Each of the four listed vectors lies in \(U\), because the coordinates of each sum to \(0\):

\begin{equation*} 1 + 0 + (-1) = 0,\qquad 0 + 1 + (-1) = 0,\qquad 1 + (-1) + 0 = 0,\qquad 1 + 1 + (-2) = 0. \end{equation*}

By 2.6 the span of a list is the smallest subspace containing all the vectors in the list, so the span of our list is contained in \(U\).

Conversely, \(U \subseteq \operatorname{span}\bigl((1,0,-1),(0,1,-1),(1,-1,0),(1,1,-2)\bigr)\). Suppose \((x,y,z) \in U\). Then \(z = -x-y\), so

\begin{equation*} (x,y,z) = (x,\,y,\,-x-y) = x(1,0,-1) + y(0,1,-1) + 0\,(1,-1,0) + 0\,(1,1,-2), \end{equation*}

which exhibits \((x,y,z)\) as a linear combination of the four vectors in the list.

The two inclusions give

\begin{equation*} \operatorname{span}\bigl((1,0,-1),(0,1,-1),(1,-1,0),(1,1,-2)\bigr) = U, \end{equation*}

as required. (Of course the answer is far from unique: any four distinct vectors of \(U\) that include two vectors spanning the plane \(U\) will do. Note also that a list of four vectors spanning this \(2\)-dimensional set must be linearly dependent, which is why the last two entries were free to be redundant.)

The statement is true; we prove it.

Suppose \(v_1, v_2, v_3, v_4\) spans \(V\), and write

\begin{equation*} W = \operatorname{span}(v_1 - v_2,\, v_2 - v_3,\, v_3 - v_4,\, v_4). \end{equation*}

By 2.6, \(W\) is a subspace of \(V\). We show that \(v_1, v_2, v_3, v_4\) all lie in \(W\).

Working from the last vector backwards:

\begin{equation*} \begin{aligned} v_4 &= v_4, \\ v_3 &= (v_3 - v_4) + v_4, \\ v_2 &= (v_2 - v_3) + (v_3 - v_4) + v_4, \\ v_1 &= (v_1 - v_2) + (v_2 - v_3) + (v_3 - v_4) + v_4. \end{aligned} \end{equation*}

Each right-hand side is a linear combination (with all coefficients equal to \(1\) or \(0\)) of the vectors \(v_1 - v_2,\ v_2 - v_3,\ v_3 - v_4,\ v_4\), so \(v_1, v_2, v_3, v_4 \in W\).

Now \(W\) is a subspace of \(V\) containing \(v_1, v_2, v_3, v_4\). By 2.6, \(\operatorname{span}(v_1, v_2, v_3, v_4)\) is the smallest such subspace, so

\begin{equation*} V = \operatorname{span}(v_1, v_2, v_3, v_4) \subseteq W. \end{equation*}

Since \(W \subseteq V\) as well, we conclude \(W = V\); that is, the list \(v_1 - v_2,\ v_2 - v_3,\ v_3 - v_4,\ v_4\) spans \(V\).

(Alternatively, one can verify the inclusion directly: given \(u \in V\), write \(u = a_1 v_1 + a_2 v_2 + a_3 v_3 + a_4 v_4\) and substitute the displayed expressions for \(v_1, v_2, v_3, v_4\) to obtain

\begin{equation*} u = a_1 (v_1 - v_2) + (a_1 + a_2)(v_2 - v_3) + (a_1 + a_2 + a_3)(v_3 - v_4) + (a_1 + a_2 + a_3 + a_4) v_4, \end{equation*}

which displays \(u\) as a linear combination of the new list.)

Write \(A = \operatorname{span}(v_1, \dots, v_m)\) and \(B = \operatorname{span}(w_1, \dots, w_m)\). By 2.6 both \(A\) and \(B\) are subspaces of \(V\), and each is the smallest subspace containing the vectors of its defining list. So it suffices to show that every \(w_k\) lies in \(A\) and that every \(v_k\) lies in \(B\).

Each \(w_k\) lies in \(A\). Indeed, by definition

\begin{equation*} w_k = v_1 + \cdots + v_k = 1\,v_1 + \cdots + 1\,v_k + 0\,v_{k+1} + \cdots + 0\,v_m, \end{equation*}

which is a linear combination of \(v_1, \dots, v_m\). Hence \(w_k \in A\) for each \(k \in \{1,\dots,m\}\). Since \(A\) is a subspace containing \(w_1, \dots, w_m\), the minimality in 2.6 gives \(B \subseteq A\).

Each \(v_k\) lies in \(B\). For \(k = 1\) we have \(v_1 = w_1 \in B\). For \(k \in \{2, \dots, m\}\),

\begin{equation*} w_k - w_{k-1} = (v_1 + \cdots + v_k) - (v_1 + \cdots + v_{k-1}) = v_k, \end{equation*}

so \(v_k = w_k - w_{k-1}\) is a linear combination of \(w_1, \dots, w_m\) (take coefficient \(1\) on \(w_k\), coefficient \(-1\) on \(w_{k-1}\), and \(0\) elsewhere). Hence \(v_k \in B\) for each \(k\). Since \(B\) is a subspace containing \(v_1, \dots, v_m\), minimality in 2.6 gives \(A \subseteq B\).

Combining the two inclusions, \(A = B\), that is,

\begin{equation*} \operatorname{span}(v_1, \dots, v_m) = \operatorname{span}(w_1, \dots, w_m). \end{equation*}

(If one prefers an explicit computation rather than an appeal to 2.6: given scalars \(a_1, \dots, a_m\), the identity \(v_k = w_k - w_{k-1}\) with the convention \(w_0 = 0\) yields

\begin{equation*} \sum_{k=1}^{m} a_k v_k = \sum_{k=1}^{m} a_k (w_k - w_{k-1}) = \sum_{k=1}^{m-1} (a_k - a_{k+1}) w_k + a_m w_m, \end{equation*}

after summation by parts, and conversely \(\sum_{k=1}^m b_k w_k = \sum_{k=1}^m (b_k + b_{k+1} + \cdots + b_m) v_k\). Each display converts a linear combination of one list into a linear combination of the other.)

Throughout, \(V\) is a vector space over \(\mathbf{F}\). Recall the definition 2.15: a list \(v_1, \dots, v_m\) is linearly independent if the only choice of \(a_1, \dots, a_m \in \mathbf{F}\) with \(a_1 v_1 + \cdots + a_m v_m = 0\) is \(a_1 = \cdots = a_m = 0\).

(a) Let \(v \in V\) and consider the list of length one consisting of \(v\).

Suppose first that \(v \neq 0\). Let \(a \in \mathbf{F}\) satisfy \(a v = 0\). If \(a \neq 0\), then \(a\) is invertible in the field \(\mathbf{F}\), and

\begin{equation*} v = 1\,v = \bigl(a^{-1} a\bigr) v = a^{-1} (a v) = a^{-1}\, 0 = 0 \end{equation*}

by 1.31, contradicting \(v \neq 0\). Hence \(a = 0\), and the list \(v\) is linearly independent.

Conversely, suppose \(v = 0\). Then \(1 \cdot v = 1 \cdot 0 = 0\) by 1.31, and the scalar \(1\) is not \(0\) (in a field \(1 \neq 0\)). So there is a choice of scalar that is not \(0\) producing the zero vector, and the list \(v\) is linearly dependent (2.17).

Thus the list \(v\) is linearly independent if and only if \(v \neq 0\).

(b) Let \(v, w \in V\) and consider the list \(v, w\).

Suppose the list \(v, w\) is linearly dependent. By 2.17 there exist \(a, b \in \mathbf{F}\), not both \(0\), with

\begin{equation*} a v + b w = 0. \end{equation*}

If \(a \neq 0\), then multiplying by \(a^{-1}\) gives \(v + a^{-1} b\, w = 0\), hence

\begin{equation*} v = -\bigl(a^{-1} b\bigr) w, \end{equation*}

so \(v\) is a scalar multiple of \(w\). If instead \(a = 0\), then \(b \neq 0\), and \(b w = 0\) forces \(w = b^{-1}(bw) = b^{-1} 0 = 0\); then \(w = 0 = 0\,v\) by 1.30, so \(w\) is a scalar multiple of \(v\). In either case one of the two vectors is a scalar multiple of the other.

Conversely, suppose one of the two vectors is a scalar multiple of the other. If \(v = \lambda w\) for some \(\lambda \in \mathbf{F}\), then

\begin{equation*} 1 \cdot v + (-\lambda) w = v - \lambda w = 0, \end{equation*}

and the coefficients \(1, -\lambda\) are not both \(0\) (since \(1 \neq 0\)); hence the list \(v, w\) is linearly dependent. Symmetrically, if \(w = \mu v\) for some \(\mu \in \mathbf{F}\), then \((-\mu) v + 1 \cdot w = 0\) with coefficients not both \(0\), so again the list is linearly dependent.

Taking the contrapositive of both implications: the list \(v, w\) is linearly independent if and only if neither of \(v, w\) is a scalar multiple of the other.

(These are exactly the assertions recorded without proof in 2.16(c) and 2.16(d). One direction of part (a) can also be read off the linear dependence lemma 2.19: if the list \(v\) of length one is linearly dependent, then 2.19 supplies \(k = 1\) with \(v \in \operatorname{span}( ) = \{0\}\), so \(v = 0\). The converse direction still has to be checked by hand, as above.)

The answer is \(t = 2\).

Verification. With \(t = 2\) we exhibit an explicit nontrivial linear combination equal to \(0\):

\begin{equation*} (-3)(3,1,4) + 2(2,-3,5) + 1\,(5,9,2) = (-9 + 4 + 5,\ -3 - 6 + 9,\ -12 + 10 + 2) = (0,0,0). \end{equation*}

Since the scalars \(-3, 2, 1\) are not all \(0\), the list \((3,1,4), (2,-3,5), (5,9,2)\) is linearly dependent, i.e. not linearly independent (2.17). This settles the exercise.

How the value was found. Suppose \(a,b,c \in \mathbf{R}\) satisfy

\begin{equation*} a(3,1,4) + b(2,-3,5) + c(5,9,t) = (0,0,0), \end{equation*}

which is the system

\begin{equation*} \begin{aligned} 3a + 2b + 5c &= 0, \\ a - 3b + 9c &= 0, \\ 4a + 5b + tc &= 0. \end{aligned} \end{equation*}

The list fails to be linearly independent exactly when this system has a solution with \((a,b,c) \neq (0,0,0)\).

First note that \((3,1,4), (2,-3,5)\) is linearly independent. Both vectors are nonzero, so a scalar relating one to the other would be nonzero and hence invertible; it is enough to rule out \((2,-3,5) = \lambda(3,1,4)\). Comparing second coordinates forces \(\lambda = -3\), but then the first coordinate would be \(3\lambda = -9 \neq 2\). So neither vector is a scalar multiple of the other, and Exercise 2A.4(b) (recorded in the book as 2.16(d)) shows that the list of length two is linearly independent. Consequently, in any nontrivial solution we must have \(c \neq 0\); otherwise \(a(3,1,4) + b(2,-3,5) = 0\) would force \(a = b = 0\) too. So we may scale to \(c = 1\) and ask when \((5,9,t)\) is a linear combination of the first two vectors.

Solving the first two equations with \(c = 1\): from \(a = 3b - 9\) (the second equation) substituted into the first,

\begin{equation*} 3(3b - 9) + 2b + 5 = 11b - 22 = 0, \end{equation*}

so \(b = 2\) and \(a = -3\). The third equation then reads \(4(-3) + 5(2) + t = t - 2 = 0\), i.e. \(t = 2\). Thus \(t = 2\) is in fact the only value of \(t\) for which the list fails to be linearly independent; for every other \(t\) the list is linearly independent.

(Equivalently, the \(3 \times 3\) determinant with these three vectors as rows equals

\begin{equation*} 3\bigl((-3)t - 45\bigr) - 1\bigl(2t - 25\bigr) + 4\bigl(18 + 15\bigr) = -11t + 22, \end{equation*}

which vanishes precisely when \(t = 2\). Determinants are not needed for the solution above, and are only introduced in Chapter 9 of the book.)

We prove both implications.

Suppose \(c = 8\). Then

\begin{equation*} 2(2,3,1) + 3(1,-1,2) + (-1)(7,3,8) = (4 + 3 - 7,\ 6 - 3 - 3,\ 2 + 6 - 8) = (0,0,0), \end{equation*}

and the scalars \(2, 3, -1\) are not all \(0\). By 2.17 the list \((2,3,1), (1,-1,2), (7,3,8)\) is linearly dependent. (This is exactly the computation displayed in 2.18.)

Conversely, suppose the list \((2,3,1), (1,-1,2), (7,3,c)\) is linearly dependent. By 2.17 there exist \(a, b, d \in \mathbf{F}\), not all \(0\), with

\begin{equation*} a(2,3,1) + b(1,-1,2) + d(7,3,c) = (0,0,0). \end{equation*}

We first observe that the list \((2,3,1), (1,-1,2)\) is linearly independent. Both vectors are nonzero, so if either were a scalar multiple of the other, the scalar would be nonzero and could be inverted; it therefore suffices to rule out \((1,-1,2) = \lambda(2,3,1)\) for \(\lambda \in \mathbf{F}\). Comparing third coordinates gives \(\lambda = 2\), while comparing first coordinates gives \(2\lambda = 1\); these are incompatible, since \(\lambda = 2\) yields \(2\lambda = 4 \neq 1\). Hence neither vector is a scalar multiple of the other, and by Exercise 2A.4(b) (recorded in the book as 2.16(d)) the list of length two is linearly independent.

Now \(d \neq 0\). For if \(d = 0\), the displayed equation becomes \(a(2,3,1) + b(1,-1,2) = (0,0,0)\), and linear independence of \((2,3,1),(1,-1,2)\) forces \(a = b = 0\); then \(a, b, d\) would all be \(0\), contrary to assumption.

Since \(d \neq 0\) we may divide by \(d\) and set \(\alpha = -a/d\) and \(\beta = -b/d\), obtaining

\begin{equation*} (7,3,c) = \alpha (2,3,1) + \beta (1,-1,2), \end{equation*}

that is,

\begin{equation*} 2\alpha + \beta = 7, \qquad 3\alpha - \beta = 3, \qquad \alpha + 2\beta = c. \end{equation*}

Adding the first two equations gives \(5\alpha = 10\), so \(\alpha = 2\), and then \(\beta = 7 - 2\alpha = 3\). Substituting into the third equation,

\begin{equation*} c = \alpha + 2\beta = 2 + 6 = 8. \end{equation*}

Therefore the list \((2,3,1), (1,-1,2), (7,3,c)\) is linearly dependent in \(\mathbf{F}^3\) if and only if \(c = 8\), verifying the claim asserted in 2.18.

The point of the exercise is that linear independence depends on the scalar field, not just on the vectors: the same list of two complex numbers is independent over \(\mathbf{R}\) and dependent over \(\mathbf{C}\).

(a) Regard \(\mathbf{C}\) as a vector space over \(\mathbf{R}\), so the allowed scalars are real numbers. Suppose \(a, b \in \mathbf{R}\) satisfy

\begin{equation*} a(1+i) + b(1-i) = 0. \end{equation*}

Expanding and collecting real and imaginary parts,

\begin{equation*} (a + b) + (a - b)i = 0. \end{equation*}

Since \(a + b\) and \(a - b\) are real numbers, and a complex number is \(0\) if and only if its real and imaginary parts are both \(0\), we get the system

\begin{equation*} a + b = 0, \qquad a - b = 0. \end{equation*}

Adding these gives \(2a = 0\), so \(a = 0\); subtracting gives \(2b = 0\), so \(b = 0\). Thus the only real scalars producing \(0\) are \(a = b = 0\), and by 2.15 the list \(1+i,\ 1-i\) is linearly independent in \(\mathbf{C}\) viewed as a vector space over \(\mathbf{R}\).

(b) Now regard \(\mathbf{C}\) as a vector space over \(\mathbf{C}\), so complex scalars are allowed. Observe that

\begin{equation*} (-i)(1+i) = -i - i^2 = -i + 1 = 1 - i, \end{equation*}

so \(1 - i\) is a scalar multiple of \(1 + i\), with the scalar \(-i \in \mathbf{C}\). By Exercise 2A.4(b) (recorded in the book as 2.16(d)) the list \(1+i,\ 1-i\) is therefore linearly dependent.

Explicitly, taking the scalars \(-i\) and \(1\), which are not both \(0\), we have

\begin{equation*} (-i)(1+i) + (-1)(1-i) = (1 - i) - (1 - i) = 0, \end{equation*}

a nontrivial linear combination equal to \(0\), so the list is linearly dependent by 2.17.

The contrast is instructive: the scalar \(-i\) that witnesses the dependence in (b) is not available in (a), and indeed \(1+i,\ 1-i\) is a spanning list (in fact a basis) of the two-dimensional real vector space \(\mathbf{C}\), whereas over \(\mathbf{C}\) the space \(\mathbf{C}\) is one-dimensional, so by 2.22 no list of length two in it can be linearly independent.

Suppose \(a_1, a_2, a_3, a_4 \in \mathbb{F}\) satisfy

\begin{equation*} a_1(v_1 - v_2) + a_2(v_2 - v_3) + a_3(v_3 - v_4) + a_4 v_4 = 0 . \end{equation*}

Expanding and collecting the coefficient of each \(v_j\) gives

\begin{equation*} a_1 v_1 + (a_2 - a_1) v_2 + (a_3 - a_2) v_3 + (a_4 - a_3) v_4 = 0 . \end{equation*}

Because \(v_1, v_2, v_3, v_4\) is linearly independent, every coefficient in the equation above must equal \(0\). Hence

\begin{equation*} a_1 = 0, \qquad a_2 - a_1 = 0, \qquad a_3 - a_2 = 0, \qquad a_4 - a_3 = 0 . \end{equation*}

Reading these in order, \(a_1 = 0\), then \(a_2 = a_1 = 0\), then \(a_3 = a_2 = 0\), then \(a_4 = a_3 = 0\). Thus the only choice of \(a_1, a_2, a_3, a_4\) making the original linear combination equal \(0\) is \(a_1 = a_2 = a_3 = a_4 = 0\), which is exactly the statement that

\begin{equation*} v_1 - v_2,\quad v_2 - v_3,\quad v_3 - v_4,\quad v_4 \end{equation*}

is linearly independent (see the definition of linear independence, 2.15).

The statement is true; here is a proof. (Implicitly \(m \geq 2\), so that the list makes sense.)

Suppose \(a_1, a_2, \ldots, a_m \in \mathbb{F}\) satisfy

\begin{equation*} a_1(5v_1 - 4v_2) + a_2 v_2 + a_3 v_3 + \cdots + a_m v_m = 0 . \end{equation*}

Collecting the coefficient of each \(v_j\) turns this into

\begin{equation*} (5a_1) v_1 + (a_2 - 4a_1) v_2 + a_3 v_3 + \cdots + a_m v_m = 0 . \end{equation*}

Because \(v_1, \ldots, v_m\) is linearly independent, all of these coefficients equal \(0\) (2.15). Thus

\begin{equation*} 5a_1 = 0, \qquad a_2 - 4a_1 = 0, \qquad a_3 = \cdots = a_m = 0 . \end{equation*}

Here \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), so \(5 \neq 0\) in \(\mathbb{F}\) and hence \(5a_1 = 0\) forces \(a_1 = 0\). Then \(a_2 = 4a_1 = 0\). Therefore \(a_1 = a_2 = \cdots = a_m = 0\), and the list

\begin{equation*} 5v_1 - 4v_2,\quad v_2,\quad v_3,\quad \ldots,\quad v_m \end{equation*}

is linearly independent.

Remark. Nothing is special about the numbers \(5\) and \(-4\) beyond \(5 \neq 0\): the same computation shows that if \(b, c \in \mathbb{F}\) with \(b \neq 0\), then \(bv_1 + cv_2, v_2, v_3, \ldots, v_m\) is linearly independent. If instead the first coefficient were \(0\), the claim would fail, since the list \(-4v_2, v_2, v_3, \ldots, v_m\) is linearly dependent.

The statement is true; here is a proof.

Suppose \(a_1, \ldots, a_m \in \mathbb{F}\) satisfy

\begin{equation*} a_1(\lambda v_1) + a_2(\lambda v_2) + \cdots + a_m(\lambda v_m) = 0 . \end{equation*}

By the commutativity and associativity of scalar multiplication, the left side equals

\begin{equation*} (a_1 \lambda) v_1 + (a_2 \lambda) v_2 + \cdots + (a_m \lambda) v_m . \end{equation*}

Thus \((a_1\lambda) v_1 + \cdots + (a_m \lambda) v_m = 0\). Because \(v_1, \ldots, v_m\) is linearly independent, the definition 2.15 forces

\begin{equation*} a_1 \lambda = a_2 \lambda = \cdots = a_m \lambda = 0 . \end{equation*}

Since \(\lambda \neq 0\), it has a multiplicative inverse in \(\mathbb{F}\); multiplying each equation by \(\lambda^{-1}\) gives \(a_k = 0\) for every \(k \in \{1, \ldots, m\}\).

Hence the only scalars \(a_1, \ldots, a_m\) with \(a_1(\lambda v_1) + \cdots + a_m(\lambda v_m) = 0\) are \(a_1 = \cdots = a_m = 0\), so \(\lambda v_1, \ldots, \lambda v_m\) is linearly independent.

Remark. The hypothesis \(\lambda \neq 0\) cannot be dropped when \(m \geq 1\): if \(\lambda = 0\), then \(\lambda v_1, \ldots, \lambda v_m\) is the list \(0, \ldots, 0\), which is linearly dependent because every list containing the vector \(0\) is linearly dependent (2.18).

The statement is false. Here is a counterexample.

Take \(V = \mathbb{R}^2\) and \(m = 2\), with

\begin{equation*} v_1 = (1,0), \quad v_2 = (0,1), \qquad w_1 = (-1,0), \quad w_2 = (0,-1). \end{equation*}

The list \(v_1, v_2\) is linearly independent: if \(a_1(1,0) + a_2(0,1) = (0,0)\), then \((a_1, a_2) = (0,0)\), so \(a_1 = a_2 = 0\). The list \(w_1, w_2\) is linearly independent by the same computation, since \(b_1(-1,0) + b_2(0,-1) = (-b_1, -b_2)\), which equals \((0,0)\) only when \(b_1 = b_2 = 0\).

However

\begin{equation*} v_1 + w_1 = (0,0), \qquad v_2 + w_2 = (0,0), \end{equation*}

so the list \(v_1 + w_1, v_2 + w_2\) is the list \((0,0), (0,0)\). This list is linearly dependent, because

\begin{equation*} 1 \cdot (0,0) + 0 \cdot (0,0) = (0,0) \end{equation*}

exhibits a linear combination equal to \(0\) with scalars not all \(0\) (indeed, every list containing the vector \(0\) is linearly dependent, by 2.18).

Thus linear independence of \(v_1, \ldots, v_m\) and of \(w_1, \ldots, w_m\) does not imply linear independence of \(v_1 + w_1, \ldots, v_m + w_m\).

Remark. The same idea works whenever \(V\) actually contains a linearly independent list \(v_1, \ldots, v_m\) of the given length \(m \geq 1\) (for instance in \(\mathbb{F}^m\), or in any \(V\) with \(\dim V \geq m\)): set \(w_k = -v_k\). Then \(w_1, \ldots, w_m\) is linearly independent (by Exercise 2A.10 with \(\lambda = -1\)), while each \(v_k + w_k = 0\), so the summed list is linearly dependent. In particular no hypothesis on \(V\) or \(m\) can rescue the statement, apart from the degenerate cases \(m = 0\) and \(V = \{0\}\), where there is no linearly independent list of length \(m \geq 1\) to begin with.

Suppose \(v_1 + w, \ldots, v_m + w\) is linearly dependent. By the definition of linear dependence (2.17), there exist \(a_1, \ldots, a_m \in \mathbb{F}\), not all \(0\), such that

\begin{equation*} a_1(v_1 + w) + a_2(v_2 + w) + \cdots + a_m(v_m + w) = 0 . \end{equation*}

Expanding and separating the terms involving \(w\) gives

\begin{equation*} a_1 v_1 + \cdots + a_m v_m + (a_1 + \cdots + a_m) w = 0 . \end{equation*}

Put \(s = a_1 + \cdots + a_m\), so the equation above reads

\begin{equation*} a_1 v_1 + \cdots + a_m v_m + s w = 0 . \end{equation*}

First suppose \(s = 0\). Then the equation becomes \(a_1 v_1 + \cdots + a_m v_m = 0\) with \(a_1, \ldots, a_m\) not all \(0\), contradicting the linear independence of \(v_1, \ldots, v_m\) (2.15). Hence \(s \neq 0\).

Because \(s \neq 0\), we may divide by \(s\): solving the displayed equation for \(w\) gives

\begin{equation*} w = -\frac{a_1}{s} v_1 - \frac{a_2}{s} v_2 - \cdots - \frac{a_m}{s} v_m . \end{equation*}

The right side is a linear combination of \(v_1, \ldots, v_m\) with coefficients in \(\mathbb{F}\). Therefore \(w \in \operatorname{span}(v_1, \ldots, v_m)\), as desired.

We prove the two implications separately.

First suppose \(v_1, \ldots, v_m, w\) is linearly independent; we show \(w \notin \operatorname{span}(v_1, \ldots, v_m)\). Suppose instead that \(w \in \operatorname{span}(v_1, \ldots, v_m)\). Then there exist \(c_1, \ldots, c_m \in \mathbb{F}\) with

\begin{equation*} w = c_1 v_1 + \cdots + c_m v_m . \end{equation*}

Moving \(w\) to the other side gives

\begin{equation*} c_1 v_1 + \cdots + c_m v_m + (-1) w = 0 . \end{equation*}

This is a linear combination of \(v_1, \ldots, v_m, w\) equal to \(0\) in which the last scalar is \(-1 \neq 0\), so the scalars are not all \(0\). Hence \(v_1, \ldots, v_m, w\) is linearly dependent (2.17), contradicting our assumption. Therefore \(w \notin \operatorname{span}(v_1, \ldots, v_m)\).

Conversely, suppose \(w \notin \operatorname{span}(v_1, \ldots, v_m)\); we show \(v_1, \ldots, v_m, w\) is linearly independent. Let \(a_1, \ldots, a_m, b \in \mathbb{F}\) satisfy

\begin{equation*} a_1 v_1 + \cdots + a_m v_m + b w = 0 . \end{equation*}

If \(b \neq 0\), then \(b\) is invertible in \(\mathbb{F}\) and we may solve for \(w\):

\begin{equation*} w = -\frac{a_1}{b} v_1 - \cdots - \frac{a_m}{b} v_m , \end{equation*}

which exhibits \(w\) as a linear combination of \(v_1, \ldots, v_m\) and hence puts \(w\) in \(\operatorname{span}(v_1, \ldots, v_m)\), contrary to hypothesis. Thus \(b = 0\), and the displayed equation becomes

\begin{equation*} a_1 v_1 + \cdots + a_m v_m = 0 . \end{equation*}

Because \(v_1, \ldots, v_m\) is linearly independent, this forces \(a_1 = \cdots = a_m = 0\) (2.15). Hence \(a_1 = \cdots = a_m = b = 0\), which is exactly what is required for \(v_1, \ldots, v_m, w\) to be linearly independent.

Combining the two implications gives the stated equivalence.

Remark. The case \(m = 0\) is included: then \(v_1, \ldots, v_m\) is the empty list, \(\operatorname{span}(\,) = \{0\}\), and the equivalence says that the one-term list \(w\) is linearly independent if and only if \(w \neq 0\), which agrees with 2.16(c).

Note first the two ways of passing between the lists:

\begin{equation*} w_k = v_1 + \cdots + v_k \quad (1 \leq k \leq m), \qquad v_1 = w_1, \qquad v_k = w_k - w_{k-1} \quad (2 \leq k \leq m). \end{equation*}

The second set of formulas follows immediately from the first, since \(w_k - w_{k-1} = (v_1 + \cdots + v_k) - (v_1 + \cdots + v_{k-1}) = v_k\).

Suppose first that \(v_1, \ldots, v_m\) is linearly independent, and let \(a_1, \ldots, a_m \in \mathbb{F}\) satisfy

\begin{equation*} a_1 w_1 + a_2 w_2 + \cdots + a_m w_m = 0 . \end{equation*}

Substituting \(w_k = v_1 + \cdots + v_k\) and collecting the terms involving each \(v_j\), we see that \(v_j\) occurs in \(w_k\) exactly when \(k \geq j\), so the left side equals

\begin{equation*} \sum_{j=1}^{m} \Bigl( \sum_{k=j}^{m} a_k \Bigr) v_j . \end{equation*}

Since \(v_1, \ldots, v_m\) is linearly independent, every coefficient vanishes (2.15):

\begin{equation*} \sum_{k=j}^{m} a_k = 0 \qquad \text{for each } j \in \{1, \ldots, m\}. \end{equation*}

Taking \(j = m\) gives \(a_m = 0\). For \(j < m\), subtracting the equation for \(j+1\) from the equation for \(j\) gives

\begin{equation*} a_j = \sum_{k=j}^{m} a_k - \sum_{k=j+1}^{m} a_k = 0 - 0 = 0 . \end{equation*}

Hence \(a_1 = \cdots = a_m = 0\), so \(w_1, \ldots, w_m\) is linearly independent.

Conversely, suppose \(w_1, \ldots, w_m\) is linearly independent, and let \(b_1, \ldots, b_m \in \mathbb{F}\) satisfy

\begin{equation*} b_1 v_1 + b_2 v_2 + \cdots + b_m v_m = 0 . \end{equation*}

Substituting \(v_1 = w_1\) and \(v_k = w_k - w_{k-1}\) for \(k \geq 2\), the left side equals

\begin{equation*} \sum_{k=1}^{m} b_k w_k - \sum_{k=2}^{m} b_k w_{k-1} = \sum_{k=1}^{m} b_k w_k - \sum_{k=1}^{m-1} b_{k+1} w_k = \sum_{k=1}^{m-1} (b_k - b_{k+1}) w_k + b_m w_m . \end{equation*}

Since \(w_1, \ldots, w_m\) is linearly independent, all these coefficients vanish:

\begin{equation*} b_k - b_{k+1} = 0 \quad \text{for } k \in \{1, \ldots, m-1\}, \qquad b_m = 0 . \end{equation*}

The first family of equations says \(b_1 = b_2 = \cdots = b_m\), and the last says \(b_m = 0\); hence \(b_1 = \cdots = b_m = 0\). Thus \(v_1, \ldots, v_m\) is linearly independent.

This proves both implications, so \(v_1, \ldots, v_m\) is linearly independent if and only if \(w_1, \ldots, w_m\) is.

Remark. Exercise 2A.3 shows that these two lists always have the same span. The present exercise is the companion statement for linear independence; together they say that passing from \(v_1, \ldots, v_m\) to its list of partial sums preserves both properties, and hence carries bases to bases.

Recall from the discussion following 2.12 that

\begin{equation*} \mathcal{P}_4(\mathbf{F}) = \operatorname{span}(1, z, z^2, z^3, z^4), \end{equation*}

where \(z^k\) denotes the function \(z \mapsto z^k\) from \(\mathbf{F}\) to \(\mathbf{F}\). Indeed, if \(p \in \mathcal{P}_4(\mathbf{F})\), then \(\deg p \le 4\) by 2.12, so by 2.10 and 2.11 there exist \(a_0, a_1, a_2, a_3, a_4 \in \mathbf{F}\) with \(p(z) = a_0 + a_1 z + a_2 z^2 + a_3 z^3 + a_4 z^4\) for every \(z \in \mathbf{F}\) (take all the coefficients to be \(0\) if \(p = 0\), whose degree is \(-\infty\); otherwise take the coefficients supplied by 2.11 and pad with zeros). This exhibits \(p\) as a linear combination of \(1, z, z^2, z^3, z^4\); conversely every such linear combination is a polynomial of degree at most \(4\).

Thus \(1, z, z^2, z^3, z^4\) is a list of length \(5\) that spans \(\mathcal{P}_4(\mathbf{F})\). In particular \(\mathcal{P}_4(\mathbf{F})\) is finite-dimensional (2.9).

Now 2.22 states that in a finite-dimensional vector space the length of every linearly independent list is less than or equal to the length of every spanning list. Hence every linearly independent list in \(\mathcal{P}_4(\mathbf{F})\) has length at most \(5\).

A list of six polynomials that is linearly independent in \(\mathcal{P}_4(\mathbf{F})\) would have length \(6 > 5\), which is impossible. Therefore no such list exists.

Consider the list \(1, z, z^2, z^3, z^4\) of elements of \(\mathcal{P}_4(\mathbf{F})\), where \(z^k\) denotes the function \(z \mapsto z^k\). This list has length \(5\), and it is linearly independent by 2.16(b): if \(a_0, a_1, a_2, a_3, a_4 \in \mathbf{F}\) satisfy

\begin{equation*} a_0 + a_1 z + a_2 z^2 + a_3 z^3 + a_4 z^4 = 0 \end{equation*}

as an element of \(\mathcal{P}(\mathbf{F})\), meaning that \(a_0 + a_1 z + a_2 z^2 + a_3 z^3 + a_4 z^4 = 0\) for every \(z \in \mathbf{F}\), then \(a_0 = a_1 = a_2 = a_3 = a_4 = 0\) (this is the uniqueness of the coefficients of a polynomial, which follows from 4.8). Although 2.16(b) is stated for \(\mathcal{P}(\mathbf{F})\), the displayed condition defining linear independence involves only the vectors in the list and the scalar field \(\mathbf{F}\), not the ambient vector space; since \(1, z, z^2, z^3, z^4\) all lie in the subspace \(\mathcal{P}_4(\mathbf{F})\), this list is linearly independent in \(\mathcal{P}_4(\mathbf{F})\) as well.

Suppose now that some list \(q_1, q_2, q_3, q_4\) of four polynomials spanned \(\mathcal{P}_4(\mathbf{F})\). Then \(\mathcal{P}_4(\mathbf{F})\) would be a finite-dimensional vector space (2.9) with a spanning list of length \(4\). By 2.22, the length of every linearly independent list in \(\mathcal{P}_4(\mathbf{F})\) would then be at most \(4\). But \(1, z, z^2, z^3, z^4\) is a linearly independent list in \(\mathcal{P}_4(\mathbf{F})\) of length \(5 > 4\), a contradiction.

Hence no list of four polynomials spans \(\mathcal{P}_4(\mathbf{F})\).

We prove the two implications separately.

First suppose there is a sequence \(v_1, v_2, \dots\) of vectors in \(V\) such that \(v_1, \dots, v_m\) is linearly independent for every positive integer \(m\). We must show \(V\) is infinite-dimensional, that is (2.13), that \(V\) is not finite-dimensional. Suppose instead that \(V\) is finite-dimensional. Then by 2.9 there is a list \(w_1, \dots, w_n\) of vectors in \(V\) that spans \(V\). Take \(m = n + 1\). By hypothesis the list \(v_1, \dots, v_{n+1}\) is linearly independent in \(V\), and it has length \(n + 1\). But 2.22 says that in a finite-dimensional vector space the length of every linearly independent list is at most the length of every spanning list, so \(n + 1 \le n\), which is false. This contradiction shows \(V\) is not finite-dimensional; that is, \(V\) is infinite-dimensional.

Conversely, suppose \(V\) is infinite-dimensional. We construct the required sequence recursively, choosing \(v_m\) at step \(m\).

Step \(1\). The list of length \(0\), namely the empty list, has span \(\{0\}\) (by the convention that \(\operatorname{span}( ) = \{0\}\)). If \(V = \{0\}\), then the empty list spans \(V\) and \(V\) would be finite-dimensional, contrary to hypothesis. Hence \(V \ne \{0\}\), so we may choose \(v_1 \in V\) with \(v_1 \ne 0\). By 2.16(c), the list \(v_1\) is linearly independent.

Step \(m + 1\). Suppose \(m \ge 1\) and vectors \(v_1, \dots, v_m \in V\) have been chosen so that \(v_1, \dots, v_m\) is linearly independent. Because \(V\) is infinite-dimensional, no list of vectors in \(V\) spans \(V\); in particular \(\operatorname{span}(v_1, \dots, v_m) \ne V\). Hence we may choose

\begin{equation*} v_{m+1} \in V \setminus \operatorname{span}(v_1, \dots, v_m). \end{equation*}

By Exercise 13 of Section 2A (a linearly independent list stays linearly independent when a vector outside its span is adjoined), the list \(v_1, \dots, v_m, v_{m+1}\) is linearly independent.

For completeness, here is the verification of that step. Suppose \(a_1, \dots, a_{m+1} \in \mathbf{F}\) satisfy \(a_1 v_1 + \cdots + a_m v_m + a_{m+1} v_{m+1} = 0\). If \(a_{m+1} \ne 0\), then

\begin{equation*} v_{m+1} = -\frac{a_1}{a_{m+1}} v_1 - \cdots - \frac{a_m}{a_{m+1}} v_m \in \operatorname{span}(v_1, \dots, v_m), \end{equation*}

contradicting the choice of \(v_{m+1}\). Hence \(a_{m+1} = 0\), and then \(a_1 v_1 + \cdots + a_m v_m = 0\), which forces \(a_1 = \cdots = a_m = 0\) because \(v_1, \dots, v_m\) is linearly independent. Thus all the scalars are \(0\), proving linear independence.

This recursion never halts, so it produces a sequence \(v_1, v_2, \dots\) of vectors in \(V\). By construction \(v_1, \dots, v_m\) is linearly independent for every positive integer \(m\), as desired.

Recall from 1.23 that \(\mathbf{F}^\infty\) is the vector space of all sequences of elements of \(\mathbf{F}\),

\begin{equation*} \mathbf{F}^\infty = \{(x_1, x_2, \dots) : x_k \in \mathbf{F} \text{ for } k = 1, 2, \dots\}, \end{equation*}

with addition and scalar multiplication defined coordinatewise.

For each positive integer \(k\) let \(e_k \in \mathbf{F}^\infty\) be the sequence whose \(k\)-th coordinate is \(1\) and whose other coordinates are all \(0\):

\begin{equation*} e_k = (0, \dots, 0, 1, 0, 0, \dots), \end{equation*}

with the \(1\) in position \(k\).

Claim: for every positive integer \(m\), the list \(e_1, \dots, e_m\) is linearly independent in \(\mathbf{F}^\infty\). To see this, suppose \(a_1, \dots, a_m \in \mathbf{F}\) and

\begin{equation*} a_1 e_1 + \cdots + a_m e_m = 0, \end{equation*}

where \(0\) here denotes the sequence all of whose coordinates are \(0\). Computing the left side coordinatewise gives

\begin{equation*} a_1 e_1 + \cdots + a_m e_m = (a_1, a_2, \dots, a_m, 0, 0, \dots). \end{equation*}

Two sequences are equal exactly when all their corresponding coordinates are equal, so \(a_1 = a_2 = \cdots = a_m = 0\). Thus \(e_1, \dots, e_m\) is linearly independent, proving the claim.

Hence \(e_1, e_2, \dots\) is a sequence of vectors in \(\mathbf{F}^\infty\) such that \(e_1, \dots, e_m\) is linearly independent for every positive integer \(m\). By Exercise 17 of Section 2A, \(\mathbf{F}^\infty\) is infinite-dimensional.

Alternatively, one can argue directly without invoking Exercise 17. Suppose \(\mathbf{F}^\infty\) were finite-dimensional. Then by 2.9 some list \(w_1, \dots, w_n\) spans \(\mathbf{F}^\infty\), and by 2.22 every linearly independent list in \(\mathbf{F}^\infty\) has length at most \(n\). But the claim above provides the linearly independent list \(e_1, \dots, e_{n+1}\), of length \(n + 1 > n\). This contradiction shows that \(\mathbf{F}^\infty\) is not finite-dimensional, that is, \(\mathbf{F}^\infty\) is infinite-dimensional (2.13).

Write \(C[0,1]\) for the set of continuous functions from \([0,1]\) to \(\mathbf{R}\). This is a subspace of the vector space \(\mathbf{R}^{[0,1]}\) of all functions from \([0,1]\) to \(\mathbf{R}\) (see 1.24 and 1.34): the zero function is continuous, and sums and real scalar multiples of continuous functions are continuous. So \(C[0,1]\) is a real vector space, with the usual pointwise operations.

For each nonnegative integer \(k\) define \(f_k \in C[0,1]\) by

\begin{equation*} f_k(x) = x^k \qquad \text{for } x \in [0,1], \end{equation*}

with the convention \(f_0(x) = 1\). Each \(f_k\) is continuous on \([0,1]\).

Claim: for every nonnegative integer \(m\), the list \(f_0, f_1, \dots, f_m\) is linearly independent in \(C[0,1]\). To prove this, suppose \(a_0, a_1, \dots, a_m \in \mathbf{R}\) and

\begin{equation*} a_0 f_0 + a_1 f_1 + \cdots + a_m f_m = 0 \end{equation*}

in \(C[0,1]\), which means

\begin{equation*} a_0 + a_1 x + \cdots + a_m x^m = 0 \qquad \text{for every } x \in [0,1]. \end{equation*}

Let \(p \in \mathcal{P}(\mathbf{R})\) be the polynomial \(p(z) = a_0 + a_1 z + \cdots + a_m z^m\) for \(z \in \mathbf{R}\). Then every point of \([0,1]\) is a zero of \(p\), so \(p\) has infinitely many zeros in \(\mathbf{R}\). If \(p\) were not the zero polynomial, then \(p\) would have some degree \(d\) with \(0 \le d \le m\), and \(p\) would have at most \(d\) zeros in \(\mathbf{R}\): for \(d \ge 1\) this is 4.8, and for \(d = 0\) the polynomial \(p\) is a nonzero constant and so has no zeros at all. Either way \(p\) would have only finitely many zeros, a contradiction. Hence \(p\) is the zero polynomial, and therefore, by the uniqueness of the coefficients of a polynomial (see the discussion preceding 2.11, which rests on 4.8), we get

\begin{equation*} a_0 = a_1 = \cdots = a_m = 0. \end{equation*}

This proves the claim.

Note carefully that the claim really is a statement about \(C[0,1]\), not about \(\mathcal{P}(\mathbf{R})\): two polynomials that agree at every point of \([0,1]\) determine the same element of \(C[0,1]\), and the argument above shows that this forces their coefficients to agree.

Now set \(g_k = f_{k-1}\) for each positive integer \(k\), so that \(g_1, g_2, \dots\) is a sequence in \(C[0,1]\) with \(g_1, \dots, g_m = f_0, \dots, f_{m-1}\) linearly independent for every positive integer \(m\), by the claim. By Exercise 17 of Section 2A, \(C[0,1]\) is infinite-dimensional.

Directly: if \(C[0,1]\) were finite-dimensional, then by 2.9 some list \(w_1, \dots, w_n\) of continuous functions would span \(C[0,1]\), and by 2.22 every linearly independent list in \(C[0,1]\) would have length at most \(n\). But \(f_0, f_1, \dots, f_n\) is linearly independent of length \(n + 1 > n\), a contradiction. Hence \(C[0,1]\) is infinite-dimensional (2.13).

Let

\begin{equation*} U = \{p \in \mathcal{P}_m(\mathbf{F}) : p(2) = 0\}. \end{equation*}

By hypothesis \(p_0, p_1, \dots, p_m\) is a list of \(m + 1\) vectors in \(U\).

First, \(U\) is a subspace of \(\mathcal{P}_m(\mathbf{F})\). Indeed the zero polynomial lies in \(U\); if \(p, q \in U\) then \((p + q)(2) = p(2) + q(2) = 0 + 0 = 0\), so \(p + q \in U\); and if \(p \in U\) and \(\lambda \in \mathbf{F}\) then \((\lambda p)(2) = \lambda p(2) = 0\), so \(\lambda p \in U\). Thus \(U\) is a subspace by 1.34, and in particular \(U\) is itself a vector space over \(\mathbf{F}\).

Next we exhibit a spanning list for \(U\) of length \(m\). For \(j \in \{1, \dots, m\}\) define \(q_j \in \mathcal{P}_m(\mathbf{F})\) by

\begin{equation*} q_j(z) = z^j - 2^j \qquad \text{for } z \in \mathbf{F}. \end{equation*}

Each \(q_j\) has degree \(j \le m\) and satisfies \(q_j(2) = 2^j - 2^j = 0\), so \(q_j \in U\). (When \(m = 0\) this is the empty list.)

Claim: \(U = \operatorname{span}(q_1, \dots, q_m)\).

Since each \(q_j \in U\) and \(U\) is a subspace, every linear combination of \(q_1, \dots, q_m\) lies in \(U\); hence \(\operatorname{span}(q_1, \dots, q_m) \subseteq U\). (For \(m = 0\) this reads \(\operatorname{span}( ) = \{0\} \subseteq U\), which also holds.)

Conversely, let \(p \in U\). By 2.11 and 2.12 there exist \(a_0, a_1, \dots, a_m \in \mathbf{F}\) with

\begin{equation*} p(z) = a_0 + a_1 z + \cdots + a_m z^m \qquad \text{for every } z \in \mathbf{F}. \end{equation*}

Because \(p(2) = 0\), for every \(z \in \mathbf{F}\) we have

\begin{equation*} \begin{aligned} p(z) &= p(z) - p(2) \\ &= (a_0 - a_0) + a_1(z - 2) + a_2(z^2 - 2^2) + \cdots + a_m(z^m - 2^m) \\ &= a_1 q_1(z) + a_2 q_2(z) + \cdots + a_m q_m(z). \end{aligned} \end{equation*}

Thus \(p = a_1 q_1 + \cdots + a_m q_m \in \operatorname{span}(q_1, \dots, q_m)\). (This is the same manipulation recorded in 4.7, inside the proof of 4.6.) When \(m = 0\) the computation degenerates to \(p = p - p(2) = 0\), so \(U = \{0\} = \operatorname{span}( )\), again as claimed.

This proves the claim, so \(U\) is spanned by a list of length \(m\); in particular \(U\) is finite-dimensional (2.9).

Now suppose, for contradiction, that \(p_0, p_1, \dots, p_m\) were linearly independent in \(\mathcal{P}_m(\mathbf{F})\). Linear independence of a list depends only on the vectors themselves and the scalar field, not on which vector space we regard them as sitting in; since \(p_0, \dots, p_m\) all lie in the subspace \(U\), the list \(p_0, \dots, p_m\) would then be a linearly independent list in the vector space \(U\). It has length \(m + 1\), while \(q_1, \dots, q_m\) spans \(U\) and has length \(m\). This contradicts 2.22, which asserts that in the finite-dimensional vector space \(U\) the length of every linearly independent list is at most the length of every spanning list, i.e. \(m + 1 \le m\).

Therefore \(p_0, p_1, \dots, p_m\) is not linearly independent in \(\mathcal{P}_m(\mathbf{F})\).

Claim: the only vector space with exactly one basis is the zero vector space \(\{0\}\) (whose unique basis is the empty list).

First, \(\{0\}\) does have exactly one basis. The empty list is linearly independent, since 2.15 explicitly declares the empty list to be linearly independent, and its span is \(\{0\}\) by the convention in 2.4 that the span of the empty list is \(\{0\}\); hence the empty list is a basis of \(\{0\}\) by 2.26. Any nonempty list of vectors in \(\{0\}\) has the form \(0, \ldots, 0\), which is not linearly independent (take the first coefficient to be \(1\) and the rest \(0\)), so no other list is a basis. Thus \(\{0\}\) has exactly one basis.

Conversely, suppose \(V\) is a vector space with exactly one basis; call it \(v_1, \ldots, v_n\). We must show \(V = \{0\}\), i.e. that \(n = 0\). Suppose instead \(n \geq 1\). Since \(v_1, \ldots, v_n\) is linearly independent, \(v_1 \neq 0\): otherwise \(1 \cdot v_1 + 0 v_2 + \cdots + 0 v_n = 0\) would be a nontrivial vanishing linear combination.

Consider the list

\begin{equation*} 2v_1,\, v_2, \ldots, v_n . \end{equation*}

It spans \(V\), because \(v_1 = \tfrac{1}{2}(2v_1) \in \operatorname{span}(2v_1, v_2, \ldots, v_n)\) and hence this span contains \(\operatorname{span}(v_1, \ldots, v_n) = V\); note \(\tfrac12 \in \mathbf{F}\) since \(\mathbf{F}\) is \(\mathbf{R}\) or \(\mathbf{C}\). It is linearly independent: if \(a_1(2v_1) + a_2 v_2 + \cdots + a_n v_n = 0\), then \((2a_1)v_1 + a_2v_2 + \cdots + a_nv_n = 0\), so linear independence of \(v_1, \ldots, v_n\) gives \(2a_1 = a_2 = \cdots = a_n = 0\), hence \(a_1 = \cdots = a_n = 0\). So \(2v_1, v_2, \ldots, v_n\) is a basis of \(V\) by 2.26.

But this basis differs from \(v_1, \ldots, v_n\), since \(2v_1 - v_1 = v_1 \neq 0\) and so \(2v_1 \neq v_1\). This contradicts the assumption that \(V\) has exactly one basis. Hence \(n = 0\), and therefore \(V = \operatorname{span}() = \{0\}\).

(Note that a vector space with exactly one basis in particular has a basis, which in this book is by definition a finite list; so such a space is automatically finite-dimensional and the argument above applies to it.)

Throughout, a list is a basis exactly when it is linearly independent and spans (2.26).

(a) The standard basis of \(\mathbf{F}^n\) is \(e_1, \ldots, e_n\), where \(e_k\) is the \(n\)-tuple with \(1\) in slot \(k\) and \(0\) elsewhere. For \(a_1, \ldots, a_n \in \mathbf{F}\) we have \(a_1 e_1 + \cdots + a_n e_n = (a_1, \ldots, a_n)\). Hence every \((x_1, \ldots, x_n) \in \mathbf{F}^n\) equals \(x_1e_1 + \cdots + x_ne_n\), so the list spans \(\mathbf{F}^n\); and if \(a_1e_1 + \cdots + a_ne_n = (0,\ldots,0)\) then \((a_1, \ldots, a_n) = (0, \ldots, 0)\), so \(a_1 = \cdots = a_n = 0\) and the list is linearly independent. Thus \(e_1, \ldots, e_n\) is a basis of \(\mathbf{F}^n\).

(b) For \((1,2), (3,5)\) in \(\mathbf{F}^2\): given \((x,y) \in \mathbf{F}^2\), the equation \(a(1,2) + b(3,5) = (x,y)\) says \(a + 3b = x\) and \(2a + 5b = y\). Doubling the first equation and subtracting the second gives \(b = 2x - y\), and then \(a = x - 3b = -5x + 3y\). Conversely \((-5x+3y)(1,2) + (2x-y)(3,5) = (-5x+3y+6x-3y,\; -10x+6y+10x-5y) = (x,y)\). So every \((x,y)\) is uniquely of the form \(a(1,2)+b(3,5)\), and \((1,2),(3,5)\) is a basis of \(\mathbf{F}^2\) by 2.28. (In particular, taking \((x,y) = (0,0)\) gives \(a = b = 0\), which is the linear independence.) This list has length two, as does the standard basis of \(\mathbf{F}^2\) from (a).

(c) For \((1,2,-4), (7,-5,6)\) in \(\mathbf{F}^3\): if \(a(1,2,-4) + b(7,-5,6) = (0,0,0)\), then \(a + 7b = 0\) and \(2a - 5b = 0\). The first gives \(a = -7b\); substituting into the second gives \(-14b - 5b = -19b = 0\), so \(b = 0\) and then \(a = 0\). Hence the list is linearly independent. It does not span \(\mathbf{F}^3\): if \(a(1,2,-4)+b(7,-5,6) = (0,0,1)\), the first two coordinates force \(a = b = 0\) as above, and then the third coordinate is \(0 \neq 1\). So \((0,0,1) \notin \operatorname{span}\big((1,2,-4),(7,-5,6)\big)\), and the list is not a basis of \(\mathbf{F}^3\).

(d) For \((1,2), (3,5), (4,13)\) in \(\mathbf{F}^2\): it spans \(\mathbf{F}^2\) because already the sublist \((1,2),(3,5)\) spans \(\mathbf{F}^2\) by (b), and enlarging a list only enlarges (weakly) its span. It is not linearly independent: applying the formulas of (b) with \((x,y) = (4,13)\) gives \(b = 2\cdot 4 - 13 = -5\) and \(a = -5 \cdot 4 + 3 \cdot 13 = 19\), so \((4,13) = 19(1,2) - 5(3,5)\), i.e.

\begin{equation*} 19(1,2) - 5(3,5) - 1\cdot(4,13) = (0,0) \end{equation*}

with coefficients not all \(0\). Hence the list is not a basis.

(e) Let \(U = \{(x,x,y) \in \mathbf{F}^3 : x, y \in \mathbf{F}\}\). Every linear combination \(a(1,1,0) + b(0,0,1) = (a,a,b)\) has its first two coordinates equal, hence lies in \(U\); thus \(\operatorname{span}\big((1,1,0),(0,0,1)\big) \subseteq U\). Conversely, every element of \(U\) satisfies \((x,x,y) = x(1,1,0) + y(0,0,1)\), so \(U \subseteq \operatorname{span}\big((1,1,0),(0,0,1)\big)\). Hence \(U\) equals that span; in particular \(U\) is a subspace of \(\mathbf{F}^3\) by 2.6 (so it is a vector space and the assertion makes sense), and the list spans \(U\). If \(a(1,1,0) + b(0,0,1) = (a,a,b) = (0,0,0)\), then \(a = b = 0\), so the list is linearly independent. Hence \((1,1,0),(0,0,1)\) is a basis of \(U\).

(f) Let \(U = \{(x,y,z) \in \mathbf{F}^3 : x + y + z = 0\}\). Both \((1,-1,0)\) and \((1,0,-1)\) lie in \(U\), since \(1 - 1 + 0 = 0\) and \(1 + 0 - 1 = 0\). Every linear combination \(a(1,-1,0) + b(1,0,-1) = (a+b, -a, -b)\) satisfies \((a+b) + (-a) + (-b) = 0\) and so lies in \(U\); combined with the spanning computation below this shows \(U = \operatorname{span}\big((1,-1,0),(1,0,-1)\big)\), so \(U\) is a subspace of \(\mathbf{F}^3\) by 2.6 and the assertion makes sense. If \(a(1,-1,0) + b(1,0,-1) = (a+b, -a, -b) = (0,0,0)\), then \(a = b = 0\); so the list is linearly independent. For spanning, let \((x,y,z) \in U\), so \(x = -y - z\). Then

\begin{equation*} (-y)(1,-1,0) + (-z)(1,0,-1) = (-y - z,\; y,\; z) = (x,y,z). \end{equation*}

Hence \((1,-1,0),(1,0,-1)\) is a basis of \(U\).

(g) By definition \(\mathcal{P}_m(\mathbf{F})\) is the set of polynomial functions on \(\mathbf{F}\) of the form \(p(z) = a_0 + a_1 z + \cdots + a_m z^m\) with \(a_0, \ldots, a_m \in \mathbf{F}\); that is exactly the statement that \(1, z, \ldots, z^m\) spans \(\mathcal{P}_m(\mathbf{F})\). This list is linearly independent by 2.16(b): if \(a_0 + a_1 z + \cdots + a_m z^m = 0\) as an element of \(\mathcal{P}(\mathbf{F})\), i.e. for every \(z \in \mathbf{F}\), then \(a_0 = a_1 = \cdots = a_m = 0\) (this follows from 4.8). Hence \(1, z, \ldots, z^m\) is a basis of \(\mathcal{P}_m(\mathbf{F})\).

Finally, the example asserts that \((7,5), (-4,9)\) and \((1,2),(3,5)\) are both bases of \(\mathbf{F}^2\). The second was verified in (b). For the first, let \((x,y) \in \mathbf{F}^2\) and consider \(a(7,5) + b(-4,9) = (x,y)\), i.e. \(7a - 4b = x\) and \(5a + 9b = y\). Multiplying the first by \(9\), the second by \(4\), and adding gives \(83a = 9x + 4y\); multiplying the first by \(5\), the second by \(7\), and subtracting gives \(-83b = 5x - 7y\). Since \(83 \neq 0\) in \(\mathbf{F}\), the unique solution is

\begin{equation*} a = \frac{9x + 4y}{83}, \qquad b = \frac{7y - 5x}{83}, \end{equation*}

and one checks directly that these values work: \(7a - 4b = \frac{63x + 28y - 28y + 20x}{83} = x\) and \(5a + 9b = \frac{45x + 20y + 63y - 45x}{83} = y\). So every vector of \(\mathbf{F}^2\) has a unique representation as a linear combination of \((7,5), (-4,9)\), which by 2.28 makes that list a basis of \(\mathbf{F}^2\).

(a) A vector \((x_1, x_2, x_3, x_4, x_5)\) lies in \(U\) precisely when \(x_1 = 3x_2\) and \(x_3 = 7x_4\); so \(x_2, x_4, x_5\) may be arbitrary and then \(x_1, x_3\) are determined. Writing \(x_2 = t\), \(x_4 = s\), \(x_5 = r\), the general element of \(U\) is

\begin{equation*} (3t, t, 7s, s, r) = t(3,1,0,0,0) + s(0,0,7,1,0) + r(0,0,0,0,1). \end{equation*}

Hence the list

\begin{equation*} (3,1,0,0,0),\quad (0,0,7,1,0),\quad (0,0,0,0,1) \end{equation*}

spans \(U\) (and each of these three vectors indeed lies in \(U\): \(3 = 3\cdot 1\) and \(0 = 7 \cdot 0\) for the first, etc.). It is linearly independent: if

\begin{equation*} a(3,1,0,0,0) + b(0,0,7,1,0) + c(0,0,0,0,1) = (3a, a, 7b, b, c) = (0,0,0,0,0), \end{equation*}

then reading off the second, fourth, and fifth coordinates gives \(a = b = c = 0\). So this list is a basis of \(U\) by 2.26.

(b) Adjoin \((1,0,0,0,0)\) and \((0,0,1,0,0)\), giving the list

\begin{equation*} (3,1,0,0,0),\ (0,0,7,1,0),\ (0,0,0,0,1),\ (1,0,0,0,0),\ (0,0,1,0,0). \end{equation*}

This spans \(\mathbf{R}^5\): given \((x_1, \ldots, x_5) \in \mathbf{R}^5\), take \(a = x_2\), \(b = x_4\), \(c = x_5\), \(d = x_1 - 3x_2\), \(e = x_3 - 7x_4\); then

\begin{equation*} a(3,1,0,0,0) + b(0,0,7,1,0) + c(0,0,0,0,1) + d(1,0,0,0,0) + e(0,0,1,0,0) \end{equation*}

\begin{equation*} = (3x_2 + x_1 - 3x_2,\ x_2,\ 7x_4 + x_3 - 7x_4,\ x_4,\ x_5) = (x_1, x_2, x_3, x_4, x_5). \end{equation*}

It is linearly independent: if the displayed combination equals \((0,0,0,0,0)\), then its coordinates are \((3a + d,\, a,\, 7b + e,\, b,\, c) = (0,0,0,0,0)\), so the second, fourth, and fifth coordinates give \(a = b = c = 0\), and then the first and third give \(d = 0\) and \(e = 0\). Hence this list of length five is a basis of \(\mathbf{R}^5\) extending the basis from (a).

(c) Take

\begin{equation*} W = \operatorname{span}\big((1,0,0,0,0), (0,0,1,0,0)\big) = \{(x_1, 0, x_3, 0, 0) : x_1, x_3 \in \mathbf{R}\}, \end{equation*}

which is a subspace of \(\mathbf{R}^5\) because every span is a subspace (2.6).

\(U + W = \mathbf{R}^5\): by (b) every vector of \(\mathbf{R}^5\) is a linear combination of the five listed vectors; the part coming from the first three vectors lies in \(U\) (they span \(U\)) and the part coming from the last two lies in \(W\).

\(U \cap W = \{0\}\): a vector of \(W\) has the form \((x_1, 0, x_3, 0, 0)\), and if it also lies in \(U\) then \(x_1 = 3 \cdot 0 = 0\) and \(x_3 = 7 \cdot 0 = 0\), so the vector is \(0\).

By 1.46 these two facts give \(\mathbf{R}^5 = U \oplus W\).

(a) A vector lies in \(U\) exactly when \(z_2 = 6z_1\) and \(z_3 = -2z_4 - 3z_5\); thus \(z_1, z_4, z_5\) are free and \(z_2, z_3\) are determined by them. Writing \(z_1 = t\), \(z_4 = s\), \(z_5 = r\), the general element of \(U\) is

\begin{equation*} (t, 6t, -2s - 3r, s, r) = t(1,6,0,0,0) + s(0,0,-2,1,0) + r(0,0,-3,0,1). \end{equation*}

So the list

\begin{equation*} (1,6,0,0,0),\quad (0,0,-2,1,0),\quad (0,0,-3,0,1) \end{equation*}

spans \(U\), and each of its vectors lies in \(U\) (for the first, \(6 \cdot 1 = 6\) and \(0 + 0 + 0 = 0\); for the second, \(6 \cdot 0 = 0\) and \(-2 + 2\cdot 1 + 0 = 0\); for the third, \(6\cdot 0 = 0\) and \(-3 + 0 + 3 \cdot 1 = 0\)). It is linearly independent: if

\begin{equation*} a(1,6,0,0,0) + b(0,0,-2,1,0) + c(0,0,-3,0,1) = (a,\, 6a,\, -2b - 3c,\, b,\, c) = 0, \end{equation*}

the first, fourth, and fifth coordinates give \(a = b = c = 0\). Hence this list is a basis of \(U\) by 2.26.

(b) Adjoin \((0,1,0,0,0)\) and \((0,0,1,0,0)\), giving

\begin{equation*} (1,6,0,0,0),\ (0,0,-2,1,0),\ (0,0,-3,0,1),\ (0,1,0,0,0),\ (0,0,1,0,0). \end{equation*}

For scalars \(a,b,c,d,e \in \mathbf{C}\),

\begin{equation*} a(1,6,0,0,0) + b(0,0,-2,1,0) + c(0,0,-3,0,1) + d(0,1,0,0,0) + e(0,0,1,0,0) \end{equation*}

\begin{equation*} = (a,\; 6a + d,\; -2b - 3c + e,\; b,\; c). \end{equation*}

This list spans \(\mathbf{C}^5\): given \((z_1, \ldots, z_5) \in \mathbf{C}^5\), choose \(a = z_1\), \(b = z_4\), \(c = z_5\), \(d = z_2 - 6z_1\), and \(e = z_3 + 2z_4 + 3z_5\); then the displayed vector equals \((z_1, z_2, z_3, z_4, z_5)\). It is linearly independent: if the displayed vector is \(0\), then the first, fourth, and fifth coordinates give \(a = b = c = 0\), and then the second and third coordinates give \(d = 0\) and \(e = 0\). Hence it is a basis of \(\mathbf{C}^5\) extending the basis of (a).

(c) Take

\begin{equation*} W = \operatorname{span}\big((0,1,0,0,0), (0,0,1,0,0)\big) = \{(0, z_2, z_3, 0, 0) : z_2, z_3 \in \mathbf{C}\}, \end{equation*}

a subspace of \(\mathbf{C}^5\) by 2.6.

\(U + W = \mathbf{C}^5\): by (b) every vector of \(\mathbf{C}^5\) is a linear combination of the five listed vectors; the contribution of the first three lies in \(U\) and the contribution of the last two lies in \(W\).

\(U \cap W = \{0\}\): if \((0, z_2, z_3, 0, 0) \in U\), then \(z_2 = 6 \cdot 0 = 0\) and \(z_3 + 2 \cdot 0 + 3 \cdot 0 = 0\) forces \(z_3 = 0\); so the vector is \(0\).

By 1.46, \(\mathbf{C}^5 = U \oplus W\).

Because \(V\) is finite-dimensional, so are its subspaces \(U\) and \(W\) (2.25). Hence \(U\) has a basis \(u_1, \ldots, u_m\) and \(W\) has a basis \(w_1, \ldots, w_n\) (2.31); here \(m\) or \(n\) may be \(0\), in which case the corresponding list is empty.

The combined list

\begin{equation*} u_1, \ldots, u_m, w_1, \ldots, w_n \end{equation*}

spans \(V\). Indeed, let \(v \in V\). Since \(V = U + W\), there are \(u \in U\) and \(w \in W\) with \(v = u + w\). Since \(u_1, \ldots, u_m\) spans \(U\), there are scalars \(a_1, \ldots, a_m\) with \(u = a_1u_1 + \cdots + a_mu_m\); similarly \(w = b_1w_1 + \cdots + b_nw_n\) for some scalars \(b_1, \ldots, b_n\). Thus

\begin{equation*} v = a_1u_1 + \cdots + a_mu_m + b_1w_1 + \cdots + b_nw_n, \end{equation*}

so \(v \in \operatorname{span}(u_1, \ldots, u_m, w_1, \ldots, w_n)\).

By 2.30, every spanning list in a vector space can be reduced to a basis of that space by deleting some of its vectors. Applying this to the spanning list above produces a basis of \(V\) each of whose vectors is one of \(u_1, \ldots, u_m, w_1, \ldots, w_n\). Every \(u_j\) lies in \(U\) and every \(w_k\) lies in \(W\), so every vector of this basis lies in \(U \cup W\). This is the desired basis of \(V\).

The assertion is false. Here is a counterexample. Define

\begin{equation*} p_0(z) = 1, \qquad p_1(z) = z, \qquad p_2(z) = z^2 + z^3, \qquad p_3(z) = z^3 . \end{equation*}

These all lie in \(\mathcal{P}_3(\mathbf{F})\), and their degrees are \(0, 1, 3, 3\) respectively, so none of them has degree \(2\). Nevertheless \(p_0, p_1, p_2, p_3\) is a basis of \(\mathcal{P}_3(\mathbf{F})\).

Spanning: since \(p_2 - p_3\) is the polynomial \(z \mapsto z^2\), the span of \(p_0, p_1, p_2, p_3\) contains the four polynomials \(1, z, z^2, z^3\), hence contains their span, which is all of \(\mathcal{P}_3(\mathbf{F})\) (Example 2.27(g)). Concretely, for \(a, b, c, d \in \mathbf{F}\),

\begin{equation*} a + bz + cz^2 + dz^3 = a\,p_0 + b\,p_1 + c\,p_2 + (d - c)\,p_3 . \end{equation*}

Linear independence: suppose \(a, b, c, d \in \mathbf{F}\) and \(a\,p_0 + b\,p_1 + c\,p_2 + d\,p_3 = 0\). Collecting terms, this says

\begin{equation*} a + bz + cz^2 + (c + d)z^3 = 0 \end{equation*}

for all \(z \in \mathbf{F}\). By 2.16(b) (which rests on 4.8), all coefficients vanish: \(a = b = c = 0\) and \(c + d = 0\); the last two give \(d = 0\). Hence \(a = b = c = d = 0\).

So \(p_0, p_1, p_2, p_3\) is linearly independent and spans \(\mathcal{P}_3(\mathbf{F})\), hence is a basis of \(\mathcal{P}_3(\mathbf{F})\) by 2.26, even though none of these polynomials has degree \(2\). The claim in the exercise is therefore false.

Write \(L\) for the list \(v_1 + v_2,\ v_2 + v_3,\ v_3 + v_4,\ v_4\). By 2.26 we must show that \(L\) is linearly independent and spans \(V\).

Linear independence. Suppose \(a, b, c, d \in \mathbf{F}\) satisfy

\begin{equation*} a(v_1 + v_2) + b(v_2 + v_3) + c(v_3 + v_4) + d v_4 = 0 . \end{equation*}

Regrouping by \(v_1, v_2, v_3, v_4\) gives

\begin{equation*} a v_1 + (a + b) v_2 + (b + c) v_3 + (c + d) v_4 = 0 . \end{equation*}

Since \(v_1, v_2, v_3, v_4\) is a basis of \(V\), it is linearly independent, so all four coefficients are \(0\):

\begin{equation*} a = 0, \qquad a + b = 0, \qquad b + c = 0, \qquad c + d = 0 . \end{equation*}

Reading these in order gives \(a = 0\), then \(b = -a = 0\), then \(c = -b = 0\), then \(d = -c = 0\). Hence \(a = b = c = d = 0\), so \(L\) is linearly independent.

Spanning. Let \(S = \operatorname{span}(v_1 + v_2,\ v_2 + v_3,\ v_3 + v_4,\ v_4)\). Working from the end of the list backwards:

\begin{equation*} v_4 \in S, \qquad v_3 = (v_3 + v_4) - v_4 \in S, \qquad v_2 = (v_2 + v_3) - v_3 \in S, \qquad v_1 = (v_1 + v_2) - v_2 \in S, \end{equation*}

where at each step we use that \(S\) is a subspace (2.6), hence closed under subtraction, together with the membership established at the previous step. Thus \(v_1, v_2, v_3, v_4 \in S\), and since \(S\) is a subspace containing these four vectors, it contains their span:

\begin{equation*} V = \operatorname{span}(v_1, v_2, v_3, v_4) \subseteq S \subseteq V . \end{equation*}

Therefore \(S = V\), i.e. \(L\) spans \(V\).

Being linearly independent and spanning \(V\), the list \(v_1 + v_2,\ v_2 + v_3,\ v_3 + v_4,\ v_4\) is a basis of \(V\).

The statement is false. Here is a counterexample.

Take \(V = \mathbf{F}^4\) and let \(v_1, v_2, v_3, v_4\) be the standard basis \(e_1, e_2, e_3, e_4\) of \(\mathbf{F}^4\), where \(e_k\) is the list of length \(4\) with \(1\) in the \(k^{\text{th}}\) slot and \(0\) elsewhere. Let

\begin{equation*} U = \operatorname{span}(e_1,\, e_2,\, e_3 + e_4) = \{(x_1, x_2, x_3, x_4) \in \mathbf{F}^4 : x_3 = x_4\}. \end{equation*}

To see that the two descriptions of \(U\) agree, note that \(e_1, e_2, e_3 + e_4\) all lie in the set on the right, which is a subspace of \(\mathbf{F}^4\), so the span is contained in it; conversely if \(x_3 = x_4\) then \((x_1, x_2, x_3, x_4) = x_1 e_1 + x_2 e_2 + x_3(e_3 + e_4)\).

Now check the three hypotheses.

First, \(v_1 = e_1 \in U\) and \(v_2 = e_2 \in U\), because the third and fourth coordinates of each are both \(0\).

Second, \(v_3 = e_3 \notin U\), because the third coordinate of \(e_3\) is \(1\) and its fourth coordinate is \(0\), and \(1 \neq 0\).

Third, \(v_4 = e_4 \notin U\), for the same reason with the roles of the third and fourth coordinates reversed.

Thus all the hypotheses hold. But \(v_1, v_2\) is not a basis of \(U\): the vector \(e_3 + e_4\) belongs to \(U\), and it is not in \(\operatorname{span}(e_1, e_2)\), since every element of \(\operatorname{span}(e_1, e_2)\) has third coordinate \(0\) while \(e_3 + e_4\) has third coordinate \(1\). Hence \(v_1, v_2\) does not span \(U\) and so is not a basis of \(U\) (see 2.26).

It is worth isolating exactly which half of the conclusion can fail. The list \(v_1, v_2\) is a sublist of the linearly independent list \(v_1, v_2, v_3, v_4\), and any sublist of a linearly independent list is linearly independent: a nontrivial vanishing linear combination of \(v_1, v_2\) becomes a nontrivial vanishing linear combination of \(v_1, v_2, v_3, v_4\) upon appending the coefficients \(0, 0\). So the linear independence half of the definition of a basis always holds under these hypotheses. What can fail, as the counterexample shows, is the spanning half: knowing merely that \(v_3\) and \(v_4\) individually avoid \(U\) says nothing about whether some combination of them, such as \(v_3 + v_4\), lies in \(U\).

Throughout, adopt the convention \(w_0 = 0\). Then the defining relations can be inverted:

\begin{equation*} v_k = w_k - w_{k-1} \qquad \text{for each } k \in \{1, \ldots, m\}, \end{equation*}

since \(w_k - w_{k-1} = (v_1 + \cdots + v_k) - (v_1 + \cdots + v_{k-1}) = v_k\), and for \(k = 1\) this reads \(v_1 = w_1 - 0 = w_1\).

Because a basis is precisely a linearly independent spanning list (2.26), it suffices to prove two things: that the two lists have the same span, and that one list is linearly independent if and only if the other is.

Step 1: the two lists have the same span.

Each \(w_k = v_1 + \cdots + v_k\) is a linear combination of \(v_1, \ldots, v_m\), so \(w_k \in \operatorname{span}(v_1, \ldots, v_m)\) for every \(k\). Since the span of a list is the smallest subspace containing all vectors in the list, this gives

\begin{equation*} \operatorname{span}(w_1, \ldots, w_m) \subseteq \operatorname{span}(v_1, \ldots, v_m). \end{equation*}

Conversely, each \(v_k = w_k - w_{k-1}\) is a linear combination of \(w_1, \ldots, w_m\) (with \(w_0 = 0\) contributing nothing), so \(v_k \in \operatorname{span}(w_1, \ldots, w_m)\) for every \(k\), giving the reverse inclusion. Hence

\begin{equation*} \operatorname{span}(v_1, \ldots, v_m) = \operatorname{span}(w_1, \ldots, w_m). \end{equation*}

In particular, one list spans \(V\) if and only if the other does.

Step 2: if \(v_1, \ldots, v_m\) is linearly independent, then so is \(w_1, \ldots, w_m\).

Suppose \(a_1, \ldots, a_m \in \mathbf{F}\) satisfy \(a_1 w_1 + \cdots + a_m w_m = 0\). Substituting \(w_k = v_1 + \cdots + v_k\) and collecting the coefficient of each \(v_j\) (the vector \(v_j\) appears in \(w_k\) exactly when \(k \geq j\)) gives

\begin{equation*} 0 = \sum_{k=1}^{m} a_k \sum_{j=1}^{k} v_j = \sum_{j=1}^{m} \Bigl( \sum_{k=j}^{m} a_k \Bigr) v_j. \end{equation*}

Since \(v_1, \ldots, v_m\) is linearly independent, every coefficient vanishes:

\begin{equation*} \sum_{k=j}^{m} a_k = 0 \qquad \text{for each } j \in \{1, \ldots, m\}. \end{equation*}

Taking \(j = m\) gives \(a_m = 0\). For \(j < m\), subtracting the equation for \(j+1\) from the equation for \(j\) gives

\begin{equation*} a_j = \sum_{k=j}^{m} a_k - \sum_{k=j+1}^{m} a_k = 0 - 0 = 0. \end{equation*}

Thus \(a_1 = \cdots = a_m = 0\), so \(w_1, \ldots, w_m\) is linearly independent.

Step 3: if \(w_1, \ldots, w_m\) is linearly independent, then so is \(v_1, \ldots, v_m\).

Suppose \(b_1, \ldots, b_m \in \mathbf{F}\) satisfy \(b_1 v_1 + \cdots + b_m v_m = 0\). Using \(v_k = w_k - w_{k-1}\) and the convention \(w_0 = 0\), and also setting \(b_{m+1} = 0\), we compute

\begin{equation*} \begin{aligned} 0 &= \sum_{k=1}^{m} b_k (w_k - w_{k-1}) \\ &= \sum_{k=1}^{m} b_k w_k - \sum_{k=1}^{m-1} b_{k+1} w_k \\ &= \sum_{k=1}^{m} (b_k - b_{k+1}) w_k, \end{aligned} \end{equation*}

where the last equality uses \(b_{m+1} = 0\) so that the \(k = m\) term is \(b_m w_m\). Since \(w_1, \ldots, w_m\) is linearly independent, every coefficient vanishes: \(b_k = b_{k+1}\) for each \(k \in \{1, \ldots, m\}\). Reading these backwards from \(b_{m+1} = 0\) gives \(b_m = 0\), then \(b_{m-1} = 0\), and so on, yielding \(b_1 = \cdots = b_m = 0\). Thus \(v_1, \ldots, v_m\) is linearly independent.

Conclusion.

Suppose \(v_1, \ldots, v_m\) is a basis of \(V\). Then it spans \(V\) and is linearly independent. By Step 1, \(w_1, \ldots, w_m\) spans \(V\); by Step 2, it is linearly independent. Hence \(w_1, \ldots, w_m\) is a basis of \(V\).

Conversely, suppose \(w_1, \ldots, w_m\) is a basis of \(V\). By Step 1, \(v_1, \ldots, v_m\) spans \(V\); by Step 3, it is linearly independent. Hence \(v_1, \ldots, v_m\) is a basis of \(V\).

Note that no finite-dimensionality hypothesis on \(V\) was used anywhere; the argument is a purely formal change of coordinates, the linear map sending \((a_1, \ldots, a_m)\) to its list of partial sums being invertible over any field, with the successive-differences map as its inverse.

We must show that the concatenated list spans \(V\) and is linearly independent (2.26).

The list spans \(V\).

Let \(v \in V\). Since \(V = U \oplus W\), in particular \(V = U + W\), so there exist \(u \in U\) and \(w \in W\) with

\begin{equation*} v = u + w. \end{equation*}

Because \(u_1, \ldots, u_m\) is a basis of \(U\), it spans \(U\), so there exist \(a_1, \ldots, a_m \in \mathbf{F}\) with \(u = a_1 u_1 + \cdots + a_m u_m\). Because \(w_1, \ldots, w_n\) is a basis of \(W\), it spans \(W\), so there exist \(b_1, \ldots, b_n \in \mathbf{F}\) with \(w = b_1 w_1 + \cdots + b_n w_n\). Hence

\begin{equation*} v = a_1 u_1 + \cdots + a_m u_m + b_1 w_1 + \cdots + b_n w_n, \end{equation*}

exhibiting \(v\) as a linear combination of \(u_1, \ldots, u_m, w_1, \ldots, w_n\). Since \(v \in V\) was arbitrary, the list spans \(V\).

The list is linearly independent.

Suppose \(a_1, \ldots, a_m, b_1, \ldots, b_n \in \mathbf{F}\) are such that

\begin{equation*} a_1 u_1 + \cdots + a_m u_m + b_1 w_1 + \cdots + b_n w_n = 0. \end{equation*}

Set

\begin{equation*} u = a_1 u_1 + \cdots + a_m u_m \quad \text{and} \quad w = b_1 w_1 + \cdots + b_n w_n. \end{equation*}

Then \(u \in U\) and \(w \in W\) (each is a linear combination of vectors in the relevant subspace, and subspaces are closed under addition and scalar multiplication), and \(u + w = 0\). Because the sum \(U + W\) is a direct sum, 1.45 tells us that the only way to write \(0\) as a sum of a vector in \(U\) and a vector in \(W\) is with both summands equal to \(0\). Hence

\begin{equation*} u = 0 \quad \text{and} \quad w = 0. \end{equation*}

Now \(a_1 u_1 + \cdots + a_m u_m = 0\) and \(u_1, \ldots, u_m\) is linearly independent (being a basis of \(U\)), so \(a_1 = \cdots = a_m = 0\). Likewise \(b_1 w_1 + \cdots + b_n w_n = 0\) and \(w_1, \ldots, w_n\) is linearly independent, so \(b_1 = \cdots = b_n = 0\).

Thus all the scalars are \(0\), and the list \(u_1, \ldots, u_m, w_1, \ldots, w_n\) is linearly independent.

Having shown that the list spans \(V\) and is linearly independent, we conclude that it is a basis of \(V\).

Two remarks. First, no finite-dimensionality hypothesis on \(V\) was needed: the hypotheses that \(U\) and \(W\) have the finite bases named already force \(V = U \oplus W\) to be finite-dimensional, and the proof itself uses only the definitions. Second, in place of 1.45 one could invoke 1.46 with a slightly longer argument: from \(u + w = 0\) we get \(u = -w \in U \cap W = \{0\}\), so \(u = 0\) and then \(w = -u = 0\).

Recall the definitions from Exercise 8 in Section 1B. As a set, \(V_{\mathbf{C}} = V \times V\), and the ordered pair \((u, v)\) is written \(u + iv\). Thus two elements \(u + iv\) and \(u^{\prime} + iv^{\prime}\) of \(V_{\mathbf{C}}\) are equal precisely when \(u = u^{\prime}\) and \(v = v^{\prime}\). Addition is defined by

\begin{equation*} (u_1 + iv_1) + (u_2 + iv_2) = (u_1 + u_2) + i(v_1 + v_2) \end{equation*}

and complex scalar multiplication by

\begin{equation*} (a + bi)(u + iv) = (au - bv) + i(av + bu) \end{equation*}

for all \(a, b \in \mathbf{R}\) and all \(u, v \in V\). The vector \(v \in V\) is regarded as the element \(v + i0\) of \(V_{\mathbf{C}}\), and this is how the list \(v_1, \ldots, v_n\) is to be read below. The zero vector of \(V_{\mathbf{C}}\) is \(0 + i0\).

A single computation drives everything. For \(a, b \in \mathbf{R}\) and \(v \in V\), the scalar multiplication rule with the second component equal to \(0\) gives

\begin{equation*} (a + bi)(v + i0) = (av - b \cdot 0) + i(a \cdot 0 + bv) = (av) + i(bv). \end{equation*}

Consequently, for real numbers \(a_1, \ldots, a_n, b_1, \ldots, b_n\), repeated use of the addition rule yields

\begin{equation*} \sum_{k=1}^{n} (a_k + b_k i)(v_k + i0) = \sum_{k=1}^{n} \bigl( (a_k v_k) + i(b_k v_k) \bigr) = \Bigl( \sum_{k=1}^{n} a_k v_k \Bigr) + i \Bigl( \sum_{k=1}^{n} b_k v_k \Bigr). \end{equation*}

Call this identity \((\ast)\). Note that every complex number has the form \(a + bi\) with \(a, b \in \mathbf{R}\), so \((\ast)\) covers all complex linear combinations of \(v_1 + i0, \ldots, v_n + i0\).

The list spans \(V_{\mathbf{C}}\) over \(\mathbf{C}\).

Let \(u + iv\) be an arbitrary element of \(V_{\mathbf{C}}\), where \(u, v \in V\). Since \(v_1, \ldots, v_n\) spans the real vector space \(V\), there exist real numbers \(a_1, \ldots, a_n\) and \(b_1, \ldots, b_n\) with

\begin{equation*} u = a_1 v_1 + \cdots + a_n v_n \quad \text{and} \quad v = b_1 v_1 + \cdots + b_n v_n. \end{equation*}

By \((\ast)\),

\begin{equation*} \sum_{k=1}^{n} (a_k + b_k i)(v_k + i0) = u + iv. \end{equation*}

Thus \(u + iv\) lies in the complex span of \(v_1, \ldots, v_n\). Since \(u + iv \in V_{\mathbf{C}}\) was arbitrary, the list spans \(V_{\mathbf{C}}\).

The list is linearly independent over \(\mathbf{C}\).

Suppose \(\lambda_1, \ldots, \lambda_n \in \mathbf{C}\) satisfy

\begin{equation*} \lambda_1 (v_1 + i0) + \cdots + \lambda_n (v_n + i0) = 0 + i0. \end{equation*}

Write \(\lambda_k = a_k + b_k i\) with \(a_k, b_k \in \mathbf{R}\). By \((\ast)\) the left side equals

\begin{equation*} \Bigl( \sum_{k=1}^{n} a_k v_k \Bigr) + i \Bigl( \sum_{k=1}^{n} b_k v_k \Bigr). \end{equation*}

Two elements of \(V_{\mathbf{C}} = V \times V\) are equal exactly when their two components agree, so

\begin{equation*} \sum_{k=1}^{n} a_k v_k = 0 \quad \text{and} \quad \sum_{k=1}^{n} b_k v_k = 0, \end{equation*}

these being equations in \(V\). Because \(v_1, \ldots, v_n\) is a basis of the real vector space \(V\), it is linearly independent over \(\mathbf{R}\), and the \(a_k\) and \(b_k\) are real. Hence

\begin{equation*} a_1 = \cdots = a_n = 0 \quad \text{and} \quad b_1 = \cdots = b_n = 0, \end{equation*}

so \(\lambda_k = a_k + b_k i = 0\) for every \(k\). Thus \(v_1, \ldots, v_n\) is linearly independent in \(V_{\mathbf{C}}\) over \(\mathbf{C}\).

Since the list spans \(V_{\mathbf{C}}\) and is linearly independent there, it is a basis of the complex vector space \(V_{\mathbf{C}}\) (2.26).

The moral is that complexification does not change the number of basis vectors, only the field of scalars: the real degrees of freedom that were split between the real and imaginary components of \(u + iv\) get recombined into the real and imaginary parts of the complex coefficients. Anticipating the language of Section 2C, this says \(\dim_{\mathbf{C}} V_{\mathbf{C}} = \dim_{\mathbf{R}} V\). The model case is \(V = \mathbf{R}^n\), whose complexification is \(\mathbf{C}^n\): the standard basis of \(\mathbf{R}^n\) over \(\mathbf{R}\) is also the standard basis of \(\mathbf{C}^n\) over \(\mathbf{C}\).

Recall that a line in \(\mathbb{R}^2\) is a set of the form

\begin{equation*} L = \{w + \lambda v : \lambda \in \mathbb{R}\} \end{equation*}

for some \(w \in \mathbb{R}^2\) and some \(v \in \mathbb{R}^2\) with \(v \neq 0\).

First observe that a line containing the origin is exactly the span of a single nonzero vector. Indeed, suppose \(L = \{w + \lambda v : \lambda \in \mathbb{R}\}\) with \(v \neq 0\) and \(0 \in L\). Then \(w = -\lambda_0 v\) for some \(\lambda_0 \in \mathbb{R}\), so

\begin{equation*} L = \{(\lambda - \lambda_0)v : \lambda \in \mathbb{R}\} = \{\mu v : \mu \in \mathbb{R}\} = \operatorname{span}(v). \end{equation*}

Conversely, if \(v \neq 0\) then \(\operatorname{span}(v) = \{0 + \lambda v : \lambda \in \mathbb{R}\}\) is a line containing the origin.

Each listed set is a subspace. The set \(\{0\}\) and the whole space \(\mathbb{R}^2\) are subspaces. A line containing the origin equals \(\operatorname{span}(v)\) for some \(v \neq 0\), and the span of a list of vectors in \(\mathbb{R}^2\) is a subspace of \(\mathbb{R}^2\) (by 2.6).

Each subspace is one of the listed sets. Suppose \(U\) is a subspace of \(\mathbb{R}^2\). Since \(\mathbb{R}^2\) is finite-dimensional, \(U\) is finite-dimensional (by 2.25), and \(\dim U \leq \dim \mathbb{R}^2 = 2\) by 2.37 and 2.36. So \(\dim U \in \{0, 1, 2\}\).

If \(\dim U = 0\), then the empty list is a basis of \(U\); the span of the empty list is \(\{0\}\) by 2.4, so \(U = \{0\}\).

If \(\dim U = 1\), then \(U\) has a basis consisting of a single vector \(v\), and \(v \neq 0\) because a list containing \(0\) is not linearly independent. Hence \(U = \operatorname{span}(v)\), which by the observation above is a line in \(\mathbb{R}^2\) containing the origin.

If \(\dim U = 2 = \dim \mathbb{R}^2\), then \(U = \mathbb{R}^2\) by 2.39.

Thus the subspaces of \(\mathbb{R}^2\) are precisely \(\{0\}\), the lines through the origin, and \(\mathbb{R}^2\).

Recall the definitions: a line in \(\mathbb{R}^3\) is a set of the form \(\{w + \lambda v : \lambda \in \mathbb{R}\}\) with \(v \neq 0\), and a plane in \(\mathbb{R}^3\) is a set of the form \(\{w + \lambda_1 v_1 + \lambda_2 v_2 : \lambda_1, \lambda_2 \in \mathbb{R}\}\) where \(v_1, v_2\) is a linearly independent list in \(\mathbb{R}^3\).

Step 1: lines through the origin are exactly the \(1\)-dimensional subspaces. If \(L = \{w + \lambda v : \lambda \in \mathbb{R}\}\) with \(v \neq 0\) and \(0 \in L\), then \(w = -\lambda_0 v\) for some \(\lambda_0\), so \(L = \{\mu v : \mu \in \mathbb{R}\} = \operatorname{span}(v)\), a subspace of dimension \(1\) (the list \(v\) is linearly independent because \(v \neq 0\), and it spans \(L\)). Conversely, a \(1\)-dimensional subspace has a basis consisting of one vector \(v\), necessarily nonzero, and equals \(\operatorname{span}(v) = \{0 + \lambda v : \lambda \in \mathbb{R}\}\), a line containing the origin.

Step 2: planes through the origin are exactly the \(2\)-dimensional subspaces. Suppose \(P = \{w + \lambda_1 v_1 + \lambda_2 v_2 : \lambda_1, \lambda_2 \in \mathbb{R}\}\) with \(v_1, v_2\) linearly independent, and \(0 \in P\). Then \(w = -(\mu_1 v_1 + \mu_2 v_2)\) for some \(\mu_1, \mu_2 \in \mathbb{R}\), and hence

\begin{equation*} P = \{(\lambda_1 - \mu_1)v_1 + (\lambda_2 - \mu_2)v_2 : \lambda_1, \lambda_2 \in \mathbb{R}\} = \operatorname{span}(v_1, v_2). \end{equation*}

This is a subspace (by 2.6) with basis \(v_1, v_2\), so \(\dim P = 2\). Conversely, a \(2\)-dimensional subspace \(U\) has a basis \(v_1, v_2\); this list is linearly independent and \(U = \operatorname{span}(v_1, v_2) = \{0 + \lambda_1 v_1 + \lambda_2 v_2 : \lambda_1, \lambda_2 \in \mathbb{R}\}\), a plane containing the origin.

Step 3: the classification. Each of the four listed kinds of set is a subspace: \(\{0\}\) and \(\mathbb{R}^3\) obviously, and lines and planes through the origin by Steps 1 and 2.

Conversely, let \(U\) be a subspace of \(\mathbb{R}^3\). Since \(\mathbb{R}^3\) is finite-dimensional, so is \(U\) (by 2.25), and \(\dim U \leq \dim \mathbb{R}^3 = 3\) by 2.37 and 2.36. Hence \(\dim U \in \{0,1,2,3\}\).

If \(\dim U = 0\), then \(U = \{0\}\), since the empty list is a basis of \(U\) and its span is \(\{0\}\).

If \(\dim U = 1\), then \(U\) is a line containing the origin, by Step 1.

If \(\dim U = 2\), then \(U\) is a plane containing the origin, by Step 2.

If \(\dim U = 3 = \dim \mathbb{R}^3\), then \(U = \mathbb{R}^3\) by 2.39.

Thus the subspaces of \(\mathbb{R}^3\) are precisely \(\{0\}\), the lines through the origin, the planes through the origin, and \(\mathbb{R}^3\).

Throughout, write \(\mathcal{P}_4 = \mathcal{P}_4(\mathbb{F})\); recall \(\dim \mathcal{P}_4 = 5\) by 2.36. Note first that \(U\) is a subspace of \(\mathcal{P}_4\): the zero polynomial vanishes at \(6\), and if \(p(6) = q(6) = 0\) and \(\lambda \in \mathbb{F}\) then \((p + q)(6) = 0\) and \((\lambda p)(6) = 0\).

A convenient basis of \(\mathcal{P}_4\). Consider the list

\begin{equation*} 1,\ (x-6),\ (x-6)^2,\ (x-6)^3,\ (x-6)^4 . \end{equation*}

It is linearly independent: suppose \(c_0 + c_1(x-6) + c_2(x-6)^2 + c_3(x-6)^3 + c_4(x-6)^4 = 0\) as a polynomial. Expanding, the coefficient of \(x^4\) is \(c_4\), so \(c_4 = 0\); then the coefficient of \(x^3\) is \(c_3\), so \(c_3 = 0\); continuing downward gives \(c_2 = c_1 = c_0 = 0\). This list has length \(5 = \dim \mathcal{P}_4\), so it is a basis of \(\mathcal{P}_4\) by 2.38.

(a) Every \(p \in \mathcal{P}_4\) therefore has a unique expansion

\begin{equation*} p = c_0 + c_1(x-6) + c_2(x-6)^2 + c_3(x-6)^3 + c_4(x-6)^4 , \end{equation*}

and evaluating at \(6\) gives \(p(6) = c_0\). Hence \(p \in U\) if and only if \(c_0 = 0\), i.e. if and only if \(p\) lies in the span of

\begin{equation*} (x-6),\ (x-6)^2,\ (x-6)^3,\ (x-6)^4 . \end{equation*}

These four polynomials lie in \(U\) and span \(U\), and they are linearly independent (being a sublist of a linearly independent list). So they form a basis of \(U\), and \(\dim U = 4\).

(b) Adjoining the constant polynomial \(1\) gives

\begin{equation*} (x-6),\ (x-6)^2,\ (x-6)^3,\ (x-6)^4,\ 1 , \end{equation*}

which is (after reordering) the basis of \(\mathcal{P}_4\) established above.

(c) Take \(W = \operatorname{span}(1)\), the subspace of constant polynomials. Then \(\mathcal{P}_4 = U \oplus W\).

Indeed, for any \(p \in \mathcal{P}_4\) let \(c\) denote the constant polynomial whose value is \(p(6)\), and write

\begin{equation*} p = (p - c) + c . \end{equation*}

Here \((p - c)(6) = p(6) - p(6) = 0\), so \(p - c \in U\), while \(c \in W\). Hence \(\mathcal{P}_4 = U + W\). Moreover, if \(q \in U \cap W\), then \(q\) is a constant polynomial, say \(q = c\), with \(q(6) = 0\); thus \(c = 0\) and \(q = 0\). So \(U \cap W = \{0\}\), and therefore the sum is direct by 1.46. Hence \(\mathcal{P}_4(\mathbb{F}) = U \oplus W\).

Write \(\mathcal{P}_4 = \mathcal{P}_4(\mathbb{R})\); recall \(\dim \mathcal{P}_4 = 5\) by 2.36. The set \(U\) is a subspace of \(\mathcal{P}_4\): the second derivative of the zero polynomial is \(0\), so the zero polynomial lies in \(U\), and differentiation is additive and commutes with scalar multiplication, so \((p + q)^{\prime\prime}(6) = p^{\prime\prime}(6) + q^{\prime\prime}(6)\) and \((\lambda p)^{\prime\prime}(6) = \lambda p^{\prime\prime}(6)\).

A convenient basis of \(\mathcal{P}_4\). The list

\begin{equation*} 1,\ (x-6),\ (x-6)^2,\ (x-6)^3,\ (x-6)^4 \end{equation*}

is linearly independent: if \(c_0 + c_1(x-6) + \cdots + c_4(x-6)^4 = 0\), then expanding and reading off the coefficient of \(x^4\) gives \(c_4 = 0\), then the coefficient of \(x^3\) gives \(c_3 = 0\), and continuing downward gives \(c_2 = c_1 = c_0 = 0\). Having length \(5 = \dim \mathcal{P}_4\), it is a basis of \(\mathcal{P}_4\) by 2.38.

(a) Write \(p \in \mathcal{P}_4\) in this basis:

\begin{equation*} p = c_0 + c_1(x-6) + c_2(x-6)^2 + c_3(x-6)^3 + c_4(x-6)^4 . \end{equation*}

Differentiating twice,

\begin{equation*} p^{\prime\prime} = 2c_2 + 6c_3(x-6) + 12c_4(x-6)^2 , \end{equation*}

so \(p^{\prime\prime}(6) = 2c_2\). Hence \(p \in U\) if and only if \(c_2 = 0\), i.e. if and only if \(p\) lies in the span of

\begin{equation*} 1,\ (x-6),\ (x-6)^3,\ (x-6)^4 . \end{equation*}

Each of these four polynomials lies in \(U\) (their second derivatives are \(0\), \(0\), \(6(x-6)\), and \(12(x-6)^2\), all of which vanish at \(6\)), they span \(U\) by the computation above, and they are linearly independent as a sublist of a linearly independent list. So they form a basis of \(U\), and \(\dim U = 4\).

(b) Adjoin \((x-6)^2\). The resulting list

\begin{equation*} 1,\ (x-6),\ (x-6)^2,\ (x-6)^3,\ (x-6)^4 \end{equation*}

is the basis of \(\mathcal{P}_4\) exhibited above.

(c) Take \(W = \operatorname{span}\bigl((x-6)^2\bigr)\). For any \(p \in \mathcal{P}_4\) set \(\lambda = \tfrac{1}{2}p^{\prime\prime}(6)\) and write

\begin{equation*} p = \Bigl(p - \lambda (x-6)^2\Bigr) + \lambda (x-6)^2 . \end{equation*}

The second summand lies in \(W\). The first lies in \(U\), since the second derivative of \(\lambda(x-6)^2\) is the constant \(2\lambda = p^{\prime\prime}(6)\), so

\begin{equation*} \bigl(p - \lambda(x-6)^2\bigr)^{\prime\prime}(6) = p^{\prime\prime}(6) - p^{\prime\prime}(6) = 0 . \end{equation*}

Hence \(\mathcal{P}_4 = U + W\). If \(q \in U \cap W\), then \(q = c(x-6)^2\) for some \(c \in \mathbb{R}\) and \(q^{\prime\prime}(6) = 2c = 0\), so \(c = 0\) and \(q = 0\). Thus \(U \cap W = \{0\}\), and by 1.46 the sum is direct: \(\mathcal{P}_4(\mathbb{R}) = U \oplus W\).

Write \(\mathcal{P}_4 = \mathcal{P}_4(\mathbb{F})\), so \(\dim \mathcal{P}_4 = 5\) by 2.36. The set \(U\) is a subspace: the zero polynomial satisfies the condition, and if \(p(2) = p(5)\) and \(q(2) = q(5)\) then \((p+q)(2) = (p+q)(5)\) and \((\lambda p)(2) = (\lambda p)(5)\) for every \(\lambda \in \mathbb{F}\).

A convenient basis of \(\mathcal{P}_4\). Consider

\begin{equation*} 1,\quad x-2,\quad (x-2)(x-5),\quad x(x-2)(x-5),\quad x^2(x-2)(x-5). \end{equation*}

These are monic polynomials of degrees \(0,1,2,3,4\). If a linear combination \(c_0 \cdot 1 + c_1(x-2) + c_2(x-2)(x-5) + c_3 x(x-2)(x-5) + c_4 x^2(x-2)(x-5)\) equals the zero polynomial, then reading off the coefficient of \(x^4\) gives \(c_4 = 0\); with that, the coefficient of \(x^3\) gives \(c_3 = 0\); then \(x^2\) gives \(c_2 = 0\), then \(x^1\) gives \(c_1 = 0\), and finally \(c_0 = 0\). So the list is linearly independent; having length \(5 = \dim\mathcal{P}_4\), it is a basis of \(\mathcal{P}_4\) by 2.38.

(a) Write \(p \in \mathcal{P}_4\) in this basis:

\begin{equation*} p = c_0 + c_1(x-2) + c_2(x-2)(x-5) + c_3 x(x-2)(x-5) + c_4 x^2(x-2)(x-5). \end{equation*}

Every term after the second vanishes at both \(2\) and \(5\), so

\begin{equation*} p(2) = c_0, \qquad p(5) = c_0 + 3c_1 . \end{equation*}

Hence \(p(2) = p(5)\) if and only if \(3c_1 = 0\), i.e. (since \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\)) if and only if \(c_1 = 0\). Therefore \(U\) is exactly the span of

\begin{equation*} 1,\quad (x-2)(x-5),\quad x(x-2)(x-5),\quad x^2(x-2)(x-5), \end{equation*}

each of which does lie in \(U\). Being a sublist of a linearly independent list, this list is linearly independent, so it is a basis of \(U\) and \(\dim U = 4\).

(b) Adjoin \(x-2\). The resulting list

\begin{equation*} 1,\quad x-2,\quad (x-2)(x-5),\quad x(x-2)(x-5),\quad x^2(x-2)(x-5) \end{equation*}

is the basis of \(\mathcal{P}_4\) exhibited above.

(c) Take \(W = \operatorname{span}(x-2)\). Given \(p \in \mathcal{P}_4\), set \(\lambda = \tfrac{1}{3}\bigl(p(5) - p(2)\bigr)\) and write

\begin{equation*} p = \bigl(p - \lambda(x-2)\bigr) + \lambda(x-2). \end{equation*}

The second summand lies in \(W\). For the first, put \(q = p - \lambda(x-2)\); then \(q(2) = p(2)\) and

\begin{equation*} q(5) = p(5) - 3\lambda = p(5) - \bigl(p(5) - p(2)\bigr) = p(2) = q(2), \end{equation*}

so \(q \in U\). Hence \(\mathcal{P}_4 = U + W\). If \(r \in U \cap W\), then \(r = c(x-2)\) for some \(c \in \mathbb{F}\), and \(r(2) = r(5)\) gives \(0 = 3c\), so \(c = 0\) and \(r = 0\). Thus \(U \cap W = \{0\}\), and by 1.46 the sum is direct: \(\mathcal{P}_4(\mathbb{F}) = U \oplus W\).

Write \(\mathcal{P}_4 = \mathcal{P}_4(\mathbb{F})\), so \(\dim \mathcal{P}_4 = 5\) by 2.36. As in the previous exercises, \(U\) is a subspace of \(\mathcal{P}_4\): it contains \(0\), and the conditions \(p(2) = p(5)\) and \(p(5) = p(6)\) are preserved under addition and scalar multiplication.

A convenient basis of \(\mathcal{P}_4\). Consider

\begin{equation*} 1,\quad x-2,\quad (x-2)(x-5),\quad (x-2)(x-5)(x-6),\quad x(x-2)(x-5)(x-6). \end{equation*}

These are monic of degrees \(0,1,2,3,4\). If a linear combination with coefficients \(c_0, \dots, c_4\) is the zero polynomial, then the coefficient of \(x^4\) gives \(c_4 = 0\), then the coefficient of \(x^3\) gives \(c_3 = 0\), then \(x^2\) gives \(c_2 = 0\), then \(x\) gives \(c_1 = 0\), and finally \(c_0 = 0\). So the list is linearly independent, and having length \(5 = \dim\mathcal{P}_4\) it is a basis of \(\mathcal{P}_4\) by 2.38.

(a) Write \(p \in \mathcal{P}_4\) in this basis:

\begin{equation*} p = c_0 + c_1(x-2) + c_2(x-2)(x-5) + c_3(x-2)(x-5)(x-6) + c_4 x(x-2)(x-5)(x-6). \end{equation*}

The last two terms vanish at each of \(2, 5, 6\). Evaluating,

\begin{equation*} p(2) = c_0, \qquad p(5) = c_0 + 3c_1, \qquad p(6) = c_0 + 4c_1 + 4c_2 . \end{equation*}

Thus \(p \in U\) if and only if \(3c_1 = 0\) and \(4c_1 + 4c_2 = 0\), which (since \(3\) and \(4\) are invertible in \(\mathbb{F}\)) holds if and only if \(c_1 = c_2 = 0\). Hence \(U\) is precisely the span of

\begin{equation*} 1,\quad (x-2)(x-5)(x-6),\quad x(x-2)(x-5)(x-6), \end{equation*}

all three of which do belong to \(U\) (the constant \(1\) takes the same value everywhere; the other two vanish at \(2\), \(5\), and \(6\)). This list is linearly independent as a sublist of a linearly independent list, so it is a basis of \(U\) and \(\dim U = 3\).

(b) Adjoin \(x-2\) and \((x-2)(x-5)\). The resulting list

\begin{equation*} 1,\quad (x-2)(x-5)(x-6),\quad x(x-2)(x-5)(x-6),\quad x-2,\quad (x-2)(x-5) \end{equation*}

is, after reordering, the basis of \(\mathcal{P}_4\) exhibited above.

(c) Take \(W = \operatorname{span}\bigl(x-2,\ (x-2)(x-5)\bigr)\).

Since the basis of \(U\) from (a) together with the two adjoined polynomials spans \(\mathcal{P}_4\), and every vector in that spanning list lies in \(U\) or in \(W\), we get \(\mathcal{P}_4 = U + W\).

Now suppose \(q \in U \cap W\). Because \(q \in U\), part (a) gives scalars \(a_0, a_3, a_4\) with

\begin{equation*} q = a_0 + a_3 (x-2)(x-5)(x-6) + a_4\, x(x-2)(x-5)(x-6); \end{equation*}

because \(q \in W\), there are scalars \(b_1, b_2\) with \(q = b_1(x-2) + b_2(x-2)(x-5)\). Subtracting these two expressions expresses the zero polynomial as a linear combination of the basis of \(\mathcal{P}_4\) from (b); by the uniqueness in the criterion for a basis (2.28), all coefficients are \(0\). In particular \(a_0 = a_3 = a_4 = 0\), so \(q = 0\).

Hence \(U \cap W = \{0\}\), and by 1.46 the sum is direct: \(\mathcal{P}_4(\mathbb{F}) = U \oplus W\).

(As a check on dimensions, \(\dim U + \dim W = 3 + 2 = 5 = \dim \mathcal{P}_4\), consistent with 2.43.)

Write \(\mathcal{P}_4 = \mathcal{P}_4(\mathbb{R})\), so \(\dim \mathcal{P}_4 = 5\) by 2.36. Because the integral is linear, \(U\) is a subspace of \(\mathcal{P}_4\): the zero polynomial has integral \(0\), and

\begin{equation*} \int_{-1}^{1}(p + q) = \int_{-1}^{1} p + \int_{-1}^{1} q, \qquad \int_{-1}^{1} \lambda p = \lambda \int_{-1}^{1} p . \end{equation*}

The basic computation we need is, for an integer \(k \geq 0\),

\begin{equation*} \int_{-1}^{1} x^{k}\,dx = \frac{1 - (-1)^{k+1}}{k+1} = \begin{cases} \dfrac{2}{k+1}, & k \text{ even},\\[4pt] 0, & k \text{ odd}. \end{cases} \end{equation*}

In particular \(\int_{-1}^{1} 1 = 2\), \(\int_{-1}^{1} x = 0\), \(\int_{-1}^{1} x^2 = \tfrac{2}{3}\), \(\int_{-1}^{1} x^3 = 0\), and \(\int_{-1}^{1} x^4 = \tfrac{2}{5}\).

A convenient basis of \(\mathcal{P}_4\). Consider

\begin{equation*} 1,\quad x,\quad x^2 - \tfrac{1}{3},\quad x^3,\quad x^4 - \tfrac{1}{5}. \end{equation*}

These are monic of degrees \(0,1,2,3,4\), so the list is linearly independent: if a linear combination with coefficients \(c_0, \dots, c_4\) is the zero polynomial, the coefficient of \(x^4\) forces \(c_4 = 0\), then the coefficient of \(x^3\) forces \(c_3 = 0\), then \(x^2\) forces \(c_2 = 0\), then \(x\) forces \(c_1 = 0\), and finally \(c_0 = 0\). Having length \(5 = \dim\mathcal{P}_4\), it is a basis of \(\mathcal{P}_4\) by 2.38.

(a) By the computation above,

\begin{equation*} \int_{-1}^{1}\Bigl(x^2 - \tfrac{1}{3}\Bigr) = \tfrac{2}{3} - \tfrac{2}{3} = 0, \qquad \int_{-1}^{1}\Bigl(x^4 - \tfrac{1}{5}\Bigr) = \tfrac{2}{5} - \tfrac{2}{5} = 0, \end{equation*}

and \(\int_{-1}^{1} x = \int_{-1}^{1} x^3 = 0\), while \(\int_{-1}^{1} 1 = 2\). So if

\begin{equation*} p = c_0 + c_1 x + c_2\Bigl(x^2 - \tfrac{1}{3}\Bigr) + c_3 x^3 + c_4\Bigl(x^4 - \tfrac{1}{5}\Bigr), \end{equation*}

then \(\int_{-1}^{1} p = 2c_0\). Hence \(p \in U\) if and only if \(c_0 = 0\), i.e. if and only if \(p\) lies in the span of

\begin{equation*} x,\quad x^2 - \tfrac{1}{3},\quad x^3,\quad x^4 - \tfrac{1}{5}. \end{equation*}

These four polynomials lie in \(U\) and span \(U\), and they are linearly independent as a sublist of a linearly independent list. So they form a basis of \(U\), and \(\dim U = 4\).

(b) Adjoin the constant polynomial \(1\):

\begin{equation*} 1,\quad x,\quad x^2 - \tfrac{1}{3},\quad x^3,\quad x^4 - \tfrac{1}{5} \end{equation*}

is the basis of \(\mathcal{P}_4\) exhibited above.

(c) Take \(W = \operatorname{span}(1)\), the subspace of constant polynomials. Given \(p \in \mathcal{P}_4\), set \(c = \tfrac{1}{2}\int_{-1}^{1} p\) and write

\begin{equation*} p = (p - c) + c . \end{equation*}

The second summand is a constant, hence in \(W\), and

\begin{equation*} \int_{-1}^{1}(p - c) = \int_{-1}^{1} p - 2c = 0 , \end{equation*}

so \(p - c \in U\). Hence \(\mathcal{P}_4 = U + W\). If \(q \in U \cap W\), then \(q\) is a constant, say \(q = a\), and \(0 = \int_{-1}^{1} a = 2a\), so \(a = 0\) and \(q = 0\). Thus \(U \cap W = \{0\}\), and by 1.46 the sum is direct: \(\mathcal{P}_4(\mathbb{R}) = U \oplus W\).

Write \(U = \operatorname{span}(v_1 + w, \dots, v_m + w)\). Since \(U\) is spanned by a finite list, \(U\) is finite-dimensional and \(\dim U\) makes sense.

If \(m = 1\) the claim reads \(\dim U \ge 0\), which is true because dimensions are nonnegative. So assume \(m \ge 2\).

Consider the list of \(m - 1\) vectors

\begin{equation*} v_1 - v_m,\ v_2 - v_m,\ \dots,\ v_{m-1} - v_m. \end{equation*}

First, every vector in this list belongs to \(U\): for each \(k \in \{1, \dots, m-1\}\),

\begin{equation*} v_k - v_m = (v_k + w) - (v_m + w), \end{equation*}

which is a linear combination of \(v_1 + w, \dots, v_m + w\) and hence lies in \(U\).

Second, this list is linearly independent. Suppose \(a_1, \dots, a_{m-1} \in \mathbb{F}\) satisfy

\begin{equation*} a_1 (v_1 - v_m) + \dots + a_{m-1}(v_{m-1} - v_m) = 0. \end{equation*}

Regrouping, this says

\begin{equation*} a_1 v_1 + \dots + a_{m-1} v_{m-1} + \bigl(-(a_1 + \dots + a_{m-1})\bigr) v_m = 0 . \end{equation*}

Because \(v_1, \dots, v_m\) is linearly independent, every coefficient in this equation is \(0\); in particular \(a_1 = \dots = a_{m-1} = 0\). Hence \(v_1 - v_m, \dots, v_{m-1} - v_m\) is linearly independent.

So \(U\) contains a linearly independent list of length \(m - 1\). A basis of \(U\) is a spanning list of \(U\) of length \(\dim U\), and by 2.22 the length of a linearly independent list in \(U\) is at most the length of a spanning list of \(U\). Therefore

\begin{equation*} m - 1 \le \dim U = \dim \operatorname{span}(v_1 + w, \dots, v_m + w), \end{equation*}

as desired.

Two remarks. The bound is sharp: if \(w = -v_1\) and \(m \ge 2\), then the list \(v_1 + w, \dots, v_m + w\) begins with \(0\) and one checks (by the same regrouping computation) that the remaining \(m-1\) vectors \(v_2 - v_1, \dots, v_m - v_1\) are linearly independent, so the span has dimension exactly \(m - 1\). On the other hand, if \(w \notin \operatorname{span}(v_1, \dots, v_m)\), or more generally if \(w\) is not of the form \(-\sum b_k v_k\) with \(\sum b_k = 1\), then the dimension is \(m\).

Each \(p_k\) has degree \(k \le m\), so \(p_0, p_1, \dots, p_m\) is a list of vectors in \(\mathcal{P}_m(\mathbb{F})\). This list has length \(m + 1\), and \(\dim \mathcal{P}_m(\mathbb{F}) = m + 1\) by 2.36. Hence by 2.38 (a linearly independent list of the right length is a basis), it suffices to prove that \(p_0, p_1, \dots, p_m\) is linearly independent.

Suppose \(a_0, a_1, \dots, a_m \in \mathbb{F}\) are such that

\begin{equation*} a_0 p_0 + a_1 p_1 + \dots + a_m p_m = 0, \end{equation*}

meaning that the left side is the zero function on \(\mathbb{F}\). Suppose, for the sake of contradiction, that not all the \(a_k\) equal \(0\). Let \(k\) be the largest index in \(\{0, 1, \dots, m\}\) with \(a_k \ne 0\). Then

\begin{equation*} a_0 p_0 + a_1 p_1 + \dots + a_k p_k = 0 . \end{equation*}

For each \(j\), write \(p_j(z) = c_{j,0} + c_{j,1} z + \dots + c_{j,j} z^{\,j}\) with \(c_{j,j} \ne 0\) (possible because \(p_j\) has degree \(j\)). Collecting coefficients, the left side of the displayed equation is the polynomial whose coefficient of \(z^k\) equals

\begin{equation*} a_k c_{k,k}, \end{equation*}

because for \(j < k\) the polynomial \(p_j\) has degree \(j < k\) and so contributes nothing to the coefficient of \(z^k\). Since \(a_k \ne 0\) and \(c_{k,k} \ne 0\), we get \(a_k c_{k,k} \ne 0\).

But the left side of the displayed equation is the zero function, all of whose coefficients are \(0\). As noted immediately after 4.8, the coefficients of a polynomial are uniquely determined (if a polynomial had two different lists of coefficients, subtracting the two representations would produce a polynomial with a nonzero coefficient and yet infinitely many zeros, contradicting 4.8 together with the fact that \(\mathbb{F}\), being \(\mathbb{R}\) or \(\mathbb{C}\), is infinite). Hence the coefficient of \(z^k\) on the left side must be \(0\), i.e. \(a_k c_{k,k} = 0\), a contradiction.

Therefore all the \(a_k\) equal \(0\), so \(p_0, p_1, \dots, p_m\) is linearly independent, and therefore it is a basis of \(\mathcal{P}_m(\mathbb{F})\).

Note that this argument covers the case \(k = 0\) as well: there the equation reads \(a_0 p_0 = 0\) with \(p_0\) the nonzero constant \(c_{0,0}\), so \(a_0 c_{0,0} = 0\) forces \(a_0 = 0\).

Each \(p_k\) is a product of a polynomial of degree \(k\) and a polynomial of degree \(m - k\), hence is a polynomial of degree \(m\); in particular \(p_k \in \mathcal{P}_m(\mathbb{F})\). Thus \(p_0, \dots, p_m\) is a list of \(m + 1\) vectors in \(\mathcal{P}_m(\mathbb{F})\), and \(\dim \mathcal{P}_m(\mathbb{F}) = m + 1\) by 2.36. So by 2.38 it suffices to show that this list is linearly independent.

Suppose \(a_0, \dots, a_m \in \mathbb{F}\) are such that

\begin{equation*} a_0 p_0 + a_1 p_1 + \dots + a_m p_m = 0 , \end{equation*}

where \(0\) denotes the zero function. Suppose, for the sake of contradiction, that not all the \(a_k\) equal \(0\), and let \(j\) be the smallest index with \(a_j \ne 0\). Then

\begin{equation*} \sum_{k=j}^{m} a_k x^k (1 - x)^{m-k} = 0 \end{equation*}

for every \(x \in \mathbb{F}\).

Now expand the left side as a polynomial and look at the coefficient of \(x^{\,j}\). By the binomial theorem,

\begin{equation*} x^k (1 - x)^{m-k} = \sum_{i=0}^{m-k} \binom{m-k}{i} (-1)^i x^{k+i}, \end{equation*}

so every monomial occurring in \(x^k (1-x)^{m-k}\) has degree at least \(k\). Consequently the terms with \(k > j\) contribute nothing to the coefficient of \(x^{\,j}\), and the term with \(k = j\) contributes \(a_j \binom{m-j}{0} (-1)^0 = a_j\). Hence the left side, written in standard form, has \(x^{\,j}\)-coefficient equal to \(a_j \ne 0\).

But the left side of the displayed equation is the zero function, all of whose coefficients are \(0\). As noted immediately after 4.8, the coefficients of a polynomial are uniquely determined: if a polynomial had two different lists of coefficients, subtracting the two representations would give a polynomial with a nonzero coefficient and yet infinitely many zeros, which is impossible because such a polynomial has some degree \(d \ge 0\), and a polynomial of degree \(d \ge 1\) has at most \(d\) zeros by 4.8 while a polynomial of degree \(0\) is a nonzero constant and so has none, whereas \(\mathbb{F}\) (being \(\mathbb{R}\) or \(\mathbb{C}\)) is infinite. Hence the \(x^{\,j}\)-coefficient of the left side is \(0\), i.e. \(a_j = 0\), contradicting the choice of \(j\).

Therefore all the \(a_k\) equal \(0\), so \(p_0, \dots, p_m\) is linearly independent. Being a linearly independent list of length \(\dim \mathcal{P}_m(\mathbb{F})\), it is a basis of \(\mathcal{P}_m(\mathbb{F})\) by 2.38.

Remark: the actual Bernstein basis polynomials of degree \(m\) are \(\binom{m}{k} x^k (1-x)^{m-k}\); multiplying each \(p_k\) by the nonzero scalar \(\binom{m}{k}\) clearly does not affect linear independence, so those also form a basis of \(\mathcal{P}_m(\mathbb{F})\).

Here \(\mathbb{C}^6\) is regarded as a vector space over \(\mathbb{C}\), so \(\dim \mathbb{C}^6 = 6\) by 2.36 and in particular \(\mathbb{C}^6\) is finite-dimensional. Every subspace of a finite-dimensional vector space is finite-dimensional (2.25), and its dimension is at most the dimension of the whole space (2.37). Since \(U \cap W\) and \(U + W\) are subspaces of \(\mathbb{C}^6\) (the intersection by Exercise 10 in Section 1C, the sum by 1.40), both are finite-dimensional and \(\dim(U + W) \le 6\).

By 2.43,

\begin{equation*} \dim(U \cap W) = \dim U + \dim W - \dim(U + W) \ge 4 + 4 - 6 = 2 . \end{equation*}

Let \(n = \dim(U \cap W) \ge 2\) and let \(u_1, \dots, u_n\) be a basis of \(U \cap W\). Then \(u_1, u_2\) are two vectors in \(U \cap W\), and the list \(u_1, u_2\) is linearly independent, because any sublist of a linearly independent list is linearly independent (if a nontrivial linear combination of \(u_1, u_2\) equalled \(0\), then padding it with zero coefficients on \(u_3, \dots, u_n\) would give a nontrivial linear combination of \(u_1, \dots, u_n\) equal to \(0\)).

Finally, a linearly independent list of length two consists of two vectors neither of which is a scalar multiple of the other. Indeed, if \(u_2 = \lambda u_1\) for some \(\lambda \in \mathbb{C}\), then

\begin{equation*} \lambda u_1 + (-1) u_2 = 0 \end{equation*}

is a linear combination equal to \(0\) whose coefficients are not both \(0\) (the coefficient of \(u_2\) is \(-1\)), contradicting linear independence. Symmetrically, if \(u_1 = \lambda u_2\) then \((-1) u_1 + \lambda u_2 = 0\) gives the same contradiction.

Thus \(u_1\) and \(u_2\) are two vectors in \(U \cap W\) such that neither is a scalar multiple of the other.

The space \(\mathbb{R}^8\) has dimension eight (by 2.36), so it is finite-dimensional and hence all of its subspaces are finite-dimensional (by 2.25); in particular \(U \cap W\), which is a subspace of \(\mathbb{R}^8\) by Exercise 10 in Section 1C, is finite-dimensional.

Apply 2.43 to \(U\) and \(W\):

\begin{equation*} \dim(U + W) = \dim U + \dim W - \dim(U \cap W). \end{equation*}

By hypothesis \(U + W = \mathbb{R}^8\), so \(\dim(U + W) = 8\), and \(\dim U = 3\), \(\dim W = 5\). Hence

\begin{equation*} 8 = 3 + 5 - \dim(U \cap W), \end{equation*}

which gives \(\dim(U \cap W) = 0\).

A vector space of dimension \(0\) equals \(\{0\}\): a basis of \(U \cap W\) is then a list of length \(0\), i.e. the empty list, and the span of the empty list is \(\{0\}\), so \(U \cap W = \{0\}\).

By 1.46, \(U + W\) is a direct sum. Since \(U + W = \mathbb{R}^8\), we conclude

\begin{equation*} \mathbb{R}^8 = U \oplus W , \end{equation*}

as desired.

(Note that the numerical hypothesis \(\dim U + \dim W = 8 = \dim \mathbb{R}^8\) is exactly what makes the argument work; without it, \(U + W = \mathbb{R}^8\) would not force the sum to be direct.)

The space \(\mathbb{R}^9\) has dimension nine (by 2.36), so it is finite-dimensional. Both \(U + W\) and \(U \cap W\) are subspaces of \(\mathbb{R}^9\) (by 1.40 and by Exercise 10 in Section 1C), hence are finite-dimensional (by 2.25) with dimension at most nine (by 2.37). In particular

\begin{equation*} \dim(U + W) \le 9 . \end{equation*}

By 2.43,

\begin{equation*} \dim(U \cap W) = \dim U + \dim W - \dim(U + W) \ge 5 + 5 - 9 = 1 . \end{equation*}

Thus \(\dim(U \cap W) \ge 1\). If we had \(U \cap W = \{0\}\), then the empty list would be a basis of \(U \cap W\) and so \(\dim(U \cap W) = 0\), contradicting the inequality above. Hence \(U \cap W \ne \{0\}\).

Equivalently: since \(\dim(U \cap W) \ge 1\), the subspace \(U \cap W\) has a basis of length at least one, and the vectors in a basis are nonzero (a list containing \(0\) is linearly dependent), so \(U \cap W\) contains a nonzero vector.

Every subspace of the finite-dimensional space \(V\) is finite-dimensional (by 2.25) with dimension at most \(\dim V = 10\) (by 2.37). Moreover the intersection of two subspaces of \(V\) is again a subspace of \(V\) (Exercise 10 in Section 1C), and the sum of two subspaces of \(V\) is again a subspace of \(V\) (by 1.40). So all the dimensions below are defined, and every sum appearing below has dimension at most \(10\).

Step 1: apply 2.43 to \(V_1\) and \(V_2\). Since \(V_1 + V_2\) is a subspace of \(V\), we have \(\dim(V_1 + V_2) \le 10\), so

\begin{equation*} \dim(V_1 \cap V_2) = \dim V_1 + \dim V_2 - \dim(V_1 + V_2) \ge 7 + 7 - 10 = 4 . \end{equation*}

Step 2: apply 2.43 to the subspaces \(V_1 \cap V_2\) and \(V_3\). Note that

\begin{equation*} (V_1 \cap V_2) \cap V_3 = V_1 \cap V_2 \cap V_3 , \end{equation*}

and \((V_1 \cap V_2) + V_3\) is a subspace of \(V\), so its dimension is at most \(10\). Therefore

\begin{equation*} \begin{aligned} \dim(V_1 \cap V_2 \cap V_3) &= \dim(V_1 \cap V_2) + \dim V_3 - \dim\bigl((V_1 \cap V_2) + V_3\bigr) \\ &\ge 4 + 7 - 10 \\ &= 1 . \end{aligned} \end{equation*}

Hence \(\dim(V_1 \cap V_2 \cap V_3) \ge 1\). If \(V_1 \cap V_2 \cap V_3\) were \(\{0\}\), its dimension would be \(0\) (the empty list is a basis of \(\{0\}\)), contradicting the inequality above. Therefore

\begin{equation*} V_1 \cap V_2 \cap V_3 \ne \{0\} , \end{equation*}

as desired.

Remark: the same telescoping argument shows in general that if \(V_1, \dots, V_n\) are subspaces of an \(N\)-dimensional space \(V\), then

\begin{equation*} \dim(V_1 \cap \dots \cap V_n) \ge \dim V_1 + \dots + \dim V_n - (n-1) N , \end{equation*}

which here gives \(7 + 7 + 7 - 2 \cdot 10 = 1 > 0\).

The key is the following observation: if \(A\) and \(B\) are subspaces of the finite-dimensional space \(V\), then

\begin{equation*} \dim(A \cap B) \geq \dim A + \dim B - \dim V. \end{equation*}

Indeed, \(A + B\) is a subspace of \(V\) (by 1.40), so \(\dim(A+B) \leq \dim V\) by 2.37; combining this with the formula for the dimension of a sum (2.43),

\begin{equation*} \dim(A \cap B) = \dim A + \dim B - \dim(A + B) \geq \dim A + \dim B - \dim V. \end{equation*}

(All the subspaces appearing here are finite-dimensional by 2.25, so every dimension written is a well-defined nonnegative integer and 2.43 applies.)

Now apply this twice. First, with \(A = V_1\) and \(B = V_2\),

\begin{equation*} \dim(V_1 \cap V_2) \geq \dim V_1 + \dim V_2 - \dim V. \end{equation*}

Next, \(V_1 \cap V_2\) is a subspace of \(V\), so we may apply the observation again with \(A = V_1 \cap V_2\) and \(B = V_3\):

\begin{equation*} \begin{aligned} \dim(V_1 \cap V_2 \cap V_3) &= \dim\bigl((V_1 \cap V_2) \cap V_3\bigr)\\ &\geq \dim(V_1 \cap V_2) + \dim V_3 - \dim V\\ &\geq \bigl(\dim V_1 + \dim V_2 - \dim V\bigr) + \dim V_3 - \dim V\\ &= \dim V_1 + \dim V_2 + \dim V_3 - 2 \dim V. \end{aligned} \end{equation*}

By hypothesis the last quantity is strictly positive. Hence \(\dim(V_1 \cap V_2 \cap V_3) \geq 1\), and a subspace of dimension at least \(1\) contains a nonzero vector (the subspace \(\{0\}\) has dimension \(0\), its basis being the empty list). Therefore \(V_1 \cap V_2 \cap V_3 \neq \{0\}\).

First note that \(m < n\). Indeed \(m \leq n\) by 2.37, and if \(m = n\) then \(U = V\) by 2.39, contrary to hypothesis. Thus \(n - m \geq 1\), and in particular \(n \geq 1\).

Let \(u_1, \dots, u_m\) be a basis of \(U\) (this is the empty list when \(m = 0\), i.e. when \(U = \{0\}\)). A basis of \(U\) is a linearly independent list in \(V\), so by 2.32 it extends to a basis

\begin{equation*} u_1, \dots, u_m, v_1, \dots, v_{n-m} \end{equation*}

of \(V\); the extended list has length \(n\) because every basis of \(V\) has length \(\dim V = n\) (by 2.34).

For each \(k \in \{1, \dots, n-m\}\) define

\begin{equation*} W_k = \operatorname{span}(u_1, \dots, u_m, v_1, \dots, v_{k-1}, v_{k+1}, \dots, v_{n-m}), \end{equation*}

that is, the span of the whole basis with \(v_k\) deleted. This produces \(n - m\) subspaces of \(V\).

Each \(W_k\) has dimension \(n-1\). The list spanning \(W_k\) has length \(n - 1\), and it is a sublist of a linearly independent list, hence is itself linearly independent. A linearly independent list that spans \(W_k\) is a basis of \(W_k\), so \(\dim W_k = n - 1\).

The intersection contains \(U\). Each of \(u_1, \dots, u_m\) appears in the spanning list of \(W_k\), so \(U = \operatorname{span}(u_1, \dots, u_m) \subseteq W_k\) for every \(k\). Hence \(U \subseteq W_1 \cap \dots \cap W_{n-m}\).

The intersection is contained in \(U\). Suppose \(v \in W_1 \cap \dots \cap W_{n-m}\). Because \(u_1, \dots, u_m, v_1, \dots, v_{n-m}\) is a basis of \(V\), we may write

\begin{equation*} v = a_1 u_1 + \dots + a_m u_m + b_1 v_1 + \dots + b_{n-m} v_{n-m} \end{equation*}

for scalars \(a_1, \dots, a_m, b_1, \dots, b_{n-m}\), and by 2.28 this representation is unique. Fix \(k\). Since \(v \in W_k\), we can also write \(v\) as a linear combination of \(u_1, \dots, u_m, v_1, \dots, v_{k-1}, v_{k+1}, \dots, v_{n-m}\), which is a representation of \(v\) in the basis of \(V\) whose coefficient of \(v_k\) equals \(0\). Uniqueness now forces \(b_k = 0\). As \(k\) was arbitrary, \(b_1 = \dots = b_{n-m} = 0\), so

\begin{equation*} v = a_1 u_1 + \dots + a_m u_m \in U. \end{equation*}

Combining the two inclusions gives \(W_1 \cap \dots \cap W_{n-m} = U\), with \(W_1, \dots, W_{n-m}\) being \(n - m\) subspaces of \(V\) each of dimension \(n - 1\), as required.

For each \(k \in \{1, \dots, m\}\) put \(n_k = \dim V_k\) and choose a basis

\begin{equation*} v_{k,1}, \dots, v_{k,n_k} \end{equation*}

of \(V_k\); such a basis exists because \(V_k\) is finite-dimensional (by 2.31), and it has length \(n_k\) by the definition of dimension. (If \(V_k = \{0\}\) then \(n_k = 0\) and the list is empty; nothing below is affected.)

Let \(L\) denote the list obtained by concatenating these \(m\) lists:

\begin{equation*} v_{1,1}, \dots, v_{1,n_1}, \; v_{2,1}, \dots, v_{2,n_2}, \; \dots, \; v_{m,1}, \dots, v_{m,n_m}. \end{equation*}

This is a finite list of vectors of \(V\), of length \(n_1 + \dots + n_m\).

Claim: \(L\) spans \(V_1 + \dots + V_m\). By definition of the sum of subspaces (1.36), every element of \(V_1 + \dots + V_m\) has the form \(u_1 + \dots + u_m\) with \(u_k \in V_k\) for each \(k\). Since \(v_{k,1}, \dots, v_{k,n_k}\) spans \(V_k\), we may write

\begin{equation*} u_k = a_{k,1} v_{k,1} + \dots + a_{k,n_k} v_{k,n_k} \end{equation*}

for suitable scalars. Summing over \(k\) exhibits \(u_1 + \dots + u_m\) as a linear combination of the vectors in \(L\). Hence \(V_1 + \dots + V_m \subseteq \operatorname{span} L\). Conversely each vector of \(L\) lies in some \(V_k\) and hence in the subspace \(V_1 + \dots + V_m\) (a sum of subspaces is a subspace, by 1.40), so \(\operatorname{span} L \subseteq V_1 + \dots + V_m\). Therefore

\begin{equation*} \operatorname{span} L = V_1 + \dots + V_m. \end{equation*}

Because \(V_1 + \dots + V_m\) is a vector space with a finite spanning list, it is finite-dimensional by definition.

Finally, \(L\) is a spanning list of \(V_1 + \dots + V_m\) and any basis of \(V_1 + \dots + V_m\) is a linearly independent list in \(V_1 + \dots + V_m\); hence by 2.22 the length of that basis is at most the length of \(L\). In other words

\begin{equation*} \dim(V_1 + \dots + V_m) \leq n_1 + \dots + n_m = \dim V_1 + \dots + \dim V_m, \end{equation*}

as desired. (Equivalently, one can reduce \(L\) to a basis of \(V_1 + \dots + V_m\) using 2.30; the reduced list has length at most that of \(L\).)

Because \(V\) is finite-dimensional it has a basis (by 2.31), say \(v_1, \dots, v_n\); this list has length \(n = \dim V\) by the definition of dimension. For each \(k \in \{1, \dots, n\}\) set

\begin{equation*} V_k = \operatorname{span}(v_k). \end{equation*}

Each \(V_k\) is a subspace of \(V\), since a span is always a subspace.

Each \(V_k\) is one-dimensional. The list \(v_1, \dots, v_n\) is linearly independent, so no \(v_k\) equals \(0\): if \(v_k = 0\) then \(0 v_1 + \dots + 1 v_k + \dots + 0 v_n = 0\) would be a nontrivial vanishing linear combination. A one-vector list \(v_k\) with \(v_k \neq 0\) is linearly independent, because \(a v_k = 0\) with \(a \neq 0\) would give \(v_k = a^{-1}(a v_k) = 0\). Thus \(v_k\) is a linearly independent list spanning \(V_k\), hence a basis of \(V_k\), and \(\dim V_k = 1\).

The sum is all of \(V\). Given \(v \in V\), write \(v = a_1 v_1 + \dots + a_n v_n\) with \(a_1, \dots, a_n \in \mathbf{F}\), which is possible because \(v_1, \dots, v_n\) spans \(V\). Each \(a_k v_k \in V_k\), so \(v \in V_1 + \dots + V_n\). Conversely \(V_1 + \dots + V_n \subseteq V\) because each \(V_k\) is a subspace of \(V\). Hence \(V = V_1 + \dots + V_n\).

The sum is direct. By the condition for a direct sum (1.45), it suffices to show that the only way to write \(0 = w_1 + \dots + w_n\) with each \(w_k \in V_k\) is by taking every \(w_k = 0\). So suppose \(w_1 + \dots + w_n = 0\) with \(w_k \in V_k\). Each \(w_k\) has the form \(w_k = a_k v_k\) for some \(a_k \in \mathbf{F}\), so

\begin{equation*} a_1 v_1 + \dots + a_n v_n = 0. \end{equation*}

Since \(v_1, \dots, v_n\) is linearly independent, \(a_1 = \dots = a_n = 0\), and therefore \(w_1 = \dots = w_n = 0\).

Hence \(V = V_1 \oplus \dots \oplus V_n\) with each \(V_k\) one-dimensional, as required.

Why one might guess it. The table preceding these exercises sets up a dictionary between finite sets and finite-dimensional vector spaces: the set \(S\) corresponds to the space \(V\), the number of elements \(\#S\) corresponds to \(\dim V\), the union \(S_1 \cup S_2\) (the smallest subset containing \(S_1\) and \(S_2\)) corresponds to the sum \(V_1 + V_2\) (the smallest subspace containing \(V_1\) and \(V_2\), by 1.40), and intersection corresponds to intersection. Under this dictionary the inclusion-exclusion formula for two sets,

\begin{equation*} \#(S_1 \cup S_2) = \#S_1 + \#S_2 - \#(S_1 \cap S_2), \end{equation*}

becomes exactly 2.43:

\begin{equation*} \dim(V_1 + V_2) = \dim V_1 + \dim V_2 - \dim(V_1 \cap V_2), \end{equation*}

which is a true theorem. For three finite sets the counting formula reads

\begin{equation*} \begin{aligned} \#(S_1 \cup S_2 \cup S_3) = {}& \#S_1 + \#S_2 + \#S_3\\ & - \#(S_1 \cap S_2) - \#(S_1 \cap S_3) - \#(S_2 \cap S_3)\\ & + \#(S_1 \cap S_2 \cap S_3), \end{aligned} \end{equation*}

so translating term by term through the same dictionary produces the displayed guess. Since the dictionary gave the correct answer in the two-subspace case, it is natural to expect it to work again.

The guess is false. Take \(V = \mathbf{R}^2\) and

\begin{equation*} V_1 = \operatorname{span}\bigl((1,0)\bigr), \qquad V_2 = \operatorname{span}\bigl((0,1)\bigr), \qquad V_3 = \operatorname{span}\bigl((1,1)\bigr). \end{equation*}

Each of these is a one-dimensional subspace of \(\mathbf{R}^2\) (each is the span of a single nonzero vector), so \(\dim V_1 = \dim V_2 = \dim V_3 = 1\).

The sum contains \((1,0)\) and \((0,1)\), which span \(\mathbf{R}^2\), so \(V_1 + V_2 + V_3 = \mathbf{R}^2\) and

\begin{equation*} \dim(V_1 + V_2 + V_3) = 2. \end{equation*}

Each pairwise intersection is \(\{0\}\). For instance, a vector in \(V_1 \cap V_3\) has the form \((a, 0)\) and also the form \((b, b)\); equating second coordinates gives \(b = 0\), so the vector is \((0,0)\). The same argument (comparing the coordinate that vanishes) handles \(V_1 \cap V_2\) and \(V_2 \cap V_3\): an element of \(V_1 \cap V_2\) is \((a,0) = (0,b)\), forcing \(a = b = 0\), and an element of \(V_2 \cap V_3\) is \((0,a) = (b,b)\), forcing \(b = 0\). Hence all three pairwise intersections, and therefore also \(V_1 \cap V_2 \cap V_3\), equal \(\{0\}\), and all four of these intersections have dimension \(0\).

The right side of the proposed formula therefore equals

\begin{equation*} 1 + 1 + 1 - 0 - 0 - 0 + 0 = 3, \end{equation*}

while the left side equals \(2\). Since \(2 \neq 3\), the formula is false, and these three subspaces of \(\mathbf{R}^2\) form a counterexample.

The analogy breaks because \(\dim\) does not behave like a measure on subspaces: three distinct lines through the origin in the plane pairwise overlap only in \(\{0\}\), yet they cannot be independent, and the “overlap” responsible for this is invisible to the pairwise intersections. Exercise 2C.20 shows what the correct three-subspace identity looks like: it involves the spaces \((V_1 + V_2) \cap V_3\) and its permutations, not merely the pairwise intersections.

Let \(V\) be the ambient finite-dimensional vector space. Every subspace appearing below is a subspace of \(V\) (sums of subspaces are subspaces by 1.40, and an intersection of subspaces is a subspace, as is immediate from the conditions in 1.34), hence is finite-dimensional by 2.25. So all the dimensions written are well-defined nonnegative integers and 2.43 may be applied freely.

Step 1: a two-term grouping. Apply 2.43 to the two subspaces \(V_1 + V_2\) and \(V_3\), whose sum is \(V_1 + V_2 + V_3\):

\begin{equation*} \dim(V_1 + V_2 + V_3) = \dim(V_1 + V_2) + \dim V_3 - \dim\bigl((V_1 + V_2) \cap V_3\bigr). \end{equation*}

Apply 2.43 again, this time to \(V_1\) and \(V_2\):

\begin{equation*} \dim(V_1 + V_2) = \dim V_1 + \dim V_2 - \dim(V_1 \cap V_2). \end{equation*}

Substituting gives

\begin{equation*} \dim(V_1 + V_2 + V_3) = \dim V_1 + \dim V_2 + \dim V_3 - \dim(V_1 \cap V_2) - \dim\bigl((V_1 + V_2) \cap V_3\bigr). \end{equation*}

Call this identity (1).

Step 2: the other two groupings. Addition of subspaces is commutative and associative, so \(V_1 + V_2 + V_3\) equals \((V_1 + V_3) + V_2\) and also \((V_2 + V_3) + V_1\). Repeating the computation of Step 1 with these groupings yields

\begin{equation*} \dim(V_1 + V_2 + V_3) = \dim V_1 + \dim V_2 + \dim V_3 - \dim(V_1 \cap V_3) - \dim\bigl((V_1 + V_3) \cap V_2\bigr) \end{equation*}

(call this identity (2)) and

\begin{equation*} \dim(V_1 + V_2 + V_3) = \dim V_1 + \dim V_2 + \dim V_3 - \dim(V_2 \cap V_3) - \dim\bigl((V_2 + V_3) \cap V_1\bigr) \end{equation*}

(call this identity (3)). Explicitly, for (2): \(\dim(V_1 + V_2 + V_3) = \dim(V_1 + V_3) + \dim V_2 - \dim((V_1 + V_3) \cap V_2)\) and \(\dim(V_1 + V_3) = \dim V_1 + \dim V_3 - \dim(V_1 \cap V_3)\); and similarly for (3).

Step 3: average the three identities. Each of (1), (2), (3) has left side \(\dim(V_1 + V_2 + V_3)\). Adding them and dividing by \(3\) leaves the left side unchanged, and gives

\begin{equation*} \begin{aligned} \dim(V_1 + V_2 + V_3) = {}& \dim V_1 + \dim V_2 + \dim V_3\\ & - \frac{\dim(V_1 \cap V_2) + \dim(V_1 \cap V_3) + \dim(V_2 \cap V_3)}{3}\\ & - \frac{\dim\bigl((V_1 + V_2) \cap V_3\bigr) + \dim\bigl((V_1 + V_3) \cap V_2\bigr) + \dim\bigl((V_2 + V_3) \cap V_1\bigr)}{3}, \end{aligned} \end{equation*}

which is precisely the claimed formula.

Remark on the apparent non-integrality. Nothing is amiss. Write

\begin{equation*} c_1 = \dim(V_1 \cap V_2) + \dim\bigl((V_1 + V_2) \cap V_3\bigr), \end{equation*}

and let \(c_2, c_3\) be the analogous quantities from (2) and (3). Identities (1)-(3) say that \(c_1 = c_2 = c_3\), each being equal to the integer \(\dim V_1 + \dim V_2 + \dim V_3 - \dim(V_1 + V_2 + V_3)\). The two fractions in the statement together form \((c_1 + c_2 + c_3)/3\), the average of three equal integers, which is that same integer. The formula only looks non-integral because the pairing of each intersection term with its matching sum-intersection term has been broken up in the display.

First suppose \(b = c = 0\), so that

\begin{equation*} T(x,y,z) = (2x - 4y + 3z,\; 6x). \end{equation*}

To verify additivity, let \(u = (x_1, y_1, z_1)\) and \(v = (x_2, y_2, z_2)\) be elements of \(\mathbb{R}^3\). Then

\begin{equation*} \begin{aligned} T(u + v) &= T(x_1 + x_2,\; y_1 + y_2,\; z_1 + z_2) \\ &= \bigl(2(x_1 + x_2) - 4(y_1 + y_2) + 3(z_1 + z_2),\; 6(x_1 + x_2)\bigr) \\ &= (2x_1 - 4y_1 + 3z_1,\; 6x_1) + (2x_2 - 4y_2 + 3z_2,\; 6x_2) \\ &= Tu + Tv, \end{aligned} \end{equation*}

where the third equality holds by expanding each product and regrouping the resulting real numbers, using that addition in \(\mathbb{R}^2\) is coordinatewise. To verify homogeneity, let \(\lambda \in \mathbb{R}\) and \(v = (x,y,z) \in \mathbb{R}^3\). Then

\begin{equation*} \begin{aligned} T(\lambda v) &= T(\lambda x, \lambda y, \lambda z) \\ &= (2\lambda x - 4\lambda y + 3 \lambda z,\; 6 \lambda x) \\ &= \lambda(2x - 4y + 3z,\; 6x) \\ &= \lambda (Tv). \end{aligned} \end{equation*}

Hence \(T\) satisfies the two conditions in 3.1, so \(T\) is linear.

Now suppose, conversely, that \(T\) is linear; we must show \(b = c = 0\).

Because \(T\) is linear, 3.10 gives \(T(0,0,0) = (0,0)\). On the other hand, directly from the formula defining \(T\),

\begin{equation*} T(0,0,0) = (2 \cdot 0 - 4 \cdot 0 + 3 \cdot 0 + b,\; 6 \cdot 0 + c \cdot 0 \cdot 0 \cdot 0) = (b, 0). \end{equation*}

Comparing the first coordinates of these two expressions gives \(b = 0\).

To show that \(c = 0\), apply homogeneity with \(\lambda = 2\) and \(v = (1,1,1)\). From the formula defining \(T\),

\begin{equation*} T(1,1,1) = (2 - 4 + 3 + b,\; 6 + c) = (1 + b,\; 6 + c) \end{equation*}

and

\begin{equation*} T(2,2,2) = (4 - 8 + 6 + b,\; 12 + 8c) = (2 + b,\; 12 + 8c), \end{equation*}

since \(c \cdot 2 \cdot 2 \cdot 2 = 8c\). Homogeneity requires \(T(2,2,2) = 2\,T(1,1,1)\), and the second coordinate of \(2\,T(1,1,1)\) equals \(12 + 2c\). Thus

\begin{equation*} 12 + 8c = 12 + 2c, \end{equation*}

which gives \(6c = 0\) and hence \(c = 0\).

Therefore \(T\) is linear if and only if \(b = c = 0\).

First suppose \(b = c = 0\), so that

\begin{equation*} Tp = \left(3p(4) + 5p^{\prime}(6),\; \int_{-1}^{2} x^3 p(x)\,dx\right). \end{equation*}

Let \(p, q \in \mathcal{P}(\mathbb{R})\). Evaluation at a point is additive, and differentiation is additive (that is, \((p+q)^{\prime} = p^{\prime} + q^{\prime}\)), so the first coordinate of \(T(p+q)\) is

\begin{equation*} 3(p+q)(4) + 5(p+q)^{\prime}(6) = \bigl(3p(4) + 5p^{\prime}(6)\bigr) + \bigl(3q(4) + 5q^{\prime}(6)\bigr). \end{equation*}

The integral of a sum is the sum of the integrals, so the second coordinate of \(T(p+q)\) is

\begin{equation*} \int_{-1}^{2} x^3\bigl(p(x) + q(x)\bigr) dx = \int_{-1}^{2} x^3 p(x)\,dx + \int_{-1}^{2} x^3 q(x)\,dx. \end{equation*}

Since addition in \(\mathbb{R}^2\) is coordinatewise, these two displays give \(T(p+q) = Tp + Tq\).

Similarly, if \(\lambda \in \mathbb{R}\) and \(p \in \mathcal{P}(\mathbb{R})\), then \((\lambda p)(4) = \lambda p(4)\), \((\lambda p)^{\prime} = \lambda p^{\prime}\), and \(\int_{-1}^{2} x^3 \lambda p(x)\,dx = \lambda \int_{-1}^{2} x^3 p(x)\,dx\). Hence each coordinate of \(T(\lambda p)\) equals \(\lambda\) times the corresponding coordinate of \(Tp\), so \(T(\lambda p) = \lambda(Tp)\). Thus \(T\) is linear.

Now suppose, conversely, that \(T\) is linear; we must show \(b = c = 0\).

To show \(b = 0\), let \(p\) be the constant polynomial \(1\), so \(p^{\prime} = 0\). Then

\begin{equation*} 3p(4) + 5p^{\prime}(6) + b\,p(1)p(2) = 3 + b, \end{equation*}

while \(2p\) is the constant polynomial \(2\), whose first coordinate under \(T\) is

\begin{equation*} 3 \cdot 2 + 5 \cdot 0 + b \cdot 2 \cdot 2 = 6 + 4b. \end{equation*}

Homogeneity of \(T\) requires \(T(2p) = 2\,(Tp)\); comparing first coordinates gives

\begin{equation*} 6 + 4b = 2(3 + b) = 6 + 2b, \end{equation*}

so \(2b = 0\) and hence \(b = 0\).

To show \(c = 0\), let \(q\) be the constant polynomial \(\pi/2\). First note that

\begin{equation*} \int_{-1}^{2} x^3\,dx = \left[\frac{x^4}{4}\right]_{-1}^{2} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}. \end{equation*}

Since \(q(0) = \pi/2\) and \(\sin(\pi/2) = 1\), the second coordinate of \(Tq\) is

\begin{equation*} \int_{-1}^{2} x^3 \cdot \frac{\pi}{2}\,dx + c \sin\frac{\pi}{2} = \frac{15\pi}{8} + c. \end{equation*}

Since \(2q\) is the constant polynomial \(\pi\) and \(\sin \pi = 0\), the second coordinate of \(T(2q)\) is

\begin{equation*} \int_{-1}^{2} x^3 \cdot \pi\,dx + c \sin \pi = \frac{15\pi}{4}. \end{equation*}

Homogeneity of \(T\) requires \(T(2q) = 2\,(Tq)\); comparing second coordinates gives

\begin{equation*} \frac{15\pi}{4} = 2\left(\frac{15\pi}{8} + c\right) = \frac{15\pi}{4} + 2c, \end{equation*}

so \(2c = 0\) and hence \(c = 0\).

Therefore \(T\) is linear if and only if \(b = c = 0\).

Let \(e_1, \ldots, e_n\) denote the standard basis of \(\mathbb{F}^n\) (see 2.27); thus \(e_k\) is the element of \(\mathbb{F}^n\) whose \(k^{\text{th}}\) coordinate equals \(1\) and whose other coordinates equal \(0\).

For each \(k \in \{1, \ldots, n\}\) the vector \(Te_k\) lies in \(\mathbb{F}^m\), so it has \(m\) coordinates; define \(A_{j,k} \in \mathbb{F}\), for \(j = 1, \ldots, m\), to be the \(j^{\text{th}}\) coordinate of \(Te_k\). In other words, the scalars \(A_{j,k}\) are defined by

\begin{equation*} Te_k = (A_{1,k},\, A_{2,k},\, \ldots,\, A_{m,k}) \end{equation*}

for each \(k = 1, \ldots, n\). This defines \(A_{j,k}\) for all \(j = 1, \ldots, m\) and all \(k = 1, \ldots, n\), as required.

Now let \((x_1, \ldots, x_n) \in \mathbb{F}^n\) be arbitrary. Because addition and scalar multiplication in \(\mathbb{F}^n\) are defined coordinatewise, we have

\begin{equation*} (x_1, \ldots, x_n) = x_1 e_1 + \cdots + x_n e_n. \end{equation*}

The additivity of \(T\), applied repeatedly (formally, by induction on the number of summands), shows that \(T\) takes a sum of \(n\) vectors to the sum of their images; combining this with the homogeneity of \(T\) gives

\begin{equation*} T(x_1, \ldots, x_n) = T(x_1 e_1 + \cdots + x_n e_n) = x_1 (Te_1) + \cdots + x_n (Te_n). \end{equation*}

Substituting the coordinates of each \(Te_k\) and again using that the vector space operations on \(\mathbb{F}^m\) are coordinatewise, we obtain

\begin{equation*} \begin{aligned} T(x_1, \ldots, x_n) &= x_1 (A_{1,1}, \ldots, A_{m,1}) + \cdots + x_n (A_{1,n}, \ldots, A_{m,n}) \\ &= (A_{1,1}x_1 + \cdots + A_{1,n}x_n,\; \ldots,\; A_{m,1}x_1 + \cdots + A_{m,n}x_n), \end{aligned} \end{equation*}

because for each \(j\) the \(j^{\text{th}}\) coordinate of the middle expression equals \(A_{j,1}x_1 + \cdots + A_{j,n}x_n\).

Since \((x_1, \ldots, x_n)\) was an arbitrary element of \(\mathbb{F}^n\), the scalars \(A_{j,k}\) constructed above have the required property.

Conversely, every choice of scalars \(A_{j,k}\) produces a linear map by the formula above (this is the second to last item of Example 3.3), so the maps of this form are exactly the elements of \(\mathcal{L}(\mathbb{F}^n, \mathbb{F}^m)\).

Suppose \(a_1, \ldots, a_m \in \mathbb{F}\) are such that

\begin{equation*} a_1 v_1 + \cdots + a_m v_m = 0. \end{equation*}

We must show that \(a_1 = \cdots = a_m = 0\).

Apply \(T\) to both sides of the equation above. The homogeneity of \(T\) gives \(T(a_k v_k) = a_k (Tv_k)\) for each \(k\), and the additivity of \(T\), applied repeatedly, shows that \(T\) takes a sum of \(m\) vectors to the sum of their images. Hence

\begin{equation*} T(a_1 v_1 + \cdots + a_m v_m) = a_1 (Tv_1) + \cdots + a_m (Tv_m). \end{equation*}

On the other hand, the left side equals \(T(0)\), which equals \(0\) by 3.10. Therefore

\begin{equation*} a_1 (Tv_1) + \cdots + a_m (Tv_m) = 0. \end{equation*}

Because \(Tv_1, \ldots, Tv_m\) is a linearly independent list in \(W\), the only choice of scalars making such a linear combination equal \(0\) is the choice in which all the scalars are \(0\). Thus \(a_1 = \cdots = a_m = 0\).

We have shown that the only way to write \(0\) as a linear combination of \(v_1, \ldots, v_m\) is with all coefficients equal to \(0\). Hence \(v_1, \ldots, v_m\) is linearly independent, as desired.

Note that the converse of this result is false: if \(T = 0\) and \(v_1, \ldots, v_m\) is any linearly independent list with \(m \geq 1\), then \(Tv_1, \ldots, Tv_m\) is the list \(0, \ldots, 0\), which is not linearly independent.

Recall the operations, defined in 3.5: for \(S, T \in \mathcal{L}(V,W)\) and \(\lambda \in \mathbb{F}\),

\begin{equation*} (S+T)(v) = Sv + Tv \quad \text{and} \quad (\lambda T)(v) = \lambda(Tv) \end{equation*}

for all \(v \in V\). Throughout, two functions from \(V\) to \(W\) are equal precisely when they agree at every \(v \in V\), so every identity below is verified by evaluating both sides at an arbitrary \(v \in V\). All manipulations on the right-hand sides take place in the vector space \(W\), whose vector space properties we use freely.

Step 1: the operations produce elements of \(\mathcal{L}(V,W)\).

Let \(S, T \in \mathcal{L}(V,W)\) and \(\lambda \in \mathbb{F}\). For \(u, v \in V\),

\begin{equation*} \begin{aligned} (S+T)(u+v) &= S(u+v) + T(u+v) = (Su + Sv) + (Tu + Tv) \\ &= (Su + Tu) + (Sv + Tv) = (S+T)(u) + (S+T)(v), \end{aligned} \end{equation*}

using the additivity of \(S\) and \(T\) and then the commutativity and associativity of addition in \(W\). For \(\mu \in \mathbb{F}\) and \(v \in V\),

\begin{equation*} (S+T)(\mu v) = S(\mu v) + T(\mu v) = \mu(Sv) + \mu(Tv) = \mu(Sv + Tv) = \mu\bigl((S+T)(v)\bigr), \end{equation*}

using homogeneity of \(S\) and \(T\) and a distributive property in \(W\). Hence \(S + T \in \mathcal{L}(V,W)\).

Similarly, for \(u, v \in V\),

\begin{equation*} (\lambda T)(u+v) = \lambda\bigl(T(u+v)\bigr) = \lambda(Tu + Tv) = \lambda(Tu) + \lambda(Tv) = (\lambda T)(u) + (\lambda T)(v), \end{equation*}

and for \(\mu \in \mathbb{F}\) and \(v \in V\),

\begin{equation*} (\lambda T)(\mu v) = \lambda\bigl(T(\mu v)\bigr) = \lambda\bigl(\mu (Tv)\bigr) = (\lambda\mu)(Tv) = (\mu\lambda)(Tv) = \mu\bigl(\lambda(Tv)\bigr) = \mu\bigl((\lambda T)(v)\bigr), \end{equation*}

using associativity of scalar multiplication in \(W\) and commutativity of multiplication in \(\mathbb{F}\). Hence \(\lambda T \in \mathcal{L}(V,W)\). So \(\mathcal{L}(V,W)\) is closed under both operations, and they are genuine operations on \(\mathcal{L}(V,W)\).

Step 2: commutativity. For \(S, T \in \mathcal{L}(V,W)\) and \(v \in V\),

\begin{equation*} (S+T)(v) = Sv + Tv = Tv + Sv = (T+S)(v), \end{equation*}

so \(S + T = T + S\).

Step 3: associativity. For \(R, S, T \in \mathcal{L}(V,W)\) and \(v \in V\),

\begin{equation*} \bigl((R+S)+T\bigr)(v) = (Rv + Sv) + Tv = Rv + (Sv + Tv) = \bigl(R+(S+T)\bigr)(v), \end{equation*}

so \((R+S)+T = R+(S+T)\). For \(\lambda, \mu \in \mathbb{F}\), \(T \in \mathcal{L}(V,W)\), and \(v \in V\),

\begin{equation*} \bigl((\lambda\mu)T\bigr)(v) = (\lambda\mu)(Tv) = \lambda\bigl(\mu(Tv)\bigr) = \lambda\bigl((\mu T)(v)\bigr) = \bigl(\lambda(\mu T)\bigr)(v), \end{equation*}

so \((\lambda\mu)T = \lambda(\mu T)\).

Step 4: additive identity. Let \(0\) denote the zero linear map from \(V\) to \(W\) defined in Example 3.3, so \(0v = 0\) for all \(v \in V\); it belongs to \(\mathcal{L}(V,W)\). For every \(T \in \mathcal{L}(V,W)\) and \(v \in V\),

\begin{equation*} (T + 0)(v) = Tv + 0 = Tv, \end{equation*}

so \(T + 0 = T\). Thus \(\mathcal{L}(V,W)\) has an additive identity.

Step 5: additive inverses. Let \(T \in \mathcal{L}(V,W)\) and set \(-T = (-1)T\), which lies in \(\mathcal{L}(V,W)\) by Step 1. For every \(v \in V\),

\begin{equation*} \bigl(T + (-T)\bigr)(v) = Tv + (-1)(Tv) = 0, \end{equation*}

using 1.32 in \(W\) (namely that \((-1)w\) is the additive inverse of \(w\)). Hence \(T + (-T) = 0\), so every element of \(\mathcal{L}(V,W)\) has an additive inverse.

Step 6: multiplicative identity. For \(T \in \mathcal{L}(V,W)\) and \(v \in V\),

\begin{equation*} (1T)(v) = 1(Tv) = Tv, \end{equation*}

so \(1T = T\).

Step 7: distributive properties. For \(\lambda \in \mathbb{F}\), \(S, T \in \mathcal{L}(V,W)\), and \(v \in V\),

\begin{equation*} \bigl(\lambda(S+T)\bigr)(v) = \lambda(Sv + Tv) = \lambda(Sv) + \lambda(Tv) = (\lambda S)(v) + (\lambda T)(v) = (\lambda S + \lambda T)(v), \end{equation*}

so \(\lambda(S+T) = \lambda S + \lambda T\). For \(\lambda, \mu \in \mathbb{F}\), \(T \in \mathcal{L}(V,W)\), and \(v \in V\),

\begin{equation*} \bigl((\lambda + \mu)T\bigr)(v) = (\lambda+\mu)(Tv) = \lambda(Tv) + \mu(Tv) = (\lambda T)(v) + (\mu T)(v) = (\lambda T + \mu T)(v), \end{equation*}

so \((\lambda + \mu)T = \lambda T + \mu T\).

All the conditions in the definition of a vector space (1.20) have now been verified. Hence \(\mathcal{L}(V,W)\), with the operations of 3.5, is a vector space over \(\mathbb{F}\), and its additive identity is the zero linear map.

Recall from 3.7 that if \(T \in \mathcal{L}(U,V)\) and \(S \in \mathcal{L}(V,W)\), then the product \(ST \in \mathcal{L}(U,W)\) is defined by \((ST)(u) = S(Tu)\) for all \(u \in U\). In each part below, the two sides of the claimed equation are functions with the same domain, so it suffices to show that they agree at every point of that domain.

Associativity. Suppose \(T_1\), \(T_2\), \(T_3\) are linear maps for which the products below make sense; that is, suppose there are vector spaces \(V_1, V_2, V_3, V_4\) with

\begin{equation*} T_3 \in \mathcal{L}(V_1, V_2), \qquad T_2 \in \mathcal{L}(V_2, V_3), \qquad T_1 \in \mathcal{L}(V_3, V_4). \end{equation*}

Then \(T_2 T_3 \in \mathcal{L}(V_1, V_3)\) and \(T_1 T_2 \in \mathcal{L}(V_2, V_4)\), so both \((T_1T_2)T_3\) and \(T_1(T_2T_3)\) are maps from \(V_1\) to \(V_4\). Let \(u \in V_1\). Applying the definition of the product twice to each side,

\begin{equation*} \bigl((T_1T_2)T_3\bigr)(u) = (T_1T_2)(T_3 u) = T_1\bigl(T_2(T_3 u)\bigr) \end{equation*}

and

\begin{equation*} \bigl(T_1(T_2T_3)\bigr)(u) = T_1\bigl((T_2T_3)(u)\bigr) = T_1\bigl(T_2(T_3 u)\bigr). \end{equation*}

The right sides agree, so \((T_1T_2)T_3 = T_1(T_2T_3)\). (Note that no linearity was used here; this is just associativity of composition of functions.)

Identity. Suppose \(T \in \mathcal{L}(V,W)\). Let \(I\) denote the identity operator on \(V\) in the product \(TI\), and the identity operator on \(W\) in the product \(IT\); both products then make sense and are maps from \(V\) to \(W\). For every \(v \in V\),

\begin{equation*} (TI)(v) = T(Iv) = Tv \qquad \text{and} \qquad (IT)(v) = I(Tv) = Tv, \end{equation*}

since the identity operator takes each vector to itself. Hence \(TI = T\) and \(IT = T\), so \(TI = IT = T\).

Distributive properties. Suppose \(T, T_1, T_2 \in \mathcal{L}(U,V)\) and \(S, S_1, S_2 \in \mathcal{L}(V,W)\). All four products appearing below are maps from \(U\) to \(W\), and by 3.5 so are the sums \(S_1T + S_2T\) and \(ST_1 + ST_2\).

For the first identity, let \(u \in U\). Then

\begin{equation*} \begin{aligned} \bigl((S_1 + S_2)T\bigr)(u) &= (S_1 + S_2)(Tu) \\ &= S_1(Tu) + S_2(Tu) \\ &= (S_1T)(u) + (S_2T)(u) \\ &= (S_1T + S_2T)(u), \end{aligned} \end{equation*}

where the first and third equalities use the definition of a product of linear maps and the second and fourth use the definition of the sum of two linear maps. Hence \((S_1 + S_2)T = S_1T + S_2T\).

For the second identity, let \(u \in U\). Then

\begin{equation*} \begin{aligned} \bigl(S(T_1 + T_2)\bigr)(u) &= S\bigl((T_1 + T_2)(u)\bigr) \\ &= S(T_1 u + T_2 u) \\ &= S(T_1 u) + S(T_2 u) \\ &= (ST_1)(u) + (ST_2)(u) \\ &= (ST_1 + ST_2)(u), \end{aligned} \end{equation*}

where the third equality uses the additivity of \(S\) (this is the one place where linearity, rather than just the definitions of sum and composition, is needed). Hence \(S(T_1 + T_2) = ST_1 + ST_2\).

This establishes all three properties asserted in 3.8.

Suppose \(\dim V = 1\) and \(T \in \mathcal{L}(V)\).

Because \(\dim V = 1\), the space \(V\) has a basis consisting of exactly one vector; call it \(u\). Thus

\begin{equation*} V = \operatorname{span}(u) = \{au : a \in \mathbb{F}\}, \end{equation*}

and \(u \neq 0\) (a list containing \(0\) is not linearly independent, so a basis vector cannot be \(0\)).

Since \(T\) maps \(V\) into \(V\), we have \(Tu \in V = \operatorname{span}(u)\). Hence there exists a scalar \(\lambda \in \mathbb{F}\) such that

\begin{equation*} Tu = \lambda u. \end{equation*}

We claim this \(\lambda\) works for every vector in \(V\). Indeed, let \(v \in V\) be arbitrary. Because \(u\) spans \(V\), there exists \(a \in \mathbb{F}\) with \(v = au\). Then, using the homogeneity of \(T\) and then the associativity and commutativity of scalar multiplication,

\begin{equation*} Tv = T(au) = a(Tu) = a(\lambda u) = (a\lambda)u = (\lambda a) u = \lambda(au) = \lambda v. \end{equation*}

Thus \(Tv = \lambda v\) for all \(v \in V\), which is exactly the statement that \(T\) is multiplication by the scalar \(\lambda\).

Finally, note that \(\lambda\) is uniquely determined by \(T\). If \(\mu \in \mathbb{F}\) also satisfies \(Tv = \mu v\) for all \(v \in V\), then in particular \(\lambda u = \mu u\), so \((\lambda - \mu)u = 0\). If \(\lambda - \mu \neq 0\), multiplying both sides by \((\lambda - \mu)^{-1}\) would give \(u = 0\), contradicting \(u \neq 0\). Hence \(\lambda = \mu\).

Conversely, for each \(\lambda \in \mathbb{F}\) the map \(v \mapsto \lambda v\) is a linear map from \(V\) to \(V\) (its additivity and homogeneity are immediate from the vector space properties of \(V\)). Combining this with the above, when \(\dim V = 1\) the map \(\lambda \mapsto \lambda I\) is a one-to-one correspondence between \(\mathbb{F}\) and \(\mathcal{L}(V)\).

Define \(\varphi \colon \mathbb{R}^2 \to \mathbb{R}\) by

\begin{equation*} \varphi(x, y) = \begin{cases} \dfrac{x^3}{x^2 + y^2} & \text{if } (x,y) \neq (0,0),\\ 0 & \text{if } (x,y) = (0,0). \end{cases} \end{equation*}

This is a well-defined function, since the denominator \(x^2 + y^2\) vanishes only at \((0,0)\), where the second clause applies instead.

Homogeneity. Let \(a \in \mathbb{R}\) and \(v = (x,y) \in \mathbb{R}^2\). If \(a = 0\) or \(v = (0,0)\), then \(av = (0,0)\), so \(\varphi(av) = 0\) and also \(a\varphi(v) = 0\) (in the first case because the scalar is \(0\), in the second because \(\varphi(0,0) = 0\)); hence \(\varphi(av) = a\varphi(v)\). Now suppose \(a \neq 0\) and \((x,y) \neq (0,0)\). Then \(av = (ax, ay) \neq (0,0)\), so

\begin{equation*} \varphi(av) = \frac{(ax)^3}{(ax)^2 + (ay)^2} = \frac{a^3 x^3}{a^2(x^2+y^2)} = a \cdot \frac{x^3}{x^2+y^2} = a\varphi(v). \end{equation*}

Thus \(\varphi(av) = a\varphi(v)\) for all \(a \in \mathbb{R}\) and all \(v \in \mathbb{R}^2\).

Failure of additivity. Take \(u = (1,0)\) and \(v = (0,1)\). Then

\begin{equation*} \varphi(u) = \frac{1}{1} = 1, \qquad \varphi(v) = \frac{0}{1} = 0, \qquad \varphi(u+v) = \varphi(1,1) = \frac{1}{2}. \end{equation*}

Hence \(\varphi(u+v) = \tfrac12 \neq 1 = \varphi(u) + \varphi(v)\), so \(\varphi\) is not additive and therefore is not a linear map (see 3.1).

Thus \(\varphi\) satisfies the homogeneity condition of 3.1 for every scalar and every vector, yet \(\varphi\) is not linear: homogeneity alone does not imply linearity.

Define \(\varphi \colon \mathbb{C} \to \mathbb{C}\) to be complex conjugation:

\begin{equation*} \varphi(z) = \overline{z}, \end{equation*}

so that \(\varphi(a + bi) = a - bi\) for all \(a, b \in \mathbb{R}\).

Additivity. Let \(w = a + bi\) and \(z = c + di\) with \(a, b, c, d \in \mathbb{R}\). Then \(w + z = (a+c) + (b+d)i\), so

\begin{equation*} \varphi(w + z) = (a + c) - (b + d)i = (a - bi) + (c - di) = \varphi(w) + \varphi(z). \end{equation*}

Thus \(\varphi\) is additive.

Failure of homogeneity. Here \(\mathbb{C}\) is regarded as a vector space over \(\mathbb{C}\), so the scalars include \(i\). Take \(\lambda = i\) and \(z = 1\). Then

\begin{equation*} \varphi(\lambda z) = \varphi(i) = \overline{i} = -i, \qquad \lambda\varphi(z) = i \cdot \overline{1} = i. \end{equation*}

Since \(-i \neq i\), the homogeneity condition of 3.1 fails, so \(\varphi\) is not a linear map from the complex vector space \(\mathbb{C}\) to itself.

More generally, for any \(z \neq 0\) we have \(\varphi(iz) = \overline{iz} = \overline{i}\,\overline{z} = -i\overline{z}\) while \(i\varphi(z) = i\overline{z}\). Their difference is \(i\overline{z} - (-i\overline{z}) = 2i\overline{z}\), which is nonzero because \(z \neq 0\) implies \(\overline{z} \neq 0\). So homogeneity fails for every nonzero \(z\) when \(\lambda = i\).

Together with the previous exercise, this shows that neither of the two conditions in the definition 3.1 of a linear map follows from the other. Note that the parenthetical remark in the statement is essential: if \(\mathbb{C}\) were regarded as a vector space over \(\mathbb{R}\), then conjugation would satisfy \(\varphi(av) = a\varphi(v)\) for all real \(a\) and would indeed be linear.

The statement is false. Here is a counterexample.

Let \(q \in \mathcal{P}(\mathbb{R})\) be given by \(q(x) = x^2\), and let \(T \colon \mathcal{P}(\mathbb{R}) \to \mathcal{P}(\mathbb{R})\) be defined by \(Tp = q \circ p\), so that

\begin{equation*} (Tp)(x) = \bigl(p(x)\bigr)^2 \qquad \text{for all } x \in \mathbb{R}. \end{equation*}

(This really does map \(\mathcal{P}(\mathbb{R})\) into \(\mathcal{P}(\mathbb{R})\): the square of a polynomial is a polynomial.)

Homogeneity fails. Let \(p\) be the constant polynomial \(p(x) = 1\) and take \(\lambda = 2\). Then \((2p)(x) = 2\) for all \(x\), so

\begin{equation*} \bigl(T(2p)\bigr)(x) = \bigl((2p)(x)\bigr)^2 = 4, \end{equation*}

whereas

\begin{equation*} \bigl(2(Tp)\bigr)(x) = 2\bigl(p(x)\bigr)^2 = 2. \end{equation*}

Thus \(T(2p)\) is the constant polynomial \(4\) and \(2(Tp)\) is the constant polynomial \(2\); these are different elements of \(\mathcal{P}(\mathbb{R})\), so \(T(2p) \neq 2(Tp)\) and the homogeneity condition of 3.1 fails.

Additivity fails too. With the same \(p\), we have \(p + p = 2p\), so

\begin{equation*} T(p + p) = T(2p) = 4 \neq 2 = Tp + Tp . \end{equation*}

Hence \(T\) is not a linear map, and the assertion of the exercise is false.

Remark. Contrast this with the last bullet point of 3.3, where \(T\) is defined by \((Tp)(x) = p(q(x))\), that is, \(Tp = p \circ q\). That map is linear, because the composition happens on the inside: for all \(x \in \mathbb{R}\),

\begin{equation*} \bigl(T(p_1 + p_2)\bigr)(x) = (p_1 + p_2)(q(x)) = p_1(q(x)) + p_2(q(x)) = (Tp_1)(x) + (Tp_2)(x), \end{equation*}

and \(\bigl(T(\lambda p)\bigr)(x) = (\lambda p)(q(x)) = \lambda\, p(q(x)) = \lambda (Tp)(x)\). The failure in the present exercise comes from the fact that \(q\) is applied after \(p\), so the vector-space operations of \(\mathcal{P}(\mathbb{R})\) get fed into the nonlinear function \(q\).

Indeed, the map \(p \mapsto q \circ p\) is linear precisely when \(q(x) = cx\) for some \(c \in \mathbb{R}\). If \(q(x) = cx\), then \(q \circ p = cp\) for every \(p\), which is linear. Conversely, suppose \(q\) is not of this form. If \(q(0) \neq 0\), then \(T\) sends the zero polynomial to the nonzero constant polynomial \(q(0)\), so \(T\) is not linear by 3.10. If \(q(0) = 0\) but \(q(x) \neq cx\) for every \(c\), then in particular \(q\) is not the polynomial \(q(1)x\), so there is a real number \(\lambda\) with \(q(\lambda) \neq \lambda\, q(1)\); taking \(p\) to be the constant polynomial \(1\) gives \(T(\lambda p) = q(\lambda) \neq \lambda\, q(1) = \lambda (Tp)\) as constant polynomials, so again \(T\) is not linear.

If \(V = \{0\}\) then \(\mathcal{L}(V)\) contains only the zero map, which equals \(0 \cdot I\), and both conditions hold trivially. So assume \(V \neq \{0\}\).

The easy direction. Suppose \(T = \lambda I\) for some \(\lambda \in \mathbf{F}\). Let \(S \in \mathcal{L}(V)\) and \(v \in V\). Then

\begin{equation*} (ST)v = S(\lambda v) = \lambda (Sv) = (\lambda I)(Sv) = (TS)v, \end{equation*}

where the second equality is the homogeneity of \(S\). Since \(v \in V\) was arbitrary, \(ST = TS\).

The interesting direction. Now suppose \(ST = TS\) for every \(S \in \mathcal{L}(V)\). We first show that every nonzero vector of \(V\) is mapped by \(T\) into its own span.

Claim: if \(v \in V\) and \(v \neq 0\), then \(Tv \in \operatorname{span}(v)\).

Suppose not. Then \(Tv \notin \operatorname{span}(v)\), and hence the list \(v, Tv\) is linearly independent: if \(av + b(Tv) = 0\) with \(b \neq 0\) then \(Tv = -(a/b)v \in \operatorname{span}(v)\), a contradiction, so \(b = 0\); then \(av = 0\) with \(v \neq 0\) forces \(a = 0\).

Because \(V\) is finite-dimensional, the linearly independent list \(v, Tv\) extends to a basis \(v, Tv, u_3, \dots, u_n\) of \(V\) by 2.32. By the linear map lemma 3.4 there exists \(S \in \mathcal{L}(V)\) with

\begin{equation*} Sv = v, \qquad S(Tv) = 0, \qquad Su_k = 0 \text{ for } k = 3, \dots, n. \end{equation*}

Then

\begin{equation*} (ST)v = S(Tv) = 0, \qquad (TS)v = T(Sv) = Tv . \end{equation*}

But \(Tv \neq 0\), since \(Tv\) belongs to a linearly independent list. Hence \((ST)v \neq (TS)v\), contradicting \(ST = TS\). This proves the claim.

Thus for each nonzero \(v \in V\) there is a scalar \(a_v \in \mathbf{F}\) with

\begin{equation*} Tv = a_v v . \end{equation*}

(The scalar \(a_v\) is unique, because \(v \neq 0\).)

It remains to show that \(a_v\) does not depend on \(v\). Let \(v, w \in V\) be nonzero.

Case 1: \(v, w\) is linearly dependent. Since both are nonzero, there is \(c \in \mathbf{F}\) with \(c \neq 0\) and \(w = cv\). Then

\begin{equation*} a_w w = Tw = T(cv) = c\,Tv = c\,a_v v = a_v (cv) = a_v w . \end{equation*}

Because \(w \neq 0\), this gives \(a_w = a_v\).

Case 2: \(v, w\) is linearly independent. Then \(v + w \neq 0\), and

\begin{equation*} a_{v+w} v + a_{v+w} w = a_{v+w}(v+w) = T(v+w) = Tv + Tw = a_v v + a_w w . \end{equation*}

Rearranging,

\begin{equation*} (a_{v+w} - a_v)v + (a_{v+w} - a_w)w = 0 . \end{equation*}

Linear independence of \(v, w\) forces \(a_{v+w} = a_v\) and \(a_{v+w} = a_w\), so \(a_v = a_w\).

In both cases \(a_v = a_w\). Hence there is a single scalar \(\lambda \in \mathbf{F}\) with \(a_v = \lambda\) for every nonzero \(v \in V\) (such a \(v\) exists because \(V \neq \{0\}\)). Therefore \(Tv = \lambda v\) for every nonzero \(v \in V\), and also \(T0 = 0 = \lambda \cdot 0\) by 3.10. Thus \(T = \lambda I\), so \(T\) is a scalar multiple of the identity.

First note that \(T\) is a well-defined function from \(V\) to \(W\): each \(v \in V\) satisfies exactly one of the conditions \(v \in U\) and \(v \notin U\), so exactly one clause of the definition applies.

Because \(S \neq 0\), there exists \(u \in U\) with

\begin{equation*} Su \neq 0 . \end{equation*}

Because \(U \neq V\) and \(U \subseteq V\), there exists

\begin{equation*} w \in V \text{ with } w \notin U . \end{equation*}

Consider the vector \(u + w \in V\). We claim \(u + w \notin U\). Indeed, if \(u + w \in U\), then since \(u \in U\) and \(U\) is a subspace (hence closed under additive inverses and addition),

\begin{equation*} w = (u + w) - u \in U, \end{equation*}

contradicting \(w \notin U\). So \(u + w \notin U\), and therefore, by the definition of \(T\),

\begin{equation*} T(u + w) = 0 . \end{equation*}

On the other hand, \(u \in U\) gives \(Tu = Su\), and \(w \notin U\) gives \(Tw = 0\), so

\begin{equation*} Tu + Tw = Su + 0 = Su \neq 0 . \end{equation*}

Hence

\begin{equation*} T(u + w) = 0 \neq Su = Tu + Tw, \end{equation*}

so \(T\) fails the additivity condition in the definition 3.1 of a linear map. Therefore \(T\) is not a linear map from \(V\) to \(W\).

Remark. The point of this exercise is that “extend by zero” is not a legitimate way to extend a linear map from a proper subspace \(U\) to all of \(V\): the set \(V \setminus U\) is not closed under addition with elements of \(U\), so the two clauses of the definition interfere with each other. The next exercise shows that when \(V\) is finite-dimensional a genuine linear extension does always exist; it is obtained by extending a basis of \(U\) to a basis of \(V\) and defining the extension to be \(0\) on the new basis vectors only, not on all of \(V \setminus U\).

Suppose \(U\) is a subspace of \(V\) and \(S \in \mathcal{L}(U, W)\).

Since \(V\) is finite-dimensional, so is \(U\), by 2.25; hence \(U\) has a basis

\begin{equation*} u_1, \dots, u_m \end{equation*}

by 2.31 (if \(U = \{0\}\) this is the empty list, so \(m = 0\)). In particular \(u_1, \dots, u_m\) is linearly independent in \(U\), hence also linearly independent as a list in \(V\). Because \(V\) is finite-dimensional, 2.32 says this list extends to a basis

\begin{equation*} u_1, \dots, u_m, v_1, \dots, v_n \end{equation*}

of \(V\).

By the linear map lemma 3.4 applied to this basis of \(V\) and to the vectors \(Su_1, \dots, Su_m, 0, \dots, 0 \in W\), there exists a (unique) linear map \(T \in \mathcal{L}(V, W)\) such that

\begin{equation*} Tu_k = Su_k \quad \text{for } k = 1, \dots, m, \qquad Tv_j = 0 \quad \text{for } j = 1, \dots, n . \end{equation*}

We check that \(T\) extends \(S\). Let \(u \in U\). Because \(u_1, \dots, u_m\) is a basis of \(U\), there are scalars \(c_1, \dots, c_m \in \mathbf{F}\) with

\begin{equation*} u = c_1 u_1 + \dots + c_m u_m . \end{equation*}

Using the linearity of \(T\) and then the linearity of \(S\),

\begin{equation*} \begin{aligned} Tu &= T(c_1 u_1 + \dots + c_m u_m)\\ &= c_1 (Tu_1) + \dots + c_m (Tu_m)\\ &= c_1 (Su_1) + \dots + c_m (Su_m)\\ &= S(c_1 u_1 + \dots + c_m u_m)\\ &= Su . \end{aligned} \end{equation*}

(When \(m = 0\) this reads \(T0 = 0 = S0\), which holds by 3.10.)

Thus \(T \in \mathcal{L}(V, W)\) and \(Tu = Su\) for all \(u \in U\), so \(S\) has been extended to a linear map on all of \(V\), as desired.

Remark. The extension is generally far from unique: any choice of vectors \(w_1, \dots, w_n \in W\) in place of the zeros above yields, via 3.4, another linear extension of \(S\). Compare exercise 3A.12, which shows that the naive extension “\(Tv = 0\) for every \(v \notin U\)” never works when \(U \neq V\) and \(S \neq 0\).

Recall that \(\mathcal{L}(V, W)\) is a vector space with the operations of 3.5, by 3.6. Recall also that a vector space is called infinite-dimensional (2.13) when it is not finite-dimensional, that is, when no finite list of its vectors spans it.

We prove the result by exhibiting, for every positive integer \(m\), a linearly independent list of length \(m\) in \(\mathcal{L}(V, W)\).

Step 1: \(W\) contains linearly independent lists of every length.

We show by induction on \(k\) that for every nonnegative integer \(k\) there is a linearly independent list \(w_1, \dots, w_k\) of vectors in \(W\). For \(k = 0\) the empty list is linearly independent. Suppose \(w_1, \dots, w_k\) is a linearly independent list in \(W\). Then \(\operatorname{span}(w_1, \dots, w_k) \neq W\), for otherwise the finite list \(w_1, \dots, w_k\) would span \(W\), making \(W\) finite-dimensional (2.9) and contradicting our hypothesis. Choose

\begin{equation*} w_{k+1} \in W \setminus \operatorname{span}(w_1, \dots, w_k) . \end{equation*}

Then \(w_1, \dots, w_{k+1}\) is linearly independent: suppose \(a_1, \dots, a_{k+1} \in \mathbf{F}\) satisfy

\begin{equation*} a_1 w_1 + \dots + a_k w_k + a_{k+1} w_{k+1} = 0 . \end{equation*}

If \(a_{k+1} \neq 0\), then

\begin{equation*} w_{k+1} = -\frac{a_1}{a_{k+1}} w_1 - \dots - \frac{a_k}{a_{k+1}} w_k \in \operatorname{span}(w_1, \dots, w_k), \end{equation*}

a contradiction. Hence \(a_{k+1} = 0\), and then \(a_1 w_1 + \dots + a_k w_k = 0\) forces \(a_1 = \dots = a_k = 0\) by the linear independence of \(w_1, \dots, w_k\). This completes the induction.

Step 2: transporting these lists into \(\mathcal{L}(V, W)\).

Since \(V\) is finite-dimensional with \(\dim V > 0\), it has a basis \(v_1, \dots, v_p\) with \(p \geq 1\) (2.31). Fix a positive integer \(m\), and by Step 1 choose a linearly independent list \(w_1, \dots, w_m\) in \(W\).

For each \(j = 1, \dots, m\), the linear map lemma 3.4 provides a linear map \(T_j \in \mathcal{L}(V, W)\) with

\begin{equation*} T_j v_1 = w_j \qquad \text{and} \qquad T_j v_k = 0 \text{ for } k = 2, \dots, p . \end{equation*}

We claim \(T_1, \dots, T_m\) is linearly independent in \(\mathcal{L}(V, W)\). Suppose \(a_1, \dots, a_m \in \mathbf{F}\) and

\begin{equation*} a_1 T_1 + \dots + a_m T_m = 0, \end{equation*}

where the \(0\) on the right is the zero map from \(V\) to \(W\). Evaluating both sides at \(v_1\) and using the definition 3.5 of addition and scalar multiplication in \(\mathcal{L}(V, W)\), we get

\begin{equation*} a_1 (T_1 v_1) + \dots + a_m (T_m v_1) = 0, \qquad \text{that is,} \qquad a_1 w_1 + \dots + a_m w_m = 0 . \end{equation*}

Since \(w_1, \dots, w_m\) is linearly independent, \(a_1 = \dots = a_m = 0\). Hence \(T_1, \dots, T_m\) is linearly independent.

Step 3: conclusion.

Suppose, for a contradiction, that \(\mathcal{L}(V, W)\) were finite-dimensional. Then by 2.9 there would be a finite list \(R_1, \dots, R_d\) spanning \(\mathcal{L}(V, W)\). By Step 2 there is a linearly independent list in \(\mathcal{L}(V, W)\) of length \(d + 1\). But 2.22 says the length of a linearly independent list is at most the length of a spanning list, so \(d + 1 \leq d\), a contradiction.

Therefore \(\mathcal{L}(V, W)\) is not finite-dimensional; that is, \(\mathcal{L}(V, W)\) is infinite-dimensional.

Remark. The hypothesis \(\dim V > 0\) is needed: if \(V = \{0\}\), then the only linear map from \(V\) to \(W\) is the zero map, so \(\mathcal{L}(V, W) = \{0\}\) is finite-dimensional (of dimension \(0\)) no matter how large \(W\) is. Intuitively, Step 2 says that a single nonzero vector \(v_1\) of \(V\) suffices to encode an arbitrary vector of \(W\) inside a linear map, and \(W\) has room for arbitrarily long independent lists.

Because \(v_1, \dots, v_m\) is linearly dependent, there exist scalars \(a_1, \dots, a_m \in \mathbf{F}\), not all zero, such that

\begin{equation*} a_1 v_1 + \dots + a_m v_m = 0 . \end{equation*}

Fix an index \(j\) with \(a_j \neq 0\). Because \(W \neq \{0\}\), we may also fix a vector \(w \in W\) with \(w \neq 0\).

Now define the list \(w_1, \dots, w_m\) in \(W\) by

\begin{equation*} w_j = w \quad \text{and} \quad w_k = 0 \ \text{ for } k \neq j . \end{equation*}

We claim no \(T \in \mathcal{L}(V, W)\) satisfies \(Tv_k = w_k\) for each \(k = 1, \dots, m\). Suppose, seeking a contradiction, that some \(T \in \mathcal{L}(V, W)\) did satisfy these \(m\) equations. Applying \(T\) to the dependence relation above and using additivity and homogeneity of \(T\), we get

\begin{equation*} T(a_1 v_1 + \dots + a_m v_m) = a_1 Tv_1 + \dots + a_m Tv_m = a_1 w_1 + \dots + a_m w_m = a_j w , \end{equation*}

where the last equality holds because \(w_k = 0\) for every \(k \neq j\). On the other hand, \(a_1 v_1 + \dots + a_m v_m = 0\), and every linear map takes \(0\) to \(0\) (by 3.10), so the left side equals \(0\). Hence

\begin{equation*} a_j w = 0 . \end{equation*}

Since \(a_j \neq 0\), we may multiply by \(1/a_j\) to conclude \(w = 0\), contradicting the choice of \(w \neq 0\).

Thus no \(T \in \mathcal{L}(V, W)\) satisfies \(Tv_k = w_k\) for each \(k = 1, \dots, m\), as desired.

Remark: this exercise is the exact converse of the linear map lemma 3.4 in the following sense. If \(v_1, \dots, v_m\) is a basis of \(V\), then 3.4 says that every choice of \(w_1, \dots, w_m \in W\) is achieved by some (indeed exactly one) \(T \in \mathcal{L}(V, W)\). This exercise shows that if the list \(v_1, \dots, v_m\) is linearly dependent, then — provided \(W\) contains a nonzero vector — some choice of \(w_1, \dots, w_m\) is achieved by no \(T \in \mathcal{L}(V,W)\) at all. So linear independence of \(v_1, \dots, v_m\) is exactly what is needed for arbitrary prescription of the values \(Tv_k\) to be possible.

Let \(n = \dim V\); by hypothesis \(n \geq 2\) and \(n < \infty\). Choose a basis \(v_1, \dots, v_n\) of \(V\).

By the linear map lemma 3.4, a linear map on \(V\) may be defined by specifying its values arbitrarily on the basis \(v_1, \dots, v_n\). Use this to define \(S, T \in \mathcal{L}(V)\) by

\begin{equation*} \begin{aligned} S v_1 &= v_1, & S v_k &= 0 \quad \text{for } k = 2, \dots, n, \\ T v_1 &= v_2, & T v_k &= 0 \quad \text{for } k = 2, \dots, n . \end{aligned} \end{equation*}

Note that these definitions make sense precisely because \(n \geq 2\): we need a second basis vector \(v_2\) for \(T\) to aim at.

Now evaluate both compositions at \(v_1\). First,

\begin{equation*} (ST)v_1 = S(Tv_1) = S v_2 = 0 , \end{equation*}

since \(2 \leq n\) means \(v_2\) is one of the basis vectors sent to \(0\) by \(S\). Second,

\begin{equation*} (TS)v_1 = T(S v_1) = T v_1 = v_2 . \end{equation*}

Since \(v_1, \dots, v_n\) is a basis, it is in particular linearly independent, so \(v_2 \neq 0\). Therefore \((ST)v_1 = 0 \neq v_2 = (TS)v_1\), and the two operators \(ST\) and \(TS\) disagree at the vector \(v_1\). Hence \(ST \neq TS\), as desired.

Remark: the hypothesis \(\dim V > 1\) cannot be dropped. If \(\dim V \leq 1\) then every operator on \(V\) is a scalar multiple of the identity (this is trivial when \(V = \{0\}\), and is Exercise 7 of this section when \(\dim V = 1\)), and any two scalar multiples of the identity commute. Thus \(\mathcal{L}(V)\) is a commutative algebra exactly when \(\dim V \leq 1\).

First note that \(\{0\}\) and \(\mathcal{L}(V)\) really are two-sided ideals: both are subspaces of \(\mathcal{L}(V)\), the products \(T0\) and \(0T\) equal \(0\) for every \(T \in \mathcal{L}(V)\), and \(\mathcal{L}(V)\) trivially contains all products of its elements. So the content of the exercise is that there are no others.

Suppose then that \(\mathcal{E}\) is a two-sided ideal of \(\mathcal{L}(V)\) with \(\mathcal{E} \neq \{0\}\). We show \(\mathcal{E} = \mathcal{L}(V)\). The strategy is to produce the identity operator \(I\) inside \(\mathcal{E}\); once \(I \in \mathcal{E}\), every \(T \in \mathcal{L}(V)\) satisfies \(T = TI\) by the identity property in 3.8, and \(TI \in \mathcal{E}\) by the ideal property applied to the element \(I\) of \(\mathcal{E}\) and the operator \(T\); so \(\mathcal{L}(V) \subseteq \mathcal{E}\) and hence \(\mathcal{E} = \mathcal{L}(V)\).

Because \(\mathcal{E} \neq \{0\}\), there is some \(E \in \mathcal{E}\) with \(E \neq 0\). Since \(E\) is not the zero operator, there exists \(u \in V\) with

\begin{equation*} Eu \neq 0 . \end{equation*}

In particular \(V \neq \{0\}\), so \(n := \dim V \geq 1\); and \(n < \infty\) because \(V\) is finite-dimensional. Fix a basis \(v_1, \dots, v_n\) of \(V\).

Two families of auxiliary operators. For each \(j \in \{1, \dots, n\}\), the linear map lemma 3.4 lets us define \(A_j \in \mathcal{L}(V)\) by its values on the basis:

\begin{equation*} A_j v_j = u \quad \text{and} \quad A_j v_i = 0 \ \text{ for } i \neq j . \end{equation*}

Next, since \(Eu \neq 0\), the one-term list \(Eu\) is linearly independent in \(V\). By 2.32 it extends to a basis

\begin{equation*} u_1, u_2, \dots, u_n \quad \text{of } V, \qquad \text{where } u_1 = Eu . \end{equation*}

(The extended basis has \(n\) terms because every basis of \(V\) has length \(\dim V = n\).) For each \(j \in \{1, \dots, n\}\), use 3.4 again to define \(B_j \in \mathcal{L}(V)\) by

\begin{equation*} B_j u_1 = v_j \quad \text{and} \quad B_j u_i = 0 \ \text{ for } i = 2, \dots, n . \end{equation*}

The products lie in \(\mathcal{E}\). Fix \(j\). Since \(E \in \mathcal{E}\) and \(A_j \in \mathcal{L}(V)\), the ideal property gives \(E A_j \in \mathcal{E}\). Since \(E A_j \in \mathcal{E}\) and \(B_j \in \mathcal{L}(V)\), the ideal property gives \(B_j (E A_j) \in \mathcal{E}\). By associativity (3.8) we may write this operator unambiguously as \(B_j E A_j\). Thus

\begin{equation*} B_j E A_j \in \mathcal{E} \qquad \text{for each } j = 1, \dots, n . \end{equation*}

Computing these products on the basis. Fix \(j\) and let \(i \in \{1, \dots, n\}\). If \(i \neq j\) then \(A_j v_i = 0\), so

\begin{equation*} (B_j E A_j) v_i = B_j E (0) = 0 , \end{equation*}

using 3.10 (linear maps take \(0\) to \(0\)). If \(i = j\) then

\begin{equation*} (B_j E A_j) v_j = B_j E (u) = B_j (Eu) = B_j u_1 = v_j . \end{equation*}

Assembling the identity. Because \(\mathcal{E}\) is a subspace of \(\mathcal{L}(V)\), it is closed under addition, so the finite sum

\begin{equation*} R := B_1 E A_1 + B_2 E A_2 + \dots + B_n E A_n \end{equation*}

belongs to \(\mathcal{E}\). (Finite-dimensionality of \(V\) is what makes this legitimate: the sum has only finitely many terms. It was also used earlier, to produce the finite basis \(v_1, \dots, v_n\) and to invoke 2.32.) For each \(i \in \{1, \dots, n\}\), the computation above gives

\begin{equation*} R v_i = \sum_{j=1}^{n} (B_j E A_j) v_i = (B_i E A_i) v_i = v_i , \end{equation*}

since all terms with \(j \neq i\) vanish. So the operators \(R\) and \(I\) agree on the basis \(v_1, \dots, v_n\); by the uniqueness half of the linear map lemma 3.4 (or simply because every \(v \in V\) is a linear combination of the \(v_i\) and both maps are linear), \(R = I\).

Therefore \(I \in \mathcal{E}\), and by the first paragraph \(\mathcal{E} = \mathcal{L}(V)\).

Conclusion: every two-sided ideal of \(\mathcal{L}(V)\) is either \(\{0\}\) or all of \(\mathcal{L}(V)\). (If \(V = \{0\}\) then \(\mathcal{L}(V) = \{0\}\) and the two listed ideals coincide, so the statement holds vacuously in that degenerate case.)

Remark: the hypothesis that \(V\) is finite-dimensional is essential. If \(V\) is infinite-dimensional, then the set of operators \(T \in \mathcal{L}(V)\) whose range is finite-dimensional is a two-sided ideal — it is a subspace, and composing a finite-rank operator with any operator on either side again has finite-dimensional range — and it is neither \(\{0\}\) (it contains, for instance, any nonzero operator with one-dimensional range) nor all of \(\mathcal{L}(V)\) (it does not contain \(I\)). The argument above breaks down precisely at the assembling step, where the sum \(\sum_j B_j E A_j\) would have to be infinite.

The fundamental theorem of linear maps (3.21) tells us where to look: if \(T\) is defined on a finite-dimensional space \(V\), then \(\dim V = \dim \operatorname{null} T + \dim \operatorname{range} T = 3 + 2 = 5\). So take \(V = \mathbf{F}^5\) and \(W = \mathbf{F}^2\), and define \(T \in \mathcal{L}(\mathbf{F}^5, \mathbf{F}^2)\) by

\begin{equation*} T(z_1, z_2, z_3, z_4, z_5) = (z_1, z_2). \end{equation*}

This map is linear: each coordinate of the output is a coordinate of the input, so additivity and homogeneity are immediate from the coordinatewise definitions of addition and scalar multiplication on \(\mathbf{F}^5\) and \(\mathbf{F}^2\).

Null space. We have \(T(z_1, \dots, z_5) = (0,0)\) if and only if \(z_1 = z_2 = 0\). Hence

\begin{equation*} \operatorname{null} T = \{(0, 0, z_3, z_4, z_5) : z_3, z_4, z_5 \in \mathbf{F}\}. \end{equation*}

The list \((0,0,1,0,0), (0,0,0,1,0), (0,0,0,0,1)\) spans this set and is linearly independent (a linear combination with coefficients \(z_3, z_4, z_5\) equals \((0,0,z_3,z_4,z_5)\), which is \(0\) only when all three scalars are \(0\)). Thus it is a basis of \(\operatorname{null} T\) and \(\dim \operatorname{null} T = 3\).

Range. For any \((w_1, w_2) \in \mathbf{F}^2\) we have \(T(w_1, w_2, 0, 0, 0) = (w_1, w_2)\), so \(\operatorname{range} T = \mathbf{F}^2\) and therefore \(\dim \operatorname{range} T = 2\).

As a check, \(\dim \mathbf{F}^5 = 5 = 3 + 2 = \dim \operatorname{null} T + \dim \operatorname{range} T\), in agreement with 3.21.

Recall that \((ST)^2\) denotes the composition \(ST \circ ST = STST\), and that the zero element of \(\mathcal{L}(V)\) is the map sending every vector to \(0\). So we must show \((STST)v = 0\) for every \(v \in V\).

Let \(v \in V\). Put \(u = S(Tv)\). Then \(u \in \operatorname{range} S\), and since \(\operatorname{range} S \subseteq \operatorname{null} T\) we get \(u \in \operatorname{null} T\), i.e.

\begin{equation*} T\bigl(S(Tv)\bigr) = Tu = 0 . \end{equation*}

Therefore

\begin{equation*} (ST)^2 v = S\Bigl(T\bigl(S(Tv)\bigr)\Bigr) = S(0) = 0, \end{equation*}

where the last equality holds because every linear map sends \(0\) to \(0\): indeed \(S(0) = S(0 + 0) = S(0) + S(0)\) by additivity, and adding \(-S(0)\) to both sides gives \(S(0) = 0\).

Since \(v \in V\) was arbitrary, \((ST)^2\) is the zero map, that is, \((ST)^2 = 0\).

(a) The list \(v_1, \dots, v_m\) spans \(V\) if and only if \(T\) is surjective.

To see this, compute the range of \(T\). By definition,

\begin{equation*} \operatorname{range} T = \{z_1 v_1 + \cdots + z_m v_m : z_1, \dots, z_m \in \mathbf{F}\}, \end{equation*}

which is exactly the set of all linear combinations of \(v_1, \dots, v_m\), i.e. \(\operatorname{range} T = \operatorname{span}(v_1, \dots, v_m)\). Now \(T\) is surjective means \(\operatorname{range} T = V\) (3.19), and \(v_1, \dots, v_m\) spans \(V\) means \(\operatorname{span}(v_1, \dots, v_m) = V\). Since the two sets in question are the same set, the two conditions are equivalent.

(b) The list \(v_1, \dots, v_m\) is linearly independent if and only if \(T\) is injective.

Compute the null space:

\begin{equation*} \operatorname{null} T = \{(z_1, \dots, z_m) \in \mathbf{F}^m : z_1 v_1 + \cdots + z_m v_m = 0\}. \end{equation*}

By definition, \(v_1, \dots, v_m\) is linearly independent exactly when the only choice of scalars \(z_1, \dots, z_m \in \mathbf{F}\) making \(z_1 v_1 + \cdots + z_m v_m = 0\) is \(z_1 = \cdots = z_m = 0\); that is, exactly when \(\operatorname{null} T = \{0\}\). By 3.15, \(\operatorname{null} T = \{0\}\) if and only if \(T\) is injective. Hence linear independence of the list is equivalent to injectivity of \(T\).

Remark. Combining the two parts: \(v_1, \dots, v_m\) is a basis of \(V\) if and only if \(T\) is both injective and surjective, i.e. if and only if \(T\) is an isomorphism from \(\mathbf{F}^m\) onto \(V\).

Write \(E = \{T \in \mathcal{L}(\mathbf{R}^5, \mathbf{R}^4) : \dim \operatorname{null} T > 2\}\). We show \(E\) is not closed under addition, which is enough to conclude that \(E\) is not a subspace.

Define \(S, T \in \mathcal{L}(\mathbf{R}^5, \mathbf{R}^4)\) by

\begin{equation*} S(x_1, x_2, x_3, x_4, x_5) = (x_1, x_2, 0, 0), \qquad T(x_1, x_2, x_3, x_4, x_5) = (0, 0, x_3, x_4). \end{equation*}

Both are linear, since each output coordinate is either \(0\) or one of the input coordinates.

\(S\) lies in \(E\): we have \(S(x_1, \dots, x_5) = 0\) if and only if \(x_1 = x_2 = 0\), so

\begin{equation*} \operatorname{null} S = \{(0, 0, x_3, x_4, x_5) : x_3, x_4, x_5 \in \mathbf{R}\}, \end{equation*}

which has basis \((0,0,1,0,0), (0,0,0,1,0), (0,0,0,0,1)\) and hence \(\dim \operatorname{null} S = 3 > 2\).

\(T\) lies in \(E\): similarly \(T(x_1, \dots, x_5) = 0\) if and only if \(x_3 = x_4 = 0\), so

\begin{equation*} \operatorname{null} T = \{(x_1, x_2, 0, 0, x_5) : x_1, x_2, x_5 \in \mathbf{R}\}, \end{equation*}

which has basis \((1,0,0,0,0), (0,1,0,0,0), (0,0,0,0,1)\) and hence \(\dim \operatorname{null} T = 3 > 2\).

But \(S + T\) is given by

\begin{equation*} (S+T)(x_1, x_2, x_3, x_4, x_5) = (x_1, x_2, x_3, x_4), \end{equation*}

so \((S+T)(x_1, \dots, x_5) = 0\) if and only if \(x_1 = x_2 = x_3 = x_4 = 0\). Thus

\begin{equation*} \operatorname{null}(S+T) = \{(0,0,0,0,x_5) : x_5 \in \mathbf{R}\}, \end{equation*}

which is spanned by the single nonzero vector \((0,0,0,0,1)\), so \(\dim \operatorname{null}(S+T) = 1\). Since \(1\) is not greater than \(2\), we have \(S + T \notin E\).

Hence \(E\) contains \(S\) and \(T\) but not \(S + T\), so \(E\) is not closed under addition and therefore is not a subspace of \(\mathcal{L}(\mathbf{R}^5, \mathbf{R}^4)\).

Remark. \(E\) does contain the zero map (whose null space is all of \(\mathbf{R}^5\), of dimension \(5\)) and \(E\) is closed under scalar multiplication (for \(\lambda \neq 0\), \(\operatorname{null}(\lambda T) = \operatorname{null} T\); and \(0 T = 0 \in E\)). So failure of closure under addition is the only way this set can fail to be a subspace, and the example above exhibits that failure.

Define \(T \in \mathcal{L}(\mathbf{R}^4)\) by

\begin{equation*} T(x_1, x_2, x_3, x_4) = (0, 0, x_1, x_2). \end{equation*}

This map is linear, since each output coordinate is either the constant \(0\) or one of the input coordinates.

Null space. We have \(T(x_1, x_2, x_3, x_4) = (0,0,0,0)\) if and only if \(x_1 = x_2 = 0\). Hence

\begin{equation*} \operatorname{null} T = \{(0, 0, x_3, x_4) : x_3, x_4 \in \mathbf{R}\}. \end{equation*}

Range. As \((x_1, x_2, x_3, x_4)\) ranges over \(\mathbf{R}^4\), the vector \((0, 0, x_1, x_2)\) ranges over all vectors whose first two coordinates are \(0\): given any \(a, b \in \mathbf{R}\) we have \(T(a, b, 0, 0) = (0, 0, a, b)\). Hence

\begin{equation*} \operatorname{range} T = \{(0, 0, a, b) : a, b \in \mathbf{R}\}. \end{equation*}

The two sets described are literally the same subset of \(\mathbf{R}^4\), so \(\operatorname{range} T = \operatorname{null} T\), as desired.

Sanity check with 3.21: \(\dim \operatorname{null} T = \dim \operatorname{range} T = 2\), and indeed \(\dim \mathbf{R}^4 = 4 = 2 + 2\). This is why such a \(T\) can exist on \(\mathbf{R}^4\) but, as the next exercise shows, not on \(\mathbf{R}^5\).

Suppose, for contradiction, that \(T \in \mathcal{L}(\mathbf{R}^5)\) satisfies \(\operatorname{range} T = \operatorname{null} T\).

The space \(\mathbf{R}^5\) is finite-dimensional with \(\dim \mathbf{R}^5 = 5\), so the fundamental theorem of linear maps (3.21) applies and gives

\begin{equation*} 5 = \dim \mathbf{R}^5 = \dim \operatorname{null} T + \dim \operatorname{range} T . \end{equation*}

Since \(\operatorname{range} T\) and \(\operatorname{null} T\) are the same subspace of \(\mathbf{R}^5\), they have the same dimension; write \(k = \dim \operatorname{null} T = \dim \operatorname{range} T\), a nonnegative integer. The displayed equation becomes

\begin{equation*} 5 = k + k = 2k . \end{equation*}

Thus \(5\) would be an even integer, which is false; equivalently, \(k = 5/2\) is not an integer, contradicting the fact that the dimension of a finite-dimensional vector space is a nonnegative integer.

This contradiction shows no such \(T\) exists.

Remark. The same argument shows more generally that if \(V\) is finite-dimensional with \(\dim V\) odd, then no \(T \in \mathcal{L}(V)\) satisfies \(\operatorname{range} T = \operatorname{null} T\). When \(\dim V\) is even, say \(\dim V = 2k\), such a map does exist: take a basis \(v_1, \dots, v_{2k}\) of \(V\) and use the linear map lemma (3.4) to define \(T\) by \(Tv_j = v_{k+j}\) for \(j = 1, \dots, k\) and \(Tv_j = 0\) for \(j = k+1, \dots, 2k\); then both \(\operatorname{range} T\) and \(\operatorname{null} T\) equal \(\operatorname{span}(v_{k+1}, \dots, v_{2k})\). Exercise 3B.5 is the case \(\dim V = 4\).

Write \(E = \{T \in \mathcal{L}(V,W) : T \text{ is not injective}\}\). We produce \(S, T \in E\) with \(S + T \notin E\); since a subspace must be closed under addition, this shows \(E\) is not a subspace of \(\mathcal{L}(V,W)\).

Set \(n = \dim V\) and \(m = \dim W\), so \(2 \leq n \leq m\). Choose a basis \(v_1, \dots, v_n\) of \(V\) and a basis \(w_1, \dots, w_m\) of \(W\). Since \(n \leq m\), the vectors \(w_1, \dots, w_n\) are among the basis vectors of \(W\), and in particular \(w_1, \dots, w_n\) is a linearly independent list in \(W\).

By the linear map lemma (3.4), there are unique \(S, T \in \mathcal{L}(V, W)\) satisfying

\begin{equation*} \begin{aligned} S v_1 &= 0, & S v_j &= w_j \quad (j = 2, \dots, n),\\ T v_1 &= w_1, & T v_j &= 0 \quad (j = 2, \dots, n). \end{aligned} \end{equation*}

\(S\) is not injective. The vector \(v_1\) is nonzero (it belongs to a basis, and a list containing \(0\) is linearly dependent) and \(S v_1 = 0\), so \(\operatorname{null} S \neq \{0\}\); by 3.15, \(S\) is not injective. Hence \(S \in E\).

\(T\) is not injective. Here we use the hypothesis \(n = \dim V \geq 2\), which guarantees that the basis vector \(v_2\) exists. It is nonzero and \(T v_2 = 0\), so \(\operatorname{null} T \neq \{0\}\) and \(T\) is not injective by 3.15. Hence \(T \in E\).

\(S + T\) is injective. For each \(j \in \{1, \dots, n\}\) we have

\begin{equation*} (S+T)v_j = S v_j + T v_j = w_j, \end{equation*}

since exactly one of the two summands is \(w_j\) and the other is \(0\). Now suppose \(v \in \operatorname{null}(S+T)\). Write \(v = a_1 v_1 + \cdots + a_n v_n\) with \(a_1, \dots, a_n \in \mathbf{F}\), which is possible because \(v_1, \dots, v_n\) is a basis of \(V\). Then, by linearity,

\begin{equation*} 0 = (S+T)v = a_1 (S+T)v_1 + \cdots + a_n (S+T)v_n = a_1 w_1 + \cdots + a_n w_n . \end{equation*}

Because \(w_1, \dots, w_n\) is linearly independent, \(a_1 = \cdots = a_n = 0\), hence \(v = 0\). Thus \(\operatorname{null}(S+T) = \{0\}\), so \(S+T\) is injective by 3.15, and therefore \(S + T \notin E\).

We have exhibited \(S, T \in E\) whose sum is not in \(E\), so \(E\) is not closed under addition and consequently is not a subspace of \(\mathcal{L}(V, W)\).

Remark on the hypotheses. The condition \(\dim V \leq \dim W\) is needed: if \(\dim V > \dim W\) then by 3.22 no linear map from \(V\) to \(W\) is injective, so \(E\) would be all of \(\mathcal{L}(V, W)\), which is a subspace of itself. The condition \(\dim V \geq 2\) is also needed: if \(\dim V = 1\), then a linear map \(T\) on \(V\) fails to be injective exactly when \(T = 0\) (for \(v_1\) a basis of \(V\), \(T\) is determined by \(T v_1\), and \(\operatorname{null} T \neq \{0\}\) forces \(T v_1 = 0\)), so \(E = \{0\}\), again a subspace. If \(\dim V = 0\) then every map is injective and \(E\) is empty, hence not a subspace, but the argument above would not apply.

Write \(n = \dim V\) and \(m = \dim W\), so that \(n \geq m \geq 2\). Let \(v_1, \dots, v_n\) be a basis of \(V\) and let \(w_1, \dots, w_m\) be a basis of \(W\). Denote by

\begin{equation*} E = \{T \in \mathcal{L}(V, W) : T \text{ is not surjective}\}. \end{equation*}

We exhibit two elements of \(E\) whose sum is not in \(E\); since a subspace must be closed under addition, this shows \(E\) is not a subspace of \(\mathcal{L}(V,W)\).

By the linear map lemma (3.4) there is a unique \(S \in \mathcal{L}(V, W)\) with

\begin{equation*} S v_k = \begin{cases} w_k & \text{if } 1 \leq k \leq m - 1, \\ 0 & \text{if } m \leq k \leq n, \end{cases} \end{equation*}

and a unique \(T \in \mathcal{L}(V, W)\) with

\begin{equation*} T v_k = \begin{cases} w_m & \text{if } k = m, \\ 0 & \text{otherwise.} \end{cases} \end{equation*}

Note that both definitions make sense: \(m - 1 \geq 1\) because \(m \geq 2\), and the index \(m\) is a legitimate index for the basis of \(V\) because \(m \leq n\).

Neither \(S\) nor \(T\) is surjective. Indeed, every \(v \in V\) has the form \(v = a_1 v_1 + \dots + a_n v_n\) for some scalars \(a_1, \dots, a_n\), so by linearity \(Sv = a_1 w_1 + \dots + a_{m-1} w_{m-1}\). Hence

\begin{equation*} \operatorname{range} S \subseteq \operatorname{span}(w_1, \dots, w_{m-1}), \end{equation*}

and the reverse inclusion holds too because \(w_k = S v_k\) for \(1 \leq k \leq m - 1\), so in fact \(\operatorname{range} S = \operatorname{span}(w_1, \dots, w_{m-1})\). As \(w_1, \dots, w_{m-1}\) is linearly independent (being a sublist of a basis), this subspace has dimension \(m - 1 < m = \dim W\); since equal subspaces have equal dimensions, \(\operatorname{range} S \neq W\). The same computation gives

\begin{equation*} \operatorname{range} T = \operatorname{span}(w_m), \end{equation*}

of dimension \(1 < 2 \leq m = \dim W\), so \(\operatorname{range} T \neq W\). Thus \(S \in E\) and \(T \in E\).

Now consider \(S + T\). For \(1 \leq k \leq m - 1\) we have \((S + T) v_k = w_k + 0 = w_k\), and \((S + T) v_m = 0 + w_m = w_m\). Hence \(w_1, \dots, w_m\) all lie in \(\operatorname{range}(S + T)\). Since \(\operatorname{range}(S + T)\) is a subspace of \(W\) (by 3.18) containing the basis \(w_1, \dots, w_m\) of \(W\), we get \(\operatorname{range}(S + T) = W\). Therefore \(S + T\) is surjective, i.e. \(S + T \notin E\).

So \(E\) is not closed under addition, and consequently \(E\) is not a subspace of \(\mathcal{L}(V, W)\).

Both hypotheses are genuinely used. If \(\dim W = 1\), then a map \(T \in \mathcal{L}(V,W)\) fails to be surjective exactly when \(\operatorname{range} T = \{0\}\), i.e. exactly when \(T = 0\); so in that case \(E = \{0\}\), which is a subspace. If \(\dim V < \dim W\), then no linear map from \(V\) to \(W\) is surjective (by 3.24), so \(E = \mathcal{L}(V, W)\), again a subspace.

Suppose \(a_1, \dots, a_n \in \mathbf{F}\) satisfy

\begin{equation*} a_1 T v_1 + \dots + a_n T v_n = 0. \end{equation*}

Because \(T\) is linear (it commutes with scalar multiplication and preserves sums), the left side equals \(T(a_1 v_1 + \dots + a_n v_n)\). Thus

\begin{equation*} T(a_1 v_1 + \dots + a_n v_n) = 0, \end{equation*}

which says that \(a_1 v_1 + \dots + a_n v_n \in \operatorname{null} T\).

Since \(T\) is injective, 3.15 gives \(\operatorname{null} T = \{0\}\). Hence

\begin{equation*} a_1 v_1 + \dots + a_n v_n = 0. \end{equation*}

Now the linear independence of \(v_1, \dots, v_n\) forces \(a_1 = \dots = a_n = 0\).

We have shown that the only choice of scalars making \(a_1 T v_1 + \dots + a_n T v_n = 0\) is \(a_1 = \dots = a_n = 0\). By definition, \(T v_1, \dots, T v_n\) is linearly independent in \(W\).

We prove the two inclusions

\begin{equation*} \operatorname{span}(T v_1, \dots, T v_n) \subseteq \operatorname{range} T \quad \text{and} \quad \operatorname{range} T \subseteq \operatorname{span}(T v_1, \dots, T v_n). \end{equation*}

First inclusion. Each \(T v_k\) lies in \(\operatorname{range} T\) by definition of the range. By 3.18, \(\operatorname{range} T\) is a subspace of \(W\), hence is closed under scalar multiplication and addition; therefore every linear combination \(a_1 T v_1 + \dots + a_n T v_n\) lies in \(\operatorname{range} T\). Since the span of a list is precisely the set of its linear combinations, \(\operatorname{span}(T v_1, \dots, T v_n) \subseteq \operatorname{range} T\).

Second inclusion. Let \(w \in \operatorname{range} T\). Then \(w = Tv\) for some \(v \in V\). Because \(v_1, \dots, v_n\) spans \(V\), there are scalars \(a_1, \dots, a_n \in \mathbf{F}\) with

\begin{equation*} v = a_1 v_1 + \dots + a_n v_n. \end{equation*}

Applying \(T\) and using linearity,

\begin{equation*} w = Tv = T(a_1 v_1 + \dots + a_n v_n) = a_1 T v_1 + \dots + a_n T v_n, \end{equation*}

which lies in \(\operatorname{span}(T v_1, \dots, T v_n)\).

The two inclusions give \(\operatorname{span}(T v_1, \dots, T v_n) = \operatorname{range} T\); that is, the list \(T v_1, \dots, T v_n\) spans \(\operatorname{range} T\).

Note in particular that if \(V\) is finite-dimensional then \(\operatorname{range} T\) has a finite spanning list, so \(\operatorname{range} T\) is finite-dimensional and \(\dim \operatorname{range} T \leq \dim V\).

By 3.13, \(\operatorname{null} T\) is a subspace of \(V\); since \(V\) is finite-dimensional, \(\operatorname{null} T\) is finite-dimensional by 2.25, so it has a basis \(u_1, \dots, u_m\) (by 2.31). This list is linearly independent in \(V\), so by 2.32 it extends to a basis

\begin{equation*} u_1, \dots, u_m, w_1, \dots, w_n \end{equation*}

of \(V\). (Either \(m\) or \(n\) may be \(0\); the argument below is unaffected.) Define

\begin{equation*} U = \operatorname{span}(w_1, \dots, w_n), \end{equation*}

which is a subspace of \(V\) because a span is always a subspace.

Step 1: \(U \cap \operatorname{null} T = \{0\}\).

Certainly \(0\) lies in both \(U\) and \(\operatorname{null} T\). Conversely, suppose \(v \in U \cap \operatorname{null} T\). Since \(v \in U\) there are scalars \(b_1, \dots, b_n\) with \(v = b_1 w_1 + \dots + b_n w_n\), and since \(v \in \operatorname{null} T\) and \(u_1, \dots, u_m\) is a basis of \(\operatorname{null} T\) there are scalars \(a_1, \dots, a_m\) with \(v = a_1 u_1 + \dots + a_m u_m\). Subtracting,

\begin{equation*} a_1 u_1 + \dots + a_m u_m - b_1 w_1 - \dots - b_n w_n = 0. \end{equation*}

The list \(u_1, \dots, u_m, w_1, \dots, w_n\) is linearly independent, so every coefficient here vanishes: \(a_1 = \dots = a_m = 0\) and \(b_1 = \dots = b_n = 0\). Hence \(v = 0\), proving \(U \cap \operatorname{null} T = \{0\}\).

Step 2: \(\operatorname{range} T = \{Tu : u \in U\}\).

The inclusion \(\{Tu : u \in U\} \subseteq \operatorname{range} T\) is immediate, since every \(Tu\) with \(u \in U \subseteq V\) belongs to \(\operatorname{range} T\).

For the reverse inclusion, let \(x \in \operatorname{range} T\), say \(x = Tv\) with \(v \in V\). Write \(v\) in the basis of \(V\):

\begin{equation*} v = a_1 u_1 + \dots + a_m u_m + b_1 w_1 + \dots + b_n w_n \end{equation*}

for some scalars \(a_j, b_k\). Put \(u = b_1 w_1 + \dots + b_n w_n\), so \(u \in U\). Since \(u_1, \dots, u_m \in \operatorname{null} T\), linearity of \(T\) gives

\begin{equation*} x = Tv = a_1 T u_1 + \dots + a_m T u_m + T u = 0 + \dots + 0 + Tu = Tu. \end{equation*}

Thus \(x \in \{Tu : u \in U\}\).

Combining the two inclusions, \(\operatorname{range} T = \{Tu : u \in U\}\), and with Step 1 this completes the proof.

Remark. The two conclusions say exactly that the restriction \(T|_U\) is injective (its null space is \(U \cap \operatorname{null} T = \{0\}\), by 3.15) and has the same range as \(T\). Counting dimensions, the list \(w_1, \dots, w_n\) is linearly independent (a sublist of a basis) and spans \(U\), hence is a basis of \(U\) and \(\dim U = n\); since \(n = \dim V - m = \dim V - \dim \operatorname{null} T\), the fundamental theorem of linear maps (3.21) gives \(\dim U = \dim \operatorname{range} T\), which is consistent with \(T|_U\) being an injective map of \(U\) onto \(\operatorname{range} T\).

The strategy is to compute \(\dim \operatorname{null} T\), apply the fundamental theorem of linear maps (3.21) to get \(\dim \operatorname{range} T = 2\), and then conclude \(\operatorname{range} T = \mathbf{F}^2\) from 2.39.

Step 1: \(\dim \operatorname{null} T = 2\).

A vector \((x_1, x_2, x_3, x_4) \in \mathbf{F}^4\) lies in \(\operatorname{null} T\) precisely when \(x_1 = 5x_2\) and \(x_3 = 7x_4\), that is, precisely when it has the form

\begin{equation*} (5x_2, x_2, 7x_4, x_4) = x_2 (5, 1, 0, 0) + x_4 (0, 0, 7, 1) \end{equation*}

for some \(x_2, x_4 \in \mathbf{F}\). Conversely every such combination satisfies the two equations. Hence

\begin{equation*} \operatorname{null} T = \operatorname{span}\big((5,1,0,0), (0,0,7,1)\big). \end{equation*}

The list \((5,1,0,0), (0,0,7,1)\) is linearly independent: if

\begin{equation*} a (5,1,0,0) + b (0,0,7,1) = (5a, a, 7b, b) = (0,0,0,0), \end{equation*}

then looking at the second coordinate gives \(a = 0\) and at the fourth gives \(b = 0\). So this list is a basis of \(\operatorname{null} T\) and \(\dim \operatorname{null} T = 2\).

Step 2: \(\dim \operatorname{range} T = 2\).

Since \(\mathbf{F}^4\) is finite-dimensional with \(\dim \mathbf{F}^4 = 4\), the fundamental theorem of linear maps (3.21) gives

\begin{equation*} 4 = \dim \mathbf{F}^4 = \dim \operatorname{null} T + \dim \operatorname{range} T = 2 + \dim \operatorname{range} T, \end{equation*}

so \(\dim \operatorname{range} T = 2\).

Step 3: \(T\) is surjective.

By 3.18, \(\operatorname{range} T\) is a subspace of \(\mathbf{F}^2\), and \(\dim \mathbf{F}^2 = 2 = \dim \operatorname{range} T\). A subspace of full dimension equals the whole space (2.39), so \(\operatorname{range} T = \mathbf{F}^2\). That is, \(T\) is surjective.

The space \(\mathbf{R}^8\) is finite-dimensional with \(\dim \mathbf{R}^8 = 8\), and by hypothesis

\begin{equation*} \dim \operatorname{null} T = \dim U = 3. \end{equation*}

The fundamental theorem of linear maps (3.21) applied to \(T \in \mathcal{L}(\mathbf{R}^8, \mathbf{R}^5)\) gives

\begin{equation*} 8 = \dim \mathbf{R}^8 = \dim \operatorname{null} T + \dim \operatorname{range} T = 3 + \dim \operatorname{range} T, \end{equation*}

so

\begin{equation*} \dim \operatorname{range} T = 5. \end{equation*}

By 3.18, \(\operatorname{range} T\) is a subspace of \(\mathbf{R}^5\), and \(\dim \mathbf{R}^5 = 5 = \dim \operatorname{range} T\). Since a subspace whose dimension equals the dimension of the whole (finite-dimensional) space is the whole space (2.39), we conclude

\begin{equation*} \operatorname{range} T = \mathbf{R}^5. \end{equation*}

Hence \(T\) is surjective.

Write

\begin{equation*} N = \{(x_1, x_2, x_3, x_4, x_5) \in \mathbf{F}^5 : x_1 = 3x_2 \text{ and } x_3 = x_4 = x_5\}. \end{equation*}

Step 1: \(\dim N = 2\).

A vector of \(\mathbf{F}^5\) lies in \(N\) exactly when it has the form

\begin{equation*} (3x_2, x_2, x_3, x_3, x_3) = x_2 (3, 1, 0, 0, 0) + x_3 (0, 0, 1, 1, 1) \end{equation*}

for some \(x_2, x_3 \in \mathbf{F}\); conversely every such combination satisfies \(x_1 = 3x_2\) and \(x_3 = x_4 = x_5\). Hence

\begin{equation*} N = \operatorname{span}\big((3,1,0,0,0), (0,0,1,1,1)\big). \end{equation*}

This spanning list is linearly independent: if

\begin{equation*} a (3,1,0,0,0) + b (0,0,1,1,1) = (3a, a, b, b, b) = 0, \end{equation*}

then the second coordinate gives \(a = 0\) and the third gives \(b = 0\). So \((3,1,0,0,0), (0,0,1,1,1)\) is a basis of \(N\) and \(\dim N = 2\).

Step 2: no linear map can have \(N\) as its null space.

Suppose, for contradiction, that \(T \in \mathcal{L}(\mathbf{F}^5, \mathbf{F}^2)\) satisfies \(\operatorname{null} T = N\). Since \(\mathbf{F}^5\) is finite-dimensional with \(\dim \mathbf{F}^5 = 5\), the fundamental theorem of linear maps (3.21) gives

\begin{equation*} 5 = \dim \mathbf{F}^5 = \dim \operatorname{null} T + \dim \operatorname{range} T = 2 + \dim \operatorname{range} T, \end{equation*}

so \(\dim \operatorname{range} T = 3\).

But \(\operatorname{range} T\) is a subspace of \(\mathbf{F}^2\) (by 3.18), and \(\mathbf{F}^2\) is finite-dimensional of dimension \(2\), so 2.37 yields

\begin{equation*} \dim \operatorname{range} T \leq \dim \mathbf{F}^2 = 2. \end{equation*}

This contradicts \(\dim \operatorname{range} T = 3\).

Therefore no such linear map exists.

Remark. The same counting shows more generally that a subspace \(N \subseteq \mathbf{F}^5\) can be the null space of some \(T \in \mathcal{L}(\mathbf{F}^5, \mathbf{F}^2)\) only when \(\dim N \geq 3\). Conversely, if \(\dim N = k \geq 3\) then such a \(T\) does exist: extend a basis \(n_1, \dots, n_k\) of \(N\) to a basis \(n_1, \dots, n_k, w_1, \dots, w_{5-k}\) of \(\mathbf{F}^5\), and since \(5 - k \leq 2\) we may pick a linearly independent list \(e_1, \dots, e_{5-k}\) in \(\mathbf{F}^2\). The linear map lemma (3.4) supplies \(T\) with \(T n_i = 0\) and \(T w_j = e_j\), and one checks from the linear independence of \(e_1, \dots, e_{5-k}\) that \(\operatorname{null} T = N\).

Let \(T\) be such a map. (The argument below uses nothing about the target space, so we may as well suppose \(T \in \mathcal{L}(V, W)\) for some vector space \(W\); the case of an operator \(T \in \mathcal{L}(V)\) is the case \(W = V\).) By hypothesis \(\operatorname{null} T\) and \(\operatorname{range} T\) are both finite-dimensional. We produce a finite list that spans \(V\), which by definition makes \(V\) finite-dimensional.

Let \(u_1, \dots, u_m\) be a basis of \(\operatorname{null} T\) and let \(w_1, \dots, w_n\) be a basis of \(\operatorname{range} T\) (these exist because both spaces are finite-dimensional, by 2.31). For each \(k \in \{1, \dots, n\}\) we have \(w_k \in \operatorname{range} T\), so we may choose \(v_k \in V\) with

\begin{equation*} T v_k = w_k . \end{equation*}

We claim that the list

\begin{equation*} u_1, \dots, u_m, v_1, \dots, v_n \end{equation*}

spans \(V\). To see this, let \(v \in V\). Then \(Tv \in \operatorname{range} T\), so there exist \(b_1, \dots, b_n \in \mathbf{F}\) with

\begin{equation*} T v = b_1 w_1 + \dots + b_n w_n = b_1 T v_1 + \dots + b_n T v_n = T(b_1 v_1 + \dots + b_n v_n), \end{equation*}

where the last equality uses the linearity of \(T\). Hence

\begin{equation*} T\bigl(v - (b_1 v_1 + \dots + b_n v_n)\bigr) = 0, \end{equation*}

so \(v - (b_1 v_1 + \dots + b_n v_n) \in \operatorname{null} T\). Because \(u_1, \dots, u_m\) spans \(\operatorname{null} T\), there exist \(a_1, \dots, a_m \in \mathbf{F}\) with

\begin{equation*} v - (b_1 v_1 + \dots + b_n v_n) = a_1 u_1 + \dots + a_m u_m . \end{equation*}

Therefore

\begin{equation*} v = a_1 u_1 + \dots + a_m u_m + b_1 v_1 + \dots + b_n v_n , \end{equation*}

which exhibits \(v\) as a linear combination of \(u_1, \dots, u_m, v_1, \dots, v_n\).

Thus the finite list \(u_1, \dots, u_m, v_1, \dots, v_n\) spans \(V\), and hence \(V\) is finite-dimensional.

Note that the conclusion is exactly what one expects from the fundamental theorem of linear maps (3.21): once \(V\) is known to be finite-dimensional, that theorem gives \(\dim V = \dim \operatorname{null} T + \dim \operatorname{range} T\), and indeed the spanning list constructed above has length \(m + n\). (In fact that list is a basis of \(V\): if \(a_1 u_1 + \dots + a_m u_m + b_1 v_1 + \dots + b_n v_n = 0\), applying \(T\) gives \(b_1 w_1 + \dots + b_n w_n = 0\), so all the \(b_k\) are \(0\) because \(w_1, \dots, w_n\) is linearly independent, and then all the \(a_j\) are \(0\) because \(u_1, \dots, u_m\) is linearly independent.)

First suppose there exists an injective \(T \in \mathcal{L}(V, W)\). Because \(T\) is injective, \(\operatorname{null} T = \{0\}\) (by 3.15), so \(\dim \operatorname{null} T = 0\). The fundamental theorem of linear maps (3.21) now gives

\begin{equation*} \dim V = \dim \operatorname{null} T + \dim \operatorname{range} T = \dim \operatorname{range} T . \end{equation*}

Since \(\operatorname{range} T\) is a subspace of \(W\) (by 3.18) and \(W\) is finite-dimensional, we have \(\dim \operatorname{range} T \le \dim W\) (by 2.37). Combining the two displays gives \(\dim V \le \dim W\).

Conversely, suppose \(\dim V \le \dim W\). Write \(n = \dim V\) and \(m = \dim W\), so \(n \le m\). Let \(v_1, \dots, v_n\) be a basis of \(V\) and let \(w_1, \dots, w_m\) be a basis of \(W\) (by 2.31). Because \(n \le m\), the vectors \(w_1, \dots, w_n\) are among \(w_1, \dots, w_m\); in particular \(w_1, \dots, w_n\) is a linearly independent list in \(W\), being a sublist of a linearly independent list.

By the linear map lemma (3.4) there exists \(T \in \mathcal{L}(V, W)\) with

\begin{equation*} T v_k = w_k \quad \text{for each } k = 1, \dots, n . \end{equation*}

We claim \(T\) is injective. By 3.15 it suffices to show \(\operatorname{null} T = \{0\}\). Suppose \(v \in \operatorname{null} T\). Because \(v_1, \dots, v_n\) is a basis of \(V\), we may write \(v = c_1 v_1 + \dots + c_n v_n\) with \(c_1, \dots, c_n \in \mathbf{F}\). Then

\begin{equation*} 0 = Tv = c_1 T v_1 + \dots + c_n T v_n = c_1 w_1 + \dots + c_n w_n . \end{equation*}

Because \(w_1, \dots, w_n\) is linearly independent, \(c_1 = \dots = c_n = 0\), and hence \(v = 0\). Thus \(\operatorname{null} T = \{0\}\) and \(T\) is injective, as desired.

(If \(n = 0\), so that \(V = \{0\}\), the lists above are empty and the map \(T\) produced is the zero map, which is indeed injective on \(\{0\}\); the argument as written covers this degenerate case.)

First suppose there exists a surjective \(T \in \mathcal{L}(V, W)\). Surjectivity means \(\operatorname{range} T = W\) (by 3.19). The fundamental theorem of linear maps (3.21) gives

\begin{equation*} \dim V = \dim \operatorname{null} T + \dim \operatorname{range} T = \dim \operatorname{null} T + \dim W \ge \dim W , \end{equation*}

where the last inequality holds because \(\dim \operatorname{null} T \ge 0\). Thus \(\dim V \ge \dim W\).

Conversely, suppose \(\dim V \ge \dim W\). Write \(n = \dim V\) and \(m = \dim W\), so \(n \ge m\). Let \(v_1, \dots, v_n\) be a basis of \(V\) and let \(w_1, \dots, w_m\) be a basis of \(W\) (by 2.31). By the linear map lemma (3.4) there exists \(T \in \mathcal{L}(V, W)\) with

\begin{equation*} T v_k = \begin{cases} w_k & \text{if } 1 \le k \le m, \\ 0 & \text{if } m < k \le n. \end{cases} \end{equation*}

(This is legitimate precisely because \(m \le n\): we have prescribed a value in \(W\) for each of the \(n\) basis vectors of \(V\).)

We claim \(T\) is surjective. Indeed, \(\operatorname{range} T\) is a subspace of \(W\) (by 3.18) that contains \(w_1, \dots, w_m\), so it contains \(\operatorname{span}(w_1, \dots, w_m) = W\) (using 2.6, that the span is the smallest subspace containing the list). Hence \(\operatorname{range} T = W\), so \(T\) is surjective.

Alternatively, one can see surjectivity directly: given \(w \in W\), write \(w = c_1 w_1 + \dots + c_m w_m\); then \(w = T(c_1 v_1 + \dots + c_m v_m)\).

(If \(m = 0\), so that \(W = \{0\}\), the map \(T\) is the zero map, whose range is \(\{0\} = W\); so \(T\) is surjective and the argument covers this degenerate case as well.)

First suppose there exists \(T \in \mathcal{L}(V, W)\) with \(\operatorname{null} T = U\). The fundamental theorem of linear maps (3.21) gives

\begin{equation*} \dim V = \dim \operatorname{null} T + \dim \operatorname{range} T = \dim U + \dim \operatorname{range} T . \end{equation*}

Because \(\operatorname{range} T\) is a subspace of the finite-dimensional space \(W\) (by 3.18), we have \(\dim \operatorname{range} T \le \dim W\) (by 2.37). Hence

\begin{equation*} \dim V \le \dim U + \dim W , \end{equation*}

which rearranges to \(\dim U \ge \dim V - \dim W\), as desired.

Conversely, suppose \(\dim U \ge \dim V - \dim W\). Let \(u_1, \dots, u_j\) be a basis of \(U\), where \(j = \dim U\); note \(U\) is finite-dimensional because it is a subspace of the finite-dimensional space \(V\) (by 2.25). Extend this linearly independent list to a basis

\begin{equation*} u_1, \dots, u_j, v_1, \dots, v_n \end{equation*}

of \(V\) (by 2.32). Then \(\dim V = j + n\), so

\begin{equation*} n = \dim V - \dim U \le \dim W , \end{equation*}

where the inequality is exactly the hypothesis rearranged. Let \(w_1, \dots, w_m\) be a basis of \(W\); since \(n \le m\), the list \(w_1, \dots, w_n\) is a well-defined sublist of it and hence is linearly independent in \(W\).

By the linear map lemma (3.4) there exists \(T \in \mathcal{L}(V, W)\) with

\begin{equation*} T u_i = 0 \quad \text{for } i = 1, \dots, j, \qquad T v_k = w_k \quad \text{for } k = 1, \dots, n . \end{equation*}

We show \(\operatorname{null} T = U\).

First, \(U \subseteq \operatorname{null} T\): each \(u_i\) lies in \(\operatorname{null} T\), and \(\operatorname{null} T\) is a subspace of \(V\) (by 3.13), so it contains \(\operatorname{span}(u_1, \dots, u_j) = U\).

Second, \(\operatorname{null} T \subseteq U\): suppose \(v \in \operatorname{null} T\) and write

\begin{equation*} v = a_1 u_1 + \dots + a_j u_j + b_1 v_1 + \dots + b_n v_n \end{equation*}

with all coefficients in \(\mathbf{F}\), which is possible because \(u_1, \dots, u_j, v_1, \dots, v_n\) is a basis of \(V\). Applying \(T\) gives

\begin{equation*} 0 = Tv = b_1 w_1 + \dots + b_n w_n . \end{equation*}

Because \(w_1, \dots, w_n\) is linearly independent, \(b_1 = \dots = b_n = 0\). Hence \(v = a_1 u_1 + \dots + a_j u_j \in U\).

Thus \(\operatorname{null} T = U\), completing the proof.

First suppose there exists \(S \in \mathcal{L}(W, V)\) with \(ST = I\), the identity operator on \(V\). If \(u, v \in V\) satisfy \(Tu = Tv\), then applying \(S\) gives

\begin{equation*} u = (ST)u = S(Tu) = S(Tv) = (ST)v = v . \end{equation*}

Hence \(T\) is injective. (This direction uses no finite-dimensionality at all.)

Now suppose \(T\) is injective; we construct \(S\). Let \(w_1, \dots, w_n\) be a basis of \(\operatorname{range} T\); this exists because \(\operatorname{range} T\) is a subspace of the finite-dimensional space \(W\) (by 3.18 and 2.25). Extend it to a basis

\begin{equation*} w_1, \dots, w_n, w_{n+1}, \dots, w_m \end{equation*}

of \(W\) (by 2.32; here is where finite-dimensionality of \(W\) is used). For each \(k \in \{1, \dots, n\}\), the vector \(w_k\) lies in \(\operatorname{range} T\), so we may choose \(v_k \in V\) with

\begin{equation*} T v_k = w_k . \end{equation*}

By the linear map lemma (3.4) there exists \(S \in \mathcal{L}(W, V)\) with

\begin{equation*} S w_k = \begin{cases} v_k & \text{if } 1 \le k \le n, \\ 0 & \text{if } n < k \le m. \end{cases} \end{equation*}

We claim \(ST = I\). Let \(v \in V\). Then \(Tv \in \operatorname{range} T\), so there exist \(c_1, \dots, c_n \in \mathbf{F}\) with

\begin{equation*} T v = c_1 w_1 + \dots + c_n w_n . \end{equation*}

Applying \(S\) and using linearity gives

\begin{equation*} (ST)v = S(Tv) = c_1 S w_1 + \dots + c_n S w_n = c_1 v_1 + \dots + c_n v_n . \end{equation*}

On the other hand,

\begin{equation*} T(c_1 v_1 + \dots + c_n v_n) = c_1 T v_1 + \dots + c_n T v_n = c_1 w_1 + \dots + c_n w_n = Tv . \end{equation*}

Because \(T\) is injective, the equation \(T(c_1 v_1 + \dots + c_n v_n) = Tv\) forces

\begin{equation*} c_1 v_1 + \dots + c_n v_n = v . \end{equation*}

Combining the last two displays gives \((ST)v = v\). Since \(v \in V\) was arbitrary, \(ST\) is the identity operator on \(V\), as desired.

Remark: it is worth noting where each hypothesis enters. Finite-dimensionality of \(W\) was used only to extend a basis of \(\operatorname{range} T\) to a basis of \(W\), i.e. to split off a complement of \(\operatorname{range} T\) inside \(W\). (Injectivity of \(T\) in fact forces \(V\) to be finite-dimensional as well: an injective linear map sends a linearly independent list in \(V\) to a linearly independent list in \(W\), so no linearly independent list in \(V\) has length greater than \(\dim W\); so we may choose a linearly independent list \(u_1, \dots, u_k\) in \(V\) of maximal length. That list spans \(V\): if some \(v \in V\) were not in \(\operatorname{span}(u_1, \dots, u_k)\), then \(u_1, \dots, u_k, v\) would be linearly independent, contradicting maximality. Indeed, if \(u_1, \dots, u_k, v\) were linearly dependent, the linear dependence lemma (2.19) would supply a term lying in the span of the terms preceding it; that term cannot be \(v\), since \(v \notin \operatorname{span}(u_1, \dots, u_k)\), and it cannot be some \(u_j\), since \(u_j \in \operatorname{span}(u_1, \dots, u_{j-1})\) would make \(u_1, \dots, u_k\) linearly dependent. Hence \(V\) is spanned by a finite list, so \(V\) is finite-dimensional. We did not need this fact, and the proof above deliberately avoids it.)

First suppose there exists \(S \in \mathcal{L}(W, V)\) with \(TS = I\), the identity operator on \(W\). Let \(w \in W\). Then

\begin{equation*} w = (TS)w = T(Sw) , \end{equation*}

so \(w \in \operatorname{range} T\). Since \(w \in W\) was arbitrary, \(\operatorname{range} T = W\), i.e. \(T\) is surjective. (This direction uses no finite-dimensionality.)

Now suppose \(T\) is surjective; we construct \(S\). Because \(W\) is finite-dimensional it has a basis \(w_1, \dots, w_m\) (by 2.31). Since \(\operatorname{range} T = W\), for each \(k \in \{1, \dots, m\}\) we may choose \(v_k \in V\) with

\begin{equation*} T v_k = w_k . \end{equation*}

By the linear map lemma (3.4) there exists \(S \in \mathcal{L}(W, V)\) with

\begin{equation*} S w_k = v_k \quad \text{for each } k = 1, \dots, m . \end{equation*}

(Finite-dimensionality of \(W\) is used exactly here, to have a finite basis on which to prescribe \(S\).)

We claim \(TS = I\). For each \(k\),

\begin{equation*} (TS) w_k = T(S w_k) = T v_k = w_k , \end{equation*}

so the linear maps \(TS\) and \(I\) (both in \(\mathcal{L}(W)\)) agree on the basis \(w_1, \dots, w_m\). Hence they are equal: explicitly, given \(w \in W\), write \(w = c_1 w_1 + \dots + c_m w_m\) and compute

\begin{equation*} (TS)w = c_1 (TS) w_1 + \dots + c_m (TS) w_m = c_1 w_1 + \dots + c_m w_m = w . \end{equation*}

(One can also invoke the uniqueness assertion of 3.4: a linear map on \(W\) is determined by its values on a basis.)

Thus \(TS\) is the identity operator on \(W\), as desired.

Write

\begin{equation*} E = \{v \in V : Tv \in U\} . \end{equation*}

Step 1: \(E\) is a subspace of \(V\). Because \(T\) is linear, \(T0 = 0\) (by 3.10), and \(0 \in U\) because \(U\) is a subspace of \(W\); hence \(0 \in E\). If \(v_1, v_2 \in E\), then \(Tv_1, Tv_2 \in U\), so

\begin{equation*} T(v_1 + v_2) = Tv_1 + Tv_2 \in U \end{equation*}

because \(U\) is closed under addition; hence \(v_1 + v_2 \in E\), so \(E\) is closed under addition. If \(v \in E\) and \(\lambda \in \mathbf{F}\), then

\begin{equation*} T(\lambda v) = \lambda Tv \in U \end{equation*}

because \(U\) is closed under scalar multiplication; hence \(\lambda v \in E\), so \(E\) is closed under scalar multiplication. Since \(E\) contains \(0\) and is closed under addition and scalar multiplication, \(E\) is a subspace of \(V\) (by 1.34).

Step 2: the dimension formula. Because \(V\) is finite-dimensional and \(E\) is a subspace of \(V\), the space \(E\) is finite-dimensional (by 2.25). Consider the restriction of \(T\) to \(E\),

\begin{equation*} T|_E : E \to W, \qquad (T|_E)v = Tv \quad \text{for } v \in E . \end{equation*}

This is a linear map from \(E\) to \(W\), since it satisfies additivity and homogeneity as inherited from \(T\); that is, \(T|_E \in \mathcal{L}(E, W)\). Applying the fundamental theorem of linear maps (3.21) to \(T|_E\), whose domain \(E\) is finite-dimensional, gives

\begin{equation*} \dim E = \dim \operatorname{null} (T|_E) + \dim \operatorname{range} (T|_E) . \end{equation*}

It remains to identify the two spaces on the right.

First we show \(\operatorname{null}(T|_E) = \operatorname{null} T\). If \(v \in \operatorname{null}(T|_E)\), then \(v \in E \subseteq V\) and \(Tv = 0\), so \(v \in \operatorname{null} T\). Conversely, if \(v \in \operatorname{null} T\), then \(Tv = 0 \in U\), so \(v \in E\), and therefore \((T|_E)v = Tv = 0\), giving \(v \in \operatorname{null}(T|_E)\). Hence the two null spaces are equal, and in particular

\begin{equation*} \dim \operatorname{null}(T|_E) = \dim \operatorname{null} T . \end{equation*}

Next we show \(\operatorname{range}(T|_E) = U \cap \operatorname{range} T\). If \(w \in \operatorname{range}(T|_E)\), then \(w = Tv\) for some \(v \in E\); then \(w \in \operatorname{range} T\) by definition, and \(w = Tv \in U\) by the definition of \(E\), so \(w \in U \cap \operatorname{range} T\). Conversely, suppose \(w \in U \cap \operatorname{range} T\). Since \(w \in \operatorname{range} T\), there exists \(v \in V\) with \(Tv = w\). Since \(w \in U\), we have \(Tv \in U\), so \(v \in E\); hence \(w = (T|_E)v \in \operatorname{range}(T|_E)\). Hence the two sets are equal, and in particular

\begin{equation*} \dim \operatorname{range}(T|_E) = \dim (U \cap \operatorname{range} T) . \end{equation*}

Substituting these two identifications into the displayed application of 3.21 yields

\begin{equation*} \dim \{v \in V : Tv \in U\} = \dim E = \dim \operatorname{null} T + \dim (U \cap \operatorname{range} T), \end{equation*}

as desired.

Remark: taking \(U = \{0\}\) recovers \(E = \operatorname{null} T\) and the trivial identity \(\dim \operatorname{null} T = \dim \operatorname{null} T\), while taking \(U = W\) recovers \(E = V\) and the fundamental theorem of linear maps itself, since then \(U \cap \operatorname{range} T = \operatorname{range} T\).

Here \(ST \in \mathcal{L}(U, W)\), so \(\operatorname{null} ST\) is a subspace of \(U\) (by 3.13). Since \(U\) is finite-dimensional, \(\operatorname{null} ST\) is finite-dimensional (by 2.25).

The idea is to apply the fundamental theorem of linear maps to \(T\) restricted to \(\operatorname{null} ST\). Define

\begin{equation*} R = T|_{\operatorname{null} ST} \in \mathcal{L}(\operatorname{null} ST,\ V), \end{equation*}

so \(Ru = Tu\) for every \(u \in \operatorname{null} ST\). This is a linear map because \(T\) is linear and \(\operatorname{null} ST\) is a subspace of the domain \(U\) of \(T\).

Step 1: \(\operatorname{null} R \subseteq \operatorname{null} T\). If \(u \in \operatorname{null} R\) then \(u \in \operatorname{null} ST\) and \(Tu = Ru = 0\), so \(u \in \operatorname{null} T\). Both \(\operatorname{null} R\) and \(\operatorname{null} T\) are subspaces of \(U\) (by 3.13), and \(\operatorname{null} T\) is finite-dimensional (by 2.25, as \(U\) is finite-dimensional); since \(\operatorname{null} R\) is contained in \(\operatorname{null} T\), it is a subspace of the vector space \(\operatorname{null} T\), so 2.37 gives

\begin{equation*} \dim \operatorname{null} R \le \dim \operatorname{null} T. \end{equation*}

Step 2: \(\operatorname{range} R \subseteq \operatorname{null} S\). Suppose \(v \in \operatorname{range} R\), say \(v = Ru = Tu\) with \(u \in \operatorname{null} ST\). Then

\begin{equation*} Sv = S(Tu) = (ST)u = 0, \end{equation*}

so \(v \in \operatorname{null} S\). Both \(\operatorname{range} R\) and \(\operatorname{null} S\) are subspaces of the finite-dimensional space \(V\) (by 3.18 and 3.13), hence both are finite-dimensional (by 2.25); as \(\operatorname{range} R \subseteq \operatorname{null} S\), it is a subspace of \(\operatorname{null} S\), so 2.37 gives

\begin{equation*} \dim \operatorname{range} R \le \dim \operatorname{null} S. \end{equation*}

Step 3: combine. Applying the fundamental theorem of linear maps (3.21) to \(R\), whose domain \(\operatorname{null} ST\) is finite-dimensional, we get

\begin{equation*} \dim \operatorname{null} ST = \dim \operatorname{null} R + \dim \operatorname{range} R. \end{equation*}

Substituting the two inequalities from Steps 1 and 2 yields

\begin{equation*} \dim \operatorname{null} ST \le \dim \operatorname{null} T + \dim \operatorname{null} S, \end{equation*}

as desired.

Intuitively: a vector killed by \(ST\) is either killed already by \(T\), or else \(T\) sends it into \(\operatorname{null} S\); the first possibility accounts for at most \(\dim \operatorname{null} T\) dimensions and the second for at most \(\dim \operatorname{null} S\).

Again \(ST \in \mathcal{L}(U, W)\). Because \(U\) and \(V\) are finite-dimensional, the fundamental theorem of linear maps (3.21) tells us that \(\operatorname{range} T\), \(\operatorname{range} S\), and \(\operatorname{range} ST\) are all finite-dimensional, so all three dimensions appearing above make sense. We prove the two inequalities separately.

Claim 1: \(\dim \operatorname{range} ST \le \dim \operatorname{range} S\). Suppose \(w \in \operatorname{range} ST\). Then \(w = (ST)u = S(Tu)\) for some \(u \in U\), and \(Tu \in V\), so \(w \in \operatorname{range} S\). Thus

\begin{equation*} \operatorname{range} ST \subseteq \operatorname{range} S. \end{equation*}

Both are subspaces of \(W\) (by 3.18), and \(\operatorname{range} S\) is finite-dimensional, so 2.37 applied inside the vector space \(\operatorname{range} S\) gives \(\dim \operatorname{range} ST \le \dim \operatorname{range} S\).

Claim 2: \(\dim \operatorname{range} ST \le \dim \operatorname{range} T\). Restrict \(S\) to \(\operatorname{range} T\): define

\begin{equation*} R = S|_{\operatorname{range} T} \in \mathcal{L}(\operatorname{range} T,\ W), \end{equation*}

which makes sense because \(\operatorname{range} T\) is a subspace of \(V\) (by 3.18), and which is linear because \(S\) is. We compute the range of \(R\):

\begin{equation*} \operatorname{range} R = \{Sv : v \in \operatorname{range} T\} = \{S(Tu) : u \in U\} = \{(ST)u : u \in U\} = \operatorname{range} ST, \end{equation*}

where the second equality holds because \(v\) ranges over \(\operatorname{range} T\) exactly when \(v = Tu\) for some \(u \in U\).

Now \(\operatorname{range} T\) is finite-dimensional, so applying 3.21 to \(R\) gives

\begin{equation*} \dim \operatorname{range} T = \dim \operatorname{null} R + \dim \operatorname{range} R \ge \dim \operatorname{range} R = \dim \operatorname{range} ST. \end{equation*}

Since \(\dim \operatorname{range} ST\) is less than or equal to each of \(\dim \operatorname{range} S\) and \(\dim \operatorname{range} T\), it is less than or equal to their minimum, as desired.

(a) First we turn the hypothesis \(ST = 0\) into a statement about ranges and null spaces. Let \(v \in V\). Then \(S(Tv) = (ST)v = 0\), so \(Tv \in \operatorname{null} S\). As \(v\) was arbitrary,

\begin{equation*} \operatorname{range} T \subseteq \operatorname{null} S. \end{equation*}

Both \(\operatorname{range} T\) and \(\operatorname{null} S\) are subspaces of the five-dimensional space \(V\) (by 3.18 and 3.13), so both are finite-dimensional (by 2.25); since \(\operatorname{range} T \subseteq \operatorname{null} S\), it is a subspace of \(\operatorname{null} S\) and 2.37 gives \(\dim \operatorname{range} T \le \dim \operatorname{null} S\). Combining this with the fundamental theorem of linear maps (3.21) applied to \(S\), namely \(\dim \operatorname{null} S = 5 - \dim \operatorname{range} S\), we obtain

\begin{equation*} \dim \operatorname{range} S + \dim \operatorname{range} T \le 5. \end{equation*}

Call this the rank bound.

Now apply Exercise 23 of this section to the composition \(TS\) (with \(S\) as the inner map and \(T\) as the outer map; its hypotheses hold because \(V\) is finite-dimensional and both maps act on \(V\)):

\begin{equation*} \dim \operatorname{range} TS \le \min\{\dim \operatorname{range} T, \dim \operatorname{range} S\}. \end{equation*}

If we had \(\min\{\dim \operatorname{range} T, \dim \operatorname{range} S\} \ge 3\), then both \(\dim \operatorname{range} S \ge 3\) and \(\dim \operatorname{range} T \ge 3\), giving \(\dim \operatorname{range} S + \dim \operatorname{range} T \ge 6\), which contradicts the rank bound. Hence the minimum is at most \(2\), and therefore

\begin{equation*} \dim \operatorname{range} TS \le 2, \end{equation*}

as desired.

(b) Define \(S, T \in \mathcal{L}(\mathbf{F}^5)\) by

\begin{equation*} S(x_1, x_2, x_3, x_4, x_5) = (x_1, x_2, 0, 0, 0), \qquad T(x_1, x_2, x_3, x_4, x_5) = (0, 0, x_1, x_2, 0). \end{equation*}

Both maps are linear, since each coordinate of the output is a linear function of the input coordinates.

For every \((x_1, \dots, x_5) \in \mathbf{F}^5\) we have

\begin{equation*} (ST)(x_1, \dots, x_5) = S(0, 0, x_1, x_2, 0) = (0, 0, 0, 0, 0), \end{equation*}

because the first two coordinates of \(T(x_1, \dots, x_5)\) are \(0\). Thus \(ST = 0\).

On the other hand,

\begin{equation*} (TS)(x_1, \dots, x_5) = T(x_1, x_2, 0, 0, 0) = (0, 0, x_1, x_2, 0), \end{equation*}

so

\begin{equation*} \operatorname{range} TS = \{(0, 0, x_1, x_2, 0) : x_1, x_2 \in \mathbf{F}\} = \operatorname{span}\big((0,0,1,0,0),\ (0,0,0,1,0)\big). \end{equation*}

The two spanning vectors listed above are linearly independent, so \(\dim \operatorname{range} TS = 2\).

(Consistently with the rank bound, here \(\dim \operatorname{range} S = \dim \operatorname{range} T = 2\) and \(2 + 2 \le 5\).)

The easy direction. Suppose there exists \(E \in \mathcal{L}(W)\) with \(T = ES\). Let \(v \in \operatorname{null} S\). Then

\begin{equation*} Tv = (ES)v = E(Sv) = E0 = 0, \end{equation*}

using \(E0 = 0\) (by 3.10). Hence \(v \in \operatorname{null} T\). Thus \(\operatorname{null} S \subseteq \operatorname{null} T\). Note that this direction does not use the finite-dimensionality of \(W\).

The other direction. Now suppose \(\operatorname{null} S \subseteq \operatorname{null} T\); we must manufacture \(E \in \mathcal{L}(W)\) with \(ES = T\). The point is that the hypothesis makes the assignment sending \(Sv\) to \(Tv\) unambiguous on \(\operatorname{range} S\); finite-dimensionality of \(W\) then lets us extend that assignment to all of \(W\).

By 3.18, \(\operatorname{range} S\) is a subspace of \(W\), hence is finite-dimensional (by 2.25). Let

\begin{equation*} w_1, \dots, w_m \end{equation*}

be a basis of \(\operatorname{range} S\) (the empty list if \(\operatorname{range} S = \{0\}\)). For each \(k \in \{1, \dots, m\}\), since \(w_k \in \operatorname{range} S\) we may choose \(v_k \in V\) with

\begin{equation*} S v_k = w_k. \end{equation*}

The list \(w_1, \dots, w_m\) is linearly independent in \(W\), so by 2.32 it extends to a basis

\begin{equation*} w_1, \dots, w_m, w_{m+1}, \dots, w_n \end{equation*}

of \(W\). By the linear map lemma (3.4) there is a unique \(E \in \mathcal{L}(W)\) with

\begin{equation*} E w_k = T v_k \quad \text{for } k = 1, \dots, m, \qquad E w_k = 0 \quad \text{for } k = m+1, \dots, n. \end{equation*}

We claim \(ES = T\). Let \(v \in V\). Since \(Sv \in \operatorname{range} S\) and \(w_1, \dots, w_m\) is a basis of \(\operatorname{range} S\), there are scalars \(a_1, \dots, a_m \in \mathbf{F}\) with

\begin{equation*} Sv = a_1 w_1 + \dots + a_m w_m. \end{equation*}

Put \(u = a_1 v_1 + \dots + a_m v_m \in V\). Then by linearity of \(S\),

\begin{equation*} Su = a_1 S v_1 + \dots + a_m S v_m = a_1 w_1 + \dots + a_m w_m = Sv, \end{equation*}

so \(S(v - u) = 0\), that is, \(v - u \in \operatorname{null} S\). By hypothesis \(v - u \in \operatorname{null} T\), so \(Tv = Tu\). Therefore

\begin{equation*} (ES)v = E(Sv) = E(a_1 w_1 + \dots + a_m w_m) = a_1 E w_1 + \dots + a_m E w_m = a_1 T v_1 + \dots + a_m T v_m = Tu = Tv. \end{equation*}

Since \(v \in V\) was arbitrary, \(ES = T\), as desired.

(If \(\operatorname{range} S = \{0\}\) the argument still works, read with empty sums: then \(S = 0\), so \(\operatorname{null} S = V \subseteq \operatorname{null} T\) forces \(T = 0\), and \(E = 0\) works.)

The easy direction. Suppose there exists \(E \in \mathcal{L}(V)\) with \(S = TE\). Let \(w \in \operatorname{range} S\), say \(w = Sv\) with \(v \in V\). Then

\begin{equation*} w = Sv = (TE)v = T(Ev), \end{equation*}

and \(Ev \in V\), so \(w \in \operatorname{range} T\). Hence \(\operatorname{range} S \subseteq \operatorname{range} T\). This direction does not use finite-dimensionality of \(V\).

The other direction. Suppose \(\operatorname{range} S \subseteq \operatorname{range} T\). Because \(V\) is finite-dimensional, it has a basis

\begin{equation*} v_1, \dots, v_n. \end{equation*}

For each \(k \in \{1, \dots, n\}\) we have \(S v_k \in \operatorname{range} S \subseteq \operatorname{range} T\), so we may choose \(u_k \in V\) with

\begin{equation*} T u_k = S v_k. \end{equation*}

(This is exactly where the containment of ranges is used: it guarantees that each \(S v_k\) is hit by \(T\).)

By the linear map lemma (3.4), there exists a (unique) linear map \(E \in \mathcal{L}(V)\) with

\begin{equation*} E v_k = u_k \quad \text{for } k = 1, \dots, n. \end{equation*}

We claim \(TE = S\). Both \(TE\) and \(S\) are linear maps from \(V\) to \(W\), and for each \(k\),

\begin{equation*} (TE) v_k = T(E v_k) = T u_k = S v_k. \end{equation*}

Thus \(TE\) and \(S\) agree on the basis \(v_1, \dots, v_n\). Now let \(v \in V\) be arbitrary and write \(v = a_1 v_1 + \dots + a_n v_n\) with \(a_1, \dots, a_n \in \mathbf{F}\). By linearity,

\begin{equation*} (TE)v = a_1 (TE)v_1 + \dots + a_n (TE)v_n = a_1 S v_1 + \dots + a_n S v_n = Sv. \end{equation*}

Hence \(S = TE\), as desired.

(Equivalently, one may invoke the uniqueness clause of 3.4: two linear maps on \(V\) that agree on a basis of \(V\) are equal.)

Recall that \(P^2\) means \(P \circ P\), so the hypothesis says \(P(Pv) = Pv\) for every \(v \in V\). By 3.13 and 3.18, \(\operatorname{null} P\) and \(\operatorname{range} P\) are subspaces of \(V\), so the sum \(\operatorname{null} P + \operatorname{range} P\) makes sense. Note that no finite-dimensionality is assumed or needed.

Step 1: \(V = \operatorname{null} P + \operatorname{range} P\). Let \(v \in V\). Write

\begin{equation*} v = (v - Pv) + Pv. \end{equation*}

Clearly \(Pv \in \operatorname{range} P\). For the first summand, linearity of \(P\) and the hypothesis give

\begin{equation*} P(v - Pv) = Pv - P(Pv) = Pv - P^2 v = Pv - Pv = 0, \end{equation*}

so \(v - Pv \in \operatorname{null} P\). Hence \(v \in \operatorname{null} P + \operatorname{range} P\). Since \(v\) was arbitrary and both summands are subspaces of \(V\), we conclude \(V = \operatorname{null} P + \operatorname{range} P\).

Step 2: \(\operatorname{null} P \cap \operatorname{range} P = \{0\}\). Suppose \(v \in \operatorname{null} P \cap \operatorname{range} P\). Because \(v \in \operatorname{range} P\), there exists \(u \in V\) with \(v = Pu\). Because \(v \in \operatorname{null} P\), we have \(Pv = 0\). Therefore

\begin{equation*} v = Pu = P^2 u = P(Pu) = Pv = 0. \end{equation*}

Hence the intersection contains only \(0\); the reverse containment \(\{0\} \subseteq \operatorname{null} P \cap \operatorname{range} P\) holds because both are subspaces.

Conclusion. By 1.46, a sum of two subspaces is a direct sum if and only if their intersection is \(\{0\}\). Steps 1 and 2 therefore give

\begin{equation*} V = \operatorname{null} P \oplus \operatorname{range} P, \end{equation*}

as desired.

Concretely, Step 1 exhibits the decomposition explicitly: the \(\operatorname{range} P\) component of \(v\) is \(Pv\) and the \(\operatorname{null} P\) component is \(v - Pv\), and by directness this is the only such decomposition. Such a \(P\) is called a projection onto \(\operatorname{range} P\) along \(\operatorname{null} P\); note also that \(P\) acts as the identity on \(\operatorname{range} P\), since \(Pu = v\) implies \(Pv = P^2 u = Pu = v\).

Throughout, \(\mathcal{P}_m(\mathbf{R})\) denotes the space of polynomials with real coefficients of degree at most \(m\); the list \(1, x, \dots, x^m\) is a basis of it, so \(\dim \mathcal{P}_m(\mathbf{R}) = m + 1\). We use the standard convention \(\deg 0 = -\infty\), and the word nonconstant means \(\deg p \ge 1\). The strategy is to show that \(D\) maps \(\mathcal{P}_m(\mathbf{R})\) onto \(\mathcal{P}_{m-1}(\mathbf{R})\) for every \(m \ge 1\); since every polynomial lies in some \(\mathcal{P}_{m-1}(\mathbf{R})\), surjectivity of \(D\) follows.

Step 1: \(D1\) is a constant polynomial. The hypothesis says nothing directly about constant polynomials, so we must extract this from linearity. The polynomial \(x\) is nonconstant with \(\deg x = 1\), so \(\deg Dx = 0\); thus \(Dx = c\) for some nonzero \(c \in \mathbf{R}\). Likewise \(x + 1\) is nonconstant with \(\deg(x+1) = 1\), so \(\deg D(x+1) = 0\); thus \(D(x+1) = c^{\prime}\) for some nonzero \(c^{\prime} \in \mathbf{R}\). By linearity \(D(x+1) = Dx + D1\), hence

\begin{equation*} D1 = D(x+1) - Dx = c^{\prime} - c, \end{equation*}

a constant polynomial. Consequently, for every \(a \in \mathbf{R}\) the polynomial \(D(a \cdot 1) = a\,D1\) is constant.

Step 2: \(D\) maps \(\mathcal{P}_m(\mathbf{R})\) into \(\mathcal{P}_{m-1}(\mathbf{R})\) for every \(m \ge 1\). Let \(p \in \mathcal{P}_m(\mathbf{R})\). If \(p\) is nonconstant, then

\begin{equation*} \deg Dp = (\deg p) - 1 \le m - 1, \end{equation*}

so \(Dp \in \mathcal{P}_{m-1}(\mathbf{R})\). If \(p\) is constant, then \(Dp\) is constant by Step 1, so \(Dp \in \mathcal{P}_0(\mathbf{R}) \subseteq \mathcal{P}_{m-1}(\mathbf{R})\) because \(m - 1 \ge 0\). Either way \(Dp \in \mathcal{P}_{m-1}(\mathbf{R})\).

Step 3: \(\operatorname{null} D \subseteq \mathcal{P}_0(\mathbf{R})\). Suppose \(p \in \mathcal{P}(\mathbf{R})\) is nonconstant. Then \(\deg Dp = (\deg p) - 1 \ge 0\), so \(Dp\) has a well-defined nonnegative degree and hence \(Dp \ne 0\) (the zero polynomial has degree \(-\infty\)). Thus no nonconstant polynomial lies in \(\operatorname{null} D\), that is, \(\operatorname{null} D \subseteq \mathcal{P}_0(\mathbf{R})\).

Step 4: the restriction of \(D\) is surjective onto \(\mathcal{P}_{m-1}(\mathbf{R})\). Fix an integer \(m \ge 1\) and let

\begin{equation*} D_m = D|_{\mathcal{P}_m(\mathbf{R})}. \end{equation*}

By Step 2 we may regard \(D_m \in \mathcal{L}\big(\mathcal{P}_m(\mathbf{R}),\ \mathcal{P}_{m-1}(\mathbf{R})\big)\); it is linear because \(D\) is and \(\mathcal{P}_m(\mathbf{R})\) is a subspace of \(\mathcal{P}(\mathbf{R})\). Its null space is

\begin{equation*} \operatorname{null} D_m = \operatorname{null} D \cap \mathcal{P}_m(\mathbf{R}) \subseteq \mathcal{P}_0(\mathbf{R}) \end{equation*}

by Step 3, so \(\dim \operatorname{null} D_m \le \dim \mathcal{P}_0(\mathbf{R}) = 1\) by 2.37. Since \(\mathcal{P}_m(\mathbf{R})\) is finite-dimensional, the fundamental theorem of linear maps (3.21) gives

\begin{equation*} \dim \operatorname{range} D_m = \dim \mathcal{P}_m(\mathbf{R}) - \dim \operatorname{null} D_m \ge (m+1) - 1 = m = \dim \mathcal{P}_{m-1}(\mathbf{R}). \end{equation*}

But \(\operatorname{range} D_m\) is a subspace of \(\mathcal{P}_{m-1}(\mathbf{R})\) (by 3.18), so \(\dim \operatorname{range} D_m \le \dim \mathcal{P}_{m-1}(\mathbf{R})\) by 2.37; hence equality holds. A subspace of full dimension equals the whole space (2.39), so

\begin{equation*} \operatorname{range} D_m = \mathcal{P}_{m-1}(\mathbf{R}). \end{equation*}

Step 5: conclusion. Let \(q \in \mathcal{P}(\mathbf{R})\) be arbitrary. Set \(m = 1\) if \(q = 0\), and \(m = (\deg q) + 1\) otherwise; in either case \(m \ge 1\) and \(q \in \mathcal{P}_{m-1}(\mathbf{R})\). By Step 4 there exists \(p \in \mathcal{P}_m(\mathbf{R})\) with

\begin{equation*} Dp = D_m p = q. \end{equation*}

Hence \(q \in \operatorname{range} D\). Since \(q\) was arbitrary, \(\operatorname{range} D = \mathcal{P}(\mathbf{R})\), that is, \(D\) is surjective (by 3.19).

Note that \(D\) need not be the differentiation map: for instance \(Dp = p^{\prime} + p^{\prime\prime\prime}\) also satisfies the degree hypothesis, and the proof above shows it too is surjective. What drives the argument is only that lowering the degree by exactly one forces the restriction \(D_m\) to have at most a one-dimensional null space while its target is one dimension smaller than its domain.

Choose \(m \in \{0, 1, 2, \dots\}\) with \(p \in \mathcal{P}_m(\mathbb{R})\); such an \(m\) exists because \(p\) is a polynomial (take \(m = \deg p\) if \(p \neq 0\), and \(m = 0\) if \(p = 0\)).

Define

\begin{equation*} T \colon \mathcal{P}_{m+1}(\mathbb{R}) \to \mathcal{P}_m(\mathbb{R}), \qquad Tq = 5q^{\prime\prime} + 3q^{\prime}. \end{equation*}

This makes sense: if \(q \in \mathcal{P}_{m+1}(\mathbb{R})\) then \(q^{\prime}\) and \(q^{\prime\prime}\) both lie in \(\mathcal{P}_m(\mathbb{R})\) (differentiation lowers degree by at least one), and \(\mathcal{P}_m(\mathbb{R})\) is closed under sums and scalar multiples. Moreover \(T\) is linear, because differentiation is a linear operation on polynomials: for \(q, r \in \mathcal{P}_{m+1}(\mathbb{R})\) and \(\lambda \in \mathbb{R}\),

\begin{equation*} T(q + r) = 5(q+r)^{\prime\prime} + 3(q+r)^{\prime} = (5q^{\prime\prime} + 3q^{\prime}) + (5r^{\prime\prime} + 3r^{\prime}) = Tq + Tr, \end{equation*}

and \(T(\lambda q) = 5\lambda q^{\prime\prime} + 3 \lambda q^{\prime} = \lambda\, Tq\).

Step 1: compute \(\operatorname{null} T\).

Suppose \(q \in \mathcal{P}_{m+1}(\mathbb{R})\) satisfies \(5q^{\prime\prime} + 3q^{\prime} = 0\). Put \(r = q^{\prime}\), a polynomial, so that

\begin{equation*} 5r^{\prime} + 3r = 0, \qquad \text{i.e.} \qquad r = -\tfrac{5}{3} r^{\prime}. \end{equation*}

If \(r \neq 0\), let \(d = \deg r \geq 0\). Then \(\deg r^{\prime} < d\) (indeed \(r^{\prime} = 0\) if \(d = 0\), and \(\deg r^{\prime} = d - 1\) if \(d \geq 1\)), so \(\deg(-\tfrac53 r^{\prime}) < d = \deg r\), contradicting \(r = -\tfrac53 r^{\prime}\). Hence \(r = 0\), i.e. \(q^{\prime} = 0\), so \(q\) is a constant polynomial.

Conversely every constant polynomial \(q\) satisfies \(5q^{\prime\prime} + 3q^{\prime} = 0\). Therefore

\begin{equation*} \operatorname{null} T = \{\text{constant polynomials}\} = \mathcal{P}_0(\mathbb{R}), \end{equation*}

which has dimension \(1\).

Step 2: conclude surjectivity.

By the fundamental theorem of linear maps (3.21),

\begin{equation*} \dim \operatorname{range} T = \dim \mathcal{P}_{m+1}(\mathbb{R}) - \dim \operatorname{null} T = (m + 2) - 1 = m + 1, \end{equation*}

using \(\dim \mathcal{P}_k(\mathbb{R}) = k + 1\) (2.36). But \(\operatorname{range} T\) is a subspace of \(\mathcal{P}_m(\mathbb{R})\) (3.18), and \(\dim \mathcal{P}_m(\mathbb{R}) = m + 1\) as well. A subspace whose dimension equals the dimension of the whole (finite-dimensional) space is the whole space (2.39), so

\begin{equation*} \operatorname{range} T = \mathcal{P}_m(\mathbb{R}). \end{equation*}

Since \(p \in \mathcal{P}_m(\mathbb{R}) = \operatorname{range} T\), there exists \(q \in \mathcal{P}_{m+1}(\mathbb{R}) \subseteq \mathcal{P}(\mathbb{R})\) with \(Tq = p\), that is, \(5q^{\prime\prime} + 3q^{\prime} = p\), as desired.

Write \(U = \{au : a \in \mathbb{F}\} = \operatorname{span}(u)\). Then \(U\) is a subspace of \(V\) (the span of a list of vectors in \(V\) is a subspace of \(V\), by 2.6), and \(\operatorname{null} \varphi\) is a subspace of \(V\) (3.13). By 1.46 it therefore suffices to prove two things: that \(\operatorname{null}\varphi + U = V\), and that \((\operatorname{null}\varphi) \cap U = \{0\}\).

The hypothesis \(u \notin \operatorname{null}\varphi\) says precisely that

\begin{equation*} \varphi(u) \neq 0 . \end{equation*}

Note that this already forces \(\varphi \neq 0\), so the hypothesis \(\varphi \neq 0\) is only there to guarantee that such a \(u\) can be found; it is not needed again below.

The sum is all of \(V\). Let \(v \in V\) and set

\begin{equation*} a = \frac{\varphi(v)}{\varphi(u)} \in \mathbb{F}, \end{equation*}

which makes sense because \(\varphi(u) \neq 0\). Then, by linearity of \(\varphi\),

\begin{equation*} \varphi(v - au) = \varphi(v) - a\,\varphi(u) = \varphi(v) - \frac{\varphi(v)}{\varphi(u)}\,\varphi(u) = 0 , \end{equation*}

so \(v - au \in \operatorname{null}\varphi\). Hence

\begin{equation*} v = (v - au) + au \in \operatorname{null}\varphi + U . \end{equation*}

As \(v \in V\) was arbitrary, \(V = \operatorname{null}\varphi + U\) (the reverse inclusion is automatic, since \(\operatorname{null}\varphi\) and \(U\) are subsets of \(V\)).

The intersection is trivial. Suppose \(w \in (\operatorname{null}\varphi) \cap U\). Since \(w \in U\), we have \(w = au\) for some \(a \in \mathbb{F}\). Since \(w \in \operatorname{null}\varphi\),

\begin{equation*} 0 = \varphi(w) = \varphi(au) = a\,\varphi(u). \end{equation*}

Because \(\varphi(u) \neq 0\) and \(\mathbb{F}\) is a field, \(a = 0\), and hence \(w = 0u = 0\). Thus \((\operatorname{null}\varphi) \cap U = \{0\}\).

By 1.46 the sum \(\operatorname{null}\varphi + U\) is a direct sum, and combined with the first part we conclude

\begin{equation*} V = \operatorname{null} \varphi \oplus \{au : a \in \mathbb{F}\}, \end{equation*}

as desired.

Note first that \(X\) is automatically finite-dimensional, being a subspace of the finite-dimensional space \(V\) (2.25), so \(\dim X\) makes sense; \(\dim Y\) makes sense by hypothesis. Write \(n = \dim V\) and \(m = \dim X\), so \(0 \le m \le n\) by 2.37.

Suppose first that such a \(T\) exists, say \(T \in \mathcal{L}(V,W)\) with \(\operatorname{null} T = X\) and \(\operatorname{range} T = Y\). Since \(V\) is finite-dimensional, the fundamental theorem of linear maps (3.21) gives

\begin{equation*} \dim V = \dim \operatorname{null} T + \dim \operatorname{range} T = \dim X + \dim Y . \end{equation*}

This is the desired equality.

Conversely, suppose \(\dim X + \dim Y = \dim V\), i.e. \(\dim Y = n - m\).

Choose a basis \(x_1, \dots, x_m\) of \(X\). This is a linearly independent list in \(V\), so it extends to a basis

\begin{equation*} x_1, \dots, x_m, v_1, \dots, v_{n-m} \end{equation*}

of \(V\) (2.32); the extension has exactly \(n - m\) extra vectors because every basis of \(V\) has length \(n\) (2.34). Choose also a basis \(y_1, \dots, y_{n-m}\) of \(Y\), which has length \(n - m\) since \(\dim Y = n - m\).

By the linear map lemma (3.4) there is a unique \(T \in \mathcal{L}(V, W)\) with

\begin{equation*} T x_k = 0 \quad (k = 1, \dots, m), \qquad T v_j = y_j \quad (j = 1, \dots, n-m), \end{equation*}

where we regard \(y_1, \dots, y_{n-m}\) as vectors in \(W\) (they lie in the subspace \(Y\) of \(W\)). If \(m = n\), so that \(Y = \{0\}\) and there are no \(v_j\)’s, this simply defines \(T = 0\); if \(m = 0\), there are no \(x_k\)’s. Both degenerate cases are covered by what follows.

Range of \(T\). Since \(x_1, \dots, x_m, v_1, \dots, v_{n-m}\) spans \(V\) and \(T\) is linear, every \(v \in V\) can be written \(v = \sum_k a_k x_k + \sum_j b_j v_j\), and then

\begin{equation*} Tv = \sum_{j=1}^{n-m} b_j y_j . \end{equation*}

Hence \(\operatorname{range} T = \operatorname{span}(y_1, \dots, y_{n-m}) = Y\): the displayed formula shows \(\operatorname{range} T \subseteq \operatorname{span}(y_1,\dots,y_{n-m})\), and each \(y_j = Tv_j\) lies in \(\operatorname{range} T\), which is a subspace (3.18), so the reverse inclusion holds too.

Null space of \(T\). Each \(x_k\) satisfies \(T x_k = 0\), so \(x_k \in \operatorname{null} T\); since \(\operatorname{null} T\) is a subspace (3.13) containing \(x_1, \dots, x_m\), it contains their span, namely \(X\). Thus \(X \subseteq \operatorname{null} T\). On the other hand, by 3.21 together with the computation of the range just made,

\begin{equation*} \dim \operatorname{null} T = \dim V - \dim \operatorname{range} T = n - (n - m) = m = \dim X . \end{equation*}

So \(X\) is a subspace of the finite-dimensional space \(\operatorname{null} T\) with \(\dim X = \dim \operatorname{null} T\), whence \(X = \operatorname{null} T\) by 2.39.

Therefore this \(T\) satisfies \(\operatorname{null} T = X\) and \(\operatorname{range} T = Y\), completing the proof of the converse and of the equivalence.

Recall the definition from Exercise 17 in Section 3A: a subspace \(\mathcal{E}\) of \(\mathcal{L}(V)\) is a two-sided ideal of \(\mathcal{L}(V)\) if \(TE \in \mathcal{E}\) and \(ET \in \mathcal{E}\) for all \(E \in \mathcal{E}\) and all \(T \in \mathcal{L}(V)\). That exercise asserts that when \(V\) is finite-dimensional the only two-sided ideals of \(\mathcal{L}(V)\) are \(\{0\}\) and \(\mathcal{L}(V)\). For completeness we prove this first.

Lemma (Exercise 3A.17). If \(V\) is finite-dimensional and \(\mathcal{E}\) is a two-sided ideal of \(\mathcal{L}(V)\) with \(\mathcal{E} \neq \{0\}\), then \(\mathcal{E} = \mathcal{L}(V)\).

Proof of the lemma. We may assume \(V \neq \{0\}\) (otherwise \(\mathcal{L}(V) = \{0\}\) and there is nothing to prove). Let \(n = \dim V \ge 1\) and fix a basis \(v_1, \dots, v_n\) of \(V\). Pick \(E \in \mathcal{E}\) with \(E \neq 0\); then there is \(u \in V\) with \(Eu \neq 0\).

For each \(k \in \{1, \dots, n\}\), the linear map lemma (3.4) provides \(A_k \in \mathcal{L}(V)\) with

\begin{equation*} A_k v_k = u, \qquad A_k v_j = 0 \ \text{ for } j \neq k . \end{equation*}

Because \(Eu \neq 0\), the one-element list \(Eu\) is linearly independent, so it extends to a basis \(Eu, w_2, \dots, w_n\) of \(V\) (2.32). Again by 3.4, for each \(k\) there is \(B_k \in \mathcal{L}(V)\) with

\begin{equation*} B_k (Eu) = v_k, \qquad B_k w_i = 0 \ \text{ for } i = 2, \dots, n . \end{equation*}

Now compute \(B_k E A_k\) on the basis: \(B_k E A_k v_k = B_k E u = v_k\), while \(B_k E A_k v_j = B_k E 0 = 0\) for \(j \neq k\). Summing over \(k\), the operator \(\sum_{k=1}^n B_k E A_k\) sends each \(v_j\) to \(v_j\), so by 3.4 (uniqueness) it equals the identity \(I\).

Since \(\mathcal{E}\) is a two-sided ideal, \(B_k E A_k \in \mathcal{E}\) for each \(k\) (multiply \(E \in \mathcal{E}\) on the right by \(A_k\), then on the left by \(B_k\)), and \(\mathcal{E}\) is a subspace, so \(I = \sum_k B_k E A_k \in \mathcal{E}\). Finally, for any \(T \in \mathcal{L}(V)\) we get \(T = TI \in \mathcal{E}\). Hence \(\mathcal{E} = \mathcal{L}(V)\), proving the lemma.

Now let \(\varphi \colon \mathcal{L}(V) \to \mathbb{F}\) be linear with \(\varphi(ST) = \varphi(S)\varphi(T)\) for all \(S, T \in \mathcal{L}(V)\).

Claim: \(\operatorname{null}\varphi\) is a two-sided ideal of \(\mathcal{L}(V)\). Indeed, \(\operatorname{null}\varphi\) is a subspace of \(\mathcal{L}(V)\) because \(\varphi\) is linear (3.13). If \(E \in \operatorname{null}\varphi\) and \(T \in \mathcal{L}(V)\), then

\begin{equation*} \varphi(TE) = \varphi(T)\varphi(E) = \varphi(T)\cdot 0 = 0, \qquad \varphi(ET) = \varphi(E)\varphi(T) = 0\cdot \varphi(T) = 0, \end{equation*}

so \(TE \in \operatorname{null}\varphi\) and \(ET \in \operatorname{null}\varphi\). This proves the claim.

By the lemma, \(\operatorname{null}\varphi = \{0\}\) or \(\operatorname{null}\varphi = \mathcal{L}(V)\).

Suppose, for contradiction, that \(\operatorname{null}\varphi = \{0\}\). Then \(\varphi\) is injective (3.15). But \(\mathcal{L}(V)\) is finite-dimensional with

\begin{equation*} \dim \mathcal{L}(V) = (\dim V)^2 \ge 2^2 = 4 > 1 = \dim \mathbb{F}, \end{equation*}

using \(\dim \mathcal{L}(V, W) = (\dim V)(\dim W)\) (3.72 applied with \(W = V\)) and \(\dim V > 1\). Since \(\dim \mathcal{L}(V) > \dim \mathbb{F}\), no linear map from \(\mathcal{L}(V)\) to \(\mathbb{F}\) is injective (3.22). This contradicts the injectivity of \(\varphi\).

(If one prefers to avoid the dimension formula 3.72, here is a direct version of the same contradiction. Fix a basis \(v_1, \dots, v_n\) of \(V\) with \(n \ge 2\), and let \(P \in \mathcal{L}(V)\) be defined by \(Pv_1 = v_1\) and \(Pv_j = 0\) for \(j \ge 2\). The operators \(I\) and \(P\) are linearly independent in \(\mathcal{L}(V)\): if \(aI + bP = 0\), applying this to \(v_2\) gives \(av_2 = 0\), so \(a = 0\), and then applying \(bP\) to \(v_1\) gives \(bv_1 = 0\), so \(b = 0\). But \(\varphi(I)\) and \(\varphi(P)\) lie in the one-dimensional space \(\mathbb{F}\), so they are linearly dependent: there exist \(a, b \in \mathbb{F}\), not both \(0\), with \(a\varphi(I) + b\varphi(P) = 0\), i.e. \(\varphi(aI + bP) = 0\). Injectivity of \(\varphi\) then forces \(aI + bP = 0\), contradicting the linear independence of \(I\) and \(P\).)

Hence \(\operatorname{null}\varphi = \mathcal{L}(V)\), which says exactly that \(\varphi(T) = 0\) for every \(T \in \mathcal{L}(V)\), i.e. \(\varphi = 0\).

Remark. The hypothesis \(\dim V > 1\) cannot be dropped: if \(\dim V = 1\) then every \(T \in \mathcal{L}(V)\) is \(T = \lambda_T I\) for a unique scalar \(\lambda_T\), and \(\varphi(T) = \lambda_T\) is a nonzero linear map satisfying \(\varphi(ST) = \varphi(S)\varphi(T)\).

Throughout we use the definitions from Exercise 8 in Section 1B. As a set, \(V_{\mathbb{C}} = V \times V\), and the ordered pair \((u,v)\) is written \(u + iv\). Thus two elements \(u_1 + iv_1\) and \(u_2 + iv_2\) of \(V_{\mathbb{C}}\) are equal if and only if \(u_1 = u_2\) and \(v_1 = v_2\); in particular \(u + iv = 0\) if and only if \(u = 0\) and \(v = 0\). Addition and complex scalar multiplication are given by

\begin{equation*} (u_1 + iv_1) + (u_2 + iv_2) = (u_1 + u_2) + i(v_1 + v_2), \end{equation*}

\begin{equation*} (a + bi)(u + iv) = (au - bv) + i(av + bu) \qquad (a, b \in \mathbb{R}). \end{equation*}

The same formulas define the complex vector space \(W_{\mathbb{C}}\). Note also that \(T_{\mathbb{C}}\) is a genuinely well-defined function: every element of \(V_{\mathbb{C}}\) has exactly one expression as \(u + iv\) with \(u, v \in V\), so the formula \(T_{\mathbb{C}}(u+iv) = Tu + iTv\) assigns exactly one value to each input.

(a) We check additivity and complex homogeneity.

Additivity: for \(u_1, v_1, u_2, v_2 \in V\),

\begin{equation*} \begin{aligned} T_{\mathbb{C}}\big((u_1 + iv_1) + (u_2 + iv_2)\big) &= T_{\mathbb{C}}\big((u_1 + u_2) + i(v_1 + v_2)\big) \\ &= T(u_1 + u_2) + i\,T(v_1 + v_2) \\ &= (Tu_1 + Tu_2) + i(Tv_1 + Tv_2) \\ &= (Tu_1 + iTv_1) + (Tu_2 + iTv_2) \\ &= T_{\mathbb{C}}(u_1 + iv_1) + T_{\mathbb{C}}(u_2 + iv_2), \end{aligned} \end{equation*}

where the third equality uses the additivity of \(T\) and the fourth uses the definition of addition in \(W_{\mathbb{C}}\).

Homogeneity over \(\mathbb{C}\): let \(a, b \in \mathbb{R}\) and \(u, v \in V\). Then

\begin{equation*} \begin{aligned} T_{\mathbb{C}}\big((a+bi)(u+iv)\big) &= T_{\mathbb{C}}\big((au - bv) + i(av + bu)\big) \\ &= T(au - bv) + i\,T(av + bu) \\ &= (a\,Tu - b\,Tv) + i(a\,Tv + b\,Tu) \\ &= (a + bi)(Tu + iTv) \\ &= (a+bi)\,T_{\mathbb{C}}(u + iv), \end{aligned} \end{equation*}

where the third equality uses that \(T\) is linear over \(\mathbb{R}\) and the fourth is the definition of complex scalar multiplication in \(W_{\mathbb{C}}\). Hence \(T_{\mathbb{C}} \in \mathcal{L}(V_{\mathbb{C}}, W_{\mathbb{C}})\) as a complex linear map.

(b) By (a) and 3.15, \(T_{\mathbb{C}}\) is injective if and only if \(\operatorname{null} T_{\mathbb{C}} = \{0\}\), and \(T\) is injective if and only if \(\operatorname{null} T = \{0\}\). We first identify the null space of \(T_{\mathbb{C}}\):

\begin{equation*} u + iv \in \operatorname{null} T_{\mathbb{C}} \iff Tu + iTv = 0 \iff Tu = 0 \text{ and } Tv = 0 \iff u, v \in \operatorname{null} T, \end{equation*}

the middle equivalence because an element \(w_1 + iw_2\) of \(W_{\mathbb{C}}\) is \(0\) exactly when \(w_1 = 0\) and \(w_2 = 0\). In other words \(\operatorname{null} T_{\mathbb{C}} = (\operatorname{null} T)_{\mathbb{C}}\).

Now suppose \(T\) is injective, so \(\operatorname{null} T = \{0\}\). If \(u + iv \in \operatorname{null} T_{\mathbb{C}}\), then by the display \(u \in \operatorname{null} T\) and \(v \in \operatorname{null} T\), so \(u = v = 0\) and hence \(u + iv = 0\). Thus \(\operatorname{null} T_{\mathbb{C}} = \{0\}\) and \(T_{\mathbb{C}}\) is injective.

Conversely suppose \(T_{\mathbb{C}}\) is injective, and let \(u \in \operatorname{null} T\). Then \(T_{\mathbb{C}}(u + i0) = Tu + i\,T0 = 0 + i0 = 0\), so \(u + i0 = 0\), which gives \(u = 0\). Thus \(\operatorname{null} T = \{0\}\) and \(T\) is injective.

(c) Suppose first that \(\operatorname{range} T = W\). Let \(w_1 + iw_2 \in W_{\mathbb{C}}\) be arbitrary, with \(w_1, w_2 \in W\). Choose \(u, v \in V\) with \(Tu = w_1\) and \(Tv = w_2\). Then

\begin{equation*} T_{\mathbb{C}}(u + iv) = Tu + iTv = w_1 + iw_2 . \end{equation*}

Hence every element of \(W_{\mathbb{C}}\) is in \(\operatorname{range} T_{\mathbb{C}}\), so \(\operatorname{range} T_{\mathbb{C}} = W_{\mathbb{C}}\).

Conversely suppose \(\operatorname{range} T_{\mathbb{C}} = W_{\mathbb{C}}\). Let \(w \in W\). Then \(w + i0 \in W_{\mathbb{C}}\), so there exist \(u, v \in V\) with

\begin{equation*} Tu + iTv = T_{\mathbb{C}}(u + iv) = w + i0 . \end{equation*}

Comparing the two coordinates gives \(Tu = w\) (and \(Tv = 0\)). Hence \(w \in \operatorname{range} T\). As \(w \in W\) was arbitrary, \(\operatorname{range} T = W\).

(The two computations in (c) together show more precisely that \(\operatorname{range} T_{\mathbb{C}} = (\operatorname{range} T)_{\mathbb{C}}\), viewing \((\operatorname{range} T)_{\mathbb{C}}\) as the set of \(w_1 + iw_2 \in W_{\mathbb{C}}\) with \(w_1, w_2 \in \operatorname{range} T\).)

Fix a basis \(v_1, \dots, v_n\) of \(V\) and a basis \(w_1, \dots, w_m\) of \(W\), and let \(A = \mathcal{M}(T)\) be the matrix of \(T\) with respect to these bases. By the definition of the matrix of a linear map,

\begin{equation*} T v_k = \sum_{j=1}^{m} A_{j,k}\, w_j \qquad \text{for each } k \in \{1, \dots, n\}. \end{equation*}

Let

\begin{equation*} K = \{\, k \in \{1, \dots, n\} : \text{column } k \text{ of } A \text{ has at least one nonzero entry} \,\}. \end{equation*}

First we bound \(\dim \operatorname{range} T\) by the size of \(K\). Because \(v_1, \dots, v_n\) spans \(V\) and \(T\) is linear, \(\operatorname{range} T = \operatorname{span}(T v_1, \dots, T v_n)\). If \(k \notin K\), then every entry \(A_{j,k}\) is \(0\), and the displayed formula gives \(T v_k = 0\); such a vector contributes nothing to a span. Hence

\begin{equation*} \operatorname{range} T = \operatorname{span}(T v_k : k \in K), \end{equation*}

a span of a list of length \(|K|\). The dimension of the span of a list is at most the length of that list (a spanning list of \(\operatorname{span}(Tv_k : k \in K)\) can be reduced to a basis of it, by 2.30, and that basis is no longer than the original list), so

\begin{equation*} \dim \operatorname{range} T \le |K|. \end{equation*}

Now we bound \(|K|\) by the number of nonzero entries of \(A\). For each \(k \in K\) choose one entry of column \(k\) that is nonzero; distinct indices \(k\) give entries in distinct columns of \(A\), hence distinct entries of \(A\). This exhibits \(|K|\) distinct nonzero entries of \(A\), so

\begin{equation*} |K| \le \#\{\text{nonzero entries of } A\}. \end{equation*}

Combining the two inequalities,

\begin{equation*} \dim \operatorname{range} T \le |K| \le \#\{\text{nonzero entries of } \mathcal{M}(T)\}, \end{equation*}

which is exactly the assertion: the matrix of \(T\) with respect to any choice of bases has at least \(\dim \operatorname{range} T\) nonzero entries.

(Note that Exercise 3C.5 shows that this bound is attained: there always exist bases for which \(\mathcal{M}(T)\) has exactly \(\dim \operatorname{range} T\) nonzero entries, all of them equal to \(1\) and sitting on the diagonal.)

Write \(n = \dim V\) and \(m = \dim W\); because \(V\) and \(W\) are nonzero we have \(n \ge 1\) and \(m \ge 1\).

First suppose there are bases \(v_1, \dots, v_n\) of \(V\) and \(w_1, \dots, w_m\) of \(W\) with respect to which every entry of \(\mathcal{M}(T)\) equals \(1\). Put \(w = w_1 + \cdots + w_m\). By the definition of the matrix of a linear map, for each \(k \in \{1, \dots, n\}\) we have

\begin{equation*} T v_k = \sum_{j=1}^{m} 1 \cdot w_j = w. \end{equation*}

The vector \(w\) is nonzero, because \(w_1, \dots, w_m\) is linearly independent and the coefficients \(1, \dots, 1\) are not all \(0\). Since \(v_1, \dots, v_n\) spans \(V\),

\begin{equation*} \operatorname{range} T = \operatorname{span}(T v_1, \dots, T v_n) = \operatorname{span}(w), \end{equation*}

and \(\operatorname{span}(w)\) has dimension \(1\) because \(w \neq 0\). Thus \(\dim \operatorname{range} T = 1\).

Conversely, suppose \(\dim \operatorname{range} T = 1\). Choose \(w \in W\) with \(w \neq 0\) and \(\operatorname{range} T = \operatorname{span}(w)\).

We first build the basis of \(W\). The list \(w\) is linearly independent in \(W\) (as \(w \ne 0\)), so it extends to a basis \(u_1, \dots, u_m\) of \(W\) with \(u_1 = w\) (by 2.32). Define

\begin{equation*} w_1 = u_1 - u_2 - \cdots - u_m, \qquad w_j = u_j \ \text{ for } j \in \{2, \dots, m\}. \end{equation*}

(When \(m = 1\) this simply says \(w_1 = u_1\).) Then \(\operatorname{span}(w_1, \dots, w_m)\) contains \(u_2, \dots, u_m\) and also contains \(w_1 + u_2 + \cdots + u_m = u_1\); hence it contains \(\operatorname{span}(u_1, \dots, u_m) = W\). So \(w_1, \dots, w_m\) is a spanning list of \(W\) of length \(m = \dim W\), and therefore a basis of \(W\) (by 2.42, a spanning list of the right length is a basis). By construction

\begin{equation*} w_1 + w_2 + \cdots + w_m = u_1 = w. \end{equation*}

Now we build the basis of \(V\). Since \(w \in \operatorname{range} T\), there is \(v_1 \in V\) with \(T v_1 = w\). By the fundamental theorem of linear maps (3.21),

\begin{equation*} \dim \operatorname{null} T = \dim V - \dim \operatorname{range} T = n - 1. \end{equation*}

Let \(z_1, \dots, z_{n-1}\) be a basis of \(\operatorname{null} T\) (an empty list if \(n = 1\)); the letters \(z\) are used to avoid any clash with the vectors \(u_1, \dots, u_m\) of \(W\) above. The list \(v_1, z_1, \dots, z_{n-1}\) is linearly independent: if \(a v_1 + c_1 z_1 + \cdots + c_{n-1} z_{n-1} = 0\), applying \(T\) gives \(a w = 0\), hence \(a = 0\) since \(w \neq 0\), and then \(c_1 = \cdots = c_{n-1} = 0\) by the linear independence of \(z_1, \dots, z_{n-1}\). Having length \(n = \dim V\), this list is a basis of \(V\) (by 2.38).

Define

\begin{equation*} v_k = v_1 + z_{k-1} \qquad \text{for } k \in \{2, \dots, n\}. \end{equation*}

Then \(\operatorname{span}(v_1, \dots, v_n)\) contains \(v_1\) and contains \(v_k - v_1 = z_{k-1}\) for each \(k \ge 2\), hence contains \(\operatorname{span}(v_1, z_1, \dots, z_{n-1}) = V\). So \(v_1, \dots, v_n\) spans \(V\) and has length \(n = \dim V\), and is therefore a basis of \(V\) (by 2.42).

Finally, \(T v_1 = w\), and for \(k \in \{2, \dots, n\}\) we have \(T v_k = T v_1 + T z_{k-1} = w + 0 = w\). Thus for every \(k\),

\begin{equation*} T v_k = w = w_1 + w_2 + \cdots + w_m = \sum_{j=1}^{m} 1 \cdot w_j, \end{equation*}

so every entry of column \(k\) of \(\mathcal{M}(T)\) equals \(1\). As this holds for each \(k\), all entries of \(\mathcal{M}(T)\) with respect to the bases \(v_1, \dots, v_n\) and \(w_1, \dots, w_m\) equal \(1\).

Throughout, all matrices are taken with respect to the fixed bases \(v_1, \dots, v_n\) of \(V\) and \(w_1, \dots, w_m\) of \(W\). Recall the defining property of the matrix of a linear map \(R \in \mathcal{L}(V, W)\): \(\mathcal{M}( R)\) is the \(m\)-by-\(n\) matrix whose entries \(\mathcal{M}( R)_{j,k}\) are the unique scalars satisfying

\begin{equation*} R v_k = \sum_{j=1}^{m} \mathcal{M}( R)_{j,k}\, w_j \qquad \text{for each } k \in \{1, \dots, n\}. \end{equation*}

Uniqueness of these scalars is exactly the statement that \(w_1, \dots, w_m\), being a basis of \(W\), represents each vector of \(W\) as a linear combination of \(w_1, \dots, w_m\) in only one way (the criterion for basis, 2.28). This uniqueness is what makes the arguments below work: to identify the entries of a matrix \(\mathcal{M}( R)\), it suffices to exhibit any one expression of \(R v_k\) as a linear combination of \(w_1, \dots, w_m\) and read off its coefficients.

(a) Let \(S, T \in \mathcal{L}(V, W)\), and set \(A = \mathcal{M}(S)\) and \(C = \mathcal{M}(T)\); both are \(m\)-by-\(n\) matrices. Fix \(k \in \{1, \dots, n\}\). Using the definition of the sum \(S + T\) of two linear maps and then the two displayed formulas above,

\begin{equation*} \begin{aligned} (S + T) v_k &= S v_k + T v_k \\ &= \sum_{j=1}^{m} A_{j,k}\, w_j + \sum_{j=1}^{m} C_{j,k}\, w_j \\ &= \sum_{j=1}^{m} (A_{j,k} + C_{j,k})\, w_j . \end{aligned} \end{equation*}

Because \(S + T \in \mathcal{L}(V, W)\) (by 3.5 and 3.6) and the representation of \((S + T) v_k\) in the basis \(w_1, \dots, w_m\) is unique, we conclude that

\begin{equation*} \mathcal{M}(S + T)_{j,k} = A_{j,k} + C_{j,k} \qquad \text{for all } j \in \{1, \dots, m\},\ k \in \{1, \dots, n\}. \end{equation*}

By the definition of matrix addition (3.34), the right side is the entry in row \(j\), column \(k\), of \(A + C\). Since \(\mathcal{M}(S+T)\) and \(A + C\) are both \(m\)-by-\(n\) and agree in every entry, \(\mathcal{M}(S + T) = \mathcal{M}(S) + \mathcal{M}(T)\). This proves 3.35.

(b) Let \(\lambda \in \mathbf{F}\) and \(T \in \mathcal{L}(V, W)\), and set \(C = \mathcal{M}(T)\). Fix \(k \in \{1, \dots, n\}\). Using the definition of the scalar multiple \(\lambda T\) and then the defining formula for \(C\),

\begin{equation*} (\lambda T) v_k = \lambda (T v_k) = \lambda \sum_{j=1}^{m} C_{j,k}\, w_j = \sum_{j=1}^{m} (\lambda C_{j,k})\, w_j , \end{equation*}

where the last equality uses distributivity and associativity of scalar multiplication in \(W\). Again \(\lambda T \in \mathcal{L}(V, W)\) (by 3.5 and 3.6), and uniqueness of the representation in the basis \(w_1, \dots, w_m\) gives

\begin{equation*} \mathcal{M}(\lambda T)_{j,k} = \lambda C_{j,k} \qquad \text{for all } j, k . \end{equation*}

By the definition of scalar multiplication of a matrix (3.36), the right side is the entry in row \(j\), column \(k\), of \(\lambda C\). Hence \(\mathcal{M}(\lambda T) = \lambda \mathcal{M}(T)\). This proves 3.38.

Take the basis of \(\mathcal{P}_3(\mathbf{R})\) to be

\begin{equation*} v_1 = x, \qquad v_2 = \tfrac{x^2}{2}, \qquad v_3 = \tfrac{x^3}{3}, \qquad v_4 = 1, \end{equation*}

and the basis of \(\mathcal{P}_2(\mathbf{R})\) to be the standard one,

\begin{equation*} w_1 = 1, \qquad w_2 = x, \qquad w_3 = x^2 . \end{equation*}

First we check that these lists really are bases.

The list \(w_1, w_2, w_3\) is the standard basis \(1, x, x^2\) of \(\mathcal{P}_2(\mathbf{R})\), so it is a basis; in particular \(\dim \mathcal{P}_2(\mathbf{R}) = 3\).

For \(v_1, v_2, v_3, v_4\): the span of this list contains \(1\), \(x\), \(\tfrac{x^2}{2}\), \(\tfrac{x^3}{3}\), hence (being closed under scalar multiplication) contains \(1, x, x^2, x^3\), and therefore contains \(\operatorname{span}(1, x, x^2, x^3) = \mathcal{P}_3(\mathbf{R})\). So \(v_1, v_2, v_3, v_4\) spans \(\mathcal{P}_3(\mathbf{R})\), and its length \(4\) equals \(\dim \mathcal{P}_3(\mathbf{R}) = 4\). By 2.42, a spanning list of the right length is a basis, so \(v_1, v_2, v_3, v_4\) is a basis of \(\mathcal{P}_3(\mathbf{R})\).

Now compute the matrix. Differentiating,

\begin{equation*} \begin{aligned} D v_1 &= (x)^{\prime} = 1 &&= 1 \cdot w_1 + 0 \cdot w_2 + 0 \cdot w_3, \\ D v_2 &= \left(\tfrac{x^2}{2}\right)^{\prime} = x &&= 0 \cdot w_1 + 1 \cdot w_2 + 0 \cdot w_3, \\ D v_3 &= \left(\tfrac{x^3}{3}\right)^{\prime} = x^2 &&= 0 \cdot w_1 + 0 \cdot w_2 + 1 \cdot w_3, \\ D v_4 &= (1)^{\prime} = 0 &&= 0 \cdot w_1 + 0 \cdot w_2 + 0 \cdot w_3 . \end{aligned} \end{equation*}

The \(k\)-th column of \(\mathcal{M}(D)\) consists of the coefficients needed to write \(D v_k\) as a linear combination of \(w_1, w_2, w_3\). Reading off the four lines above,

\begin{equation*} \mathcal{M}(D) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}, \end{equation*}

as required.

Two remarks. In Example 3.33 the same map \(D\), computed with respect to the standard bases \(1, x, x^2, x^3\) and \(1, x, x^2\), has matrix

\begin{equation*} \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}, \end{equation*}

whose only nonzero entries are the \(1, 2, 3\) just above the diagonal. Two changes produce the matrix asked for here: replacing the domain basis vectors \(x, x^2, x^3\) by \(x, \tfrac{x^2}{2}, \tfrac{x^3}{3}\) divides out the factors \(1, 2, 3\), and moving the constant polynomial \(1\) (which spans \(\operatorname{null} D\)) from the first to the last position shifts the zero column from the front to the back, which is what turns the superdiagonal into the diagonal. This is exactly the recipe of Exercise 3C.5 applied to \(D\): here \(\dim \operatorname{range} D = 3\), since \(D\) is surjective onto \(\mathcal{P}_2(\mathbf{R})\), and indeed the matrix above has \(1\) in row \(k\), column \(k\), for \(1 \le k \le 3\) and \(0\) elsewhere.

Write \(n = \dim V\), \(m = \dim W\), and \(r = \dim \operatorname{range} T\). Note \(r \le m\), since \(\operatorname{range} T\) is a subspace of \(W\) (by 2.37), and \(r \le n\) by the fundamental theorem of linear maps.

Step 1: a convenient basis of \(V\).

Since \(\operatorname{null} T\) is a subspace of the finite-dimensional space \(V\), it is finite-dimensional and has a basis \(u_1, \dots, u_d\), where \(d = \dim \operatorname{null} T\). By the fundamental theorem of linear maps (3.21),

\begin{equation*} n = \dim V = \dim \operatorname{null} T + \dim \operatorname{range} T = d + r, \qquad \text{so } d = n - r. \end{equation*}

A basis of \(\operatorname{null} T\) is a linearly independent list in \(V\), so it extends to a basis of \(V\) (by 2.32); write that extension as

\begin{equation*} u_1, \dots, u_{n-r}, \; v_1, \dots, v_r , \end{equation*}

which necessarily has \(r\) new vectors because every basis of \(V\) has length \(n\).

Step 2: the images \(Tv_1, \dots, Tv_r\) form a basis of \(\operatorname{range} T\).

They span \(\operatorname{range} T\): for \(v \in V\) write \(v = a_1 u_1 + \cdots + a_{n-r} u_{n-r} + b_1 v_1 + \cdots + b_r v_r\); applying \(T\) and using \(Tu_i = 0\) gives \(Tv = b_1 Tv_1 + \cdots + b_r Tv_r\). They are linearly independent: if \(c_1 Tv_1 + \cdots + c_r Tv_r = 0\), then \(T(c_1 v_1 + \cdots + c_r v_r) = 0\), so \(c_1 v_1 + \cdots + c_r v_r \in \operatorname{null} T\) and hence

\begin{equation*} c_1 v_1 + \cdots + c_r v_r = a_1 u_1 + \cdots + a_{n-r} u_{n-r} \end{equation*}

for some scalars \(a_i\); linear independence of the basis \(u_1, \dots, u_{n-r}, v_1, \dots, v_r\) of \(V\) then forces all \(c_k = 0\). (This is precisely the argument inside the proof of 3.21.)

Step 3: the bases.

Define

\begin{equation*} w_k = T v_k \qquad \text{for } k \in \{1, \dots, r\}. \end{equation*}

By Step 2 the list \(w_1, \dots, w_r\) is linearly independent in \(W\), so by 2.32 it extends to a basis

\begin{equation*} w_1, \dots, w_r, w_{r+1}, \dots, w_m \end{equation*}

of \(W\) (again the length must be \(m = \dim W\)).

For the domain, reorder the basis found in Step 1 by putting the \(v\)’s first: set

\begin{equation*} \tilde v_k = v_k \ \text{ for } 1 \le k \le r, \qquad \tilde v_{r+i} = u_i \ \text{ for } 1 \le i \le n - r . \end{equation*}

Reordering a basis yields a basis (spanning and linear independence do not depend on the order of the list), so \(\tilde v_1, \dots, \tilde v_n\) is a basis of \(V\).

Step 4: computing \(\mathcal{M}(T)\).

Let \(A = \mathcal{M}(T)\) with respect to \(\tilde v_1, \dots, \tilde v_n\) and \(w_1, \dots, w_m\). For \(1 \le k \le r\),

\begin{equation*} T \tilde v_k = T v_k = w_k = 0 \cdot w_1 + \cdots + 0 \cdot w_{k-1} + 1 \cdot w_k + 0 \cdot w_{k+1} + \cdots + 0 \cdot w_m , \end{equation*}

so column \(k\) of \(A\) has a \(1\) in row \(k\) and \(0\) everywhere else. For \(r < k \le n\) we have \(\tilde v_k = u_{k-r} \in \operatorname{null} T\), so

\begin{equation*} T \tilde v_k = 0 = 0 \cdot w_1 + \cdots + 0 \cdot w_m , \end{equation*}

and column \(k\) of \(A\) is entirely \(0\).

Therefore \(A_{k,k} = 1\) for \(1 \le k \le r = \dim \operatorname{range} T\), and all other entries of \(A\) are \(0\), which is what was to be proved.

(In particular \(\mathcal{M}(T)\) then has exactly \(\dim \operatorname{range} T\) nonzero entries, showing that the lower bound of Exercise 3C.1 is sharp.)

Let \(n = \dim W\). Recall that if \(w_1, \dots, w_n\) is a basis of \(W\) and \(A = \mathcal{M}(T)\) with respect to \(v_1, \dots, v_m\) and \(w_1, \dots, w_n\), then the first column of \(A\) is determined by

\begin{equation*} T v_1 = A_{1,1} w_1 + A_{2,1} w_2 + \cdots + A_{n,1} w_n . \end{equation*}

So what must be arranged is precisely this: either \(T v_1 = 0\) (first column all \(0\)), or \(T v_1 = w_1\) for the first vector \(w_1\) of the chosen basis (first column equal to \(1, 0, \dots, 0\)).

Case 1: \(T v_1 = 0\).

Since \(W\) is finite-dimensional it has a basis (by 2.31); let \(w_1, \dots, w_n\) be any basis of \(W\). Then

\begin{equation*} T v_1 = 0 = 0 \cdot w_1 + 0 \cdot w_2 + \cdots + 0 \cdot w_n , \end{equation*}

so by the uniqueness of the representation of a vector in a basis (2.28), \(A_{j,1} = 0\) for every \(j \in \{1, \dots, n\}\). Every entry in the first column is \(0\), which is allowed by the statement, since the \(1\) in the upper-left corner is only possibly there.

Case 2: \(T v_1 \neq 0\).

Set \(w_1 = T v_1\). The one-element list \(w_1\) is linearly independent in \(W\), because \(w_1 \neq 0\). Since \(W\) is finite-dimensional, this list extends to a basis

\begin{equation*} w_1, w_2, \dots, w_n \end{equation*}

of \(W\) (by 2.32); the extension has length \(n = \dim W\) because all bases of \(W\) have that length. Now

\begin{equation*} T v_1 = w_1 = 1 \cdot w_1 + 0 \cdot w_2 + \cdots + 0 \cdot w_n , \end{equation*}

so again by uniqueness of coefficients (2.28) we get \(A_{1,1} = 1\) and \(A_{j,1} = 0\) for \(j \in \{2, \dots, n\}\). Thus the first column of \(\mathcal{M}(T)\) is \(0\) except for a \(1\) in row \(1\).

In both cases a basis \(w_1, \dots, w_n\) of \(W\) has been produced with the required property, so the result holds. (The case distinction is genuinely necessary: if \(T v_1 = 0\), then no choice of basis of \(W\) can put a \(1\) in the upper-left corner, since \(T v_1 = 0\) forces every entry of the first column to vanish. Conversely if \(T v_1 \ne 0\) then the first column cannot be all \(0\), no matter which basis of \(W\) is used.)

Note that nothing was assumed about \(v_2, \dots, v_m\) or about the remaining columns of \(\mathcal{M}(T)\); only the first basis vector of \(V\) enters the argument, and the freedom used is entirely the freedom to choose the basis of \(W\).

Let \(m = \dim V\). If \(m = 0\) the matrix \(\mathcal{M}(T)\) has no columns and the assertion is vacuous, so assume \(m \ge 1\).

The key observation is that the first row of \(\mathcal{M}(T)\) records, for each domain basis vector, the coefficient of \(w_1\) in its image. Make this precise as follows. Since \(w_1, \dots, w_n\) is a basis of \(W\), every \(w \in W\) can be written uniquely as \(w = a_1 w_1 + \cdots + a_n w_n\) with \(a_1, \dots, a_n \in \mathbf{F}\) (by 2.28). Define \(\pi \colon W \to \mathbf{F}\) by

\begin{equation*} \pi(a_1 w_1 + \cdots + a_n w_n) = a_1 . \end{equation*}

Uniqueness of the coefficients makes \(\pi\) a well-defined function, and it is linear: if \(w = a_1 w_1 + \cdots + a_n w_n\) and \(w^{\prime} = a^{\prime}_1 w_1 + \cdots + a^{\prime}_n w_n\) and \(\lambda \in \mathbf{F}\), then \(w + w^{\prime} = (a_1 + a^{\prime}_1) w_1 + \cdots + (a_n + a^{\prime}_n) w_n\) and \(\lambda w = (\lambda a_1) w_1 + \cdots + (\lambda a_n) w_n\), whence \(\pi(w + w^{\prime}) = a_1 + a^{\prime}_1 = \pi(w) + \pi(w^{\prime})\) and \(\pi(\lambda w) = \lambda a_1 = \lambda \pi(w)\).

Now set

\begin{equation*} \varphi = \pi \circ T \in \mathcal{L}(V, \mathbf{F}), \end{equation*}

which is linear as a composition (indeed a product) of linear maps (3.7). By construction, for any basis \(v_1, \dots, v_m\) of \(V\), if \(A = \mathcal{M}(T)\) with respect to \(v_1, \dots, v_m\) and \(w_1, \dots, w_n\), then \(T v_k = \sum_{j=1}^{n} A_{j,k} w_j\) gives

\begin{equation*} \varphi(v_k) = \pi(T v_k) = A_{1,k} . \end{equation*}

So the first row of \(\mathcal{M}(T)\) is exactly the list \(\varphi(v_1), \dots, \varphi(v_m)\), and the task is to find a basis \(v_1, \dots, v_m\) of \(V\) with \(\varphi(v_1) \in \{0, 1\}\) and \(\varphi(v_2) = \cdots = \varphi(v_m) = 0\).

Case 1: \(\varphi = 0\).

Take \(v_1, \dots, v_m\) to be any basis of \(V\) (one exists by 2.31). Then \(A_{1,k} = \varphi(v_k) = 0\) for every \(k\), so the entire first row of \(\mathcal{M}(T)\) is \(0\), which is permitted.

Case 2: \(\varphi \neq 0\).

Then \(\operatorname{range} \varphi\) is a nonzero subspace of \(\mathbf{F}\), and \(\dim \mathbf{F} = 1\), so \(\operatorname{range} \varphi = \mathbf{F}\) and \(\dim \operatorname{range} \varphi = 1\) (by 2.37 and 2.39). By the fundamental theorem of linear maps (3.21),

\begin{equation*} \dim \operatorname{null} \varphi = \dim V - \dim \operatorname{range} \varphi = m - 1 . \end{equation*}

Because \(1 \in \operatorname{range} \varphi\), choose \(v_1 \in V\) with \(\varphi(v_1) = 1\). Let \(v_2, \dots, v_m\) be a basis of \(\operatorname{null} \varphi\), which is a list of the correct length \(m - 1\) (an empty list if \(m = 1\)).

The list \(v_1, v_2, \dots, v_m\) is linearly independent: suppose

\begin{equation*} c_1 v_1 + c_2 v_2 + \cdots + c_m v_m = 0 . \end{equation*}

Applying \(\varphi\) and using \(\varphi(v_k) = 0\) for \(k \ge 2\) gives \(c_1 \varphi(v_1) = 0\), that is, \(c_1 = 0\). The equation then reads \(c_2 v_2 + \cdots + c_m v_m = 0\), and linear independence of the basis \(v_2, \dots, v_m\) of \(\operatorname{null} \varphi\) forces \(c_2 = \cdots = c_m = 0\). Since this linearly independent list has length \(m = \dim V\), it is a basis of \(V\) (by 2.38).

With respect to this basis \(v_1, \dots, v_m\) of \(V\) and the given basis \(w_1, \dots, w_n\) of \(W\), the first row of \(\mathcal{M}(T)\) is

\begin{equation*} A_{1,1} = \varphi(v_1) = 1, \qquad A_{1,k} = \varphi(v_k) = 0 \ \text{ for } k \in \{2, \dots, m\}, \end{equation*}

so all entries in the first row are \(0\) except for a \(1\) in row \(1\), column \(1\).

In both cases the required basis of \(V\) exists, completing the proof. (As in Exercise 3C.6, the two cases are forced on us: \(\varphi = 0\) says that \(\operatorname{range} T\) lies inside \(\operatorname{span}(w_2, \dots, w_n)\), and then no basis of \(V\) can produce a nonzero entry in the first row; otherwise the first row cannot be identically \(0\).)

Fix \(j\) with \(1 \le j \le m\).

First we check that both sides of the claimed equation are matrices of the same size. The matrix \(AB\) is \(m\)-by-\(p\) (by 3.41), so its row \(j\), namely \((AB)_{j,\cdot}\), is a \(1\)-by-\(p\) matrix (by 3.44). On the other side, \(A_{j,\cdot}\) is a \(1\)-by-\(n\) matrix (by 3.44), and \(B\) is \(n\)-by-\(p\); hence the product \(A_{j,\cdot}B\) is defined and is a \(1\)-by-\(p\) matrix (by 3.41). Thus both sides are \(1\)-by-\(p\) matrices, and to prove they are equal it suffices to show that they have the same entry in column \(k\) for each \(k \in \{1,\dots,p\}\).

So fix \(k\) with \(1 \le k \le p\). The entry in row \(1\), column \(k\), of \((AB)_{j,\cdot}\) is by definition the entry in row \(j\), column \(k\), of \(AB\), which by the definition of matrix multiplication (3.41) equals

\begin{equation*} (AB)_{j,k} = \sum_{r=1}^{n} A_{j,r} B_{r,k}. \end{equation*}

Now consider the right side. The \(1\)-by-\(n\) matrix \(A_{j,\cdot}\) has entries \((A_{j,\cdot})_{1,r} = A_{j,r}\) for \(r \in \{1,\dots,n\}\). Hence, again by 3.41, the entry in row \(1\), column \(k\), of the product \(A_{j,\cdot}B\) equals

\begin{equation*} \bigl(A_{j,\cdot}B\bigr)_{1,k} = \sum_{r=1}^{n} \bigl(A_{j,\cdot}\bigr)_{1,r} B_{r,k} = \sum_{r=1}^{n} A_{j,r} B_{r,k}. \end{equation*}

The two displayed sums are identical. Thus \((AB)_{j,\cdot}\) and \(A_{j,\cdot}B\) are \(1\)-by-\(p\) matrices with the same entry in column \(k\) for every \(k \in \{1,\dots,p\}\), and therefore

\begin{equation*} (AB)_{j,\cdot} = A_{j,\cdot}\,B, \end{equation*}

as desired. In words: row \(j\) of \(AB\) equals (row \(j\) of \(A\)) times \(B\).

This is exactly the row analogue of 3.48, and the proof above is the mirror image of the proof of 3.48: there one fixes a column index \(k\) and varies the row index, here one fixes a row index \(j\) and varies the column index.

First we check sizes. Since \(a\) is \(1\)-by-\(n\) and \(B\) is \(n\)-by-\(p\), the product \(aB\) is defined and is a \(1\)-by-\(p\) matrix (by 3.41). For each \(r \in \{1,\dots,n\}\), the row \(B_{r,\cdot}\) is a \(1\)-by-\(p\) matrix (by 3.44), so each scalar multiple \(a_r B_{r,\cdot}\) is \(1\)-by-\(p\) (by 3.36) and the sum \(a_1 B_{1,\cdot} + \cdots + a_n B_{n,\cdot}\) is a \(1\)-by-\(p\) matrix (by 3.34). Thus both sides of the claimed equation are \(1\)-by-\(p\) matrices, and it suffices to prove that for each \(k \in \{1,\dots,p\}\) they have the same entry in column \(k\).

Fix \(k\) with \(1 \le k \le p\). Writing \(a_{1,r} = a_r\) for the entries of \(a\), the definition of matrix multiplication (3.41) gives

\begin{equation*} (aB)_{1,k} = \sum_{r=1}^{n} a_{1,r} B_{r,k} = \sum_{r=1}^{n} a_r B_{r,k}. \end{equation*}

Now compute the entry in column \(k\) of the right side. For each \(r\), the entry in row \(1\), column \(k\), of \(B_{r,\cdot}\) is \(B_{r,k}\), so by the definitions of scalar multiplication of a matrix (3.36) and matrix addition (3.34),

\begin{equation*} \bigl(a_1 B_{1,\cdot} + \cdots + a_n B_{n,\cdot}\bigr)_{1,k} = a_1 B_{1,k} + \cdots + a_n B_{n,k} = \sum_{r=1}^{n} a_r B_{r,k}. \end{equation*}

The two displayed expressions agree. Because \(aB\) and \(a_1 B_{1,\cdot} + \cdots + a_n B_{n,\cdot}\) are \(1\)-by-\(p\) matrices whose entries in column \(k\) agree for every \(k \in \{1,\dots,p\}\), we conclude that

\begin{equation*} aB = a_1 B_{1,\cdot} + \cdots + a_n B_{n,\cdot}. \end{equation*}

Thus \(aB\) is the linear combination of the rows \(B_{1,\cdot},\dots,B_{n,\cdot}\) of \(B\) whose coefficients \(a_1,\dots,a_n\) are the entries of \(a\). This is the row version of 3.50, whose proof it mirrors: 3.50 fixes a row index and expands over the columns of \(A\), whereas here we fix a column index and expand over the rows of \(B\).

Together with Exercise 3C.8, this yields 3.51(b): if \(C\) is \(m\)-by-\(c\) and \(R\) is \(c\)-by-\(n\) and \(j \in \{1,\dots,m\}\), then row \(j\) of \(CR\) equals \(C_{j,\cdot}R\) (by 3C.8), which by the present exercise is the linear combination of the rows of \(R\) with coefficients coming from row \(j\) of \(C\).

Take

\begin{equation*} A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \qquad B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}. \end{equation*}

Using the definition of matrix multiplication (3.41), the entry in row \(j\), column \(k\), of \(AB\) is \(\sum_{r=1}^{2} A_{j,r}B_{r,k}\). Since the only nonzero entry of \(A\) is \(A_{1,2} = 1\) and the only nonzero entry of \(B\) is \(B_{2,1} = 1\), the only nonzero term arises from \(j = 1\), \(r = 2\), \(k = 1\). Hence

\begin{equation*} AB = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}. \end{equation*}

Similarly, the entry in row \(j\), column \(k\), of \(BA\) is \(\sum_{r=1}^{2} B_{j,r}A_{r,k}\), and the only nonzero term arises from \(j = 2\), \(r = 1\), \(k = 2\). Hence

\begin{equation*} BA = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}. \end{equation*}

The matrices \(AB\) and \(BA\) differ, for example, in the entry in row \(1\), column \(1\): it is \(1\) for \(AB\) and \(0\) for \(BA\). Therefore \(AB \ne BA\), showing that matrix multiplication is not commutative.

It is worth noting why such an example must exist: if \(A\) and \(B\) are the matrices (with respect to the standard bases of \(\mathbf{F}^2\)) of \(S, T \in \mathcal{L}(\mathbf{F}^2)\), then \(AB = \mathcal{M}(ST)\) and \(BA = \mathcal{M}(TS)\) by 3.43, so a pair of noncommuting matrices is precisely a pair of linear maps on \(\mathbf{F}^2\) with \(ST \ne TS\). For the matrices above, \(S\) and \(T\) are the maps \(S(x,y) = (y,0)\) and \(T(x,y) = (0,x)\), and indeed \(ST(x,y) = (x,0)\) while \(TS(x,y) = (0,y)\).

The sizes. Suppose \(A(B+C)\) makes sense. Then \(B + C\) makes sense, so by the definition of matrix addition (3.34) the matrices \(B\) and \(C\) have the same size; say both are \(n\)-by-\(p\). Then \(B + C\) is also \(n\)-by-\(p\). Since the product \(A(B+C)\) is defined, the number of columns of \(A\) equals the number of rows of \(B + C\), so \(A\) has \(n\) columns; say \(A\) is \(m\)-by-\(n\). Now \(A\) is \(m\)-by-\(n\) and each of \(B\), \(C\) is \(n\)-by-\(p\), so both products \(AB\) and \(AC\) make sense and both are \(m\)-by-\(p\) matrices (by 3.41). Being of the same size, they can be added, so \(AB + AC\) makes sense and is \(m\)-by-\(p\), the same size as \(A(B+C)\).

Similarly, suppose \((D+E)F\) makes sense. Then \(D\) and \(E\) have the same size, say both \(m\)-by-\(n\), and \(D + E\) is \(m\)-by-\(n\); since \((D+E)F\) is defined, \(F\) has \(n\) rows, say \(F\) is \(n\)-by-\(p\). Hence \(DF\) and \(EF\) both make sense and are both \(m\)-by-\(p\), so \(DF + EF\) makes sense and is \(m\)-by-\(p\), the same size as \((D+E)F\).

The first identity. Both \(A(B+C)\) and \(AB + AC\) are \(m\)-by-\(p\) matrices, so it suffices to show that their entries in row \(j\), column \(k\), agree for all \(j \in \{1,\dots,m\}\) and \(k \in \{1,\dots,p\}\). Fix such \(j\) and \(k\). Using the definition of matrix multiplication (3.41), then the definition of matrix addition (3.34) applied to \(B + C\), then the distributive property of \(\mathbf{F}\) together with the fact that a finite sum may be split, we get

\begin{equation*} \begin{aligned} \bigl(A(B+C)\bigr)_{j,k} &= \sum_{r=1}^{n} A_{j,r}\,(B+C)_{r,k} \\ &= \sum_{r=1}^{n} A_{j,r}\,\bigl(B_{r,k} + C_{r,k}\bigr) \\ &= \sum_{r=1}^{n} \bigl(A_{j,r}B_{r,k} + A_{j,r}C_{r,k}\bigr) \\ &= \sum_{r=1}^{n} A_{j,r}B_{r,k} + \sum_{r=1}^{n} A_{j,r}C_{r,k} \\ &= (AB)_{j,k} + (AC)_{j,k} \\ &= (AB + AC)_{j,k}, \end{aligned} \end{equation*}

where the last two equalities use 3.41 and 3.34 again. Since \(j\) and \(k\) were arbitrary, \(A(B+C) = AB + AC\).

The second identity. Both \((D+E)F\) and \(DF + EF\) are \(m\)-by-\(p\) matrices. Fix \(j \in \{1,\dots,m\}\) and \(k \in \{1,\dots,p\}\). Then

\begin{equation*} \begin{aligned} \bigl((D+E)F\bigr)_{j,k} &= \sum_{r=1}^{n} (D+E)_{j,r}\,F_{r,k} \\ &= \sum_{r=1}^{n} \bigl(D_{j,r} + E_{j,r}\bigr) F_{r,k} \\ &= \sum_{r=1}^{n} \bigl(D_{j,r}F_{r,k} + E_{j,r}F_{r,k}\bigr) \\ &= \sum_{r=1}^{n} D_{j,r}F_{r,k} + \sum_{r=1}^{n} E_{j,r}F_{r,k} \\ &= (DF)_{j,k} + (EF)_{j,k} \\ &= (DF + EF)_{j,k}. \end{aligned} \end{equation*}

Since \(j\) and \(k\) were arbitrary, \((D+E)F = DF + EF\).

Note that both computations rest only on the distributive property of multiplication over addition in the field \(\mathbf{F}\); that the two one-sided distributive laws must be proved separately reflects the fact that matrix multiplication is not commutative (Exercise 3C.10).

The sizes. Suppose \((AB)C\) makes sense. Then \(AB\) makes sense, so the number of columns of \(A\) equals the number of rows of \(B\); say \(A\) is \(m\)-by-\(n\) and \(B\) is \(n\)-by-\(p\), in which case \(AB\) is \(m\)-by-\(p\) (by 3.41). Since \((AB)C\) makes sense, \(C\) has as many rows as \(AB\) has columns, so \(C\) is \(p\)-by-\(q\) for some positive integer \(q\), and \((AB)C\) is \(m\)-by-\(q\). Now \(B\) is \(n\)-by-\(p\) and \(C\) is \(p\)-by-\(q\), so \(BC\) makes sense and is \(n\)-by-\(p\) times \(p\)-by-\(q\), hence \(n\)-by-\(q\). Since \(A\) is \(m\)-by-\(n\), the product \(A(BC)\) makes sense and is \(m\)-by-\(q\). Thus both \((AB)C\) and \(A(BC)\) are defined and have the same size.

A proof with the matrices thrown out. Let \(\mathbf{F}\) be the field containing the entries of \(A\), \(B\), \(C\). Recall that a matrix determines a linear map between coordinate spaces: given an \(m\)-by-\(n\) matrix \(A\), define \(T_A \colon \mathbf{F}^{n} \to \mathbf{F}^{m}\) by

\begin{equation*} T_A(x_1,\dots,x_n) = \Bigl(\sum_{r=1}^{n} A_{1,r}x_r,\ \dots,\ \sum_{r=1}^{n} A_{m,r}x_r\Bigr). \end{equation*}

Each coordinate of \(T_A\) is a sum of scalar multiples of the coordinates of the input, so additivity and homogeneity of \(T_A\) follow at once from the corresponding identities \(\sum_r A_{j,r}(x_r + y_r) = \sum_r A_{j,r}x_r + \sum_r A_{j,r}y_r\) and \(\sum_r A_{j,r}(\lambda x_r) = \lambda \sum_r A_{j,r}x_r\) in \(\mathbf{F}\); hence \(T_A \in \mathcal{L}(\mathbf{F}^n,\mathbf{F}^m)\). Moreover, if \(e_1,\dots,e_n\) is the standard basis of \(\mathbf{F}^n\) and \(f_1,\dots,f_m\) is the standard basis of \(\mathbf{F}^m\), then

\begin{equation*} T_A e_k = (A_{1,k},\dots,A_{m,k}) = \sum_{j=1}^{m} A_{j,k} f_j, \end{equation*}

so the \(k\)th column of \(\mathcal{M}(T_A)\) is the \(k\)th column of \(A\) for each \(k\), and therefore, by the definition of the matrix of a linear map (3.31),

\begin{equation*} \mathcal{M}(T_A) = A, \end{equation*}

the matrix being taken with respect to the standard bases.

Now form \(T_C \in \mathcal{L}(\mathbf{F}^q,\mathbf{F}^p)\), \(T_B \in \mathcal{L}(\mathbf{F}^p,\mathbf{F}^n)\), and \(T_A \in \mathcal{L}(\mathbf{F}^n,\mathbf{F}^m)\) as above, so that \(\mathcal{M}(T_A) = A\), \(\mathcal{M}(T_B) = B\), \(\mathcal{M}(T_C) = C\), all with respect to standard bases. Applying 3.43 (the matrix of a product of linear maps is the product of the matrices) twice, and using standard bases throughout so that 3.43 applies as stated, we get

\begin{equation*} \mathcal{M}\bigl((T_A T_B)T_C\bigr) = \mathcal{M}(T_A T_B)\,\mathcal{M}(T_C) = \bigl(\mathcal{M}(T_A)\mathcal{M}(T_B)\bigr)C = (AB)C \end{equation*}

and

\begin{equation*} \mathcal{M}\bigl(T_A (T_B T_C)\bigr) = \mathcal{M}(T_A)\,\mathcal{M}(T_B T_C) = A\bigl(\mathcal{M}(T_B)\mathcal{M}(T_C)\bigr) = A(BC). \end{equation*}

But composition of functions is associative: for every \(u \in \mathbf{F}^q\),

\begin{equation*} \bigl((T_A T_B)T_C\bigr)(u) = (T_A T_B)(T_C u) = T_A\bigl(T_B(T_C u)\bigr) = T_A\bigl((T_B T_C)(u)\bigr) = \bigl(T_A(T_B T_C)\bigr)(u). \end{equation*}

Hence \((T_A T_B)T_C = T_A(T_B T_C)\) as linear maps from \(\mathbf{F}^q\) to \(\mathbf{F}^m\). Equal linear maps have equal matrices, so comparing the two displayed computations of matrices gives

\begin{equation*} (AB)C = A(BC). \end{equation*}

This is the promised “throw the matrices out” argument: associativity of matrix multiplication is nothing but associativity of composition of functions, transported through 3.43.

A direct verification. For completeness, here is the computation with the matrices left in. Both \((AB)C\) and \(A(BC)\) are \(m\)-by-\(q\), so fix \(j \in \{1,\dots,m\}\) and \(k \in \{1,\dots,q\}\). By 3.41 applied twice, and then interchanging the order of the two finite sums (legitimate for finite sums in \(\mathbf{F}\)),

\begin{equation*} \begin{aligned} \bigl((AB)C\bigr)_{j,k} &= \sum_{s=1}^{p} (AB)_{j,s} C_{s,k} = \sum_{s=1}^{p} \Bigl(\sum_{r=1}^{n} A_{j,r}B_{r,s}\Bigr) C_{s,k} = \sum_{s=1}^{p}\sum_{r=1}^{n} A_{j,r}B_{r,s}C_{s,k} \\ &= \sum_{r=1}^{n}\sum_{s=1}^{p} A_{j,r}B_{r,s}C_{s,k} = \sum_{r=1}^{n} A_{j,r}\Bigl(\sum_{s=1}^{p} B_{r,s}C_{s,k}\Bigr) = \sum_{r=1}^{n} A_{j,r}(BC)_{r,k} \\ &= \bigl(A(BC)\bigr)_{j,k}. \end{aligned} \end{equation*}

Since \(j\) and \(k\) were arbitrary, \((AB)C = A(BC)\).

Because \(A\) is \(n\)-by-\(n\), all the products below are defined and every matrix appearing is \(n\)-by-\(n\) (by 3.41). Moreover, matrix multiplication is associative (Exercise 3C.12), so the expression \(AAA\) is unambiguous: \((AA)A = A(AA)\). We compute using the grouping \(A^3 = (AA)A = A^2 A\).

Fix \(j\) and \(k\) with \(1 \le j,k \le n\). By the definition of matrix multiplication (3.41) applied to the product \(A^2 A\), with \(r\) as the summation index,

\begin{equation*} (A^3)_{j,k} = (A^2 A)_{j,k} = \sum_{r=1}^{n} (A^2)_{j,r}\, A_{r,k}. \end{equation*}

Applying 3.41 again to \(A^2 = AA\), this time with \(p\) as the summation index, gives, for each \(r \in \{1,\dots,n\}\),

\begin{equation*} (A^2)_{j,r} = \sum_{p=1}^{n} A_{j,p}A_{p,r}. \end{equation*}

Substituting this into the previous display and using the distributive property in \(\mathbf{F}\) to move the factor \(A_{r,k}\) inside the inner sum, we obtain

\begin{equation*} (A^3)_{j,k} = \sum_{r=1}^{n} \Bigl(\sum_{p=1}^{n} A_{j,p}A_{p,r}\Bigr) A_{r,k} = \sum_{r=1}^{n}\sum_{p=1}^{n} A_{j,p}A_{p,r}A_{r,k}. \end{equation*}

Finally, the two sums are finite sums of elements of \(\mathbf{F}\), so their order of summation may be interchanged, giving

\begin{equation*} (A^3)_{j,k} = \sum_{p=1}^{n}\sum_{r=1}^{n} A_{j,p}A_{p,r}A_{r,k}, \end{equation*}

which is the desired formula.

The pattern generalizes in the obvious way: for a positive integer \(t\) and \(1 \le j,k \le n\),

\begin{equation*} (A^t)_{j,k} = \sum_{r_1=1}^{n}\cdots\sum_{r_{t-1}=1}^{n} A_{j,r_1}A_{r_1,r_2}\cdots A_{r_{t-1},k}, \end{equation*}

as follows by induction on \(t\) using the same computation.

Let \(\Phi \colon \mathbf{F}^{m,n} \to \mathbf{F}^{n,m}\) be defined by \(\Phi(A) = A^{\mathrm{t}}\).

First, \(\Phi\) really does map \(\mathbf{F}^{m,n}\) into \(\mathbf{F}^{n,m}\): if \(A\) is an \(m\)-by-\(n\) matrix with entries in \(\mathbf{F}\), then by the definition of the transpose (3.54) the matrix \(A^{\mathrm{t}}\) is the \(n\)-by-\(m\) matrix with entries in \(\mathbf{F}\) determined by

\begin{equation*} (A^{\mathrm{t}})_{k,j} = A_{j,k} \quad \text{for } 1 \le j \le m,\ 1 \le k \le n. \end{equation*}

So \(\Phi(A) \in \mathbf{F}^{n,m}\), and \(\Phi\) is a well-defined function (the equation above prescribes each entry of \(A^{\mathrm{t}}\) exactly once).

To show \(\Phi\) is linear we verify additivity and homogeneity, using the definition of a linear map (3.1) and recalling that addition and scalar multiplication on \(\mathbf{F}^{m,n}\) and \(\mathbf{F}^{n,m}\) are entrywise (3.34 and 3.36).

Additivity. Suppose \(A, B \in \mathbf{F}^{m,n}\). Then \(A + B \in \mathbf{F}^{m,n}\), so \((A+B)^{\mathrm{t}}\) and \(A^{\mathrm{t}} + B^{\mathrm{t}}\) are both \(n\)-by-\(m\) matrices. Fix \(j \in \{1,\dots,m\}\) and \(k \in \{1,\dots,n\}\). Then

\begin{equation*} \bigl((A+B)^{\mathrm{t}}\bigr)_{k,j} = (A+B)_{j,k} = A_{j,k} + B_{j,k} = (A^{\mathrm{t}})_{k,j} + (B^{\mathrm{t}})_{k,j} = \bigl(A^{\mathrm{t}} + B^{\mathrm{t}}\bigr)_{k,j}, \end{equation*}

where the first and third equalities use 3.54 and the second and fourth use 3.34. Since every entry of an \(n\)-by-\(m\) matrix is of the form (row \(k\), column \(j\)) with \(1 \le k \le n\) and \(1 \le j \le m\), this shows

\begin{equation*} \Phi(A + B) = (A+B)^{\mathrm{t}} = A^{\mathrm{t}} + B^{\mathrm{t}} = \Phi(A) + \Phi(B). \end{equation*}

Homogeneity. Suppose \(\lambda \in \mathbf{F}\) and \(A \in \mathbf{F}^{m,n}\). Then \(\lambda A \in \mathbf{F}^{m,n}\), and \((\lambda A)^{\mathrm{t}}\) and \(\lambda A^{\mathrm{t}}\) are both \(n\)-by-\(m\) matrices. Fix \(j \in \{1,\dots,m\}\) and \(k \in \{1,\dots,n\}\). Then

\begin{equation*} \bigl((\lambda A)^{\mathrm{t}}\bigr)_{k,j} = (\lambda A)_{j,k} = \lambda A_{j,k} = \lambda (A^{\mathrm{t}})_{k,j} = \bigl(\lambda A^{\mathrm{t}}\bigr)_{k,j}, \end{equation*}

where the first and third equalities use 3.54 and the second and fourth use 3.36. Hence

\begin{equation*} \Phi(\lambda A) = (\lambda A)^{\mathrm{t}} = \lambda A^{\mathrm{t}} = \lambda\,\Phi(A). \end{equation*}

Since \(\Phi\) is additive and homogeneous, \(\Phi \in \mathcal{L}(\mathbf{F}^{m,n}, \mathbf{F}^{n,m})\), which is what we wanted to prove.

Two remarks. First, this establishes the identities \((A+B)^{\mathrm{t}} = A^{\mathrm{t}} + B^{\mathrm{t}}\) and \((\lambda A)^{\mathrm{t}} = \lambda A^{\mathrm{t}}\) quoted in the text just after 3.55. Second, \(\Phi\) is in fact an isomorphism: applying 3.54 twice gives \(\bigl((A^{\mathrm{t}})^{\mathrm{t}}\bigr)_{j,k} = (A^{\mathrm{t}})_{k,j} = A_{j,k}\), so \((A^{\mathrm{t}})^{\mathrm{t}} = A\) for every \(A \in \mathbf{F}^{m,n}\). Writing \(\Psi \colon \mathbf{F}^{n,m} \to \mathbf{F}^{m,n}\) for the transpose map in the other direction (linear by the same argument with \(m\) and \(n\) exchanged), we get \(\Psi\Phi = I\) on \(\mathbf{F}^{m,n}\) and \(\Phi\Psi = I\) on \(\mathbf{F}^{n,m}\), so \(\Phi\) is invertible with \(\Phi^{-1} = \Psi\).

First we check that both sides of the asserted equation are matrices of the same size, so that the equation has a chance of making sense. Because \(A\) is \(m\)-by-\(n\) and \(C\) is \(n\)-by-\(p\), the product \(AC\) is defined and is an \(m\)-by-\(p\) matrix (by 3.41). Hence \((AC)^{t}\) is a \(p\)-by-\(m\) matrix (by 3.54). On the other side, \(C^{t}\) is a \(p\)-by-\(n\) matrix and \(A^{t}\) is an \(n\)-by-\(m\) matrix (both by 3.54); the number of columns of \(C^{t}\) equals the number of rows of \(A^{t}\), so the product \(C^{t}A^{t}\) is defined and is a \(p\)-by-\(m\) matrix. Thus \((AC)^{t}\) and \(C^{t}A^{t}\) are both \(p\)-by-\(m\) matrices, and to prove they are equal we need only show that their entries agree in every position.

So suppose \(k \in \{1, \dots, p\}\) and \(j \in \{1, \dots, m\}\); we compute the entry in row \(k\), column \(j\), of each side.

For the left side, the definition of the transpose (3.54) gives

\begin{equation*} \bigl((AC)^{t}\bigr)_{k,j} = (AC)_{j,k}, \end{equation*}

and the definition of matrix multiplication (3.41) gives

\begin{equation*} (AC)_{j,k} = \sum_{r=1}^{n} A_{j,r} C_{r,k}. \end{equation*}

Hence

\begin{equation*} \bigl((AC)^{t}\bigr)_{k,j} = \sum_{r=1}^{n} A_{j,r} C_{r,k}. \end{equation*}

For the right side, the definition of matrix multiplication (3.41), applied to the \(p\)-by-\(n\) matrix \(C^{t}\) and the \(n\)-by-\(m\) matrix \(A^{t}\), gives

\begin{equation*} (C^{t} A^{t})_{k,j} = \sum_{r=1}^{n} (C^{t})_{k,r} (A^{t})_{r,j}. \end{equation*}

The definition of the transpose (3.54) tells us that \((C^{t})_{k,r} = C_{r,k}\) and \((A^{t})_{r,j} = A_{j,r}\). Substituting these into the equation above yields

\begin{equation*} (C^{t} A^{t})_{k,j} = \sum_{r=1}^{n} C_{r,k} A_{j,r} = \sum_{r=1}^{n} A_{j,r} C_{r,k}, \end{equation*}

where the last equality holds because multiplication in \(\mathbf{F}\) is commutative.

The two displayed computations produce the same number. Thus

\begin{equation*} \bigl((AC)^{t}\bigr)_{k,j} = (C^{t} A^{t})_{k,j} \end{equation*}

for every \(k \in \{1, \dots, p\}\) and every \(j \in \{1, \dots, m\}\). Because \((AC)^{t}\) and \(C^{t}A^{t}\) have the same size and the same entry in every position, we conclude that \((AC)^{t} = C^{t}A^{t}\), as desired.

Note the role played by the order reversal: the sum defining \((AC)_{j,k}\) pairs row \(j\) of \(A\) with column \(k\) of \(C\), and transposing turns these into column \(j\) of \(A^{t}\) and row \(k\) of \(C^{t}\). Producing that same sum as a matrix product therefore requires taking row \(k\) of \(C^{t}\) times column \(j\) of \(A^{t}\), which is exactly the entry in row \(k\), column \(j\), of \(C^{t}A^{t}\). In general \(A^{t}C^{t}\) is not even defined (its inner dimensions are \(m\) and \(p\)), which further explains why the order must be reversed.

Throughout, recall that the rank of \(A\) is by definition the column rank of \(A\) (by 3.58), which is the dimension of the span of the columns \(A_{\cdot,1}, \dots, A_{\cdot,n}\) of \(A\) in \(\mathbf{F}^{m,1}\) (by 3.52). Write

\begin{equation*} U = \operatorname{span}(A_{\cdot,1}, \dots, A_{\cdot,n}) \subseteq \mathbf{F}^{m,1}, \end{equation*}

so that the rank of \(A\) equals \(\dim U\).

First suppose the rank of \(A\) is \(1\), so \(\dim U = 1\). Every one-dimensional vector space has a basis consisting of a single nonzero vector, so there exists \(c \in \mathbf{F}^{m,1}\) with \(c \neq 0\) and \(U = \operatorname{span}( c)\). Write

\begin{equation*} c = \begin{pmatrix} c_1 \\ \vdots \\ c_m \end{pmatrix}, \end{equation*}

so \((c_1, \dots, c_m) \in \mathbf{F}^{m}\). Now let \(k \in \{1, \dots, n\}\). The column \(A_{\cdot,k}\) lies in \(U = \operatorname{span}( c)\), so there exists \(d_k \in \mathbf{F}\) with

\begin{equation*} A_{\cdot,k} = d_k c. \end{equation*}

This defines \((d_1, \dots, d_n) \in \mathbf{F}^{n}\). Comparing the entry in row \(j\) of the two sides of the equation above, and recalling that the entry in row \(j\) of the \(m\)-by-\(1\) matrix \(A_{\cdot,k}\) is \(A_{j,k}\) (by 3.44), we obtain

\begin{equation*} A_{j,k} = d_k c_j = c_j d_k \end{equation*}

for every \(j = 1, \dots, m\) and every \(k = 1, \dots, n\), as desired.

Conversely, suppose there exist \((c_1, \dots, c_m) \in \mathbf{F}^{m}\) and \((d_1, \dots, d_n) \in \mathbf{F}^{n}\) with \(A_{j,k} = c_j d_k\) for all \(j\) and \(k\). Let \(c \in \mathbf{F}^{m,1}\) be the \(m\)-by-\(1\) matrix whose entry in row \(j\) is \(c_j\). Fix \(k \in \{1, \dots, n\}\). For each \(j \in \{1, \dots, m\}\), the entry in row \(j\) of the column \(A_{\cdot,k}\) equals \(A_{j,k} = c_j d_k\), which is the entry in row \(j\) of \(d_k c\). Because these two \(m\)-by-\(1\) matrices agree in every entry,

\begin{equation*} A_{\cdot,k} = d_k c. \end{equation*}

Thus every column of \(A\) lies in \(\operatorname{span}( c)\), and hence

\begin{equation*} U = \operatorname{span}(A_{\cdot,1}, \dots, A_{\cdot,n}) \subseteq \operatorname{span}( c). \end{equation*}

The list consisting of the single vector \(c\) spans \(\operatorname{span}( c)\), and every spanning list can be reduced to a basis (by 2.30), so \(\operatorname{span}( c)\) has a basis of length at most \(1\); hence \(\dim \operatorname{span}( c) \leq 1\). Because \(U\) is a subspace of the finite-dimensional space \(\operatorname{span}( c)\), we conclude that \(\dim U \leq \dim \operatorname{span}( c) \leq 1\) (by 2.37). On the other hand, \(A \neq 0\), so some entry \(A_{j,k}\) of \(A\) is nonzero, and hence the column \(A_{\cdot,k}\) is a nonzero element of \(U\); thus \(U \neq \{0\}\) and so \(\dim U \geq 1\). Combining the two inequalities gives \(\dim U = 1\), which says exactly that the rank of \(A\) is \(1\).

Two remarks. First, the hypothesis \(A \neq 0\) is genuinely needed for the direction just proved: taking \(c_1 = \dots = c_m = 0\) and \(d_1 = \dots = d_n = 0\) produces \(A_{j,k} = c_j d_k\) for the zero matrix, whose rank is \(0\) rather than \(1\).

Second, the condition \(A_{j,k} = c_j d_k\) can be restated as a factorization. Let \(C\) be the \(m\)-by-\(1\) matrix with \(C_{j,1} = c_j\) and let \(R\) be the \(1\)-by-\(n\) matrix with \(R_{1,k} = d_k\). Then the definition of matrix multiplication (3.41) gives

\begin{equation*} (CR)_{j,k} = \sum_{r=1}^{1} C_{j,r} R_{r,k} = c_j d_k, \end{equation*}

so the condition in the exercise says precisely that \(A = CR\) for some \(m\)-by-\(1\) matrix \(C\) and some \(1\)-by-\(n\) matrix \(R\). From this point of view the forward direction is exactly the column–row factorization 3.56 applied with column rank \(c = 1\), and the reverse direction says that any nonzero matrix of the form \(CR\), with \(C\) an \(m\)-by-\(1\) matrix and \(R\) a \(1\)-by-\(n\) matrix, has all of its columns inside the single one-dimensional subspace spanned by \(C\).

Write \(A = \mathcal{M}(T) = \mathcal{M}(T, (u_1, \dots, u_n), (v_1, \dots, v_n))\). Because \(u_1, \dots, u_n\) and \(v_1, \dots, v_n\) are both bases of \(V\), we have \(\dim V = n\) and \(A\) is an \(n\)-by-\(n\) matrix. By the definition of the matrix of a linear map (3.31), the entries of \(A\) are determined by

\begin{equation*} T u_k = \sum_{j=1}^{n} A_{j,k} v_j \quad \text{for each } k = 1, \dots, n. \end{equation*}

We will prove (a) \(\Longleftrightarrow\) (b), then (b) \(\Longleftrightarrow\) (c), then (c) \(\Longleftrightarrow\) (d), then (d) \(\Longleftrightarrow\) (e); this chain shows that all five statements are equivalent.

Step 0: a dictionary between \(V\) and \(\mathbf{F}^{n,1}\).

Suppose

\begin{equation*} x = \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix} \in \mathbf{F}^{n,1}, \qquad \text{and set} \qquad u = x_1 u_1 + \dots + x_n u_n \in V. \end{equation*}

Using the linearity of \(T\) and the displayed formula for \(Tu_k\), we compute

\begin{equation*} \begin{aligned} Tu &= \sum_{k=1}^{n} x_k \, T u_k \\ &= \sum_{k=1}^{n} x_k \sum_{j=1}^{n} A_{j,k} v_j \\ &= \sum_{j=1}^{n} \Bigl( \sum_{k=1}^{n} A_{j,k} x_k \Bigr) v_j. \end{aligned} \end{equation*}

Because \(A\) is \(n\)-by-\(n\) and \(x\) is \(n\)-by-\(1\), the product \(Ax\) is an \(n\)-by-\(1\) matrix whose entry in row \(j\) is \(\sum_{k=1}^{n} A_{j,k} x_k\) (by the definition of matrix multiplication, 3.41); moreover \(Ax = x_1 A_{\cdot,1} + \dots + x_n A_{\cdot,n}\) (by 3.50). Hence the computation above says

\begin{equation*} Tu = \sum_{j=1}^{n} (Ax)_{j,1} \, v_j. \end{equation*}

Because \(v_1, \dots, v_n\) is a basis of \(V\) and hence linearly independent, the vector \(Tu\) equals \(0\) if and only if all the coefficients \((Ax)_{j,1}\) equal \(0\), that is, if and only if

\begin{equation*} x_1 A_{\cdot,1} + \dots + x_n A_{\cdot,n} = Ax = 0. \end{equation*}

Similarly, because \(u_1, \dots, u_n\) is a basis of \(V\) and hence linearly independent, we have \(u = 0\) if and only if \(x = 0\). Finally, because \(u_1, \dots, u_n\) spans \(V\), every \(u \in V\) arises from some \(x \in \mathbf{F}^{n,1}\) in this way. To summarize Step 0: the assignment \(x \mapsto u = x_1 u_1 + \dots + x_n u_n\) maps \(\mathbf{F}^{n,1}\) onto \(V\); for corresponding \(x\) and \(u\) we have \(Tu = 0\) if and only if \(Ax = 0\), and \(u = 0\) if and only if \(x = 0\).

Step 1: proof that (a) is equivalent to (b).

First suppose \(T\) is injective. Suppose \(x_1, \dots, x_n \in \mathbf{F}\) satisfy

\begin{equation*} x_1 A_{\cdot,1} + \dots + x_n A_{\cdot,n} = 0. \end{equation*}

Let \(x \in \mathbf{F}^{n,1}\) be the column with entries \(x_1, \dots, x_n\) and let \(u = x_1 u_1 + \dots + x_n u_n\). By Step 0, \(Tu = 0\), so \(u \in \operatorname{null} T\). Because \(T\) is injective, \(\operatorname{null} T = \{0\}\) (by 3.15), so \(u = 0\); hence \(x = 0\) by Step 0, meaning \(x_1 = \dots = x_n = 0\). Thus the only choice of scalars making the displayed linear combination of the columns of \(A\) equal \(0\) is the trivial one, so the columns of \(A\) are linearly independent in \(\mathbf{F}^{n,1}\). This proves (b).

Conversely, suppose the columns of \(A\) are linearly independent. Let \(u \in \operatorname{null} T\). Because \(u_1, \dots, u_n\) is a basis of \(V\), we may write \(u = x_1 u_1 + \dots + x_n u_n\) for some \(x_1, \dots, x_n \in \mathbf{F}\); let \(x \in \mathbf{F}^{n,1}\) be the corresponding column. By Step 0, \(Tu = 0\) implies

\begin{equation*} x_1 A_{\cdot,1} + \dots + x_n A_{\cdot,n} = 0. \end{equation*}

Linear independence of the columns now forces \(x_1 = \dots = x_n = 0\), so \(u = 0\). Hence \(\operatorname{null} T = \{0\}\), and therefore \(T\) is injective (by 3.15). This proves (a).

Step 2: proof that (b) is equivalent to (c).

The columns \(A_{\cdot,1}, \dots, A_{\cdot,n}\) of \(A\) form a list of length \(n\) of vectors in \(\mathbf{F}^{n,1}\), and \(\dim \mathbf{F}^{n,1} = n\) (by 3.40). If this list is linearly independent, then because its length equals \(\dim \mathbf{F}^{n,1}\) it is a basis of \(\mathbf{F}^{n,1}\) (by 2.38); in particular it spans \(\mathbf{F}^{n,1}\), giving (c). If instead this list spans \(\mathbf{F}^{n,1}\), then because its length equals \(\dim \mathbf{F}^{n,1}\) it is a basis of \(\mathbf{F}^{n,1}\) (by 2.42); in particular it is linearly independent, giving (b). Thus (b) and (c) are equivalent, each being equivalent to the statement that the columns of \(A\) form a basis of \(\mathbf{F}^{n,1}\).

Step 3: proof that (c) is equivalent to (d).

We show that each of (c) and (d) is equivalent to the statement that the rank of \(A\) equals \(n\).

Let \(U\) denote the span of the columns of \(A\) in \(\mathbf{F}^{n,1}\); by 3.52 and 3.58, \(\dim U\) is the column rank of \(A\), which is the rank of \(A\). If the columns span \(\mathbf{F}^{n,1}\), then \(U = \mathbf{F}^{n,1}\), so \(\dim U = n\) by 3.40. Conversely, if \(\dim U = n\), then \(U\) is a subspace of \(\mathbf{F}^{n,1}\) whose dimension equals \(\dim \mathbf{F}^{n,1}\), so \(U = \mathbf{F}^{n,1}\) (by 2.39), meaning the columns span \(\mathbf{F}^{n,1}\). Hence (c) holds if and only if the column rank of \(A\) equals \(n\).

The identical argument with rows in place of columns, using \(\dim \mathbf{F}^{1,n} = n\) (by 3.40), shows that (d) holds if and only if the row rank of \(A\) equals \(n\): writing \(W\) for the span of the rows of \(A\) in \(\mathbf{F}^{1,n}\), we have \(\dim W\) equal to the row rank of \(A\) (by 3.52), and \(W = \mathbf{F}^{1,n}\) if and only if \(\dim W = n\) (by 2.39 for one direction and 3.40 for the other).

By 3.57, the column rank of \(A\) equals the row rank of \(A\). Hence the column rank equals \(n\) if and only if the row rank equals \(n\), and therefore (c) and (d) are equivalent.

Step 4: proof that (d) is equivalent to (e).

The rows \(A_{1,\cdot}, \dots, A_{n,\cdot}\) of \(A\) form a list of length \(n\) of vectors in \(\mathbf{F}^{1,n}\), and \(\dim \mathbf{F}^{1,n} = n\) (by 3.40). If this list spans \(\mathbf{F}^{1,n}\), then it is a basis of \(\mathbf{F}^{1,n}\) (by 2.42) and hence is linearly independent, giving (e). If this list is linearly independent, then it is a basis of \(\mathbf{F}^{1,n}\) (by 2.38) and hence spans \(\mathbf{F}^{1,n}\), giving (d). Thus (d) and (e) are equivalent.

Having proved (a) \(\Longleftrightarrow\) (b) \(\Longleftrightarrow\) (c) \(\Longleftrightarrow\) (d) \(\Longleftrightarrow\) (e), we conclude that all five statements are equivalent, completing the proof.

Two comments are worth making. First, nothing above required \(u_1, \dots, u_n\) and \(v_1, \dots, v_n\) to be the same basis: the two bases enter only through the formula \(Tu_k = \sum_j A_{j,k} v_j\), and Step 0 used the first basis to convert vectors of \(V\) into columns and the second basis merely to detect when \(Tu = 0\). Second, because \(T\) is an operator on the finite-dimensional space \(V\), injectivity is equivalent to surjectivity (by 3.65), so each of (a) through (e) is also equivalent to the statement that \(T\) is surjective, and hence to the statement that \(T\) is invertible.

Recall the definition of invertibility (3.59): a linear map \(T \in \mathcal{L}(V,W)\) is invertible if there exists \(S \in \mathcal{L}(W,V)\) with \(ST = I\) (the identity operator on \(V\)) and \(TS = I\) (the identity operator on \(W\)); such an \(S\) is unique (3.60) and is denoted \(T^{-1}\).

Because \(T\) is invertible, the map \(T^{-1} \in \mathcal{L}(W,V)\) exists and satisfies

\begin{equation*} T^{-1} T = I_V \qquad \text{and} \qquad T T^{-1} = I_W. \end{equation*}

Now apply the definition of invertibility to the linear map \(T^{-1} \in \mathcal{L}(W,V)\). To show \(T^{-1}\) is invertible we must produce a linear map \(R \in \mathcal{L}(V,W)\) with \(R T^{-1} = I_W\) and \(T^{-1} R = I_V\). Take \(R = T\). Then \(T \in \mathcal{L}(V,W)\) is linear, and the two displayed equations above say exactly that

\begin{equation*} T T^{-1} = I_W \qquad \text{and} \qquad T^{-1} T = I_V, \end{equation*}

which are the required identities. Hence \(T^{-1}\) is invertible, and \(T\) is an inverse of \(T^{-1}\).

Finally, by the uniqueness of the inverse (3.60), the inverse of \(T^{-1}\) is the unique linear map with these two properties, so

\begin{equation*} (T^{-1})^{-1} = T, \end{equation*}

as desired.

First, \(ST \in \mathcal{L}(U,W)\): the composition of linear maps is linear (this is noted in 3.8 and is immediate from the definitions), and \(ST\) maps \(U\) into \(W\) because \(T\) maps \(U\) into \(V\) and \(S\) maps \(V\) into \(W\).

Since \(S\) and \(T\) are invertible, we have \(S^{-1} \in \mathcal{L}(W,V)\) and \(T^{-1} \in \mathcal{L}(V,U)\), so the composition \(T^{-1} S^{-1}\) is a linear map from \(W\) to \(U\); that is, \(T^{-1}S^{-1} \in \mathcal{L}(W,U)\), which is the right type of object to be an inverse of \(ST\).

Now compute, using associativity of composition of functions. For the composition in one order,

\begin{equation*} (T^{-1} S^{-1})(ST) = T^{-1}(S^{-1} S)T = T^{-1} I_V T = T^{-1} T = I_U, \end{equation*}

where we used \(S^{-1}S = I_V\) and \(T^{-1}T = I_U\). For the other order,

\begin{equation*} (ST)(T^{-1} S^{-1}) = S(T T^{-1})S^{-1} = S I_V S^{-1} = S S^{-1} = I_W, \end{equation*}

where we used \(TT^{-1} = I_V\) and \(SS^{-1} = I_W\).

Thus \(T^{-1}S^{-1}\) is a linear map from \(W\) to \(U\) whose composition with \(ST\) is the identity in both orders. By the definition of invertibility (3.59), \(ST\) is invertible, and by the uniqueness of the inverse (3.60),

\begin{equation*} (ST)^{-1} = T^{-1} S^{-1}, \end{equation*}

as desired.

Throughout, let \(n = \dim V\). We prove (a) \(\Rightarrow\) (b) \(\Rightarrow\) (c) \(\Rightarrow\) (a).

(a) \(\Rightarrow\) (b). Suppose \(T\) is invertible and let \(v_1, \dots, v_n\) be any basis of \(V\). We show \(Tv_1, \dots, Tv_n\) is linearly independent and spans \(V\).

Linear independence: suppose \(a_1, \dots, a_n \in \mathbf{F}\) satisfy \(a_1 Tv_1 + \dots + a_n Tv_n = 0\). By linearity this says \(T(a_1 v_1 + \dots + a_n v_n) = 0\). Because \(T\) is invertible it is injective (3.63), so \(\operatorname{null} T = \{0\}\) (3.15) and hence \(a_1 v_1 + \dots + a_n v_n = 0\). Since \(v_1, \dots, v_n\) is linearly independent, \(a_1 = \dots = a_n = 0\).

Spanning: let \(v \in V\). Because \(T\) is invertible, \(T^{-1} v \in V\), so there exist \(a_1, \dots, a_n \in \mathbf{F}\) with \(T^{-1} v = a_1 v_1 + \dots + a_n v_n\). Applying \(T\) and using linearity,

\begin{equation*} v = T(T^{-1} v) = a_1 Tv_1 + \dots + a_n Tv_n . \end{equation*}

Thus \(\operatorname{span}(Tv_1, \dots, Tv_n) = V\).

Hence \(Tv_1, \dots, Tv_n\) is a basis of \(V\).

(b) \(\Rightarrow\) (c). Because \(V\) is finite-dimensional, \(V\) has a basis \(v_1, \dots, v_n\) (2.31). Applying (b) to this particular basis shows that \(Tv_1, \dots, Tv_n\) is a basis of \(V\), which is exactly statement (c). (If \(V = \{0\}\), the basis is the empty list and the statement holds trivially.)

(c) \(\Rightarrow\) (a). Suppose \(v_1, \dots, v_n\) is a basis of \(V\) such that \(Tv_1, \dots, Tv_n\) is a basis of \(V\). Since \(v_1, \dots, v_n\) spans \(V\), every \(v \in V\) has the form \(v = a_1 v_1 + \dots + a_n v_n\), so

\begin{equation*} Tv = a_1 Tv_1 + \dots + a_n Tv_n \in \operatorname{span}(Tv_1, \dots, Tv_n). \end{equation*}

Conversely each \(Tv_k\) lies in \(\operatorname{range} T\), so

\begin{equation*} \operatorname{range} T = \operatorname{span}(Tv_1, \dots, Tv_n) \end{equation*}

(this is Exercise 10 in Section 3B). Because \(Tv_1, \dots, Tv_n\) is a basis of \(V\), its span is \(V\), so \(\operatorname{range} T = V\) and \(T\) is surjective. Since \(V\) is finite-dimensional and \(T\) maps \(V\) to itself, the domain and target have the same finite dimension, so injectivity is equivalent to surjectivity (3.65); thus \(T\) is also injective. Being both injective and surjective, \(T\) is invertible (3.63).

Having shown (a) \(\Rightarrow\) (b) \(\Rightarrow\) (c) \(\Rightarrow\) (a), the three statements are equivalent.

Let \(n = \dim V\), so \(n \geq 2\), and let

\begin{equation*} \mathcal{N} = \{T \in \mathcal{L}(V) : T \text{ is not invertible}\}. \end{equation*}

We show \(\mathcal{N}\) is not closed under addition, hence is not a subspace (1.34).

Because \(V\) is finite-dimensional with \(\dim V = n \geq 2\), there is a basis \(v_1, \dots, v_n\) of \(V\) (2.31), and it has at least two entries. By the linear map lemma (3.4) there exist unique \(S, T \in \mathcal{L}(V)\) satisfying

\begin{equation*} S v_1 = v_1, \quad S v_k = 0 \ \text{ for } k = 2, \dots, n, \end{equation*}

\begin{equation*} T v_1 = 0, \quad T v_k = v_k \ \text{ for } k = 2, \dots, n. \end{equation*}

Neither \(S\) nor \(T\) is invertible. Indeed \(v_2\) exists (because \(n \geq 2\)), \(v_2 \neq 0\) (an entry of a basis is nonzero, since a list containing \(0\) is linearly dependent), and \(S v_2 = 0\); thus \(\operatorname{null} S \neq \{0\}\), so \(S\) is not injective (3.15) and therefore not invertible (3.63). Likewise \(v_1 \neq 0\) and \(T v_1 = 0\), so \(T\) is not invertible. Hence \(S, T \in \mathcal{N}\).

However, \(S + T\) agrees with the identity operator on the basis \(v_1, \dots, v_n\):

\begin{equation*} (S + T)v_1 = v_1 + 0 = v_1, \qquad (S+T)v_k = 0 + v_k = v_k \ \text{ for } k = 2, \dots, n. \end{equation*}

Two linear maps that agree on a basis are equal (each \(v \in V\) is a linear combination of \(v_1, \dots, v_n\), and both maps are determined on such combinations by linearity). Therefore \(S + T = I\), the identity operator on \(V\), which is invertible (it is its own inverse). Thus \(S + T \notin \mathcal{N}\).

So \(\mathcal{N}\) contains \(S\) and \(T\) but not \(S + T\), i.e. \(\mathcal{N}\) is not closed under addition. A subspace must be closed under addition, so \(\mathcal{N}\) is not a subspace of \(\mathcal{L}(V)\).

Remark: the hypothesis \(\dim V > 1\) is needed. If \(\dim V = 1\), then a linear map \(T \in \mathcal{L}(V)\) has the form \(T v = \lambda v\) for a fixed \(\lambda \in \mathbf{F}\), and \(T\) fails to be invertible exactly when \(\lambda = 0\); thus the noninvertible maps form \(\{0\}\), which is a subspace. If \(\dim V = 0\), then \(\mathcal{L}(V) = \{0\}\) and \(0\) is invertible, so the set of noninvertible maps is empty and hence not a subspace either; but the intended content of the exercise is the case \(\dim V \geq 2\) treated above.

Let \(n = \dim V\).

First suppose such an invertible \(T \in \mathcal{L}(V)\) exists, with \(Tu = Su\) for every \(u \in U\). Let \(u \in U\) satisfy \(Su = 0\). Then \(Tu = Su = 0\). Because \(T\) is invertible it is injective (3.63), so \(\operatorname{null} T = \{0\}\) (3.15) and hence \(u = 0\). Thus \(\operatorname{null} S = \{0\}\), so \(S\) is injective (3.15).

Conversely, suppose \(S\) is injective. Because \(U\) is a subspace of the finite-dimensional space \(V\), \(U\) is finite-dimensional (2.25); let \(m = \dim U\) and let \(u_1, \dots, u_m\) be a basis of \(U\) (2.31). Extend it to a basis

\begin{equation*} u_1, \dots, u_m, v_1, \dots, v_p \end{equation*}

of \(V\) (2.32), where \(m + p = n\).

Next, \(Su_1, \dots, Su_m\) is a linearly independent list in \(V\): if \(a_1, \dots, a_m \in \mathbf{F}\) satisfy \(a_1 Su_1 + \dots + a_m Su_m = 0\), then \(S(a_1 u_1 + \dots + a_m u_m) = 0\) by linearity, so \(a_1 u_1 + \dots + a_m u_m \in \operatorname{null} S = \{0\}\) (using injectivity of \(S\) and 3.15), and linear independence of \(u_1, \dots, u_m\) forces \(a_1 = \dots = a_m = 0\).

Extend this linearly independent list to a basis of \(V\) (2.32):

\begin{equation*} Su_1, \dots, Su_m, w_1, \dots, w_q . \end{equation*}

Every basis of \(V\) has length \(n\) (2.34), so \(m + q = n = m + p\), giving \(q = p\). Thus the two bases

\begin{equation*} u_1, \dots, u_m, v_1, \dots, v_p \qquad \text{and} \qquad Su_1, \dots, Su_m, w_1, \dots, w_p \end{equation*}

have the same length, and we may define \(T \in \mathcal{L}(V)\) by the linear map lemma (3.4) via

\begin{equation*} T u_k = S u_k \ \ (k = 1, \dots, m), \qquad T v_j = w_j \ \ (j = 1, \dots, p). \end{equation*}

The map \(T\) sends the basis \(u_1, \dots, u_m, v_1, \dots, v_p\) of \(V\) to the list \(Su_1, \dots, Su_m, w_1, \dots, w_p\), which is a basis of \(V\). By the equivalence (c) \(\Rightarrow\) (a) of Exercise 3 in this section, \(T\) is invertible. (Directly: by Exercise 10 in Section 3B, \(\operatorname{range} T\) is the span of the images of a basis, which here is the span of a basis of \(V\), hence equals \(V\); so \(T\) is surjective, hence injective by 3.65 since \(V\) is finite-dimensional, hence invertible by 3.63.)

Finally, \(T\) agrees with \(S\) on \(U\). Let \(u \in U\) and write \(u = a_1 u_1 + \dots + a_m u_m\) with \(a_1, \dots, a_m \in \mathbf{F}\), possible because \(u_1, \dots, u_m\) is a basis of \(U\). Then

\begin{equation*} Tu = a_1 Tu_1 + \dots + a_m Tu_m = a_1 Su_1 + \dots + a_m Su_m = S(a_1 u_1 + \dots + a_m u_m) = Su , \end{equation*}

using linearity of \(T\) and of \(S\). Hence \(Tu = Su\) for every \(u \in U\), completing the proof.

Note that \(V\) is not assumed to be finite-dimensional; only \(W\) is.

First suppose there is an invertible \(E \in \mathcal{L}(W)\) with \(S = ET\). If \(v \in \operatorname{null} T\), then \(Sv = E(Tv) = E0 = 0\), so \(v \in \operatorname{null} S\). Conversely, if \(v \in \operatorname{null} S\), then \(E(Tv) = Sv = 0\); since \(E\) is invertible it is injective (3.63), so \(\operatorname{null} E = \{0\}\) (3.15) and hence \(Tv = 0\), i.e. \(v \in \operatorname{null} T\). Therefore \(\operatorname{null} S = \operatorname{null} T\).

Now suppose \(\operatorname{null} S = \operatorname{null} T\). We build \(E\).

Step 1: an isomorphism from \(\operatorname{range} T\) onto \(\operatorname{range} S\). Define \(\varphi \colon \operatorname{range} T \to \operatorname{range} S\) by

\begin{equation*} \varphi(Tv) = Sv \qquad \text{for } v \in V . \end{equation*}

This is well defined: every element of \(\operatorname{range} T\) has the form \(Tv\) for some \(v \in V\), and if \(Tv_1 = Tv_2\) then \(v_1 - v_2 \in \operatorname{null} T = \operatorname{null} S\), so \(Sv_1 - Sv_2 = S(v_1 - v_2) = 0\), i.e. \(Sv_1 = Sv_2\). Thus the value assigned to a point of \(\operatorname{range} T\) does not depend on the choice of \(v\) representing it.

\(\varphi\) is linear: for \(v_1, v_2 \in V\) and \(\lambda \in \mathbf{F}\),

\begin{equation*} \varphi(Tv_1 + Tv_2) = \varphi(T(v_1 + v_2)) = S(v_1 + v_2) = Sv_1 + Sv_2 = \varphi(Tv_1) + \varphi(Tv_2), \end{equation*}

\begin{equation*} \varphi(\lambda Tv_1) = \varphi(T(\lambda v_1)) = S(\lambda v_1) = \lambda Sv_1 = \lambda \varphi(Tv_1). \end{equation*}

\(\varphi\) is surjective onto \(\operatorname{range} S\), since every element of \(\operatorname{range} S\) is \(Sv = \varphi(Tv)\) for some \(v \in V\). And \(\varphi\) is injective: if \(\varphi(Tv) = 0\) then \(Sv = 0\), so \(v \in \operatorname{null} S = \operatorname{null} T\), so \(Tv = 0\). Hence \(\operatorname{null} \varphi = \{0\}\) and \(\varphi\) is injective (3.15).

Therefore \(\varphi\) is an isomorphism from \(\operatorname{range} T\) onto \(\operatorname{range} S\) (3.63, 3.69). Both spaces are subspaces of the finite-dimensional space \(W\), hence finite-dimensional (2.25), and \(\dim \operatorname{range} T = \dim \operatorname{range} S\) (3.70).

Step 2: construction of \(E\). Let \(m = \dim \operatorname{range} T\) and \(n = \dim W\), and let \(u_1, \dots, u_m\) be a basis of \(\operatorname{range} T\) (2.31). Extend it to a basis

\begin{equation*} u_1, \dots, u_m, u_{m+1}, \dots, u_n \end{equation*}

of \(W\) (2.32).

The list \(\varphi u_1, \dots, \varphi u_m\) is a basis of \(\operatorname{range} S\): it spans \(\operatorname{range} S\) because \(\varphi\) is surjective onto \(\operatorname{range} S\) and \(u_1, \dots, u_m\) spans \(\operatorname{range} T\) (so every element of \(\operatorname{range} S\) is \(\varphi\) of a linear combination of the \(u_k\), hence a linear combination of the \(\varphi u_k\)); and it is linearly independent because if \(\sum_{k=1}^{m} a_k \varphi u_k = 0\) then \(\varphi\bigl(\sum_k a_k u_k\bigr) = 0\), so \(\sum_k a_k u_k = 0\) by injectivity of \(\varphi\), so all \(a_k = 0\).

Extend \(\varphi u_1, \dots, \varphi u_m\) to a basis of \(W\) (2.32):

\begin{equation*} \varphi u_1, \dots, \varphi u_m, y_{m+1}, \dots, y_n , \end{equation*}

where the number of adjoined vectors is \(n - m\) because every basis of \(W\) has length \(n\) (2.34).

By the linear map lemma (3.4) there is a unique \(E \in \mathcal{L}(W)\) with

\begin{equation*} E u_k = \varphi u_k \ \ (k = 1, \dots, m), \qquad E u_k = y_k \ \ (k = m+1, \dots, n). \end{equation*}

Then \(E\) maps the basis \(u_1, \dots, u_n\) of \(W\) onto the basis \(\varphi u_1, \dots, \varphi u_m, y_{m+1}, \dots, y_n\) of \(W\), so \(E\) is invertible by Exercise 3 in this section, part (c) \(\Rightarrow\) (a). (Directly: \(\operatorname{range} E\) is the span of the images of a basis (Exercise 10 in Section 3B), which is the span of a basis of \(W\), hence equals \(W\); so \(E\) is surjective, hence injective by 3.65 since \(W\) is finite-dimensional, hence invertible by 3.63.)

Step 3: \(S = ET\). Let \(v \in V\). Then \(Tv \in \operatorname{range} T\), so \(Tv = c_1 u_1 + \dots + c_m u_m\) for some \(c_1, \dots, c_m \in \mathbf{F}\). Hence

\begin{equation*} E(Tv) = c_1 E u_1 + \dots + c_m E u_m = c_1 \varphi u_1 + \dots + c_m \varphi u_m = \varphi(c_1 u_1 + \dots + c_m u_m) = \varphi(Tv) = Sv , \end{equation*}

using linearity of \(E\), the definition of \(E\) on \(u_1, \dots, u_m\), linearity of \(\varphi\), and the definition of \(\varphi\). Since \(v \in V\) was arbitrary, \(S = ET\) with \(E \in \mathcal{L}(W)\) invertible, as required.

Here \(V\) is finite-dimensional; \(W\) need not be. Let \(n = \dim V\).

First suppose there is an invertible \(E \in \mathcal{L}(V)\) with \(S = TE\). Then

\begin{equation*} \operatorname{range} S = \{ T(Ev) : v \in V \} = \{ Tu : u \in \operatorname{range} E \} = \{ Tu : u \in V \} = \operatorname{range} T, \end{equation*}

where the third equality holds because \(E\) is invertible, hence surjective (3.63), so \(\operatorname{range} E = V\).

Conversely, suppose \(\operatorname{range} S = \operatorname{range} T\).

Step 1: the two null spaces have the same dimension. Since \(V\) is finite-dimensional, the fundamental theorem of linear maps (3.21) applied to \(S\) and to \(T\) gives

\begin{equation*} \dim \operatorname{null} S = \dim V - \dim \operatorname{range} S = \dim V - \dim \operatorname{range} T = \dim \operatorname{null} T . \end{equation*}

Call this common dimension \(m\), and set \(r = n - m\). (All the spaces involved are finite-dimensional: \(\operatorname{null} S\) and \(\operatorname{null} T\) are subspaces of \(V\), and \(\operatorname{range} S\), \(\operatorname{range} T\) are finite-dimensional by 3.21.)

Step 2: adapted bases. Let \(u_1, \dots, u_m\) be a basis of \(\operatorname{null} S\) (2.31), and extend it to a basis

\begin{equation*} u_1, \dots, u_m, v_1, \dots, v_r \end{equation*}

of \(V\) (2.32); the number of adjoined vectors is \(n - m = r\) because every basis of \(V\) has length \(n\) (2.34).

Claim: \(Sv_1, \dots, Sv_r\) is a basis of \(\operatorname{range} S\). Spanning: for \(v \in V\) write \(v = \sum_{k=1}^{m} b_k u_k + \sum_{j=1}^{r} a_j v_j\); then, since \(Su_k = 0\) for each \(k\),

\begin{equation*} Sv = a_1 Sv_1 + \dots + a_r Sv_r , \end{equation*}

so \(\operatorname{range} S = \operatorname{span}(Sv_1, \dots, Sv_r)\). Linear independence: if \(\sum_{j=1}^{r} a_j Sv_j = 0\), then \(\sum_j a_j v_j \in \operatorname{null} S = \operatorname{span}(u_1, \dots, u_m)\), so \(\sum_j a_j v_j = \sum_k b_k u_k\) for some scalars \(b_k\); that is,

\begin{equation*} b_1 u_1 + \dots + b_m u_m - a_1 v_1 - \dots - a_r v_r = 0 . \end{equation*}

Since \(u_1, \dots, u_m, v_1, \dots, v_r\) is linearly independent, all coefficients vanish; in particular \(a_1 = \dots = a_r = 0\). This proves the claim.

Now let \(p_1, \dots, p_m\) be a basis of \(\operatorname{null} T\); such a basis exists because \(\operatorname{null} T\) is a subspace of the finite-dimensional space \(V\) and hence finite-dimensional (2.25, 2.31), and by Step 1 it indeed has length \(\dim \operatorname{null} T = m\).

Step 3: choice of preimages. For each \(j \in \{1, \dots, r\}\) we have \(Sv_j \in \operatorname{range} S = \operatorname{range} T\), so there exists \(q_j \in V\) with

\begin{equation*} T q_j = S v_j . \end{equation*}

Step 4: the list \(p_1, \dots, p_m, q_1, \dots, q_r\) is a basis of \(V\). Its length is \(m + r = n = \dim V\), so by 2.38 it suffices to prove linear independence. Suppose

\begin{equation*} b_1 p_1 + \dots + b_m p_m + a_1 q_1 + \dots + a_r q_r = 0 . \end{equation*}

Apply \(T\). Since \(Tp_k = 0\) for each \(k\) (the \(p_k\) lie in \(\operatorname{null} T\)) and \(Tq_j = Sv_j\), we get

\begin{equation*} a_1 Sv_1 + \dots + a_r Sv_r = 0 . \end{equation*}

By the claim in Step 2, \(Sv_1, \dots, Sv_r\) is linearly independent, so \(a_1 = \dots = a_r = 0\). The displayed equation then reduces to \(b_1 p_1 + \dots + b_m p_m = 0\), and linear independence of the basis \(p_1, \dots, p_m\) of \(\operatorname{null} T\) gives \(b_1 = \dots = b_m = 0\). Hence the list is linearly independent and therefore a basis of \(V\).

Step 5: construction of \(E\) and verification. By the linear map lemma (3.4) there is a unique \(E \in \mathcal{L}(V)\) with

\begin{equation*} E u_k = p_k \ \ (k = 1, \dots, m), \qquad E v_j = q_j \ \ (j = 1, \dots, r), \end{equation*}

since \(u_1, \dots, u_m, v_1, \dots, v_r\) is a basis of \(V\). This \(E\) carries that basis onto the list \(p_1, \dots, p_m, q_1, \dots, q_r\), which by Step 4 is a basis of \(V\); hence \(E\) is invertible by Exercise 3 in this section, part (c) \(\Rightarrow\) (a). (Directly: \(\operatorname{range} E\) is the span of the images of a basis (Exercise 10 in Section 3B), which is the span of a basis of \(V\), hence equals \(V\); so \(E\) is surjective, hence injective by 3.65 since \(V\) is finite-dimensional, hence invertible by 3.63.)

Finally, \(TE\) and \(S\) agree on the basis \(u_1, \dots, u_m, v_1, \dots, v_r\) of \(V\):

\begin{equation*} (TE)u_k = T p_k = 0 = S u_k \quad (k = 1, \dots, m), \qquad (TE)v_j = T q_j = S v_j \quad (j = 1, \dots, r), \end{equation*}

using \(p_k \in \operatorname{null} T\), \(u_k \in \operatorname{null} S\), and Step 3. Two linear maps that agree on a basis are equal, because every vector of \(V\) is a linear combination of that basis and both maps are linear. Therefore \(S = TE\) with \(E \in \mathcal{L}(V)\) invertible, completing the proof.

First suppose there exist invertible \(E_1 \in \mathcal{L}(V)\) and \(E_2 \in \mathcal{L}(W)\) with \(S = E_2 T E_1\). If \(v \in V\), then

\begin{equation*} Sv = 0 \iff E_2(T E_1 v) = 0 \iff T E_1 v = 0 \iff E_1 v \in \operatorname{null} T, \end{equation*}

where the middle equivalence holds because \(E_2\) is injective (by 3.63) and \(E_2 0 = 0\). Thus

\begin{equation*} \operatorname{null} S = \{v \in V : E_1 v \in \operatorname{null} T\} = \{E_1^{-1} u : u \in \operatorname{null} T\}. \end{equation*}

Hence the restriction of \(E_1^{-1}\) to \(\operatorname{null} T\) is a linear map from \(\operatorname{null} T\) onto \(\operatorname{null} S\); it is injective because \(E_1^{-1}\) is injective. So this restriction is an isomorphism from \(\operatorname{null} T\) onto \(\operatorname{null} S\) (by 3.63). Both \(\operatorname{null} S\) and \(\operatorname{null} T\) are subspaces of the finite-dimensional vector space \(V\) and hence are finite-dimensional (by 2.25), so 3.70 applies and gives \(\dim \operatorname{null} S = \dim \operatorname{null} T\).

Now suppose \(\dim \operatorname{null} S = \dim \operatorname{null} T\). Write \(n = \dim \operatorname{null} S = \dim \operatorname{null} T\) and \(m = \dim V\). By the fundamental theorem of linear maps (3.21),

\begin{equation*} \dim \operatorname{range} S = m - n = \dim \operatorname{range} T . \end{equation*}

Set \(k = m - n\).

Choose a basis \(u_1, \dots, u_n\) of \(\operatorname{null} S\) and extend it (by 2.32) to a basis

\begin{equation*} u_1, \dots, u_n, x_1, \dots, x_k \end{equation*}

of \(V\). Similarly choose a basis \(p_1, \dots, p_n\) of \(\operatorname{null} T\) and extend it to a basis \(p_1, \dots, p_n, y_1, \dots, y_k\) of \(V\).

The list \(S x_1, \dots, S x_k\) spans \(\operatorname{range} S\), because \(\operatorname{range} S\) is spanned by \(S u_1, \dots, S u_n, S x_1, \dots, S x_k\) and each \(S u_j = 0\). Since this spanning list has length \(k = \dim \operatorname{range} S\), it is a basis of \(\operatorname{range} S\) (by 2.42). For the same reason \(T y_1, \dots, T y_k\) is a basis of \(\operatorname{range} T\).

Extend \(S x_1, \dots, S x_k\) to a basis

\begin{equation*} S x_1, \dots, S x_k, w_1, \dots, w_r \end{equation*}

of \(W\), where \(r = \dim W - k\), and extend \(T y_1, \dots, T y_k\) to a basis \(T y_1, \dots, T y_k, z_1, \dots, z_r\) of \(W\) (both extensions use 2.32, and both lists have the same length because both start with \(k\) vectors).

By the linear map lemma (3.4) there is a unique \(E_1 \in \mathcal{L}(V)\) with

\begin{equation*} E_1 u_j = p_j \quad (j = 1, \dots, n), \qquad E_1 x_i = y_i \quad (i = 1, \dots, k), \end{equation*}

and a unique \(E_2 \in \mathcal{L}(W)\) with

\begin{equation*} E_2 (T y_i) = S x_i \quad (i = 1, \dots, k), \qquad E_2 z_j = w_j \quad (j = 1, \dots, r). \end{equation*}

Each of \(E_1, E_2\) maps a basis onto a basis, so each is surjective (its range contains a spanning list of the target space) and hence invertible by 3.65.

Finally, \(S\) and \(E_2 T E_1\) agree on the basis \(u_1, \dots, u_n, x_1, \dots, x_k\) of \(V\): for each \(j\),

\begin{equation*} E_2 T E_1 u_j = E_2 T p_j = E_2 0 = 0 = S u_j \end{equation*}

since \(p_j \in \operatorname{null} T\) and \(u_j \in \operatorname{null} S\); and for each \(i\),

\begin{equation*} E_2 T E_1 x_i = E_2 T y_i = S x_i . \end{equation*}

Two linear maps that agree on a basis of \(V\) are equal, so \(S = E_2 T E_1\), as desired.

Because \(V\) is finite-dimensional and \(\operatorname{null} T\) is a subspace of \(V\), there is a subspace \(U\) of \(V\) such that

\begin{equation*} V = \operatorname{null} T \oplus U \end{equation*}

by 2.33.

The function \(T|_U\) is a linear map from \(U\) to \(W\): it satisfies additivity and homogeneity because \(T\) does, and \(U\) is closed under addition and scalar multiplication.

\(T|_U\) is injective. Suppose \(u \in U\) and \(T|_U(u) = 0\). Then \(Tu = 0\), so \(u \in \operatorname{null} T\). Thus \(u \in \operatorname{null} T \cap U = \{0\}\), where the last equality holds because the sum \(\operatorname{null} T + U\) is a direct sum (by 1.46). Hence \(u = 0\), and so \(\operatorname{null}(T|_U) = \{0\}\), which means \(T|_U\) is injective (by 3.15).

\(T|_U\) is surjective. Suppose \(w \in W\). Because \(T\) is surjective, there exists \(v \in V\) with \(Tv = w\). Write \(v = n + u\) with \(n \in \operatorname{null} T\) and \(u \in U\). Then

\begin{equation*} w = Tv = Tn + Tu = 0 + Tu = T|_U(u). \end{equation*}

Thus \(\operatorname{range}(T|_U) = W\).

Because \(T|_U\) is injective and surjective, it is invertible (by 3.63), which is to say \(T|_U\) is an isomorphism of \(U\) onto \(W\).

(a) The zero map \(0 \in \mathcal{L}(V, W)\) satisfies \(\operatorname{null} 0 = V \supseteq U\), so \(0 \in \mathcal{E}\); in particular \(\mathcal{E}\) is nonempty.

Suppose \(S, T \in \mathcal{E}\). If \(u \in U\), then \((S + T)u = Su + Tu = 0 + 0 = 0\). Thus \(U \subseteq \operatorname{null}(S + T)\), so \(S + T \in \mathcal{E}\).

Suppose \(T \in \mathcal{E}\) and \(\lambda \in \mathbf{F}\). If \(u \in U\), then \((\lambda T)u = \lambda(Tu) = \lambda 0 = 0\). Thus \(\lambda T \in \mathcal{E}\).

Because \(\mathcal{E}\) contains the additive identity of \(\mathcal{L}(V, W)\) and is closed under addition and scalar multiplication, \(\mathcal{E}\) is a subspace of \(\mathcal{L}(V, W)\) (by 1.34).

(b) The formula is

\begin{equation*} \dim \mathcal{E} = (\dim V - \dim U)(\dim W). \end{equation*}

To prove this, note first that \(U\) is finite-dimensional (by 2.25, since \(U\) is a subspace of the finite-dimensional vector space \(V\)). Following the hint, define

\begin{equation*} \Phi \colon \mathcal{L}(V, W) \to \mathcal{L}(U, W) \quad \text{by} \quad \Phi(T) = T|_U . \end{equation*}

This makes sense: the restriction of a linear map \(V \to W\) to the subspace \(U\) is a linear map \(U \to W\). Furthermore \(\Phi\) is linear, because for all \(S, T \in \mathcal{L}(V, W)\), all \(\lambda \in \mathbf{F}\), and all \(u \in U\) we have

\begin{equation*} (S + T)|_U(u) = Su + Tu = \bigl(S|_U + T|_U\bigr)(u), \qquad (\lambda T)|_U(u) = \lambda \bigl(T|_U(u)\bigr), \end{equation*}

so \(\Phi(S + T) = \Phi(S) + \Phi(T)\) and \(\Phi(\lambda T) = \lambda \Phi(T)\).

The null space of \(\Phi\) is \(\mathcal{E}\). Indeed, \(\Phi(T) = 0\) means \(Tu = 0\) for every \(u \in U\), which says exactly that \(U \subseteq \operatorname{null} T\). Thus \(\operatorname{null} \Phi = \mathcal{E}\).

The range of \(\Phi\) is all of \(\mathcal{L}(U, W)\). To see this, let \(R \in \mathcal{L}(U, W)\). Choose a basis \(u_1, \dots, u_m\) of \(U\) and extend it to a basis \(u_1, \dots, u_m, v_1, \dots, v_n\) of \(V\) (by 2.32). By the linear map lemma (3.4) there exists \(T \in \mathcal{L}(V, W)\) with

\begin{equation*} Tu_j = R u_j \quad (j = 1, \dots, m), \qquad T v_k = 0 \quad (k = 1, \dots, n). \end{equation*}

The linear maps \(T|_U\) and \(R\) agree on the basis \(u_1, \dots, u_m\) of \(U\), hence they are equal. Thus \(\Phi(T) = R\), proving that \(\Phi\) is surjective. (The degenerate cases cause no trouble: if \(U = \{0\}\) then \(m = 0\) and the list \(u_1, \dots, u_m\) is empty, and \(T = 0\) works; if \(U = V\) then \(n = 0\) and \(T = R\) works.)

Because \(\mathcal{L}(V, W)\) is finite-dimensional (by 3.72), the fundamental theorem of linear maps (3.21) applied to \(\Phi\) gives

\begin{equation*} \dim \mathcal{L}(V, W) = \dim \operatorname{null} \Phi + \dim \operatorname{range} \Phi = \dim \mathcal{E} + \dim \mathcal{L}(U, W). \end{equation*}

Now 3.72 evaluates the two outer terms: \(\dim \mathcal{L}(V, W) = (\dim V)(\dim W)\) and \(\dim \mathcal{L}(U, W) = (\dim U)(\dim W)\). Therefore

\begin{equation*} \dim \mathcal{E} = (\dim V)(\dim W) - (\dim U)(\dim W) = (\dim V - \dim U)(\dim W), \end{equation*}

as claimed.

First suppose \(S\) and \(T\) are invertible. Then \(ST\) is invertible with \((ST)^{-1} = T^{-1}S^{-1}\), because

\begin{equation*} (ST)(T^{-1}S^{-1}) = S(TT^{-1})S^{-1} = SS^{-1} = I \end{equation*}

and

\begin{equation*} (T^{-1}S^{-1})(ST) = T^{-1}(S^{-1}S)T = T^{-1}T = I. \end{equation*}

(This is Exercise 2 in this section.) Note that this direction does not require \(V\) to be finite-dimensional.

Now suppose \(ST\) is invertible. We show separately that \(S\) is surjective and \(T\) is injective, and then invoke 3.65.

\(S\) is surjective. By 3.63 the invertible map \(ST\) is surjective, so \(\operatorname{range}(ST) = V\). But every vector of the form \((ST)v = S(Tv)\) lies in \(\operatorname{range} S\), so

\begin{equation*} V = \operatorname{range}(ST) \subseteq \operatorname{range} S \subseteq V. \end{equation*}

Hence \(\operatorname{range} S = V\), so \(S\) is surjective. Because \(V\) is finite-dimensional and \(S \in \mathcal{L}(V)\), surjectivity implies invertibility (by 3.65). Thus \(S\) is invertible.

\(T\) is injective. By 3.63 the invertible map \(ST\) is injective. Suppose \(v \in V\) and \(Tv = 0\). Then \((ST)v = S(Tv) = S0 = 0\), so \(v = 0\) by the injectivity of \(ST\) (using 3.15). Hence \(\operatorname{null} T = \{0\}\), so \(T\) is injective (by 3.15). Because \(V\) is finite-dimensional and \(T \in \mathcal{L}(V)\), injectivity implies invertibility (by 3.65). Thus \(T\) is invertible.

This completes the proof of both implications.

Finite-dimensionality is genuinely needed for the implication \(ST\) invertible \(\implies\) \(S\) and \(T\) invertible: on \(V = \mathbf{F}^\infty\), let \(S\) be the backward shift \(S(x_1, x_2, \dots) = (x_2, x_3, \dots)\) and let \(T\) be the forward shift \(T(x_1, x_2, \dots) = (0, x_1, x_2, \dots)\). Then \(ST = I\) is invertible, but neither \(S\) (not injective) nor \(T\) (not surjective) is invertible.

First we show that \(U\) is invertible. Suppose \(v \in V\) and \(Uv = 0\). Then

\begin{equation*} v = Iv = (STU)v = S\bigl(T(Uv)\bigr) = S(T0) = 0. \end{equation*}

Thus \(\operatorname{null} U = \{0\}\), so \(U\) is injective (by 3.15). Because \(V\) is finite-dimensional and \(U \in \mathcal{L}(V)\), this implies \(U\) is invertible (by 3.65).

Next we show that \(S\) is invertible. Every vector of the form \((STU)v = S\bigl((TU)v\bigr)\) lies in \(\operatorname{range} S\), so

\begin{equation*} V = \operatorname{range} I = \operatorname{range}(STU) \subseteq \operatorname{range} S \subseteq V, \end{equation*}

whence \(\operatorname{range} S = V\). Thus \(S\) is surjective, and therefore invertible (by 3.65).

Now multiply the equation \(STU = I\) on the left by \(S^{-1}\) and on the right by \(U^{-1}\):

\begin{equation*} T = S^{-1}(STU)U^{-1} = S^{-1} I U^{-1} = S^{-1}U^{-1}. \end{equation*}

Because \(S^{-1}\) and \(U^{-1}\) are invertible (Exercise 1 in this section), their product \(S^{-1}U^{-1}\) is invertible (Exercise 2 in this section). Hence \(T\) is invertible, and

\begin{equation*} T^{-1} = (S^{-1}U^{-1})^{-1} = (U^{-1})^{-1}(S^{-1})^{-1} = US, \end{equation*}

using Exercise 2 for the inverse of a product and Exercise 1 for \((A^{-1})^{-1} = A\).

As a direct check: \(T(US) = S^{-1}U^{-1}US = S^{-1}S = I\) and \((US)T = US S^{-1}U^{-1} = UU^{-1} = I\).

Take \(V = \mathbf{F}^\infty\), which is not finite-dimensional. Let \(B \in \mathcal{L}(V)\) be the backward shift

\begin{equation*} B(x_1, x_2, x_3, \dots) = (x_2, x_3, x_4, \dots) \end{equation*}

and let \(F \in \mathcal{L}(V)\) be the forward shift

\begin{equation*} F(x_1, x_2, x_3, \dots) = (0, x_1, x_2, \dots). \end{equation*}

Both are linear (see Example 3.3).

Now set

\begin{equation*} S = B, \qquad T = B, \qquad U = F^2 . \end{equation*}

Then \(STU = B^2 F^2\), and for every \((x_1, x_2, \dots) \in \mathbf{F}^\infty\),

\begin{equation*} B^2 F^2 (x_1, x_2, \dots) = B^2 (0, 0, x_1, x_2, \dots) = (x_1, x_2, \dots). \end{equation*}

Thus \(STU = I\).

However \(T = B\) is not invertible: \(B(1, 0, 0, \dots) = (0, 0, \dots)\), so \(\operatorname{null} B \neq \{0\}\) and \(B\) is not injective (by 3.15), hence not invertible (by 3.63). So the conclusion of Exercise 12 fails badly here.

The formula \(T^{-1} = US\) fails too, in the sense that \(US\) is not a two-sided inverse of \(T\): we have

\begin{equation*} US = F^2 B \quad \text{and} \quad T(US) = B F^2 B = F B, \end{equation*}

because \(BF = I\). But \(FB(1, 0, 0, \dots) = F(0, 0, \dots) = (0, 0, \dots) \neq (1, 0, 0, \dots)\), so \(T(US) \neq I\).

For a version in which \(T\) is invertible but the formula still fails, take \(S = B\), \(T = I\), \(U = F\). Then \(STU = BIF = BF = I\) and \(T^{-1} = I\), yet

\begin{equation*} US(x_1, x_2, x_3, \dots) = F(x_2, x_3, \dots) = (0, x_2, x_3, \dots), \end{equation*}

so \(US \neq I = T^{-1}\).

The statement is true; here is a proof.

Suppose \(RST\) is surjective, so \(\operatorname{range}(RST) = V\).

Step 1: \(R\) is invertible. Every vector of the form \((RST)v = R\bigl((ST)v\bigr)\) lies in \(\operatorname{range} R\), so

\begin{equation*} V = \operatorname{range}(RST) \subseteq \operatorname{range} R \subseteq V . \end{equation*}

Hence \(\operatorname{range} R = V\), so \(R\) is surjective. Because \(V\) is finite-dimensional and \(R \in \mathcal{L}(V)\), this implies \(R\) is invertible (by 3.65); in particular \(R\) is injective (by 3.63).

Step 2: \(ST\) is surjective. Let \(v \in V\). Because \(RST\) is surjective, there exists \(u \in V\) with

\begin{equation*} (RST)u = Rv . \end{equation*}

That is, \(R\bigl((ST)u\bigr) = Rv\). Since \(R\) is injective, \((ST)u = v\). Thus \(v \in \operatorname{range}(ST)\), and so \(\operatorname{range}(ST) = V\).

Step 3: \(S\) is surjective. Every vector of the form \((ST)u = S(Tu)\) lies in \(\operatorname{range} S\), so

\begin{equation*} V = \operatorname{range}(ST) \subseteq \operatorname{range} S \subseteq V, \end{equation*}

giving \(\operatorname{range} S = V\).

Step 4: \(S\) is injective. Because \(V\) is finite-dimensional and \(S \in \mathcal{L}(V)\) is surjective, 3.65 gives that \(S\) is invertible and hence injective.

Therefore \(S\) is injective, as claimed.

Remark: the finite-dimensional hypothesis is essential. On \(V = \mathbf{F}^\infty\) let \(B\) be the backward shift and \(F\) the forward shift. Take \(R = S = B\) and \(T = F^2\). Then \(RST = B^2F^2 = I\), which is surjective, but \(S = B\) is not injective, since \(B(1, 0, 0, \dots) = 0\).

First note that \(V\) is finite-dimensional: the list \(Tv_1, \ldots, Tv_m\) is a finite list of vectors in \(V\) that spans \(V\), which is exactly the definition of \(V\) being finite-dimensional.

Next, \(T\) is surjective. Indeed, each \(Tv_k\) lies in \(\operatorname{range} T\), and \(\operatorname{range} T\) is a subspace of \(V\) (by 3.18), so

\begin{equation*} V = \operatorname{span}(Tv_1, \ldots, Tv_m) \subseteq \operatorname{range} T \subseteq V . \end{equation*}

Hence \(\operatorname{range} T = V\).

Now apply 3.65 with \(W = V\) (the hypothesis \(\dim V = \dim W < \infty\) holds trivially): since \(T\) is surjective, \(T\) is injective.

Finally we show \(\operatorname{span}(v_1, \ldots, v_m) = V\). The inclusion \(\operatorname{span}(v_1, \ldots, v_m) \subseteq V\) is immediate. For the reverse inclusion, let \(v \in V\). Then \(Tv \in V\), so because \(Tv_1, \ldots, Tv_m\) spans \(V\) there exist \(a_1, \ldots, a_m \in \mathbf{F}\) with

\begin{equation*} Tv = a_1 Tv_1 + \cdots + a_m Tv_m = T(a_1 v_1 + \cdots + a_m v_m), \end{equation*}

where the second equality uses the linearity of \(T\). Because \(T\) is injective, this gives

\begin{equation*} v = a_1 v_1 + \cdots + a_m v_m \in \operatorname{span}(v_1, \ldots, v_m). \end{equation*}

Thus \(V \subseteq \operatorname{span}(v_1, \ldots, v_m)\), and therefore \(v_1, \ldots, v_m\) spans \(V\).

For \(k = 1, \ldots, n\) let \(e_k \in \mathbf{F}^{n,1}\) denote the \(n\)-by-\(1\) column vector whose \(k^{\text{th}}\) entry is \(1\) and whose other entries are \(0\). The list \(e_1, \ldots, e_n\) is a basis of \(\mathbf{F}^{n,1}\); indeed, for

\begin{equation*} x = \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix} \in \mathbf{F}^{n,1} \end{equation*}

we have \(x = x_1 e_1 + \cdots + x_n e_n\), and this representation is clearly unique because the coefficients are recovered as the entries of \(x\).

Now suppose \(T \in \mathcal{L}(\mathbf{F}^{n,1}, \mathbf{F}^{m,1})\). Each \(Te_k\) is an element of \(\mathbf{F}^{m,1}\), that is, an \(m\)-by-\(1\) column vector. Define an \(m\)-by-\(n\) matrix \(A\) by declaring its \(k^{\text{th}}\) column to be \(Te_k\); explicitly,

\begin{equation*} A_{j,k} = (Te_k)_j \qquad \text{for } j = 1, \ldots, m \text{ and } k = 1, \ldots, n, \end{equation*}

where \((Te_k)_j\) denotes the entry in row \(j\) of the column vector \(Te_k\). Thus \(A_{\cdot,k} = Te_k\) for each \(k\).

Let \(x \in \mathbf{F}^{n,1}\) have entries \(x_1, \ldots, x_n\). Using \(x = x_1 e_1 + \cdots + x_n e_n\) and the linearity of \(T\),

\begin{equation*} Tx = x_1 Te_1 + \cdots + x_n Te_n = x_1 A_{\cdot,1} + \cdots + x_n A_{\cdot,n}. \end{equation*}

On the other hand, 3.50 (linear combination of columns) states precisely that

\begin{equation*} Ax = x_1 A_{\cdot,1} + \cdots + x_n A_{\cdot,n}. \end{equation*}

Comparing the two displayed equations gives \(Tx = Ax\) for every \(x \in \mathbf{F}^{n,1}\), as desired.

Remark: this is the concrete instance of 3.76, which says \(\mathcal{M}(Tv) = \mathcal{M}(T)\mathcal{M}(v)\). Here we use the standard bases \(e_1, \ldots, e_n\) of \(\mathbf{F}^{n,1}\) and the analogous standard basis of \(\mathbf{F}^{m,1}\); with respect to these bases \(\mathcal{M}(v) = v\) for every column vector \(v\), so 3.76 reads \(Tx = \mathcal{M}(T)x\), and \(A = \mathcal{M}(T)\) is the matrix constructed above.

Write \(n = \dim V\). First note that \(\mathcal{A}\) really is a linear map from \(\mathcal{L}(V)\) to \(\mathcal{L}(V)\): if \(T_1, T_2 \in \mathcal{L}(V)\) and \(\lambda \in \mathbf{F}\), then for every \(v \in V\) we have \(\big(S(T_1 + T_2)\big)v = S(T_1 v + T_2 v) = S T_1 v + S T_2 v\) and \(\big(S(\lambda T_1)\big)v = S(\lambda T_1 v) = \lambda S T_1 v\), using the linearity of \(S\); hence \(\mathcal{A}(T_1 + T_2) = \mathcal{A}(T_1) + \mathcal{A}(T_2)\) and \(\mathcal{A}(\lambda T_1) = \lambda \mathcal{A}(T_1)\). Also, by 3.72 the space \(\mathcal{L}(V)\) is finite-dimensional with

\begin{equation*} \dim \mathcal{L}(V) = (\dim V)(\dim V) = n^2 . \end{equation*}

(a) Let \(U = \operatorname{null} S\), a subspace of \(V\); because \(V\) is finite-dimensional, so is \(U\) (by 2.25). I claim

\begin{equation*} \operatorname{null} \mathcal{A} = \{T \in \mathcal{L}(V) : \operatorname{range} T \subseteq U\}. \end{equation*}

Indeed, \(T \in \operatorname{null} \mathcal{A}\) means \(ST = 0\), which means \(S(Tv) = 0\) for every \(v \in V\), which means \(Tv \in \operatorname{null} S = U\) for every \(v \in V\), which means \(\operatorname{range} T \subseteq U\).

Now define a map

\begin{equation*} \Phi : \mathcal{L}(V, U) \to \mathcal{L}(V) \end{equation*}

by letting \(\Phi( R)\) be the linear map from \(V\) to \(V\) given by \(\big(\Phi( R)\big)(v) = R v\) for \(v \in V\); in other words, \(\Phi( R)\) is \(R\) with its target space enlarged from \(U\) to \(V\). Since \(U \subseteq V\), this makes sense, and \(\Phi( R)\) is linear because \(R\) is. The map \(\Phi\) is linear, since \(\Phi(R_1 + R_2)\) and \(\Phi(R_1) + \Phi(R_2)\) agree at every \(v \in V\), and similarly for scalar multiplication.

\(\Phi\) is injective: if \(\Phi( R) = 0\), then \(Rv = 0\) for every \(v \in V\), so \(R = 0\).

\(\operatorname{range} \Phi = \operatorname{null} \mathcal{A}\): for \(R \in \mathcal{L}(V,U)\) we have \(\operatorname{range} \Phi( R) \subseteq U\), so \(\Phi( R) \in \operatorname{null} \mathcal{A}\) by the claim above. Conversely, if \(T \in \operatorname{null} \mathcal{A}\), then \(\operatorname{range} T \subseteq U\), so the formula \(Rv = Tv\) defines an element \(R \in \mathcal{L}(V, U)\), and \(\Phi( R) = T\).

Thus \(\Phi\) is an isomorphism from \(\mathcal{L}(V,U)\) onto \(\operatorname{null} \mathcal{A}\), so these two spaces have the same dimension (by 3.70, or directly because an isomorphism carries a basis to a basis). Applying 3.72 to \(\mathcal{L}(V,U)\), which is legitimate because \(V\) and \(U\) are both finite-dimensional, gives

\begin{equation*} \dim \operatorname{null} \mathcal{A} = \dim \mathcal{L}(V, U) = (\dim V)(\dim U) = (\dim V)(\dim \operatorname{null} S), \end{equation*}

as desired.

(b) Apply the fundamental theorem of linear maps (3.21) to \(\mathcal{A} \in \mathcal{L}(\mathcal{L}(V))\), which is legitimate because \(\mathcal{L}(V)\) is finite-dimensional:

\begin{equation*} \dim \mathcal{L}(V) = \dim \operatorname{null} \mathcal{A} + \dim \operatorname{range} \mathcal{A}. \end{equation*}

Using \(\dim \mathcal{L}(V) = n^2\) and part (a),

\begin{equation*} \dim \operatorname{range} \mathcal{A} = n^2 - n \, (\dim \operatorname{null} S) = n \big( n - \dim \operatorname{null} S \big). \end{equation*}

Applying 3.21 to \(S \in \mathcal{L}(V)\) gives \(n = \dim \operatorname{null} S + \dim \operatorname{range} S\), so \(n - \dim \operatorname{null} S = \dim \operatorname{range} S\). Substituting,

\begin{equation*} \dim \operatorname{range} \mathcal{A} = (\dim V)(\dim \operatorname{range} S), \end{equation*}

as desired.

Here \(\mathbf{F}\) is regarded as a vector space over \(\mathbf{F}\) in the usual way. We exhibit an explicit isomorphism; no finite-dimensionality is needed.

Define \(\Phi : \mathcal{L}(\mathbf{F}, V) \to V\) by

\begin{equation*} \Phi(T) = T1 \end{equation*}

for \(T \in \mathcal{L}(\mathbf{F}, V)\), where \(1\) denotes the multiplicative identity of \(\mathbf{F}\), viewed as a vector in the vector space \(\mathbf{F}\).

\(\Phi\) is linear. If \(T, R \in \mathcal{L}(\mathbf{F}, V)\) then, by the definition of addition in \(\mathcal{L}(\mathbf{F}, V)\),

\begin{equation*} \Phi(T + R) = (T + R)(1) = T1 + R1 = \Phi(T) + \Phi( R), \end{equation*}

and if \(\lambda \in \mathbf{F}\) then, by the definition of scalar multiplication in \(\mathcal{L}(\mathbf{F}, V)\),

\begin{equation*} \Phi(\lambda T) = (\lambda T)(1) = \lambda (T1) = \lambda \Phi(T). \end{equation*}

\(\Phi\) is injective. Suppose \(\Phi(T) = 0\), that is, \(T1 = 0\). Then for every \(\lambda \in \mathbf{F}\),

\begin{equation*} T\lambda = T(\lambda \cdot 1) = \lambda \, T1 = \lambda \cdot 0 = 0 , \end{equation*}

using the homogeneity of \(T\). Hence \(T = 0\), so \(\operatorname{null} \Phi = \{0\}\) and \(\Phi\) is injective by 3.15.

\(\Phi\) is surjective. Let \(v \in V\) and define \(T_v : \mathbf{F} \to V\) by \(T_v \lambda = \lambda v\). Then \(T_v\) is linear, because for \(\lambda, \mu, \alpha \in \mathbf{F}\),

\begin{equation*} T_v(\lambda + \mu) = (\lambda + \mu) v = \lambda v + \mu v = T_v \lambda + T_v \mu \end{equation*}

and

\begin{equation*} T_v(\alpha \lambda) = (\alpha \lambda) v = \alpha (\lambda v) = \alpha \, T_v \lambda , \end{equation*}

where we used the vector space axioms for \(V\). Thus \(T_v \in \mathcal{L}(\mathbf{F}, V)\) and \(\Phi(T_v) = T_v 1 = 1 \cdot v = v\). Hence \(\operatorname{range} \Phi = V\).

Therefore \(\Phi\) is an invertible linear map from \(\mathcal{L}(\mathbf{F}, V)\) onto \(V\) (by 3.63, a linear map is invertible if and only if it is injective and surjective), that is, \(\Phi\) is an isomorphism in the sense of 3.69. Consequently \(V\) and \(\mathcal{L}(\mathbf{F}, V)\) are isomorphic vector spaces.

Remark: when \(V\) is finite-dimensional one can argue more quickly. By 3.72,

\begin{equation*} \dim \mathcal{L}(\mathbf{F}, V) = (\dim \mathbf{F})(\dim V) = 1 \cdot \dim V = \dim V, \end{equation*}

so 3.70 gives the isomorphism. The argument above has the advantage of also covering infinite-dimensional \(V\), and of producing an explicit isomorphism.

Throughout, \(\mathcal{M}(T, (v_1, \ldots, v_n))\) denotes the matrix of \(T\) with respect to the basis \(v_1, \ldots, v_n\) used for both the domain and the target, as in Section 3D. Recall that this matrix \(A\) is determined by

\begin{equation*} Tv_k = A_{1,k} v_1 + \cdots + A_{n,k} v_n \qquad \text{for } k = 1, \ldots, n . \end{equation*}

Let \(n = \dim V\). If \(n = 0\) then \(\mathcal{L}(V) = \{0\}\), so \(T = 0 = 0 \cdot I\) and \(T\) has the same (empty) matrix with respect to the only basis of \(V\); both sides of the equivalence hold. So assume \(n \geq 1\).

Suppose first that \(T = \lambda I\) for some \(\lambda \in \mathbf{F}\). Let \(v_1, \ldots, v_n\) be any basis of \(V\). Then

\begin{equation*} Tv_k = \lambda v_k = 0 \cdot v_1 + \cdots + 0 \cdot v_{k-1} + \lambda v_k + 0 \cdot v_{k+1} + \cdots + 0 \cdot v_n , \end{equation*}

so the \(k^{\text{th}}\) column of \(\mathcal{M}(T)\) has \(\lambda\) in row \(k\) and \(0\) elsewhere. Hence \(\mathcal{M}(T, (v_1, \ldots, v_n)) = \lambda I\), the \(n\)-by-\(n\) diagonal matrix with \(\lambda\) on the diagonal. This matrix does not depend on the chosen basis, so \(T\) has the same matrix with respect to every basis of \(V\).

Conversely, suppose there is an \(n\)-by-\(n\) matrix \(A\) with \(\mathcal{M}(T, (w_1, \ldots, w_n)) = A\) for every basis \(w_1, \ldots, w_n\) of \(V\).

Step 1: \(A\) is a diagonal matrix. Fix a basis \(v_1, \ldots, v_n\) of \(V\) and fix \(k \in \{1, \ldots, n\}\). Let

\begin{equation*} u_k = 2 v_k, \qquad u_i = v_i \ \text{ for } i \neq k . \end{equation*}

Then \(u_1, \ldots, u_n\) is again a basis of \(V\): it spans \(V\) because \(v_k = \tfrac12 u_k\), and it is linearly independent because a dependence relation among the \(u_i\) becomes one among the \(v_i\) after replacing \(u_k\) by \(2v_k\). (Here we use that \(2 \neq 0\) in \(\mathbf{F}\), which holds since \(\mathbf{F}\) is \(\mathbf{R}\) or \(\mathbf{C}\).)

Reading off column \(k\) of \(A\) with respect to the basis \(v_1, \ldots, v_n\) gives

\begin{equation*} Tv_k = \sum_{j=1}^{n} A_{j,k} v_j , \end{equation*}

while reading off column \(k\) of \(A\) with respect to the basis \(u_1, \ldots, u_n\) gives

\begin{equation*} 2 \, Tv_k = T u_k = \sum_{j=1}^{n} A_{j,k} u_j = 2 A_{k,k} v_k + \sum_{j \neq k} A_{j,k} v_j . \end{equation*}

Multiplying the first display by \(2\) and subtracting the second display yields

\begin{equation*} \sum_{j \neq k} A_{j,k} v_j = 0 . \end{equation*}

Since \(v_1, \ldots, v_n\) is linearly independent, \(A_{j,k} = 0\) for every \(j \neq k\). As \(k\) was arbitrary, \(A\) is diagonal; write \(A_{k,k} = \lambda_k\), so \(A = \operatorname{diag}(\lambda_1, \ldots, \lambda_n)\).

Note the consequence: because \(A\) is the matrix of \(T\) with respect to every basis, for every basis \(w_1, \ldots, w_n\) of \(V\) and every \(k\) we have

\begin{equation*} T w_k = \lambda_k w_k . \end{equation*}

Step 2: \(\lambda_1 = \lambda_2 = \cdots = \lambda_n\). If \(n = 1\) there is nothing to prove, so suppose \(n \geq 2\) and fix \(k \in \{2, \ldots, n\}\). Keep the basis \(v_1, \ldots, v_n\) and let \(w_1, \ldots, w_n\) be the list obtained from it by interchanging \(v_1\) and \(v_k\); reordering a basis clearly yields a basis, since spanning and linear independence do not depend on the order of the list. Since \(w_1 = v_k\), the observation at the end of Step 1 applied to the basis \(w_1, \ldots, w_n\) gives

\begin{equation*} T v_k = T w_1 = \lambda_1 w_1 = \lambda_1 v_k , \end{equation*}

while the same observation applied to the basis \(v_1, \ldots, v_n\) gives \(T v_k = \lambda_k v_k\). Subtracting,

\begin{equation*} (\lambda_1 - \lambda_k) v_k = 0 . \end{equation*}

A vector in a linearly independent list is nonzero, so \(v_k \neq 0\) and therefore \(\lambda_k = \lambda_1\).

Set \(\lambda = \lambda_1\). By Steps 1 and 2, \(T v_k = \lambda v_k = (\lambda I) v_k\) for \(k = 1, \ldots, n\). Two linear maps that agree on each vector of a basis agree everywhere: if \(v = a_1 v_1 + \cdots + a_n v_n\), then

\begin{equation*} Tv = a_1 Tv_1 + \cdots + a_n Tv_n = a_1 \lambda v_1 + \cdots + a_n \lambda v_n = \lambda v = (\lambda I) v . \end{equation*}

Hence \(T = \lambda I\), a scalar multiple of the identity operator.

This follows the pattern of Example 3.67: we produce an operator on a suitable \(\mathcal{P}_m(\mathbf{R})\), show it is injective, and then invoke 3.65 to conclude that it is surjective.

Fix a nonnegative integer \(m\) and define \(T : \mathcal{P}_m(\mathbf{R}) \to \mathcal{P}_m(\mathbf{R})\) by

\begin{equation*} (Tp)(x) = (x^2 + x) p^{\prime\prime}(x) + 2x \, p^{\prime}(x) + p(3). \end{equation*}

\(T\) maps \(\mathcal{P}_m(\mathbf{R})\) into \(\mathcal{P}_m(\mathbf{R})\). If \(\deg p \leq m\), then \(\deg p^{\prime\prime} \leq m - 2\) (interpreting this as \(p^{\prime\prime} = 0\) when \(m \leq 1\)), so \((x^2+x)p^{\prime\prime}\) has degree at most \(m\); likewise \(\deg p^{\prime} \leq m-1\), so \(2x p^{\prime}\) has degree at most \(m\); and \(p(3)\) is a constant. Hence \(Tp \in \mathcal{P}_m(\mathbf{R})\).

\(T\) is linear. Differentiation is linear, evaluation at \(3\) is linear, and multiplication by a fixed polynomial is linear; so for \(p, r \in \mathcal{P}_m(\mathbf{R})\) and \(\lambda \in \mathbf{R}\),

\begin{equation*} \big(T(p + r)\big)(x) = (x^2+x)\big(p^{\prime\prime}(x) + r^{\prime\prime}(x)\big) + 2x\big(p^{\prime}(x) + r^{\prime}(x)\big) + p(3) + r(3) = (Tp)(x) + (Tr)(x), \end{equation*}

and similarly \(T(\lambda p) = \lambda \, Tp\). Thus \(T \in \mathcal{L}(\mathcal{P}_m(\mathbf{R}))\).

\(T\) is injective. Suppose \(p \in \mathcal{P}_m(\mathbf{R})\) and \(p \neq 0\); we show \(Tp \neq 0\). Let \(d = \deg p\) and write

\begin{equation*} p(x) = a_d x^d + a_{d-1}x^{d-1} + \cdots + a_0, \qquad a_d \neq 0 . \end{equation*}

If \(d = 0\), then \(p\) is the nonzero constant \(a_0\), so \(p^{\prime} = p^{\prime\prime} = 0\) and \((Tp)(x) = p(3) = a_0 \neq 0\).

Now suppose \(d \geq 1\). We compute the coefficient of \(x^d\) in \(Tp\). First,

\begin{equation*} (x^2 + x) p^{\prime\prime}(x) = d(d-1) a_d x^{d} + (\text{terms of degree} < d). \end{equation*}

When \(d = 1\) this is the assertion that \((x^2+x)p^{\prime\prime}(x) = 0\), which is true because \(p^{\prime\prime} = 0\) for a polynomial of degree \(1\), and indeed \(d(d-1) = 0\) in that case. When \(d \geq 2\) we have \(p^{\prime\prime}(x) = d(d-1) a_d x^{d-2} + (\text{terms of degree} < d-2)\); multiplying by \(x^2\) produces \(d(d-1) a_d x^d\) plus terms of degree less than \(d\), and multiplying by \(x\) produces only terms of degree at most \(d-1\). Similarly

\begin{equation*} p^{\prime}(x) = d \, a_d x^{d-1} + (\text{terms of degree} < d-1), \quad \text{so} \quad 2x\,p^{\prime}(x) = 2d \, a_d x^{d} + (\text{terms of degree} < d). \end{equation*}

Finally \(p(3)\) contributes only to the constant term, and \(d \geq 1\). Hence the coefficient of \(x^d\) in \(Tp\) equals

\begin{equation*} d(d-1) a_d + 2d \, a_d = a_d \big( d^2 - d + 2d \big) = a_d \, d(d+1). \end{equation*}

Because \(a_d \neq 0\) and \(d \geq 1\), we have \(a_d \, d(d+1) \neq 0\), so \(Tp \neq 0\).

Thus \(\operatorname{null} T = \{0\}\), so \(T\) is injective by 3.15.

\(T\) is surjective. The space \(\mathcal{P}_m(\mathbf{R})\) is finite-dimensional (the list \(1, x, \ldots, x^m\) is a basis of it), and \(T\) is an operator on it, so trivially the domain and target have equal finite dimension. By 3.65, injectivity of \(T\) implies that \(T\) is surjective.

Conclusion. Let \(q \in \mathcal{P}(\mathbf{R})\). Choose a nonnegative integer \(m\) with \(q \in \mathcal{P}_m(\mathbf{R})\) (take \(m = \deg q\) if \(q \neq 0\), and \(m = 0\) if \(q = 0\)). By the surjectivity of the corresponding \(T\), there exists \(p \in \mathcal{P}_m(\mathbf{R}) \subseteq \mathcal{P}(\mathbf{R})\) with \(Tp = q\), that is,

\begin{equation*} q(x) = (x^2 + x) p^{\prime\prime}(x) + 2x \, p^{\prime}(x) + p(3) \end{equation*}

for all \(x \in \mathbf{R}\), as desired.

Define \(T : \mathbf{F}^n \to \mathbf{F}^n\) by

\begin{equation*} T(x_1, \ldots, x_n) = \Big( \sum_{k=1}^{n} A_{1,k} x_k , \ \ldots , \ \sum_{k=1}^{n} A_{n,k} x_k \Big). \end{equation*}

\(T\) is linear. For \(x = (x_1, \ldots, x_n)\) and \(y = (y_1, \ldots, y_n)\) in \(\mathbf{F}^n\) and \(j \in \{1, \ldots, n\}\), the \(j^{\text{th}}\) coordinate of \(T(x+y)\) is

\begin{equation*} \sum_{k=1}^{n} A_{j,k}(x_k + y_k) = \sum_{k=1}^{n} A_{j,k} x_k + \sum_{k=1}^{n} A_{j,k} y_k , \end{equation*}

which is the \(j^{\text{th}}\) coordinate of \(Tx + Ty\); hence \(T(x+y) = Tx + Ty\). Similarly, for \(\lambda \in \mathbf{F}\) the \(j^{\text{th}}\) coordinate of \(T(\lambda x)\) is \(\sum_{k=1}^{n} A_{j,k}(\lambda x_k) = \lambda \sum_{k=1}^{n} A_{j,k} x_k\), so \(T(\lambda x) = \lambda \, Tx\). Thus \(T \in \mathcal{L}(\mathbf{F}^n)\).

Reformulation of (a). A list \(x_1, \ldots, x_n\) of scalars solves the homogeneous system in (a) exactly when \(T(x_1, \ldots, x_n) = (0, \ldots, 0)\), that is, exactly when \((x_1, \ldots, x_n) \in \operatorname{null} T\). Hence (a) says \(\operatorname{null} T = \{0\}\), which by 3.15 is equivalent to \(T\) being injective.

Reformulation of (b). Given \(c_1, \ldots, c_n \in \mathbf{F}\), the system in (b) has a solution exactly when there is some \(x \in \mathbf{F}^n\) with \(Tx = (c_1, \ldots, c_n)\), that is, exactly when \((c_1, \ldots, c_n) \in \operatorname{range} T\). Hence (b) says \(\operatorname{range} T = \mathbf{F}^n\), which is exactly the statement that \(T\) is surjective (3.19).

Now \(\mathbf{F}^n\) is finite-dimensional with \(\dim \mathbf{F}^n = n\), and \(T\) maps \(\mathbf{F}^n\) to \(\mathbf{F}^n\), so the hypotheses of 3.65 are satisfied (with \(V = W = \mathbf{F}^n\)). That result states

\begin{equation*} T \text{ is invertible} \iff T \text{ is injective} \iff T \text{ is surjective}. \end{equation*}

In particular \(T\) is injective if and only if \(T\) is surjective. Combining this with the two reformulations above, (a) holds if and only if (b) holds.

Remark: the hypothesis that the number of equations equals the number of variables is essential; it is what makes the domain and target of \(T\) have the same dimension, which is required for 3.65. For example, the single equation \(x_1 + x_2 = 0\) in two variables has nontrivial solutions yet every right-hand side is attained, while the system \(x_1 = c_1\), \(x_1 = c_2\) in one variable has only the trivial solution when \(c_1 = c_2 = 0\) but is unsolvable when \(c_1 \neq c_2\).

Because \(V\) has a basis of length \(n\), the space \(V\) is finite-dimensional with \(\dim V = n\). Throughout this solution every matrix of an operator on \(V\) is taken with respect to the single basis \(v_1, \dots, v_n\), so that we may write \(\mathcal{M}(S)\) for \(\mathcal{M}(S, (v_1, \dots, v_n))\) when \(S \in \mathcal{L}(V)\). Write

\begin{equation*} A = \mathcal{M}(T) = \mathcal{M}\big(T, (v_1, \dots, v_n)\big). \end{equation*}

Two facts will be used repeatedly. First, the matrix of the identity operator with respect to a single basis is the identity matrix: indeed \(I v_k = v_k\) for each \(k\), so column \(k\) of \(\mathcal{M}(I)\) has a \(1\) in row \(k\) and \(0\) elsewhere, which says exactly that \(\mathcal{M}(I) = I\) (see 3.79 and the discussion following it). Second, applying 3.81 with \(U = V = W\) and with the basis \(v_1, \dots, v_n\) used in all three roles gives

\begin{equation*} \mathcal{M}(SR) = \mathcal{M}(S)\,\mathcal{M}( R) \qquad \text{for all } S, R \in \mathcal{L}(V). \end{equation*}

Suppose first that \(T\) is invertible. Then \(T^{-1} \in \mathcal{L}(V)\) and \(T T^{-1} = T^{-1} T = I\). Taking matrices and using the two facts above,

\begin{equation*} A\,\mathcal{M}(T^{-1}) = \mathcal{M}(T)\,\mathcal{M}(T^{-1}) = \mathcal{M}(T T^{-1}) = \mathcal{M}(I) = I \end{equation*}

and likewise

\begin{equation*} \mathcal{M}(T^{-1})\,A = \mathcal{M}(T^{-1} T) = \mathcal{M}(I) = I. \end{equation*}

Because \(\mathcal{M}(T^{-1})\) is a square matrix of the same size as \(A\), the definition 3.80 shows that \(A\) is invertible, with \(A^{-1} = \mathcal{M}(T^{-1})\). (Incidentally, this is exactly the statement 3.86, whose proof the book leaves as an exercise.)

Conversely, suppose \(A\) is invertible, and let \(B = A^{-1}\), so that \(B\) is an \(n\)-by-\(n\) matrix with \(AB = BA = I\). By 3.71 the map

\begin{equation*} \mathcal{M} \colon \mathcal{L}(V) \to \mathbf{F}^{n,n} \end{equation*}

is an isomorphism (apply 3.71 with \(W = V\) and with \(v_1, \dots, v_n\) used as the basis of both the domain and the target). In particular \(\mathcal{M}\) is surjective, so there exists \(S \in \mathcal{L}(V)\) with \(\mathcal{M}(S) = B\). Now

\begin{equation*} \mathcal{M}(ST) = \mathcal{M}(S)\,\mathcal{M}(T) = BA = I = \mathcal{M}(I) \end{equation*}

and

\begin{equation*} \mathcal{M}(TS) = \mathcal{M}(T)\,\mathcal{M}(S) = AB = I = \mathcal{M}(I). \end{equation*}

Because \(\mathcal{M}\) is also injective (again by 3.71), the two displayed equations give \(ST = I\) and \(TS = I\). Hence \(T\) is invertible (with \(T^{-1} = S\)), by the definition of invertibility.

If one prefers not to invoke injectivity of \(\mathcal{M}\) twice, note that only one of the two products is needed: from \(\mathcal{M}(ST) = \mathcal{M}(I)\) we get \(ST = I\), and then 3.68 (applied with \(W = V\), legitimate since \(V\) is finite-dimensional) yields \(TS = I\) as well.

Combining the two directions gives \(\mathcal{M}(T, (v_1, \dots, v_n))\) invertible \(\iff\) \(T\) invertible, as desired.

First note that such a \(T\) exists and is unique: by the linear map lemma (3.4) there is exactly one \(T \in \mathcal{L}(V)\) with \(T v_k = u_k\) for each \(k\), since \(v_1, \dots, v_n\) is a basis of \(V\). So the statement to be proved concerns a well-defined operator.

Both sides of the asserted equation are \(n\)-by-\(n\) matrices, so it suffices to show that corresponding entries agree. Let

\begin{equation*} A = \mathcal{M}\big(T, (v_1, \dots, v_n)\big) = \mathcal{M}\big(T, (v_1, \dots, v_n), (v_1, \dots, v_n)\big), \end{equation*}

\begin{equation*} C = \mathcal{M}\big(I, (u_1, \dots, u_n), (v_1, \dots, v_n)\big). \end{equation*}

By the definition of the matrix of a linear map (3.31), applied to \(T\) with the basis \(v_1, \dots, v_n\) of the domain and the same basis \(v_1, \dots, v_n\) of the target, the entries \(A_{j,k}\) of \(A\) are the unique scalars satisfying

\begin{equation*} T v_k = A_{1,k} v_1 + \cdots + A_{n,k} v_n \qquad \text{for each } k = 1, \dots, n. \end{equation*}

Again by 3.31, applied now to the identity operator \(I \in \mathcal{L}(V)\) with the basis \(u_1, \dots, u_n\) of the domain and the basis \(v_1, \dots, v_n\) of the target, the entries \(C_{j,k}\) of \(C\) are the unique scalars satisfying

\begin{equation*} u_k = I u_k = C_{1,k} v_1 + \cdots + C_{n,k} v_n \qquad \text{for each } k = 1, \dots, n. \end{equation*}

Fix \(k \in \{1, \dots, n\}\). By hypothesis \(T v_k = u_k\), so the two displayed equations give

\begin{equation*} A_{1,k} v_1 + \cdots + A_{n,k} v_n = T v_k = u_k = C_{1,k} v_1 + \cdots + C_{n,k} v_n. \end{equation*}

Because \(v_1, \dots, v_n\) is a basis of \(V\), every vector of \(V\), in particular \(u_k\), has a unique representation as a linear combination of \(v_1, \dots, v_n\) (this is the criterion for basis, 2.28). Hence

\begin{equation*} A_{j,k} = C_{j,k} \qquad \text{for each } j = 1, \dots, n. \end{equation*}

Since \(k\) was arbitrary, all entries of \(A\) and \(C\) agree, so \(A = C\), which is the desired equation

\begin{equation*} \mathcal{M}\big(T, (v_1, \dots, v_n)\big) = \mathcal{M}\big(I, (u_1, \dots, u_n), (v_1, \dots, v_n)\big). \end{equation*}

Two remarks tie this to the surrounding theory. As the book notes just before 3.82, column \(k\) of \(\mathcal{M}(I, (u_1, \dots, u_n), (v_1, \dots, v_n))\) consists precisely of the scalars needed to write \(u_k\) in terms of \(v_1, \dots, v_n\); the content of this exercise is that these are also the scalars needed to write \(T v_k\) in terms of \(v_1, \dots, v_n\), which is immediate once \(T v_k = u_k\). Moreover, since 3.82 tells us that \(\mathcal{M}(I, (u_1, \dots, u_n), (v_1, \dots, v_n))\) is invertible, this exercise combined with Exercise 22 of this section shows that the operator \(T\) carrying the basis \(v_1, \dots, v_n\) to the basis \(u_1, \dots, u_n\) is invertible, as one also sees directly from Exercise 3 of this section.

Let \(n\) be the common size of \(A\) and \(B\), so \(A, B \in \mathbf{F}^{n,n}\). The strategy is to convert the matrix hypothesis into a statement about operators on an \(n\)-dimensional vector space, where 3.68 applies, and then convert back.

Let \(V = \mathbf{F}^n\), with standard basis \(e_1, \dots, e_n\); thus \(\dim V = n\) and \(V\) is finite-dimensional. All matrices of operators on \(V\) below are taken with respect to this single basis, used both as the basis of the domain and as the basis of the target, so we write \(\mathcal{M}( R)\) for \(\mathcal{M}(R, (e_1, \dots, e_n), (e_1, \dots, e_n))\) when \(R \in \mathcal{L}(V)\).

By 3.71 (applied with \(W = V\) and with \(e_1, \dots, e_n\) as the basis in both roles), the map

\begin{equation*} \mathcal{M} \colon \mathcal{L}(\mathbf{F}^n) \to \mathbf{F}^{n,n} \end{equation*}

is an isomorphism; in particular it is injective and surjective. By surjectivity there exist \(S, T \in \mathcal{L}(\mathbf{F}^n)\) with

\begin{equation*} \mathcal{M}(S) = A \qquad \text{and} \qquad \mathcal{M}(T) = B. \end{equation*}

(Explicitly, by the linear map lemma 3.4 one may take \(S\) to be the unique operator with \(S e_k = \sum_{j=1}^{n} A_{j,k} e_j\) for each \(k\), and similarly for \(T\) with the entries of \(B\).)

Recall also that \(\mathcal{M}(I) = I\): the identity operator sends each \(e_k\) to itself, so by 3.31 its matrix with respect to a single basis has \(1\)’s on the diagonal and \(0\)’s elsewhere, which is the identity matrix 3.79.

Now apply 3.81 with \(U = V = W = \mathbf{F}^n\) and with \(e_1, \dots, e_n\) used as all three bases. This gives \(\mathcal{M}(ST) = \mathcal{M}(S)\mathcal{M}(T)\), hence

\begin{equation*} \mathcal{M}(ST) = AB = I = \mathcal{M}(I). \end{equation*}

Because \(\mathcal{M}\) is injective, we conclude that

\begin{equation*} ST = I, \end{equation*}

where now \(I\) denotes the identity operator on \(\mathbf{F}^n\).

Since \(\mathbf{F}^n\) is finite-dimensional and \(S, T \in \mathcal{L}(\mathbf{F}^n)\) (so the two spaces involved trivially have the same dimension), 3.68 applies and yields

\begin{equation*} TS = I. \end{equation*}

For completeness, here is that argument in this special case: if \(v \in \mathbf{F}^n\) and \(Tv = 0\), then \(v = Iv = (ST)v = S(Tv) = S(0) = 0\), so \(\operatorname{null} T = \{0\}\) and \(T\) is injective by 3.15; because \(T\) is an operator on the finite-dimensional space \(\mathbf{F}^n\), injectivity forces invertibility by 3.65; multiplying \(ST = I\) on the right by \(T^{-1}\) gives \(S = T^{-1}\), and therefore \(TS = T T^{-1} = I\).

Finally convert back to matrices. Using 3.81 once more,

\begin{equation*} BA = \mathcal{M}(T)\,\mathcal{M}(S) = \mathcal{M}(TS) = \mathcal{M}(I) = I. \end{equation*}

Thus \(BA = I\), as desired. In particular \(AB = BA = I\), so \(A\) is an invertible matrix with \(A^{-1} = B\) in the sense of 3.80: the hypothesis \(AB = I\) alone already forces \(B\) to be a two-sided inverse of \(A\).

The finiteness of the size is essential to this argument, and not merely an artifact of the proof. For linear maps on infinite-dimensional spaces the analogous statement fails: with \(V = \mathbf{F}^{\infty}\), let \(T\) be the forward shift \(T(x_1, x_2, \dots) = (0, x_1, x_2, \dots)\) and let \(S\) be the backward shift \(S(x_1, x_2, x_3, \dots) = (x_2, x_3, \dots)\). Then \(ST = I\) while \(TS \neq I\), since \(TS(1, 0, 0, \dots) = 0\). This is the phenomenon behind Exercise 13 of this section; for square matrices it cannot happen because \(\mathbf{F}^{n}\) has finite dimension.

Write \(G = \{(v, Tv) : v \in V\}\) for the graph of \(T\). The key observation, used repeatedly below, is that for \(v \in V\) and \(w \in W\) we have

\begin{equation*} (v, w) \in G \iff w = Tv . \end{equation*}

Indeed, every element of \(G\) has the form \((u, Tu)\) for some \(u \in V\), and \((u, Tu) = (v,w)\) forces \(u = v\) and hence \(w = Tv\); conversely, if \(w = Tv\) then \((v,w) = (v, Tv) \in G\). So the first slot of an element of \(G\) determines the second slot. Note that this uses only that \(T\) is a function from \(V\) to \(W\), not that it is linear.

Recall (1.34) that a subset of a vector space is a subspace exactly when it contains the additive identity and is closed under addition and under scalar multiplication. The additive identity of \(V \times W\) is \((0,0)\), by 3.89 and the remark following it.

First suppose \(T\) is a linear map. Then \(T0 = 0\) by 3.10, so \((0,0) = (0, T0) \in G\). If \((u, Tu)\) and \((v, Tv)\) are elements of \(G\), then by additivity of \(T\),

\begin{equation*} (u, Tu) + (v, Tv) = (u + v, Tu + Tv) = (u + v, T(u+v)) \in G, \end{equation*}

so \(G\) is closed under addition. If \(\lambda \in \mathbf{F}\) and \((v, Tv) \in G\), then by homogeneity of \(T\),

\begin{equation*} \lambda (v, Tv) = (\lambda v, \lambda Tv) = (\lambda v, T(\lambda v)) \in G, \end{equation*}

so \(G\) is closed under scalar multiplication. Hence \(G\) is a subspace of \(V \times W\).

Conversely, suppose \(G\) is a subspace of \(V \times W\). Let \(u, v \in V\). Then \((u, Tu) \in G\) and \((v, Tv) \in G\), so closure of \(G\) under addition gives

\begin{equation*} (u + v,\; Tu + Tv) = (u, Tu) + (v, Tv) \in G . \end{equation*}

By the key observation applied to the vector \(u + v\) in the first slot, this says exactly that \(Tu + Tv = T(u+v)\). Thus \(T\) is additive.

Now let \(\lambda \in \mathbf{F}\) and \(v \in V\). Then \((v, Tv) \in G\), so closure of \(G\) under scalar multiplication gives

\begin{equation*} (\lambda v,\; \lambda Tv) = \lambda (v, Tv) \in G . \end{equation*}

By the key observation applied to the vector \(\lambda v\) in the first slot, this says exactly that \(\lambda Tv = T(\lambda v)\). Thus \(T\) is homogeneous.

Hence \(T\) satisfies both conditions in the definition of a linear map (3.1), so \(T\) is linear.

(Note that in the reverse direction we never needed the hypothesis \((0,0) \in G\); closure under addition and under scalar multiplication already forced additivity and homogeneity of \(T\).)

Fix \(k \in \{1, \ldots, m\}\) and define the \(k\)-th coordinate projection

\begin{equation*} P_k : V_1 \times \cdots \times V_m \to V_k, \qquad P_k(v_1, \ldots, v_m) = v_k . \end{equation*}

First, \(P_k\) is linear. Addition and scalar multiplication on the product are defined slotwise (3.87), so for \((u_1, \ldots, u_m)\) and \((v_1, \ldots, v_m)\) in the product and \(\lambda \in \mathbf{F}\),

\begin{equation*} \begin{aligned} P_k\big((u_1, \ldots, u_m) + (v_1, \ldots, v_m)\big) &= P_k(u_1 + v_1, \ldots, u_m + v_m) = u_k + v_k \\ &= P_k(u_1, \ldots, u_m) + P_k(v_1, \ldots, v_m), \end{aligned} \end{equation*}

and

\begin{equation*} P_k\big(\lambda(v_1, \ldots, v_m)\big) = P_k(\lambda v_1, \ldots, \lambda v_m) = \lambda v_k = \lambda P_k(v_1, \ldots, v_m). \end{equation*}

So \(P_k \in \mathcal{L}(V_1 \times \cdots \times V_m,\; V_k)\).

Second, \(P_k\) is surjective: given \(v \in V_k\), the element

\begin{equation*} (0, \ldots, 0, v, 0, \ldots, 0) \in V_1 \times \cdots \times V_m \end{equation*}

(with \(v\) in slot \(k\) and the additive identity of \(V_j\) in slot \(j\) for \(j \neq k\)) is mapped by \(P_k\) to \(v\). Hence \(\operatorname{range} P_k = V_k\).

Now apply the fundamental theorem of linear maps (3.21) to \(P_k\). Its hypothesis is satisfied because the domain \(V_1 \times \cdots \times V_m\) is finite-dimensional by assumption. The theorem asserts in particular that \(\operatorname{range} P_k\) is finite-dimensional. Since \(\operatorname{range} P_k = V_k\), we conclude that \(V_k\) is finite-dimensional.

As \(k\) was arbitrary, \(V_k\) is finite-dimensional for each \(k = 1, \ldots, m\).

(Once each \(V_k\) is known to be finite-dimensional, 3.92 applies and gives the sharper statement \(\dim V_1 + \cdots + \dim V_m = \dim(V_1 \times \cdots \times V_m)\); in particular each \(\dim V_k\) is at most \(\dim(V_1 \times \cdots \times V_m)\).)

Both objects are vector spaces: each \(\mathcal{L}(V_k, W)\) and \(\mathcal{L}(V_1 \times \cdots \times V_m, W)\) is a vector space by 3.6, and a product of vector spaces is a vector space by 3.89 (the product \(V_1 \times \cdots \times V_m\) is a vector space by 3.89 as well). Write

\begin{equation*} \mathcal{V} = V_1 \times \cdots \times V_m . \end{equation*}

For \(k = 1, \ldots, m\) define the \(k\)-th inclusion

\begin{equation*} \iota_k : V_k \to \mathcal{V}, \qquad \iota_k(v) = (0, \ldots, 0, v, 0, \ldots, 0), \end{equation*}

with \(v\) in slot \(k\) and zeros elsewhere. Because the operations on \(\mathcal{V}\) are slotwise (3.87), each \(\iota_k\) is linear. Note also the identity

\begin{equation*} (v_1, \ldots, v_m) = \iota_1(v_1) + \cdots + \iota_m(v_m) \qquad \text{for all } (v_1, \ldots, v_m) \in \mathcal{V}, \end{equation*}

which is immediate from slotwise addition.

Define

\begin{equation*} \Phi : \mathcal{L}(\mathcal{V}, W) \to \mathcal{L}(V_1, W) \times \cdots \times \mathcal{L}(V_m, W), \qquad \Phi(T) = (T \circ \iota_1, \ldots, T \circ \iota_m). \end{equation*}

This makes sense: each \(T \circ \iota_k\) is a composition of linear maps, hence lies in \(\mathcal{L}(V_k, W)\) by 3.7 (or directly from the definitions).

\(\Phi\) is linear. Let \(S, T \in \mathcal{L}(\mathcal{V}, W)\) and \(\lambda \in \mathbf{F}\). For each \(k\) and each \(v \in V_k\), the definition of addition on \(\mathcal{L}(\mathcal{V}, W)\) (3.5) gives

\begin{equation*} \big((S + T) \circ \iota_k\big)(v) = (S+T)(\iota_k v) = S(\iota_k v) + T(\iota_k v) = (S \circ \iota_k)(v) + (T \circ \iota_k)(v), \end{equation*}

so \((S+T) \circ \iota_k = S \circ \iota_k + T \circ \iota_k\). Since addition on the product space is slotwise, \(\Phi(S+T) = \Phi(S) + \Phi(T)\). Similarly \(\big((\lambda T) \circ \iota_k\big)(v) = \lambda T(\iota_k v) = \lambda (T \circ \iota_k)(v)\), so \(\Phi(\lambda T) = \lambda \Phi(T)\).

\(\Phi\) is injective. Suppose \(\Phi(T) = (0, \ldots, 0)\), that is, \(T \circ \iota_k = 0\) for every \(k\). Then for any \((v_1, \ldots, v_m) \in \mathcal{V}\), using the identity above and linearity of \(T\),

\begin{equation*} T(v_1, \ldots, v_m) = T\big(\iota_1(v_1) + \cdots + \iota_m(v_m)\big) = (T \circ \iota_1)(v_1) + \cdots + (T \circ \iota_m)(v_m) = 0 . \end{equation*}

Hence \(T = 0\), so \(\operatorname{null} \Phi = \{0\}\) and \(\Phi\) is injective by 3.15.

\(\Phi\) is surjective. Let \((T_1, \ldots, T_m) \in \mathcal{L}(V_1, W) \times \cdots \times \mathcal{L}(V_m, W)\). Define \(T : \mathcal{V} \to W\) by

\begin{equation*} T(v_1, \ldots, v_m) = T_1 v_1 + \cdots + T_m v_m . \end{equation*}

Then \(T\) is linear: for \((u_1, \ldots, u_m), (v_1, \ldots, v_m) \in \mathcal{V}\),

\begin{equation*} \begin{aligned} T\big((u_1, \ldots, u_m) + (v_1, \ldots, v_m)\big) &= T_1(u_1 + v_1) + \cdots + T_m(u_m + v_m) \\ &= (T_1 u_1 + \cdots + T_m u_m) + (T_1 v_1 + \cdots + T_m v_m), \end{aligned} \end{equation*}

using additivity of each \(T_k\) and commutativity of addition in \(W\); and for \(\lambda \in \mathbf{F}\),

\begin{equation*} T\big(\lambda(v_1, \ldots, v_m)\big) = T_1(\lambda v_1) + \cdots + T_m(\lambda v_m) = \lambda\big(T_1 v_1 + \cdots + T_m v_m\big) . \end{equation*}

So \(T \in \mathcal{L}(\mathcal{V}, W)\). Finally, for each \(k\) and each \(v \in V_k\),

\begin{equation*} (T \circ \iota_k)(v) = T(0, \ldots, 0, v, 0, \ldots, 0) = T_1 0 + \cdots + T_k v + \cdots + T_m 0 = T_k v, \end{equation*}

where we used \(T_j 0 = 0\) (3.10). Thus \(T \circ \iota_k = T_k\) for every \(k\), i.e. \(\Phi(T) = (T_1, \ldots, T_m)\).

Being linear, injective, and surjective, \(\Phi\) is invertible by 3.63, hence an isomorphism (3.69). Therefore \(\mathcal{L}(V_1 \times \cdots \times V_m, W)\) and \(\mathcal{L}(V_1, W) \times \cdots \times \mathcal{L}(V_m, W)\) are isomorphic vector spaces. No finite-dimensionality was used anywhere.

Write \(\mathcal{W} = W_1 \times \cdots \times W_m\), a vector space by 3.89. Each \(\mathcal{L}(V, W_k)\) and \(\mathcal{L}(V, \mathcal{W})\) is a vector space by 3.6, and so is the product \(\mathcal{L}(V, W_1) \times \cdots \times \mathcal{L}(V, W_m)\) by 3.89.

For \(k = 1, \ldots, m\) let

\begin{equation*} P_k : \mathcal{W} \to W_k, \qquad P_k(w_1, \ldots, w_m) = w_k \end{equation*}

be the \(k\)-th coordinate projection; it is linear because the operations on \(\mathcal{W}\) are slotwise (3.87), as verified in Exercise 2 of this section. Note that

\begin{equation*} w = \big(P_1 w, \ldots, P_m w\big) \qquad \text{for all } w \in \mathcal{W}. \end{equation*}

Call this identity the reconstruction identity.

Define

\begin{equation*} \Psi : \mathcal{L}(V, \mathcal{W}) \to \mathcal{L}(V, W_1) \times \cdots \times \mathcal{L}(V, W_m), \qquad \Psi(T) = (P_1 \circ T, \ldots, P_m \circ T). \end{equation*}

Each \(P_k \circ T\) is a composition of linear maps and hence lies in \(\mathcal{L}(V, W_k)\) by 3.7, so \(\Psi\) indeed maps into the stated space. Concretely, if \(Tv = (T_1 v, \ldots, T_m v)\) then \(\Psi(T) = (T_1, \ldots, T_m)\): the map \(\Psi\) just records the \(m\) component maps of \(T\).

\(\Psi\) is linear. For \(S, T \in \mathcal{L}(V, \mathcal{W})\), \(\lambda \in \mathbf{F}\), \(v \in V\), and each \(k\), the definition of addition and scalar multiplication on \(\mathcal{L}(V, \mathcal{W})\) (3.5) together with linearity of \(P_k\) gives

\begin{equation*} \big(P_k \circ (S+T)\big)(v) = P_k(Sv + Tv) = P_k(Sv) + P_k(Tv) = (P_k \circ S)(v) + (P_k \circ T)(v) \end{equation*}

and

\begin{equation*} \big(P_k \circ (\lambda T)\big)(v) = P_k(\lambda\, Tv) = \lambda\, P_k(Tv) = \lambda (P_k \circ T)(v). \end{equation*}

Since the operations on the product of the spaces \(\mathcal{L}(V, W_k)\) are slotwise, these identities say \(\Psi(S+T) = \Psi(S) + \Psi(T)\) and \(\Psi(\lambda T) = \lambda \Psi(T)\).

\(\Psi\) is injective. Suppose \(\Psi(T) = (0, \ldots, 0)\), so \(P_k(Tv) = 0\) for every \(k\) and every \(v \in V\). By the reconstruction identity applied to \(w = Tv\),

\begin{equation*} Tv = \big(P_1(Tv), \ldots, P_m(Tv)\big) = (0, \ldots, 0), \end{equation*}

which is the additive identity of \(\mathcal{W}\). Hence \(T = 0\), so \(\operatorname{null} \Psi = \{0\}\) and \(\Psi\) is injective by 3.15.

\(\Psi\) is surjective. Let \((T_1, \ldots, T_m) \in \mathcal{L}(V, W_1) \times \cdots \times \mathcal{L}(V, W_m)\) and define \(T : V \to \mathcal{W}\) by

\begin{equation*} Tv = (T_1 v, \ldots, T_m v). \end{equation*}

Then \(T\) is linear: for \(u, v \in V\), using additivity of each \(T_k\) and slotwise addition in \(\mathcal{W}\),

\begin{equation*} T(u+v) = \big(T_1(u+v), \ldots, T_m(u+v)\big) = (T_1 u + T_1 v, \ldots, T_m u + T_m v) = Tu + Tv, \end{equation*}

and for \(\lambda \in \mathbf{F}\),

\begin{equation*} T(\lambda v) = (T_1(\lambda v), \ldots, T_m(\lambda v)) = (\lambda T_1 v, \ldots, \lambda T_m v) = \lambda\, Tv . \end{equation*}

So \(T \in \mathcal{L}(V, \mathcal{W})\), and by construction \((P_k \circ T)(v) = T_k v\) for all \(v\) and all \(k\), i.e. \(\Psi(T) = (T_1, \ldots, T_m)\).

Thus \(\Psi\) is a linear map that is injective and surjective, hence invertible by 3.63 and therefore an isomorphism (3.69). Consequently \(\mathcal{L}(V, W_1 \times \cdots \times W_m)\) and \(\mathcal{L}(V, W_1) \times \cdots \times \mathcal{L}(V, W_m)\) are isomorphic vector spaces. Again no finite-dimensionality hypothesis was needed.

Both \(V^m\) (by 3.89) and \(\mathcal{L}(\mathbf{F}^m, V)\) (by 3.6) are vector spaces over \(\mathbf{F}\).

Let \(e_1, \ldots, e_m\) be the standard basis of \(\mathbf{F}^m\), so \(e_k\) is the list of length \(m\) with \(1\) in slot \(k\) and \(0\) elsewhere; thus

\begin{equation*} (x_1, \ldots, x_m) = x_1 e_1 + \cdots + x_m e_m \qquad \text{for all } (x_1, \ldots, x_m) \in \mathbf{F}^m . \end{equation*}

Define

\begin{equation*} \Phi : \mathcal{L}(\mathbf{F}^m, V) \to V^m, \qquad \Phi(T) = (Te_1, \ldots, Te_m). \end{equation*}

\(\Phi\) is linear. If \(S, T \in \mathcal{L}(\mathbf{F}^m, V)\) and \(\lambda \in \mathbf{F}\), then by the definition of the operations on \(\mathcal{L}(\mathbf{F}^m, V)\) (3.5) and slotwise operations on \(V^m\) (3.87),

\begin{equation*} \Phi(S+T) = \big((S+T)e_1, \ldots, (S+T)e_m\big) = (Se_1 + Te_1, \ldots, Se_m + Te_m) = \Phi(S) + \Phi(T), \end{equation*}

\begin{equation*} \Phi(\lambda T) = \big((\lambda T)e_1, \ldots, (\lambda T)e_m\big) = (\lambda\, Te_1, \ldots, \lambda\, Te_m) = \lambda\, \Phi(T). \end{equation*}

\(\Phi\) is injective. Suppose \(\Phi(T) = (0, \ldots, 0)\), i.e. \(Te_k = 0\) for each \(k\). Then for any \((x_1, \ldots, x_m) \in \mathbf{F}^m\), linearity of \(T\) gives

\begin{equation*} T(x_1, \ldots, x_m) = T(x_1 e_1 + \cdots + x_m e_m) = x_1 Te_1 + \cdots + x_m Te_m = 0 . \end{equation*}

Hence \(T = 0\), so \(\operatorname{null} \Phi = \{0\}\) and \(\Phi\) is injective by 3.15.

\(\Phi\) is surjective. Let \((v_1, \ldots, v_m) \in V^m\). Because \(e_1, \ldots, e_m\) is a basis of \(\mathbf{F}^m\), the linear map lemma (3.4) provides a (unique) \(T \in \mathcal{L}(\mathbf{F}^m, V)\) with \(Te_k = v_k\) for each \(k = 1, \ldots, m\); explicitly, \(T(x_1, \ldots, x_m) = x_1 v_1 + \cdots + x_m v_m\). Then \(\Phi(T) = (v_1, \ldots, v_m)\).

Since \(\Phi\) is linear, injective, and surjective, it is invertible by 3.63, hence an isomorphism (3.69). Therefore \(\mathcal{L}(\mathbf{F}^m, V)\) and \(V^m\) are isomorphic vector spaces.

(Nothing here requires \(V\) to be finite-dimensional. When \(V\) is finite-dimensional, both sides have dimension \(m \dim V\), by 3.92 on the one hand and by 3.72 on the other, which is consistent.)

Recall the notation \(v + U = \{v + u : u \in U\}\) from 3.95.

First we show \(U \subseteq W\). Because \(U\) is a subspace, \(0 \in U\), so

\begin{equation*} v = v + 0 \in v + U = x + W, \end{equation*}

which means there exists \(w_0 \in W\) with

\begin{equation*} v = x + w_0 . \end{equation*}

Now let \(u \in U\) be arbitrary. Then \(v + u \in v + U = x + W\), so there exists \(w \in W\) with \(v + u = x + w\), and therefore

\begin{equation*} u = (x + w) - v = (x + w) - (x + w_0) = w - w_0 . \end{equation*}

Since \(W\) is a subspace it is closed under additive inverses and addition, so \(w - w_0 \in W\). Hence \(u \in W\). As \(u \in U\) was arbitrary, \(U \subseteq W\).

The reverse inclusion is the same argument with the roles of \((v, U)\) and \((x, W)\) interchanged, which is legitimate because the hypothesis \(v + U = x + W\) is symmetric in the two pairs. Explicitly: since \(0 \in W\) we have \(x \in x + W = v + U\), so \(x = v + u_0\) for some \(u_0 \in U\). Let \(w \in W\) be arbitrary. Then \(x + w \in x + W = v + U\), so \(x + w = v + u\) for some \(u \in U\), and therefore

\begin{equation*} w = (v + u) - x = (v + u) - (v + u_0) = u - u_0 \in U, \end{equation*}

because \(U\) is a subspace. As \(w \in W\) was arbitrary, \(W \subseteq U\).

Combining the two inclusions gives \(U = W\).

(Remark: this shows that a translate determines the subspace it translates, so it is unambiguous to speak of the subspace of which a given set is a translate. The vector \(v\) is by contrast far from unique: by 3.101, \(v + U = x + U\) precisely when \(v - x \in U\). Combining the exercise with 3.101 gives the complete description: \(v + U = x + W\) if and only if \(U = W\) and \(v - x \in U\).)

Define \(\varphi : \mathbf{R}^3 \to \mathbf{R}\) by

\begin{equation*} \varphi(x, y, z) = 2x + 3y + 5z . \end{equation*}

Then \(\varphi\) is linear: for \((x,y,z), (x^{\prime},y^{\prime},z^{\prime}) \in \mathbf{R}^3\) and \(\lambda \in \mathbf{R}\),

\begin{equation*} \varphi(x + x^{\prime}, y + y^{\prime}, z + z^{\prime}) = 2(x+x^{\prime}) + 3(y+y^{\prime}) + 5(z+z^{\prime}) = \varphi(x,y,z) + \varphi(x^{\prime},y^{\prime},z^{\prime}) \end{equation*}

and \(\varphi(\lambda x, \lambda y, \lambda z) = \lambda(2x + 3y + 5z) = \lambda\, \varphi(x,y,z)\). By definition \(U = \operatorname{null} \varphi\), so \(U\) is a subspace of \(\mathbf{R}^3\) by 3.13. For \(c \in \mathbf{R}\) write

\begin{equation*} A_c = \{p \in \mathbf{R}^3 : \varphi(p) = c\}, \end{equation*}

so the set displayed in the exercise is exactly \(A_c\), and \(A_0 = U\).

The whole proof rests on the following observation. Let \(v \in \mathbf{R}^3\) and put \(c = \varphi(v)\). Then for every \(p \in \mathbf{R}^3\),

\begin{equation*} p \in v + U \iff p - v \in U \iff \varphi(p - v) = 0 \iff \varphi(p) = \varphi(v) = c \iff p \in A_c . \end{equation*}

Here the first equivalence holds because \(p \in v + U\) means \(p = v + u\) for some \(u \in U\), which is the same as saying \(p - v \in U\); the second is the definition of \(U = \operatorname{null}\varphi\); and the third uses linearity of \(\varphi\), namely \(\varphi(p - v) = \varphi(p) - \varphi(v)\). Hence

\begin{equation*} v + U = A_{\varphi(v)} \qquad \text{for every } v \in \mathbf{R}^3 . \end{equation*}

Call this the key identity.

Now suppose \(A\) is a translate of \(U\), so \(A = v + U\) for some \(v \in \mathbf{R}^3\) (3.97). Set \(c = \varphi(v) \in \mathbf{R}\). By the key identity, \(A = v + U = A_c\), which is precisely the desired conclusion: there exists \(c \in \mathbf{R}\) with

\begin{equation*} A = \{(x,y,z) \in \mathbf{R}^3 : 2x + 3y + 5z = c\}. \end{equation*}

Conversely, suppose there exists \(c \in \mathbf{R}\) with \(A = A_c\). Put

\begin{equation*} v = \left(\tfrac{c}{2},\, 0,\, 0\right) \in \mathbf{R}^3 , \end{equation*}

so that \(\varphi(v) = 2 \cdot \tfrac{c}{2} + 3 \cdot 0 + 5 \cdot 0 = c\). By the key identity, \(v + U = A_{\varphi(v)} = A_c = A\). Thus \(A\) is a translate of \(U\).

This proves both implications, so \(A\) is a translate of \(U\) if and only if \(A = A_c\) for some \(c \in \mathbf{R}\).

(Geometrically: \(U\) is the plane through the origin with normal vector \((2,3,5)\), and the translates of \(U\) are exactly the planes parallel to it, which are the level sets of \(\varphi\). This is the general fact recorded in the last bullet of 3.98. Note also that the scalar \(c\) is uniquely determined by \(A\), since \(c\) is the single value \(\varphi\) takes on the nonempty set \(A\); and by Exercise 6, \(U\) itself is determined by \(A\).)

(a) Write

\begin{equation*} A = \{x \in V : Tx = c\}. \end{equation*}

If \(A = \varnothing\) there is nothing to prove, so assume \(A \neq \varnothing\) and fix some \(x_0 \in A\), meaning \(Tx_0 = c\). We claim that

\begin{equation*} A = x_0 + \operatorname{null} T, \end{equation*}

which exhibits \(A\) as a translate of \(\operatorname{null} T\) (recall that \(\operatorname{null} T\) is a subspace of \(V\) by 3.13, and see 3.97 for the definition of translate).

First suppose \(x \in A\). Then \(Tx = c\), so by linearity

\begin{equation*} T(x - x_0) = Tx - Tx_0 = c - c = 0. \end{equation*}

Hence \(x - x_0 \in \operatorname{null} T\), and therefore

\begin{equation*} x = x_0 + (x - x_0) \in x_0 + \operatorname{null} T. \end{equation*}

Thus \(A \subseteq x_0 + \operatorname{null} T\).

Conversely, suppose \(u \in \operatorname{null} T\). Then

\begin{equation*} T(x_0 + u) = Tx_0 + Tu = c + 0 = c, \end{equation*}

so \(x_0 + u \in A\). Thus \(x_0 + \operatorname{null} T \subseteq A\).

The two inclusions give \(A = x_0 + \operatorname{null} T\), as claimed. Hence \(A\) is either empty or a translate of \(\operatorname{null} T\).

(b) The system 3.27 consists of the \(m\) equations

\begin{equation*} \sum_{k=1}^{n} A_{j,k} x_k = c_j \qquad (j = 1, \ldots, m) \end{equation*}

in the unknowns \(x_1, \ldots, x_n \in \mathbf{F}\), where the \(A_{j,k}\) and the \(c_j\) are given scalars. Define \(T \colon \mathbf{F}^n \to \mathbf{F}^m\) as in 3.25 by

\begin{equation*} T(x_1, \ldots, x_n) = \Big( \sum_{k=1}^{n} A_{1,k} x_k, \; \ldots, \; \sum_{k=1}^{n} A_{m,k} x_k \Big); \end{equation*}

this \(T\) is a linear map. Set \(c = (c_1, \ldots, c_m) \in \mathbf{F}^m\). Then an \(n\)-tuple \((x_1, \ldots, x_n)\) satisfies the system 3.27 if and only if \(T(x_1, \ldots, x_n) = c\). In other words, the solution set of the system is exactly

\begin{equation*} \{x \in \mathbf{F}^n : Tx = c\}. \end{equation*}

By part (a), this set is either empty or a translate of \(\operatorname{null} T\). Since \(\operatorname{null} T\) is a subspace of \(\mathbf{F}^n\) (by 3.13), the solution set of a system of linear equations is either the empty set or a translate of some subspace of \(\mathbf{F}^n\), as desired.

First suppose \(A\) is a translate of a subspace of \(V\), say

\begin{equation*} A = x + U \end{equation*}

for some \(x \in V\) and some subspace \(U\) of \(V\). Let \(v, w \in A\) and let \(\lambda \in \mathbf{F}\). Then \(v = x + u_1\) and \(w = x + u_2\) for some \(u_1, u_2 \in U\). Hence

\begin{equation*} \begin{aligned} \lambda v + (1 - \lambda) w &= \lambda (x + u_1) + (1 - \lambda)(x + u_2) \\ &= \big(\lambda + (1 - \lambda)\big) x + \lambda u_1 + (1 - \lambda) u_2 \\ &= x + \big(\lambda u_1 + (1 - \lambda) u_2\big). \end{aligned} \end{equation*}

Because \(U\) is a subspace, it is closed under scalar multiplication and addition, so \(\lambda u_1 + (1 - \lambda) u_2 \in U\). Thus \(\lambda v + (1 - \lambda) w \in x + U = A\), as required.

Now suppose conversely that \(A\) is a nonempty subset of \(V\) such that

\begin{equation*} \lambda v + (1 - \lambda) w \in A \quad \text{for all } v, w \in A \text{ and all } \lambda \in \mathbf{F}. \end{equation*}

Call this condition \((\ast)\). Because \(A\) is nonempty we may fix some \(x \in A\). Define

\begin{equation*} U = A - x = \{a - x : a \in A\}. \end{equation*}

Then \(A = x + U\), because \(a \mapsto a - x\) is a bijection from \(A\) onto \(U\) whose inverse is \(u \mapsto x + u\). So it suffices to show that \(U\) is a subspace of \(V\); we verify the three conditions of 1.34.

Additive identity. Since \(x \in A\), we have \(0 = x - x \in U\).

Closure under scalar multiplication. Let \(u \in U\) and \(\lambda \in \mathbf{F}\). Write \(u = a - x\) with \(a \in A\). Applying \((\ast)\) with \(v = a\) and \(w = x\) (both in \(A\)) gives

\begin{equation*} \lambda a + (1 - \lambda) x \in A. \end{equation*}

Since

\begin{equation*} \lambda a + (1 - \lambda) x = x + \lambda (a - x) = x + \lambda u, \end{equation*}

we conclude that \(x + \lambda u \in A\), which means \(\lambda u \in A - x = U\).

Closure under addition. Let \(u_1, u_2 \in U\), say \(u_1 = a - x\) and \(u_2 = b - x\) with \(a, b \in A\). Applying \((\ast)\) with \(v = a\), \(w = b\), and \(\lambda = \tfrac{1}{2}\) (legitimate since \(\mathbf{F}\) is \(\mathbf{R}\) or \(\mathbf{C}\), so \(\tfrac{1}{2} \in \mathbf{F}\)) gives

\begin{equation*} \tfrac{1}{2} a + \tfrac{1}{2} b \in A. \end{equation*}

Hence

\begin{equation*} \tfrac{1}{2} u_1 + \tfrac{1}{2} u_2 = \tfrac{1}{2}(a - x) + \tfrac{1}{2}(b - x) = \big(\tfrac{1}{2} a + \tfrac{1}{2} b\big) - x \in A - x = U. \end{equation*}

Now the closure under scalar multiplication proved above, applied with the scalar \(2\), gives

\begin{equation*} u_1 + u_2 = 2 \big( \tfrac{1}{2} u_1 + \tfrac{1}{2} u_2 \big) \in U. \end{equation*}

Thus \(U\) is a subspace of \(V\) and \(A = x + U\) is a translate of \(U\), completing the proof.

If \(A_1 \cap A_2 = \varnothing\), then the conclusion holds. So assume \(A_1 \cap A_2 \neq \varnothing\) and fix

\begin{equation*} x \in A_1 \cap A_2 . \end{equation*}

We will show that

\begin{equation*} A_1 \cap A_2 = x + (U_1 \cap U_2), \end{equation*}

which finishes the proof because \(U_1 \cap U_2\) is a subspace of \(V\) (this is Exercise 1C.10; one checks directly that \(U_1 \cap U_2\) contains \(0\) and is closed under addition and scalar multiplication, since \(U_1\) and \(U_2\) each are).

Because \(x \in A_1 = v + U_1\), we have \(x - v \in U_1\), and hence 3.101 gives

\begin{equation*} A_1 = v + U_1 = x + U_1 . \end{equation*}

Similarly \(x - w \in U_2\), so 3.101 gives

\begin{equation*} A_2 = w + U_2 = x + U_2 . \end{equation*}

Therefore

\begin{equation*} A_1 \cap A_2 = (x + U_1) \cap (x + U_2). \end{equation*}

Now let \(y \in V\). Then

\begin{equation*} \begin{aligned} y \in (x + U_1) \cap (x + U_2) &\iff y - x \in U_1 \ \text{ and } \ y - x \in U_2 \\ &\iff y - x \in U_1 \cap U_2 \\ &\iff y \in x + (U_1 \cap U_2). \end{aligned} \end{equation*}

Hence

\begin{equation*} A_1 \cap A_2 = x + (U_1 \cap U_2), \end{equation*}

which is a translate of the subspace \(U_1 \cap U_2\) of \(V\). Thus \(A_1 \cap A_2\) is either the empty set or a translate of a subspace of \(V\), as desired.

For a sequence \(x = (x_1, x_2, \ldots) \in \mathbf{F}^\infty\), write

\begin{equation*} \operatorname{supp} x = \{k \in \mathbf{Z}^+ : x_k \neq 0\}. \end{equation*}

Thus \(U = \{x \in \mathbf{F}^\infty : \operatorname{supp} x \text{ is finite}\}\).

(a) We verify the three conditions of 1.34.

The zero sequence \(0 = (0, 0, \ldots)\) has \(\operatorname{supp} 0 = \varnothing\), which is finite, so \(0 \in U\).

Suppose \(x, y \in U\). If \(k \notin \operatorname{supp} x \cup \operatorname{supp} y\) then \(x_k = y_k = 0\), so \(x_k + y_k = 0\). Hence

\begin{equation*} \operatorname{supp}(x + y) \subseteq \operatorname{supp} x \cup \operatorname{supp} y , \end{equation*}

and the right side is a union of two finite sets, hence finite. Thus \(\operatorname{supp}(x+y)\) is finite and \(x + y \in U\).

Suppose \(x \in U\) and \(\lambda \in \mathbf{F}\). If \(k \notin \operatorname{supp} x\) then \(\lambda x_k = 0\), so \(\operatorname{supp}(\lambda x) \subseteq \operatorname{supp} x\), which is finite. Thus \(\lambda x \in U\).

Hence \(U\) is a subspace of \(\mathbf{F}^\infty\).

(b) By 2.22, if a vector space is finite-dimensional then every linearly independent list in it has length at most the length of any spanning list, and in particular at most the dimension of the space. Hence to prove that \(\mathbf{F}^\infty / U\) is infinite-dimensional it suffices to produce, for every positive integer \(n\), a linearly independent list of length \(n\) in \(\mathbf{F}^\infty/U\).

We first split \(\mathbf{Z}^+\) into infinitely many pairwise disjoint infinite pieces. For each positive integer \(j\) set

\begin{equation*} S_j = \{ 2^{\,j-1}(2k - 1) : k \in \mathbf{Z}^+ \}, \end{equation*}

so \(S_j\) is the set of positive integers whose factorization into a power of \(2\) times an odd number uses exactly \(2^{\,j-1}\). Every positive integer \(m\) can be written in exactly one way as \(m = 2^{a} q\) with \(a \geq 0\) an integer and \(q\) odd; consequently \(m\) lies in \(S_{a+1}\) and in no other \(S_j\). Thus the sets \(S_1, S_2, \ldots\) are pairwise disjoint, and each \(S_j\) is infinite (it contains \(2^{\,j-1}(2k-1)\) for every \(k \in \mathbf{Z}^+\), and these are distinct).

Now for each positive integer \(j\) define \(v_j \in \mathbf{F}^\infty\) to be the sequence whose \(k\)-th coordinate \((v_j)_k\) equals \(1\) if \(k \in S_j\) and equals \(0\) if \(k \notin S_j\).

Fix a positive integer \(n\); we claim that

\begin{equation*} v_1 + U, \; \ldots, \; v_n + U \end{equation*}

is a linearly independent list in \(\mathbf{F}^\infty/U\). Suppose \(a_1, \ldots, a_n \in \mathbf{F}\) satisfy

\begin{equation*} a_1 (v_1 + U) + \cdots + a_n (v_n + U) = 0 + U . \end{equation*}

By the definition 3.102 of the operations on \(\mathbf{F}^\infty/U\), the left side equals \((a_1 v_1 + \cdots + a_n v_n) + U\). Hence by 3.101,

\begin{equation*} a_1 v_1 + \cdots + a_n v_n \in U . \end{equation*}

Set \(x = a_1 v_1 + \cdots + a_n v_n\). Because \(S_1, \ldots, S_n\) are pairwise disjoint, for each \(j \in \{1, \ldots, n\}\) and each \(k \in S_j\) exactly one of \((v_1)_k, \ldots, (v_n)_k\) equals \(1\) (namely \((v_j)_k\)) and the rest are \(0\); hence

\begin{equation*} x_k = a_j \qquad \text{for every } k \in S_j . \end{equation*}

If some \(a_j \neq 0\), then \(x_k = a_j \neq 0\) for all \(k\) in the infinite set \(S_j\), so \(\operatorname{supp} x\) is infinite and \(x \notin U\), a contradiction. Therefore \(a_1 = \cdots = a_n = 0\), which proves that \(v_1 + U, \ldots, v_n + U\) is linearly independent.

Since \(\mathbf{F}^\infty/U\) contains linearly independent lists of every finite length, it cannot be finite-dimensional. Hence \(\mathbf{F}^\infty/U\) is infinite-dimensional.

Here \(m\) is a positive integer, as the notation \(v_1, \ldots, v_m\) indicates. Throughout, put

\begin{equation*} U = \operatorname{span}(v_2 - v_1, \ldots, v_m - v_1), \end{equation*}

a subspace of \(V\) by 2.6 (when \(m = 1\) this is the span of the empty list, namely \(\{0\}\), and \(A = \{v_1\}\)).

(a) and (c). We prove the single statement \(A = v_1 + U\); this proves (a), and since \(U\) is the span of a list of length \(m - 1\), we have \(\dim U \leq m - 1 < m\) (a spanning list of length \(m-1\) contains a basis by 2.30, so the dimension is at most \(m-1\)), which proves (c).

First suppose \(a \in A\), say \(a = \lambda_1 v_1 + \cdots + \lambda_m v_m\) with \(\lambda_1 + \cdots + \lambda_m = 1\). Then

\begin{equation*} \begin{aligned} a - v_1 &= \sum_{j=1}^{m} \lambda_j v_j - \Big( \sum_{j=1}^{m} \lambda_j \Big) v_1 \\ &= \sum_{j=1}^{m} \lambda_j (v_j - v_1) \\ &= \sum_{j=2}^{m} \lambda_j (v_j - v_1), \end{aligned} \end{equation*}

where the last equality holds because the \(j = 1\) term is \(\lambda_1 (v_1 - v_1) = 0\). Hence \(a - v_1 \in U\), so \(a \in v_1 + U\). Thus \(A \subseteq v_1 + U\).

Conversely, suppose \(u \in U\), say \(u = c_2 (v_2 - v_1) + \cdots + c_m (v_m - v_1)\) with \(c_2, \ldots, c_m \in \mathbf{F}\). Then

\begin{equation*} v_1 + u = \Big( 1 - \sum_{j=2}^{m} c_j \Big) v_1 + c_2 v_2 + \cdots + c_m v_m . \end{equation*}

The coefficients of \(v_1, \ldots, v_m\) on the right side sum to

\begin{equation*} \Big( 1 - \sum_{j=2}^{m} c_j \Big) + \sum_{j=2}^{m} c_j = 1 , \end{equation*}

so \(v_1 + u \in A\). Thus \(v_1 + U \subseteq A\).

Therefore \(A = v_1 + U\), a translate of the subspace \(U\) of \(V\), and \(\dim U < m\).

(b) Suppose \(B\) is a translate of a subspace of \(V\) with \(\{v_1, \ldots, v_m\} \subseteq B\), say

\begin{equation*} B = x + W \end{equation*}

for some \(x \in V\) and some subspace \(W\) of \(V\). Since \(v_j \in B\) for each \(j\), we have

\begin{equation*} v_j - x \in W \qquad (j = 1, \ldots, m). \end{equation*}

Let \(a \in A\), say \(a = \lambda_1 v_1 + \cdots + \lambda_m v_m\) with \(\lambda_1 + \cdots + \lambda_m = 1\). Then, using \(\sum_j \lambda_j = 1\),

\begin{equation*} a - x = \sum_{j=1}^{m} \lambda_j v_j - \Big( \sum_{j=1}^{m} \lambda_j \Big) x = \sum_{j=1}^{m} \lambda_j (v_j - x). \end{equation*}

Each \(v_j - x\) lies in the subspace \(W\), so this linear combination lies in \(W\). Hence \(a - x \in W\), which means \(a \in x + W = B\).

Therefore \(A \subseteq B\), as desired.

(Remark: parts (a) and (b) together say that \(A\) is the smallest translate of a subspace containing \(v_1, \ldots, v_m\), i.e. \(A\) is the affine hull of \(v_1, \ldots, v_m\).)

Let \(n = \dim V/U\) and choose a basis of \(V/U\). Every element of \(V/U\) is a translate \(v + U\) for some \(v \in V\) (by 3.99), so we may write this basis as

\begin{equation*} v_1 + U, \; \ldots, \; v_n + U \end{equation*}

for some \(v_1, \ldots, v_n \in V\). (If \(n = 0\) this is the empty list; the argument below goes through with all sums over the empty index set interpreted as \(0\).)

By the linear map lemma 3.4 there is a unique linear map

\begin{equation*} \Psi \in \mathcal{L}(V/U, V) \quad \text{with} \quad \Psi(v_j + U) = v_j \ \text{ for } j = 1, \ldots, n . \end{equation*}

Define \(S \colon U \times (V/U) \to V\) by

\begin{equation*} S(u, x) = u + \Psi x . \end{equation*}

We show that \(S\) is an isomorphism.

\(S\) is linear. If \(u, u^{\prime} \in U\) and \(x, x^{\prime} \in V/U\), then, using the definition 3.87 of the operations on the product \(U \times (V/U)\) and the linearity of \(\Psi\),

\begin{equation*} \begin{aligned} S\big((u, x) + (u^{\prime}, x^{\prime})\big) &= S(u + u^{\prime}, x + x^{\prime}) \\ &= (u + u^{\prime}) + \Psi(x + x^{\prime}) \\ &= (u + \Psi x) + (u^{\prime} + \Psi x^{\prime}) \\ &= S(u, x) + S(u^{\prime}, x^{\prime}). \end{aligned} \end{equation*}

Similarly, for \(\lambda \in \mathbf{F}\),

\begin{equation*} S\big(\lambda (u, x)\big) = S(\lambda u, \lambda x) = \lambda u + \Psi(\lambda x) = \lambda u + \lambda \Psi x = \lambda \, S(u, x). \end{equation*}

Thus \(S \in \mathcal{L}\big(U \times (V/U), V\big)\).

\(S\) is surjective. Let \(v \in V\). Since \(v + U \in V/U\) and \(v_1 + U, \ldots, v_n + U\) spans \(V/U\), there exist \(a_1, \ldots, a_n \in \mathbf{F}\) with

\begin{equation*} v + U = a_1 (v_1 + U) + \cdots + a_n (v_n + U) = (a_1 v_1 + \cdots + a_n v_n) + U, \end{equation*}

the last equality by 3.102. By 3.101 this gives

\begin{equation*} u := v - (a_1 v_1 + \cdots + a_n v_n) \in U . \end{equation*}

Setting \(x = v + U\) we get, by linearity of \(\Psi\),

\begin{equation*} \Psi x = a_1 \Psi(v_1 + U) + \cdots + a_n \Psi(v_n + U) = a_1 v_1 + \cdots + a_n v_n , \end{equation*}

and hence

\begin{equation*} S(u, x) = u + \Psi x = \big(v - (a_1 v_1 + \cdots + a_n v_n)\big) + (a_1 v_1 + \cdots + a_n v_n) = v . \end{equation*}

Thus \(S\) is surjective.

\(S\) is injective. By 3.15 it suffices to show that \(S(u, x) = 0\) forces \((u, x) = (0, 0)\). So suppose \(u \in U\) and \(x \in V/U\) satisfy \(u + \Psi x = 0\). Write \(x = a_1 (v_1 + U) + \cdots + a_n (v_n + U)\) with \(a_1, \ldots, a_n \in \mathbf{F}\), which is possible because \(v_1 + U, \ldots, v_n + U\) spans \(V/U\). As above, \(\Psi x = a_1 v_1 + \cdots + a_n v_n\). Hence

\begin{equation*} a_1 v_1 + \cdots + a_n v_n = -u \in U . \end{equation*}

By 3.101 this implies \((a_1 v_1 + \cdots + a_n v_n) + U = 0 + U\), that is (using 3.102),

\begin{equation*} a_1 (v_1 + U) + \cdots + a_n (v_n + U) = 0 + U , \end{equation*}

which is the zero vector of \(V/U\). Because \(v_1 + U, \ldots, v_n + U\) is linearly independent, \(a_1 = \cdots = a_n = 0\). Hence \(x = 0\) and \(\Psi x = 0\), and therefore \(u = -\Psi x = 0\). Thus \((u, x) = (0, 0)\) and \(S\) is injective.

Being linear, injective, and surjective, \(S\) is an isomorphism from \(U \times (V/U)\) onto \(V\) (by 3.63, such a map is invertible, and an invertible linear map is an isomorphism by 3.69). Therefore \(V\) is isomorphic to \(U \times (V/U)\), as desired.

Note that no assumption was made about \(U\) or \(V\) being finite-dimensional; only \(\dim V/U < \infty\) was used, and it was used to obtain a finite basis of \(V/U\).

Each \(w_k + U\) belongs to \(V/U\) by 3.99. We must show that the list \(w_1 + U, \ldots, w_m + U\) spans \(V/U\) and is linearly independent.

Spanning. Let \(x \in V/U\). By 3.99 there is \(v \in V\) with \(x = v + U\). Because \(V = U \oplus W\), in particular \(V = U + W\), so there exist \(u \in U\) and \(w \in W\) with

\begin{equation*} v = u + w . \end{equation*}

Because \(w_1, \ldots, w_m\) is a basis of \(W\), there exist \(a_1, \ldots, a_m \in \mathbf{F}\) with

\begin{equation*} w = a_1 w_1 + \cdots + a_m w_m . \end{equation*}

Then

\begin{equation*} v - (a_1 w_1 + \cdots + a_m w_m) = v - w = u \in U , \end{equation*}

so 3.101 gives

\begin{equation*} v + U = (a_1 w_1 + \cdots + a_m w_m) + U . \end{equation*}

By the definition 3.102 of the operations on \(V/U\), the right side equals

\begin{equation*} a_1 (w_1 + U) + \cdots + a_m (w_m + U). \end{equation*}

Hence

\begin{equation*} x = v + U = a_1 (w_1 + U) + \cdots + a_m (w_m + U), \end{equation*}

so \(x \in \operatorname{span}(w_1 + U, \ldots, w_m + U)\). Since \(x \in V/U\) was arbitrary, the list \(w_1 + U, \ldots, w_m + U\) spans \(V/U\).

Linear independence. Suppose \(a_1, \ldots, a_m \in \mathbf{F}\) satisfy

\begin{equation*} a_1 (w_1 + U) + \cdots + a_m (w_m + U) = 0 + U , \end{equation*}

where \(0 + U = U\) is the additive identity of \(V/U\) (see 3.103). By 3.102 the left side equals \((a_1 w_1 + \cdots + a_m w_m) + U\), so 3.101 gives

\begin{equation*} a_1 w_1 + \cdots + a_m w_m \in U . \end{equation*}

On the other hand \(a_1 w_1 + \cdots + a_m w_m \in W\), because \(w_1, \ldots, w_m \in W\) and \(W\) is a subspace. Hence

\begin{equation*} a_1 w_1 + \cdots + a_m w_m \in U \cap W . \end{equation*}

Because \(V = U \oplus W\), we have \(U \cap W = \{0\}\) by 1.46. Therefore

\begin{equation*} a_1 w_1 + \cdots + a_m w_m = 0 , \end{equation*}

and the linear independence of the basis \(w_1, \ldots, w_m\) of \(W\) forces \(a_1 = \cdots = a_m = 0\). Hence \(w_1 + U, \ldots, w_m + U\) is linearly independent.

Being a linearly independent spanning list of \(V/U\), the list \(w_1 + U, \ldots, w_m + U\) is a basis of \(V/U\), as desired.

(As a check on the result, note that it gives \(\dim V/U = m = \dim W = \dim V - \dim U\) when \(V\) is finite-dimensional, in agreement with 3.105 and 1.46 together with 2.43.)

Let \(\pi \colon V \to V/U\) denote the quotient map, so \(\pi(v) = v + U\) for each \(v \in V\) (see 3.104). Recall from the proof of 3.105 that

\begin{equation*} \operatorname{null} \pi = U, \end{equation*}

because \(\pi(v) = 0 + U\) means \(v + U = 0 + U\), which by 3.101 holds if and only if \(v - 0 = v \in U\).

First we show that the list spans \(V\). Suppose \(v \in V\). Then \(v + U \in V/U\), so because \(v_1 + U, \dots, v_m + U\) spans \(V/U\) there exist \(a_1, \dots, a_m \in \mathbf{F}\) with

\begin{equation*} v + U = a_1(v_1 + U) + \dots + a_m(v_m + U) = (a_1 v_1 + \dots + a_m v_m) + U, \end{equation*}

where the last equality uses the definition 3.102 of addition and scalar multiplication on \(V/U\). By 3.101 this gives

\begin{equation*} v - (a_1 v_1 + \dots + a_m v_m) \in U. \end{equation*}

Because \(u_1, \dots, u_n\) spans \(U\), there exist \(b_1, \dots, b_n \in \mathbf{F}\) with \(v - (a_1 v_1 + \dots + a_m v_m) = b_1 u_1 + \dots + b_n u_n\). Hence

\begin{equation*} v = a_1 v_1 + \dots + a_m v_m + b_1 u_1 + \dots + b_n u_n, \end{equation*}

so \(v \in \operatorname{span}(v_1, \dots, v_m, u_1, \dots, u_n)\). Because \(v \in V\) was arbitrary, the list \(v_1, \dots, v_m, u_1, \dots, u_n\) spans \(V\).

Now we show that the list is linearly independent. Suppose \(a_1, \dots, a_m, b_1, \dots, b_n \in \mathbf{F}\) satisfy

\begin{equation*} a_1 v_1 + \dots + a_m v_m + b_1 u_1 + \dots + b_n u_n = 0. \end{equation*}

Apply the linear map \(\pi\) to both sides. Since each \(u_j \in U = \operatorname{null} \pi\), we get \(\pi(u_j) = 0\), and therefore

\begin{equation*} a_1(v_1 + U) + \dots + a_m(v_m + U) = 0 + U, \end{equation*}

the additive identity of \(V/U\). Because \(v_1 + U, \dots, v_m + U\) is a basis of \(V/U\) and hence linearly independent, we conclude \(a_1 = \dots = a_m = 0\). The displayed equation above now reduces to

\begin{equation*} b_1 u_1 + \dots + b_n u_n = 0, \end{equation*}

and the linear independence of the basis \(u_1, \dots, u_n\) of \(U\) gives \(b_1 = \dots = b_n = 0\). Thus all the coefficients vanish, so \(v_1, \dots, v_m, u_1, \dots, u_n\) is linearly independent.

Being a spanning list of \(V\) that is linearly independent, \(v_1, \dots, v_m, u_1, \dots, u_n\) is a basis of \(V\) (by the definition 2.26 of basis).

Note that this exercise in particular shows that if \(U\) and \(V/U\) are both finite-dimensional, then so is \(V\), and \(\dim V = \dim V/U + \dim U\), in agreement with 3.105.

Note that \(V\) is not assumed to be finite-dimensional, so we cannot simply invoke 3.105; instead we use 3.107.

We first determine \(\operatorname{range} \varphi\). It is a subspace of \(\mathbf{F}\) (by 3.18). Regarded as a vector space over itself, \(\mathbf{F}\) has the one-element list \(1\) as a basis, since every \(\lambda \in \mathbf{F}\) equals \(\lambda \cdot 1\) and \(\lambda \cdot 1 = 0\) forces \(\lambda = 0\). Hence \(\dim \mathbf{F} = 1\).

Because \(\varphi \neq 0\), there exists \(v \in V\) with \(\varphi(v) \neq 0\). Then for every \(\lambda \in \mathbf{F}\) we have

\begin{equation*} \varphi\!\left( \frac{\lambda}{\varphi(v)} \, v \right) = \frac{\lambda}{\varphi(v)} \, \varphi(v) = \lambda, \end{equation*}

where the division makes sense because \(\varphi(v) \neq 0\). Thus every element of \(\mathbf{F}\) is in \(\operatorname{range} \varphi\), so

\begin{equation*} \operatorname{range} \varphi = \mathbf{F}, \end{equation*}

and therefore \(\dim \operatorname{range} \varphi = 1\).

By 3.107(d), the map \(\widetilde{\varphi}\) of 3.106 is an isomorphism from \(V/(\operatorname{null} \varphi)\) onto \(\operatorname{range} \varphi = \mathbf{F}\). An isomorphism is by definition an invertible linear map (3.69), so \(\widetilde{\varphi}^{\,-1}\) is a linear map from \(\mathbf{F}\) onto \(V/(\operatorname{null} \varphi)\), and it too is invertible, hence injective and surjective by 3.63.

Now use the general fact that an isomorphism carries a basis to a basis: if \(\Phi\) is an isomorphism from one space onto another and \(e_1, \dots, e_k\) is a basis of the domain, then \(\Phi e_1, \dots, \Phi e_k\) is a basis of the target (surjectivity of \(\Phi\) gives spanning, and injectivity of \(\Phi\) together with its linearity gives linear independence, since \(a_1 \Phi e_1 + \dots + a_k \Phi e_k = 0\) implies \(\Phi(a_1 e_1 + \dots + a_k e_k) = 0\), hence \(a_1 e_1 + \dots + a_k e_k = 0\), hence every \(a_j = 0\)). It is essential here to apply this to \(\widetilde{\varphi}^{\,-1}\) rather than to \(\widetilde{\varphi}\): we possess a basis of \(\mathbf{F}\), namely the one-vector list \(1\), whereas a basis of \(V/(\operatorname{null} \varphi)\) is exactly what we are trying to produce, and \(V\) is not assumed finite-dimensional so we may not presume one exists. Transporting the basis \(1\) of \(\mathbf{F}\) through \(\widetilde{\varphi}^{\,-1}\) shows that the one-vector list \(\widetilde{\varphi}^{\,-1}(1)\) is a basis of \(V/(\operatorname{null} \varphi)\). Hence

\begin{equation*} \dim V/(\operatorname{null} \varphi) = 1, \end{equation*}

as desired.

Here is a direct argument that avoids invoking 3.107 at all. With \(v\) as above, every \(x \in V\) satisfies

\begin{equation*} x - \frac{\varphi(x)}{\varphi(v)} \, v \in \operatorname{null} \varphi, \end{equation*}

since \(\varphi\) of the left side is \(\varphi(x) - \varphi(x) = 0\). Hence by 3.101,

\begin{equation*} x + \operatorname{null} \varphi = \frac{\varphi(x)}{\varphi(v)} \, \bigl( v + \operatorname{null} \varphi \bigr), \end{equation*}

so the single vector \(v + \operatorname{null} \varphi\) spans \(V/(\operatorname{null} \varphi)\). It is also linearly independent, because \(\lambda(v + \operatorname{null} \varphi) = 0 + \operatorname{null} \varphi\) means \(\lambda v \in \operatorname{null} \varphi\), so \(\lambda \varphi(v) = 0\), so \(\lambda = 0\). Thus \(v + \operatorname{null} \varphi\) is a basis of \(V/(\operatorname{null} \varphi)\), which therefore has dimension \(1\).

Let \(\pi \colon V \to V/U\) be the quotient map, defined by \(\pi(v) = v + U\) (see 3.104); it is a linear map. As noted in the proof of 3.105, 3.101 gives

\begin{equation*} \operatorname{null} \pi = U, \end{equation*}

because \(\pi(v) = 0 + U\) if and only if \(v \in U\).

Because \(\dim V/U = 1\), the space \(V/U\) has a basis consisting of one vector; write this basis as \(v + U\) for some \(v \in V\). Thus every element of \(V/U\) can be written in exactly one way as \(\lambda(v + U)\) with \(\lambda \in \mathbf{F}\): it can be so written because \(v + U\) spans \(V/U\), and uniquely so because \(\lambda(v+U) = \mu(v+U)\) implies \((\lambda - \mu)(v + U) = 0\), which by linear independence forces \(\lambda = \mu\).

Hence we may define a function \(\alpha \colon V/U \to \mathbf{F}\) by

\begin{equation*} \alpha\bigl( \lambda (v + U) \bigr) = \lambda \quad \text{for each } \lambda \in \mathbf{F}; \end{equation*}

the uniqueness just established shows that \(\alpha\) is well defined on all of \(V/U\). (This is the map given by 3.4, the existence of a linear map taking prescribed values on a basis, applied to the basis \(v + U\) of \(V/U\) and the value \(1 \in \mathbf{F}\).)

The map \(\alpha\) is linear: if \(\lambda, \mu, c \in \mathbf{F}\) then

\begin{equation*} \alpha\bigl( \lambda(v+U) + \mu(v+U) \bigr) = \alpha\bigl( (\lambda + \mu)(v+U) \bigr) = \lambda + \mu = \alpha\bigl(\lambda(v+U)\bigr) + \alpha\bigl(\mu(v+U)\bigr) \end{equation*}

and

\begin{equation*} \alpha\bigl( c \cdot \lambda(v+U) \bigr) = \alpha\bigl( (c\lambda)(v+U) \bigr) = c\lambda = c \, \alpha\bigl(\lambda(v+U)\bigr). \end{equation*}

The map \(\alpha\) is injective: if \(\alpha(\lambda(v+U)) = 0\) then \(\lambda = 0\), so \(\lambda(v+U) = 0\); thus \(\operatorname{null} \alpha = \{0\}\), and \(\alpha\) is injective by 3.15.

Now define

\begin{equation*} \varphi = \alpha \circ \pi. \end{equation*}

Then \(\varphi\) is a linear map from \(V\) to \(\mathbf{F}\), because the composition (product) of linear maps is linear (see 3.7). Finally,

\begin{equation*} \operatorname{null} \varphi = \{ x \in V : \alpha(\pi x) = 0 \} = \{ x \in V : \pi x = 0 \} = \operatorname{null} \pi = U, \end{equation*}

where the second equality holds because \(\alpha\) is injective and \(\alpha(0) = 0\), so \(\alpha(\pi x) = 0\) if and only if \(\pi x = 0\).

Thus \(\varphi \in \mathcal{L}(V, \mathbf{F})\) satisfies \(\operatorname{null} \varphi = U\), as desired. Note that \(\varphi \neq 0\), since \(\varphi(v) = \alpha(v + U) = 1\); combined with the previous exercise, this shows that the subspaces \(U\) of \(V\) with \(\dim V/U = 1\) are exactly the null spaces of the nonzero linear functionals on \(V\).

Throughout, let \(\pi \colon V \to V/U\) denote the quotient map \(\pi(v) = v + U\) (see 3.104), which is linear, and recall from 3.101 that \(\pi(v) = 0\) if and only if \(v \in U\); thus \(\operatorname{null} \pi = U\).

(a) Suppose \(W\) is a finite-dimensional subspace of \(V\) with \(V = U + W\). Let \(\pi|_W\) denote the restriction of \(\pi\) to \(W\); then \(\pi|_W \in \mathcal{L}(W, V/U)\).

We claim \(\pi|_W\) is surjective. Indeed, let \(v + U\) be an arbitrary element of \(V/U\). Since \(V = U + W\), we may write \(v = u + w\) with \(u \in U\) and \(w \in W\). Then \(v - w = u \in U\), so 3.101 gives \(v + U = w + U = \pi|_W(w)\). Hence

\begin{equation*} \operatorname{range} \pi|_W = V/U . \end{equation*}

Because \(W\) is finite-dimensional, the fundamental theorem of linear maps (3.21) applied to \(\pi|_W\) gives

\begin{equation*} \dim W = \dim \operatorname{null} \pi|_W + \dim \operatorname{range} \pi|_W = \dim (U \cap W) + \dim V/U \geq \dim V/U, \end{equation*}

where we used \(\operatorname{null} \pi|_W = W \cap \operatorname{null} \pi = U \cap W\) and the fact that dimensions are nonnegative. This proves (a).

(b) Let \(n = \dim V/U\) and let \(v_1 + U, \dots, v_n + U\) be a basis of \(V/U\); such a basis exists because \(V/U\) is finite-dimensional (by 2.31), and each of its vectors has the form \(v_j + U\) for some \(v_j \in V\) because the quotient map \(\pi\) is surjective. The case \(n = 0\) is permitted throughout what follows, all lists then being empty: in that case \(V/U = \{0\}\), so 3.101 forces \(V = U\), and the construction below produces \(W = \operatorname{span}() = \{0\}\), for which \(\dim W = 0 = \dim V/U\) and \(V = U \oplus \{0\}\). Define

\begin{equation*} W = \operatorname{span}(v_1, \dots, v_n), \end{equation*}

which is a finite-dimensional subspace of \(V\) (by 2.6 and the definition of finite-dimensional).

The list \(v_1, \dots, v_n\) is linearly independent in \(V\): if \(a_1 v_1 + \dots + a_n v_n = 0\) for some \(a_1, \dots, a_n \in \mathbf{F}\), then applying \(\pi\) gives

\begin{equation*} a_1(v_1 + U) + \dots + a_n(v_n + U) = 0 + U, \end{equation*}

and the linear independence of the basis \(v_1 + U, \dots, v_n + U\) of \(V/U\) forces \(a_1 = \dots = a_n = 0\). Hence \(v_1, \dots, v_n\) is a basis of \(W\) and

\begin{equation*} \dim W = n = \dim V/U . \end{equation*}

Next, \(V = U + W\). Indeed, let \(v \in V\). Because \(v_1 + U, \dots, v_n + U\) spans \(V/U\), there exist \(a_1, \dots, a_n \in \mathbf{F}\) with

\begin{equation*} v + U = a_1(v_1 + U) + \dots + a_n(v_n + U) = (a_1 v_1 + \dots + a_n v_n) + U . \end{equation*}

By 3.101 this means \(v - (a_1 v_1 + \dots + a_n v_n) \in U\), so

\begin{equation*} v = \underbrace{\bigl( v - (a_1 v_1 + \dots + a_n v_n) \bigr)}_{\in\, U} + \underbrace{(a_1 v_1 + \dots + a_n v_n)}_{\in\, W} \in U + W . \end{equation*}

Finally, \(U \cap W = \{0\}\). Suppose \(w \in U \cap W\). Because \(w \in W\), we can write \(w = a_1 v_1 + \dots + a_n v_n\) for some \(a_1, \dots, a_n \in \mathbf{F}\). Because \(w \in U\), we have \(\pi(w) = 0\), that is,

\begin{equation*} a_1(v_1 + U) + \dots + a_n(v_n + U) = 0 + U . \end{equation*}

Linear independence of \(v_1 + U, \dots, v_n + U\) gives \(a_1 = \dots = a_n = 0\), hence \(w = 0\).

Because \(V = U + W\) and \(U \cap W = \{0\}\), the sum \(U + W\) is a direct sum by 1.46, so

\begin{equation*} V = U \oplus W \quad \text{and} \quad \dim W = \dim V/U, \end{equation*}

as desired.

Part (b) shows that the lower bound in part (a) is attained, and moreover it is attained by a subspace \(W\) that is a complement of \(U\) in \(V\). Combining the two parts: among all finite-dimensional subspaces \(W\) of \(V\) with \(V = U + W\), the minimum possible dimension is exactly \(\dim V/U\).

Recall that \(\pi \in \mathcal{L}(V, V/U)\) is given by \(\pi(v) = v + U\) (see 3.104), and that by 3.101 we have \(\pi(v) = 0 + U\) if and only if \(v \in U\); in other words \(\operatorname{null} \pi = U\).

First suppose there exists \(S \in \mathcal{L}(V/U, W)\) with \(T = S \circ \pi\). Let \(u \in U\). Then \(\pi(u) = u + U = 0 + U\), which is the additive identity of \(V/U\). Hence

\begin{equation*} T u = (S \circ \pi)(u) = S(\pi(u)) = S(0) = 0, \end{equation*}

the last equality because every linear map sends \(0\) to \(0\) (by 3.10). Thus \(u \in \operatorname{null} T\). Since \(u \in U\) was arbitrary, \(U \subseteq \operatorname{null} T\).

Conversely, suppose \(U \subseteq \operatorname{null} T\). We want to define \(S \colon V/U \to W\) by

\begin{equation*} S(v + U) = Tv \quad \text{for each } v \in V . \end{equation*}

This definition makes sense (that is, \(S\) is well defined). Suppose \(v, v^{\prime} \in V\) satisfy \(v + U = v^{\prime} + U\). By 3.101 this gives \(v - v^{\prime} \in U\), and since \(U \subseteq \operatorname{null} T\) we get \(T(v - v^{\prime}) = 0\), hence \(Tv = Tv^{\prime}\). So the value assigned to the element \(v + U\) of \(V/U\) does not depend on which representative \(v\) is used, and every element of \(V/U\) has the form \(v + U\) for some \(v \in V\). Thus \(S\) is a well-defined function from \(V/U\) to \(W\).

The function \(S\) is linear. For \(v, v^{\prime} \in V\), using the definition 3.102 of addition on \(V/U\) and the additivity of \(T\),

\begin{equation*} S\bigl( (v + U) + (v^{\prime} + U) \bigr) = S\bigl( (v + v^{\prime}) + U \bigr) = T(v + v^{\prime}) = Tv + Tv^{\prime} = S(v + U) + S(v^{\prime} + U). \end{equation*}

For \(\lambda \in \mathbf{F}\) and \(v \in V\), using the definition 3.102 of scalar multiplication on \(V/U\) and the homogeneity of \(T\),

\begin{equation*} S\bigl( \lambda(v + U) \bigr) = S\bigl( (\lambda v) + U \bigr) = T(\lambda v) = \lambda \, Tv = \lambda \, S(v + U). \end{equation*}

Hence \(S \in \mathcal{L}(V/U, W)\).

Finally, for each \(v \in V\),

\begin{equation*} (S \circ \pi)(v) = S(\pi(v)) = S(v + U) = Tv , \end{equation*}

so \(T = S \circ \pi\), completing the proof of the converse direction and hence of the equivalence.

Two remarks. First, the map \(S\) produced above is unique: if \(S^{\prime} \in \mathcal{L}(V/U, W)\) also satisfies \(T = S^{\prime} \circ \pi\), then for every \(v \in V\) we have \(S^{\prime}(v + U) = (S^{\prime} \circ \pi)(v) = Tv = S(v + U)\), and since every element of \(V/U\) has the form \(v + U\), we get \(S^{\prime} = S\). Second, in the special case \(U = \operatorname{null} T\) the hypothesis \(U \subseteq \operatorname{null} T\) holds automatically, and the map \(S\) constructed here is exactly the map \(\widetilde{T}\) of 3.106; the identity \(T = S \circ \pi\) then becomes 3.107(a).

Let \(V\) be a vector space over \(\mathbf{F}\) and let \(\varphi \in V^{\prime}\), so \(\varphi\) is a linear map from \(V\) to \(\mathbf{F}\). By 3.18, \(\operatorname{range} \varphi\) is a subspace of \(\mathbf{F}\).

Now \(\mathbf{F}\), regarded as a vector space over \(\mathbf{F}\), has dimension \(1\) (the list consisting of the single vector \(1\) is a basis of \(\mathbf{F}\)). Hence if \(M\) is a subspace of \(\mathbf{F}\), then \(\dim M \le 1\) by 2.37, so \(\dim M = 0\) or \(\dim M = 1\). If \(\dim M = 0\) then \(M = \{0\}\), and if \(\dim M = 1\) then \(M = \mathbf{F}\) by 2.39. Thus the only subspaces of \(\mathbf{F}\) are \(\{0\}\) and \(\mathbf{F}\).

Applying this to \(M = \operatorname{range} \varphi\) gives two cases. If \(\operatorname{range} \varphi = \{0\}\), then \(\varphi(v) = 0\) for every \(v \in V\), which means \(\varphi\) is the zero map. If \(\operatorname{range} \varphi = \mathbf{F}\), then \(\varphi\) is surjective by the definition of surjectivity (3.19). Hence every linear functional is surjective or is the zero map.

Here is the same argument written without reference to dimension, which shows that no finite-dimensionality hypothesis of any kind is needed. Suppose \(\varphi \in V^{\prime}\) is not the zero map. Then there exists \(u \in V\) with \(\varphi(u) \ne 0\). Set \(a = \varphi(u) \in \mathbf{F}\), so \(a \ne 0\) and \(a\) has a multiplicative inverse in \(\mathbf{F}\). Given any \(\lambda \in \mathbf{F}\), the vector \(v = (\lambda/a)u \in V\) satisfies

\begin{equation*} \varphi(v) = \varphi\!\left(\frac{\lambda}{a}\,u\right) = \frac{\lambda}{a}\,\varphi(u) = \frac{\lambda}{a}\cdot a = \lambda, \end{equation*}

where the second equality uses the homogeneity of the linear map \(\varphi\). Thus every \(\lambda \in \mathbf{F}\) is in \(\operatorname{range} \varphi\), so \(\varphi\) is surjective.

Note that the crucial feature of the situation is that the target space \(\mathbf{F}\) is one-dimensional; a linear map into a space of dimension at least \(2\) can of course be neither surjective nor the zero map.

Recall that \(\mathbf{R}^{[0,1]}\) denotes the vector space of all functions from the interval \([0,1]\) to \(\mathbf{R}\), with addition and scalar multiplication defined pointwise:

\begin{equation*} (f+g)(x) = f(x) + g(x), \qquad (\lambda f)(x) = \lambda\, f(x) \end{equation*}

for all \(f, g \in \mathbf{R}^{[0,1]}\), all \(\lambda \in \mathbf{R}\), and all \(x \in [0,1]\). A linear functional on this space is a linear map from \(\mathbf{R}^{[0,1]}\) to \(\mathbf{R}\).

Define \(\varphi_1, \varphi_2, \varphi_3 \colon \mathbf{R}^{[0,1]} \to \mathbf{R}\) by

\begin{equation*} \varphi_1(f) = f(0), \qquad \varphi_2(f) = f(1), \qquad \varphi_3(f) = f(0) + f(1). \end{equation*}

Each is a linear functional. Indeed, let \(f, g \in \mathbf{R}^{[0,1]}\) and \(\lambda \in \mathbf{R}\), and fix \(t \in [0,1]\). By the pointwise definitions above,

\begin{equation*} (f+g)(t) = f(t) + g(t) \qquad \text{and} \qquad (\lambda f)(t) = \lambda\, f(t). \end{equation*}

So the evaluation map \(f \mapsto f(t)\) is additive and homogeneous, hence a linear functional on \(\mathbf{R}^{[0,1]}\). Taking \(t = 0\) gives \(\varphi_1\) and taking \(t = 1\) gives \(\varphi_2\). Finally \(\varphi_3 = \varphi_1 + \varphi_2\), and a sum of linear maps is again a linear map (3.5, and 3.6, which says \(\mathcal{L}(V,W)\) is a vector space), so \(\varphi_3\) is a linear functional as well.

These three functionals are distinct. Let \(f\) be the function on \([0,1]\) defined by \(f(0) = 1\) and \(f(x) = 0\) for \(x \in (0,1]\), and let \(g\) be defined by \(g(1) = 1\) and \(g(x) = 0\) for \(x \in [0,1)\). Then

\begin{equation*} \varphi_1(f) = 1, \quad \varphi_2(f) = 0, \quad \varphi_3(f) = 1, \end{equation*}

\begin{equation*} \varphi_1(g) = 0, \quad \varphi_2(g) = 1, \quad \varphi_3(g) = 1. \end{equation*}

From the values at \(f\) we get \(\varphi_1 \ne \varphi_2\) and \(\varphi_2 \ne \varphi_3\); from the values at \(g\) we get \(\varphi_1 \ne \varphi_3\). Hence the three examples are pairwise distinct.

Two remarks. First, the same argument shows that for any \(t_1, \dots, t_k \in [0,1]\) and any \(c_1, \dots, c_k \in \mathbf{R}\), the map \(f \mapsto c_1 f(t_1) + \cdots + c_k f(t_k)\) is a linear functional on \(\mathbf{R}^{[0,1]}\); this manufactures infinitely many distinct examples. Second, the familiar functional \(f \mapsto \int_0^1 f\) is not available here, because \(\mathbf{R}^{[0,1]}\) contains all functions from \([0,1]\) to \(\mathbf{R}\), including functions that are not Riemann integrable (for instance the indicator function of the rationals in \([0,1]\)), so that formula does not define a function on all of \(\mathbf{R}^{[0,1]}\). It does define a linear functional on the subspace of continuous real-valued functions on \([0,1]\).

First note that the length-one list \(v\) is linearly independent. Indeed, if \(a \in \mathbf{F}\) and \(av = 0\) with \(a \ne 0\), then \(v = a^{-1}(av) = a^{-1}0 = 0\), contradicting the hypothesis \(v \ne 0\); hence \(av = 0\) forces \(a = 0\).

Because \(V\) is finite-dimensional, 2.32 tells us that the linearly independent list \(v\) can be extended to a basis of \(V\). Thus there is a basis \(v_1, \dots, v_n\) of \(V\) with \(v_1 = v\).

Now apply the linear map lemma (3.4): since \(v_1, \dots, v_n\) is a basis of \(V\), there exists a unique linear map \(\varphi \colon V \to \mathbf{F}\) satisfying

\begin{equation*} \varphi(v_1) = 1 \qquad \text{and} \qquad \varphi(v_j) = 0 \quad \text{for } j = 2, \dots, n. \end{equation*}

Since \(\varphi\) is a linear map from \(V\) to \(\mathbf{F}\), it is a linear functional on \(V\), that is, \(\varphi \in V^{\prime}\). And \(\varphi(v) = \varphi(v_1) = 1\), as required.

Equivalently, one can phrase the last step using the dual basis: let \(\varphi_1, \dots, \varphi_n \in V^{\prime}\) be the dual basis of \(v_1, \dots, v_n\) (3.112), which by definition satisfies \(\varphi_j(v_k) = 1\) if \(j = k\) and \(\varphi_j(v_k) = 0\) if \(j \ne k\). Then \(\varphi = \varphi_1\) works, since \(\varphi_1(v) = \varphi_1(v_1) = 1\).

Observe that once such a \(\varphi\) exists, it is far from unique when \(\dim V > 1\): for example, in the notation above, \(\varphi_1 + \varphi_2\) also sends \(v\) to \(1\). Note also that the conclusion fails for \(v = 0\), since every linear map sends \(0\) to \(0\) (3.10). Finally, finite-dimensionality is used only to guarantee that the linearly independent list \(v\) extends to a basis; the result holds for arbitrary vector spaces if one is willing to use the fact (proved with the axiom of choice, and not needed in this book) that every vector space has a basis.

Because \(V\) is finite-dimensional, so is its subspace \(U\) (2.25), and \(\dim U \le \dim V\) (2.37). Moreover \(\dim U \ne \dim V\): if \(\dim U = \dim V\), then \(U = V\) by 2.39, contradicting the hypothesis. Hence

\begin{equation*} m := \dim U < \dim V =: n . \end{equation*}

Let \(u_1, \dots, u_m\) be a basis of \(U\) (if \(U = \{0\}\) this is the empty list, and the argument below goes through unchanged). Since \(u_1, \dots, u_m\) is a linearly independent list in \(V\), it extends to a basis

\begin{equation*} u_1, \dots, u_m, v_{m+1}, \dots, v_n \end{equation*}

of \(V\) by 2.32. Because \(m < n\), the vector \(v_{m+1}\) really is present in this list.

By the linear map lemma (3.4), there is a (unique) linear map \(\varphi \colon V \to \mathbf{F}\) with

\begin{equation*} \varphi(u_j) = 0 \ \text{ for } j = 1, \dots, m, \qquad \varphi(v_{m+1}) = 1, \qquad \varphi(v_k) = 0 \ \text{ for } k = m+2, \dots, n, \end{equation*}

since the values of a linear map may be assigned arbitrarily on a basis. As a linear map from \(V\) to \(\mathbf{F}\), we have \(\varphi \in V^{\prime}\).

This \(\varphi\) has the two required properties.

First, \(\varphi\) vanishes on \(U\). Let \(u \in U\). Since \(u_1, \dots, u_m\) is a basis of \(U\), there exist \(a_1, \dots, a_m \in \mathbf{F}\) with \(u = a_1u_1 + \cdots + a_mu_m\). Then, by linearity of \(\varphi\),

\begin{equation*} \varphi(u) = a_1\varphi(u_1) + \cdots + a_m\varphi(u_m) = a_1 \cdot 0 + \cdots + a_m \cdot 0 = 0 . \end{equation*}

(In the degenerate case \(m = 0\), where \(U = \{0\}\) and the list \(u_1, \dots, u_m\) is empty, this reads: the only \(u \in U\) is \(u = 0\), and \(\varphi(0) = 0\) by 3.10. Note also that \(m < n\) holds automatically here, since \(U \ne V\) forces \(V \ne \{0\}\) and hence \(n \ge 1\).)

Second, \(\varphi \ne 0\), because \(\varphi(v_{m+1}) = 1 \ne 0\), whereas the zero linear functional takes the value \(0\) at every vector.

In the language of this section, the conclusion says precisely that \(U^0 \ne \{0\}\), and the exercise is thus the implication \(U \ne V \Rightarrow U^0 \ne \{0\}\), which is the contrapositive of one direction of 3.127(a). The proof above is a direct one, using only 2.32 and the linear map lemma, rather than the dimension formula 3.125.

Note also that the hypothesis \(U \ne V\) cannot be dropped: if \(U = V\), then a functional vanishing on all of \(U\) vanishes on all of \(V\) and is therefore the zero functional. Finally, the construction above yields slightly more than was asked: the functional \(\varphi\) produced satisfies \(\varphi(v_{m+1}) = 1\) for a vector \(v_{m+1} \notin U\), so it not only fails to be \(0\) but takes the value \(1\) somewhere outside \(U\). In this respect the proof runs parallel to that of Exercise 3F.3, with the basis of \(U\) playing the role that the single vector \(v\) played there.

First let us be clear about why the functions \(\varphi_1, \dots, \varphi_m\) are well defined. If \(v \in V\), then \(Tv \in \operatorname{range} T\), and \(w_1, \dots, w_m\) is a basis of \(\operatorname{range} T\); by 2.28, every vector in \(\operatorname{range} T\) has a unique representation as a linear combination of \(w_1, \dots, w_m\). So there is exactly one \(m\)-tuple of scalars, which we name \(\varphi_1(v), \dots, \varphi_m(v)\), with

\begin{equation*} Tv = \varphi_1(v)w_1 + \cdots + \varphi_m(v)w_m . \end{equation*}

This assigns to each \(v \in V\) exactly one scalar \(\varphi_j(v)\) for each \(j\), so each \(\varphi_j\) is a function from \(V\) to \(\mathbf{F}\). This uniqueness is the engine of the whole proof: it lets us conclude equality of coefficients from equality of vectors.

Additivity. Suppose \(u, v \in V\). Applying the defining equation to \(u\), to \(v\), and to \(u + v\), and using the additivity of \(T\), we get

\begin{equation*} \begin{aligned} \varphi_1(u+v)w_1 + \cdots + \varphi_m(u+v)w_m &= T(u+v) \\ &= Tu + Tv \\ &= \bigl(\varphi_1(u)w_1 + \cdots + \varphi_m(u)w_m\bigr) + \bigl(\varphi_1(v)w_1 + \cdots + \varphi_m(v)w_m\bigr) \\ &= \bigl(\varphi_1(u) + \varphi_1(v)\bigr)w_1 + \cdots + \bigl(\varphi_m(u) + \varphi_m(v)\bigr)w_m , \end{aligned} \end{equation*}

where the last step just collects terms. Now the first and last lines exhibit the vector \(T(u+v) \in \operatorname{range} T\) as two linear combinations of the basis \(w_1, \dots, w_m\). By the uniqueness in 2.28, the corresponding coefficients agree:

\begin{equation*} \varphi_j(u+v) = \varphi_j(u) + \varphi_j(v) \qquad \text{for each } j = 1, \dots, m . \end{equation*}

Homogeneity. Suppose \(\lambda \in \mathbf{F}\) and \(v \in V\). Using the homogeneity of \(T\),

\begin{equation*} \begin{aligned} \varphi_1(\lambda v)w_1 + \cdots + \varphi_m(\lambda v)w_m &= T(\lambda v) \\ &= \lambda(Tv) \\ &= \lambda\bigl(\varphi_1(v)w_1 + \cdots + \varphi_m(v)w_m\bigr) \\ &= \bigl(\lambda \varphi_1(v)\bigr)w_1 + \cdots + \bigl(\lambda \varphi_m(v)\bigr)w_m . \end{aligned} \end{equation*}

Again the first and last lines are two representations of the same vector of \(\operatorname{range} T\) in terms of the basis \(w_1, \dots, w_m\), so by uniqueness

\begin{equation*} \varphi_j(\lambda v) = \lambda \varphi_j(v) \qquad \text{for each } j = 1, \dots, m . \end{equation*}

Thus each \(\varphi_j\) is a function from \(V\) to \(\mathbf{F}\) that is additive and homogeneous, hence a linear map from \(V\) to \(\mathbf{F}\); that is, each \(\varphi_j\) is a linear functional on \(V\) (3.108), so \(\varphi_1, \dots, \varphi_m \in V^{\prime}\).

A remark that identifies these functionals. Let \(\psi_1, \dots, \psi_m\) be the dual basis of \(w_1, \dots, w_m\) in \((\operatorname{range} T)^{\prime}\). Then 3.114 says exactly that \(\psi_j(w) \) is the \(j\)-th coefficient of \(w\) in the basis \(w_1, \dots, w_m\), so \(\varphi_j(v) = \psi_j(Tv)\), that is,

\begin{equation*} \varphi_j = \psi_j \circ T = S^{\prime}(\psi_j), \end{equation*}

where \(S \in \mathcal{L}(V, \operatorname{range} T)\) is \(T\) with its target space narrowed to \(\operatorname{range} T\). Since the product (composition) of two linear maps is again a linear map (3.7), this gives a second, one-line proof of linearity. Note that no finite-dimensionality of \(V\) or \(W\) was needed anywhere; all that was used is that \(\operatorname{range} T\) has the finite basis \(w_1, \dots, w_m\).

We prove the two implications separately. Note that \(V\) is not assumed to be finite-dimensional, and none of what follows uses such an assumption.

First suppose there exists \(c \in \mathbf{F}\) with \(\beta = c\varphi\). Let \(v \in \operatorname{null}\varphi\), so \(\varphi(v) = 0\). Then

\begin{equation*} \beta(v) = (c\varphi)(v) = c\,\varphi(v) = c \cdot 0 = 0 , \end{equation*}

where the second equality is the definition of scalar multiplication in \(V^{\prime}\) (3.5). Hence \(v \in \operatorname{null}\beta\). This proves \(\operatorname{null} \varphi \subseteq \operatorname{null} \beta\).

Now suppose \(\operatorname{null} \varphi \subseteq \operatorname{null} \beta\); we must produce \(c \in \mathbf{F}\) with \(\beta = c\varphi\). There are two cases.

Case 1: \(\varphi = 0\). Then \(\operatorname{null}\varphi = V\), so the hypothesis gives \(V = \operatorname{null}\varphi \subseteq \operatorname{null}\beta \subseteq V\), forcing \(\operatorname{null}\beta = V\), i.e. \(\beta = 0\). Thus \(\beta = 0 = 0 \cdot \varphi\), and \(c = 0\) works (indeed every \(c \in \mathbf{F}\) works in this degenerate case).

Case 2: \(\varphi \ne 0\). By Exercise 3F.1, a nonzero linear functional is surjective, so there exists \(u \in V\) with

\begin{equation*} \varphi(u) = 1 . \end{equation*}

(Concretely: choose \(x\) with \(\varphi(x) = a \ne 0\) and set \(u = x/a\).) Define

\begin{equation*} c = \beta(u) \in \mathbf{F} . \end{equation*}

We claim \(\beta = c\varphi\). Let \(v \in V\) be arbitrary and put

\begin{equation*} w = v - \varphi(v)\,u . \end{equation*}

Then, using linearity of \(\varphi\) and \(\varphi(u) = 1\),

\begin{equation*} \varphi(w) = \varphi(v) - \varphi(v)\,\varphi(u) = \varphi(v) - \varphi(v) = 0 , \end{equation*}

so \(w \in \operatorname{null}\varphi\). By hypothesis \(w \in \operatorname{null}\beta\), that is, \(\beta(w) = 0\). Applying the linear functional \(\beta\) to \(v = w + \varphi(v)u\) therefore yields

\begin{equation*} \beta(v) = \beta(w) + \varphi(v)\,\beta(u) = 0 + \varphi(v)\, c = c\,\varphi(v) = (c\varphi)(v) . \end{equation*}

Since \(v \in V\) was arbitrary, \(\beta = c\varphi\), completing Case 2 and the proof.

Two comments. First, in Case 2 the scalar \(c\) is unique: if \(c_1\varphi = c_2\varphi\) with \(\varphi \ne 0\), pick \(u\) with \(\varphi(u) = 1\) to get \(c_1 = c_2\). Second, the geometric content is that a nonzero linear functional is determined up to a scalar by its null space: by 3.21 (when \(V\) is finite-dimensional) the null space of a nonzero \(\varphi \in V^{\prime}\) has dimension \(\dim V - 1\), i.e. it is a hyperplane, and the exercise says that \(\beta\) vanishes on that hyperplane exactly when \(\beta\) is a multiple of \(\varphi\). The decomposition \(v = w + \varphi(v)u\) used above is precisely the direct sum decomposition \(V = \operatorname{null}\varphi \oplus \operatorname{span}(u)\).

Write \(V = V_1 \times \cdots \times V_m\), whose vector space operations are defined coordinatewise (3.87 and 3.89):

\begin{equation*} (u_1, \dots, u_m) + (v_1, \dots, v_m) = (u_1 + v_1, \dots, u_m + v_m), \qquad \lambda(v_1, \dots, v_m) = (\lambda v_1, \dots, \lambda v_m). \end{equation*}

The space \(V_1{}^{\prime} \times \cdots \times V_m{}^{\prime}\) is likewise a product of vector spaces, with coordinatewise operations. Since no finite-dimensionality is assumed, we cannot argue by comparing dimensions (3.70); instead we exhibit an explicit isomorphism, that is, an invertible linear map (3.69).

For each \(j \in \{1, \dots, m\}\) define the inclusion \(\iota_j \colon V_j \to V\) by

\begin{equation*} \iota_j(v) = (0, \dots, 0, v, 0, \dots, 0), \end{equation*}

with \(v\) in the \(j\)-th slot and the zero vector of \(V_k\) in slot \(k\) for \(k \ne j\). Each \(\iota_j\) is linear, because the operations on \(V\) are coordinatewise: \(\iota_j(u + v) = \iota_j(u) + \iota_j(v)\) and \(\iota_j(\lambda v) = \lambda \iota_j(v)\), the zero coordinates behaving correctly since \(0 + 0 = 0\) and \(\lambda 0 = 0\). Note also the decomposition identity, valid for every \((v_1, \dots, v_m) \in V\):

\begin{equation*} (v_1, \dots, v_m) = \iota_1(v_1) + \cdots + \iota_m(v_m). \end{equation*}

Define

\begin{equation*} \Gamma \colon V^{\prime} \to V_1{}^{\prime} \times \cdots \times V_m{}^{\prime}, \qquad \Gamma(\varphi) = (\varphi \circ \iota_1, \dots, \varphi \circ \iota_m). \end{equation*}

This makes sense: if \(\varphi \in V^{\prime}\), then each \(\varphi \circ \iota_j\) is a composition of linear maps \(V_j \to V \to \mathbf{F}\), hence a linear map from \(V_j\) to \(\mathbf{F}\) (3.7), i.e. an element of \(V_j{}^{\prime}\).

\(\Gamma\) is linear. Let \(\varphi, \psi \in V^{\prime}\) and \(\lambda \in \mathbf{F}\). For each \(j\) and each \(v \in V_j\),

\begin{equation*} \bigl((\varphi + \psi) \circ \iota_j\bigr)(v) = (\varphi + \psi)(\iota_j(v)) = \varphi(\iota_j(v)) + \psi(\iota_j(v)) = (\varphi \circ \iota_j)(v) + (\psi \circ \iota_j)(v), \end{equation*}

using the definition of addition in \(V^{\prime}\) (3.5); hence \((\varphi + \psi) \circ \iota_j = \varphi \circ \iota_j + \psi \circ \iota_j\). Similarly \((\lambda\varphi) \circ \iota_j = \lambda(\varphi \circ \iota_j)\). Since addition and scalar multiplication in \(V_1{}^{\prime} \times \cdots \times V_m{}^{\prime}\) are coordinatewise, these identities say exactly that

\begin{equation*} \Gamma(\varphi + \psi) = \Gamma(\varphi) + \Gamma(\psi) \qquad \text{and} \qquad \Gamma(\lambda\varphi) = \lambda\,\Gamma(\varphi). \end{equation*}

\(\Gamma\) is injective. Suppose \(\Gamma(\varphi) = 0\), i.e. \(\varphi \circ \iota_j = 0\) for each \(j\). Let \((v_1, \dots, v_m) \in V\). By the decomposition identity above and linearity of \(\varphi\),

\begin{equation*} \varphi(v_1, \dots, v_m) = \varphi(\iota_1(v_1)) + \cdots + \varphi(\iota_m(v_m)) = 0 + \cdots + 0 = 0 . \end{equation*}

Hence \(\varphi = 0\), so \(\operatorname{null}\Gamma = \{0\}\) and \(\Gamma\) is injective by 3.15.

\(\Gamma\) is surjective. Let \((\varphi_1, \dots, \varphi_m) \in V_1{}^{\prime} \times \cdots \times V_m{}^{\prime}\). Define \(\varphi \colon V \to \mathbf{F}\) by

\begin{equation*} \varphi(v_1, \dots, v_m) = \varphi_1(v_1) + \cdots + \varphi_m(v_m). \end{equation*}

Then \(\varphi\) is linear: for \((u_1, \dots, u_m), (v_1, \dots, v_m) \in V\) and \(\lambda \in \mathbf{F}\),

\begin{equation*} \begin{aligned} \varphi\bigl((u_1, \dots, u_m) + (v_1, \dots, v_m)\bigr) &= \varphi_1(u_1 + v_1) + \cdots + \varphi_m(u_m + v_m) \\ &= \bigl(\varphi_1(u_1) + \cdots + \varphi_m(u_m)\bigr) + \bigl(\varphi_1(v_1) + \cdots + \varphi_m(v_m)\bigr) \\ &= \varphi(u_1, \dots, u_m) + \varphi(v_1, \dots, v_m), \end{aligned} \end{equation*}

and

\begin{equation*} \varphi\bigl(\lambda(v_1, \dots, v_m)\bigr) = \varphi_1(\lambda v_1) + \cdots + \varphi_m(\lambda v_m) = \lambda\bigl(\varphi_1(v_1) + \cdots + \varphi_m(v_m)\bigr) = \lambda\,\varphi(v_1, \dots, v_m), \end{equation*}

each step using linearity of the individual \(\varphi_j\). Thus \(\varphi \in V^{\prime}\). Moreover, for each \(j\) and each \(v \in V_j\),

\begin{equation*} (\varphi \circ \iota_j)(v) = \varphi(0, \dots, 0, v, 0, \dots, 0) = \varphi_1(0) + \cdots + \varphi_j(v) + \cdots + \varphi_m(0) = \varphi_j(v), \end{equation*}

since a linear map sends \(0\) to \(0\) (3.10). Hence \(\varphi \circ \iota_j = \varphi_j\) for each \(j\), so \(\Gamma(\varphi) = (\varphi_1, \dots, \varphi_m)\), proving surjectivity.

Being linear, injective, and surjective, \(\Gamma\) is invertible (3.63), hence an isomorphism (3.69). Therefore

\begin{equation*} (V_1 \times \cdots \times V_m)^{\prime} \quad \text{and} \quad V_1{}^{\prime} \times \cdots \times V_m{}^{\prime} \end{equation*}

are isomorphic vector spaces.

Two remarks. The surjectivity argument in fact exhibits the inverse map explicitly: \(\Gamma^{-1}(\varphi_1, \dots, \varphi_m)\) is the functional \((v_1, \dots, v_m) \mapsto \varphi_1(v_1) + \cdots + \varphi_m(v_m)\). And in the finite-dimensional case the result is also visible through dimensions: \(\dim (V_1 \times \cdots \times V_m)^{\prime} = \dim(V_1 \times \cdots \times V_m) = \dim V_1 + \cdots + \dim V_m\) by 3.111 and 3.92, while \(\dim (V_1{}^{\prime} \times \cdots \times V_m{}^{\prime}) = \dim V_1{}^{\prime} + \cdots + \dim V_m{}^{\prime} = \dim V_1 + \cdots + \dim V_m\), so 3.70 gives the isomorphism. The proof above has the advantage of requiring no dimension hypothesis and of being natural, i.e. constructed without choosing bases.

First note that both maps are linear. Each \(\varphi_j\) is a linear functional on \(V\), so for \(u, v \in V\) and \(\lambda \in \mathbf{F}\) we have

\begin{equation*} \Gamma(u + \lambda v) = (\varphi_1(u) + \lambda\varphi_1(v), \dots, \varphi_n(u) + \lambda\varphi_n(v)) = \Gamma(u) + \lambda\Gamma(v), \end{equation*}

because addition and scalar multiplication in \(\mathbf{F}^n\) are defined coordinatewise. Similarly \(\Lambda\) is linear, since

\begin{equation*} \Lambda\big((a_1,\dots,a_n) + \lambda(b_1,\dots,b_n)\big) = \sum_{k=1}^n (a_k + \lambda b_k) v_k = \sum_{k=1}^n a_k v_k + \lambda \sum_{k=1}^n b_k v_k . \end{equation*}

Thus \(\Gamma \in \mathcal{L}(V, \mathbf{F}^n)\) and \(\Lambda \in \mathcal{L}(\mathbf{F}^n, V)\), and it makes sense to ask whether they are inverses.

Now we show \(\Lambda \Gamma\) is the identity on \(V\). If \(v \in V\), then

\begin{equation*} (\Lambda \circ \Gamma)(v) = \Lambda(\varphi_1(v), \dots, \varphi_n(v)) = \varphi_1(v) v_1 + \cdots + \varphi_n(v) v_n = v, \end{equation*}

where the last equality is exactly 3.114 (the dual basis gives the coefficients for writing \(v\) as a linear combination of \(v_1, \dots, v_n\)). Hence \(\Lambda \Gamma = I_V\).

Next we show \(\Gamma \Lambda\) is the identity on \(\mathbf{F}^n\). Let \((a_1, \dots, a_n) \in \mathbf{F}^n\) and put \(v = a_1 v_1 + \cdots + a_n v_n\). For each \(j \in \{1, \dots, n\}\), linearity of \(\varphi_j\) and the defining property \(\varphi_j(v_k) = 1\) if \(k = j\) and \(\varphi_j(v_k) = 0\) if \(k \neq j\) (see 3.112) give

\begin{equation*} \varphi_j(v) = \sum_{k=1}^n a_k \varphi_j(v_k) = a_j . \end{equation*}

Therefore

\begin{equation*} (\Gamma \circ \Lambda)(a_1, \dots, a_n) = \Gamma(v) = (\varphi_1(v), \dots, \varphi_n(v)) = (a_1, \dots, a_n), \end{equation*}

so \(\Gamma \Lambda = I_{\mathbf{F}^n}\).

Because \(\Lambda \Gamma = I_V\) and \(\Gamma \Lambda = I_{\mathbf{F}^n}\), the map \(\Lambda\) is by definition (3.59) an inverse of \(\Gamma\), and \(\Gamma\) is an inverse of \(\Lambda\). Hence both maps are invertible, and since inverses are unique (3.60) we may write \(\Gamma^{-1} = \Lambda\) and \(\Lambda^{-1} = \Gamma\).

Informally: \(\Lambda\) turns a list of scalars into the vector having those scalars as coordinates with respect to \(v_1, \dots, v_n\), and \(\Gamma\) reads off the coordinates of a vector with respect to that same basis. These operations undo each other, which is precisely the content of 3.114. In particular \(\Gamma\) is an invertible linear map from \(V\) onto \(\mathbf{F}^n\), so it is an isomorphism, and \(V\) and \(\mathbf{F}^n\) are isomorphic vector spaces whenever \(\dim V = n\).

Write \(v_j = x^j\) for \(j = 0, 1, \dots, m\), so that \(v_0, v_1, \dots, v_m\) is the given basis of \(\mathcal{P}_m(\mathbf{R})\). By 3.112, the dual basis of this list is the unique list \(\varphi_0, \dots, \varphi_m\) of linear functionals on \(\mathcal{P}_m(\mathbf{R})\) satisfying

\begin{equation*} \varphi_k(v_j) = \begin{cases} 1 & \text{if } j = k, \\ 0 & \text{if } j \neq k. \end{cases} \end{equation*}

(Uniqueness holds because a linear map on \(\mathcal{P}_m(\mathbf{R})\) is determined by its values on a basis; see the linear map lemma 3.4.) So there are two things to check for the functionals \(\varphi_k(p) = p^{(k)}(0)/k!\): that each is a linear functional, and that they satisfy the displayed condition.

Step 1: each \(\varphi_k\) is a linear functional. Fix \(k \in \{0, 1, \dots, m\}\). If \(p \in \mathcal{P}_m(\mathbf{R})\) then \(p^{(k)}\) is again a polynomial function on \(\mathbf{R}\), so \(p^{(k)}(0)/k!\) is a well-defined real number; thus \(\varphi_k\) maps \(\mathcal{P}_m(\mathbf{R})\) into \(\mathbf{R}\). Differentiation is linear, hence so is its \(k\)-fold composition: \((p + \lambda q)^{(k)} = p^{(k)} + \lambda q^{(k)}\) for all \(p, q \in \mathcal{P}_m(\mathbf{R})\) and \(\lambda \in \mathbf{R}\). Evaluation at \(0\) and division by the fixed nonzero scalar \(k!\) are also linear. Hence

\begin{equation*} \varphi_k(p + \lambda q) = \frac{(p + \lambda q)^{(k)}(0)}{k!} = \frac{p^{(k)}(0) + \lambda q^{(k)}(0)}{k!} = \varphi_k(p) + \lambda \varphi_k(q), \end{equation*}

so \(\varphi_k \in \big(\mathcal{P}_m(\mathbf{R})\big)^{\prime}\).

Step 2: the duality condition. Fix \(j, k \in \{0, 1, \dots, m\}\) and compute \(\varphi_k(x^j)\), considering three cases.

If \(k > j\), then differentiating \(x^j\) more than \(j\) times gives the zero polynomial: indeed \((x^j)^{(j)}\) is the constant \(j!\), and every further derivative is \(0\). Hence \((x^j)^{(k)} = 0\), so \(\varphi_k(x^j) = 0\).

If \(k < j\), then repeated differentiation gives

\begin{equation*} (x^j)^{(k)} = j(j-1)\cdots(j-k+1)\, x^{\,j-k}, \end{equation*}

and \(j - k \geq 1\), so evaluating at \(0\) gives \(0\). Hence \(\varphi_k(x^j) = 0\).

If \(k = j\), then by the formula above with \(k\) replaced by \(j\) in the last factor, \((x^k)^{(k)} = k(k-1)\cdots 1 = k!\), the constant polynomial. Evaluating at \(0\) gives \(k!\), so

\begin{equation*} \varphi_k(x^k) = \frac{k!}{k!} = 1 . \end{equation*}

Combining the three cases, \(\varphi_k(v_j) = 1\) if \(j = k\) and \(\varphi_k(v_j) = 0\) if \(j \neq k\), which is exactly the condition in 3.112. Therefore \(\varphi_0, \varphi_1, \dots, \varphi_m\) is the dual basis of \(1, x, \dots, x^m\), as claimed.

As a consistency check, 3.114 now says that every \(p \in \mathcal{P}_m(\mathbf{R})\) satisfies

\begin{equation*} p = \varphi_0(p) \cdot 1 + \varphi_1(p) x + \cdots + \varphi_m(p) x^m = \sum_{k=0}^{m} \frac{p^{(k)}(0)}{k!} x^k, \end{equation*}

which is the Taylor expansion of \(p\) at \(0\). This is a polynomial identity, not a limiting statement, because the sum terminates.

(a) For \(j = 0, 1, \dots, m\) set \(v_j = (x-5)^j\), meaning the polynomial function \(x \mapsto (x-5)^j\). Expanding by the binomial theorem shows \(v_j \in \mathcal{P}_m(\mathbf{R})\), since \(v_j\) has degree \(j \leq m\). The list \(v_0, v_1, \dots, v_m\) has length \(m + 1\), and \(\dim \mathcal{P}_m(\mathbf{R}) = m + 1\) (see 2.36, since the standard basis \(1, x, \dots, x^m\) has length \(m+1\)). So by 2.38, it suffices to prove that the list is linearly independent.

Suppose \(a_0, a_1, \dots, a_m \in \mathbf{R}\) satisfy

\begin{equation*} a_0 + a_1(x-5) + \cdots + a_m (x-5)^m = 0, \end{equation*}

where \(0\) denotes the zero polynomial. Suppose, seeking a contradiction, that not all the \(a_j\) equal \(0\), and let \(k\) be the largest index with \(a_k \neq 0\). Then the left side equals \(\sum_{j=0}^{k} a_j (x-5)^j\). Each term \(a_j (x-5)^j\) with \(j < k\) is a polynomial of degree at most \(k - 1\), while \(a_k(x-5)^k\) has degree exactly \(k\) with leading coefficient \(a_k\) (the coefficient of \(x^k\) in \((x-5)^k\) is \(1\)). Hence the coefficient of \(x^k\) in the sum is \(a_k \neq 0\), so the sum is a polynomial of degree \(k\), and in particular is not the zero polynomial. This contradiction shows \(a_0 = a_1 = \cdots = a_m = 0\).

Thus \(1, x-5, \dots, (x-5)^m\) is a linearly independent list of length \(\dim \mathcal{P}_m(\mathbf{R})\) in \(\mathcal{P}_m(\mathbf{R})\), and therefore is a basis of \(\mathcal{P}_m(\mathbf{R})\) by 2.38.

(b) The dual basis is \(\varphi_0, \varphi_1, \dots, \varphi_m\), where

\begin{equation*} \varphi_k(p) = \frac{p^{(k)}(5)}{k!} . \end{equation*}

That is, the dual basis is obtained from the one in Exercise 9 by evaluating derivatives at \(5\) instead of at \(0\).

To verify this, first note that each \(\varphi_k\) is a linear functional on \(\mathcal{P}_m(\mathbf{R})\): differentiating \(k\) times, evaluating at the point \(5\), and dividing by the fixed nonzero scalar \(k!\) are each linear operations, so

\begin{equation*} \varphi_k(p + \lambda q) = \frac{p^{(k)}(5) + \lambda q^{(k)}(5)}{k!} = \varphi_k(p) + \lambda\varphi_k(q) \end{equation*}

for all \(p, q \in \mathcal{P}_m(\mathbf{R})\) and \(\lambda \in \mathbf{R}\).

Now fix \(j, k \in \{0, 1, \dots, m\}\) and compute \(\varphi_k\big((x-5)^j\big)\). Repeated differentiation of \((x-5)^j\) gives, for \(k \leq j\),

\begin{equation*} \big((x-5)^j\big)^{(k)} = j(j-1)\cdots(j-k+1)\,(x-5)^{\,j-k}, \end{equation*}

and \(\big((x-5)^j\big)^{(k)} = 0\) for \(k > j\) (once the exponent reaches \(0\) the polynomial is constant, and one more derivative kills it).

If \(k > j\), we get \(\varphi_k\big((x-5)^j\big) = 0\).

If \(k < j\), then \(j - k \geq 1\), so the displayed derivative vanishes at \(x = 5\); hence \(\varphi_k\big((x-5)^j\big) = 0\).

If \(k = j\), then the displayed derivative is the constant \(k(k-1)\cdots 1 = k!\), so

\begin{equation*} \varphi_k\big((x-5)^k\big) = \frac{k!}{k!} = 1 . \end{equation*}

Thus \(\varphi_k(v_j) = 1\) if \(j = k\) and \(\varphi_k(v_j) = 0\) if \(j \neq k\), which is precisely the condition in 3.112 defining the dual basis of \(v_0, \dots, v_m\). Hence \(\varphi_0, \dots, \varphi_m\) is the dual basis of \(1, x-5, \dots, (x-5)^m\).

By 3.114 this gives, for every \(p \in \mathcal{P}_m(\mathbf{R})\), the identity

\begin{equation*} p = \sum_{k=0}^{m} \frac{p^{(k)}(5)}{k!} (x-5)^k, \end{equation*}

the Taylor expansion of \(p\) centered at \(5\).

Because \(V\) has a basis of length \(n\), the space \(V\) is finite-dimensional with \(\dim V = n\). Set

\begin{equation*} \eta = \psi(v_1)\varphi_1 + \cdots + \psi(v_n)\varphi_n . \end{equation*}

Each \(\varphi_k\) belongs to \(V^{\prime}\) and each \(\psi(v_k)\) is a scalar, so \(\eta \in V^{\prime}\) because \(V^{\prime}\) is a vector space (3.110) and hence closed under linear combinations. Thus \(\psi\) and \(\eta\) are both linear functionals on \(V\), and to prove they are equal it suffices to show they agree on the basis \(v_1, \dots, v_n\): two linear maps on \(V\) that agree on a basis agree on \(\operatorname{span}(v_1, \dots, v_n) = V\), since every \(v \in V\) can be written \(v = c_1 v_1 + \cdots + c_n v_n\) and then linearity gives \(\psi(v) = \sum_k c_k \psi(v_k) = \sum_k c_k \eta(v_k) = \eta(v)\).

So let \(j \in \{1, \dots, n\}\) and evaluate \(\eta\) at \(v_j\). Using the definition of the dual basis (3.112), namely \(\varphi_k(v_j) = 1\) if \(k = j\) and \(\varphi_k(v_j) = 0\) if \(k \neq j\), we get

\begin{equation*} \eta(v_j) = \sum_{k=1}^{n} \psi(v_k)\varphi_k(v_j) = \psi(v_j), \end{equation*}

since every term with \(k \neq j\) vanishes and the term with \(k = j\) equals \(\psi(v_j) \cdot 1\).

Hence \(\eta(v_j) = \psi(v_j)\) for each \(j \in \{1, \dots, n\}\), and by the paragraph above \(\eta = \psi\). That is,

\begin{equation*} \psi = \psi(v_1)\varphi_1 + \cdots + \psi(v_n)\varphi_n , \end{equation*}

as desired.

Two remarks. First, this result is the dual counterpart of 3.114: whereas 3.114 says the dual basis supplies the coefficients for expanding a vector in the basis \(v_1, \dots, v_n\), this exercise says the original basis supplies the coefficients for expanding a linear functional in the dual basis \(\varphi_1, \dots, \varphi_n\).

Second, one can also argue via 3.116: that result tells us \(\varphi_1, \dots, \varphi_n\) is a basis of \(V^{\prime}\), so there exist unique scalars \(a_1, \dots, a_n \in \mathbf{F}\) with \(\psi = a_1\varphi_1 + \cdots + a_n\varphi_n\). Applying both sides to \(v_j\) and using \(\varphi_k(v_j) = 0\) for \(k \neq j\) gives \(\psi(v_j) = a_j\), which yields the same conclusion. The proof given first is slightly more economical because it does not need 3.116.

Throughout, recall the definition of the dual map (3.118): if \(R \in \mathcal{L}(V, W)\), then \(R^{\prime} \in \mathcal{L}(W^{\prime}, V^{\prime})\) is defined by \(R^{\prime}(\varphi) = \varphi \circ R\) for each \(\varphi \in W^{\prime}\). Both sides of each identity to be proved are functions from \(W^{\prime}\) to \(V^{\prime}\), so they are equal precisely when they take the same value at every \(\varphi \in W^{\prime}\); and those values are linear functionals on \(V\), so they in turn are equal precisely when they agree at every \(v \in V\).

(a) Let \(\varphi \in W^{\prime}\) and let \(v \in V\). Then

\begin{equation*} \begin{aligned} \big((S+T)^{\prime}(\varphi)\big)(v) &= \big(\varphi \circ (S+T)\big)(v) \\ &= \varphi\big((S+T)v\big) \\ &= \varphi(Sv + Tv) \\ &= \varphi(Sv) + \varphi(Tv) \\ &= (\varphi \circ S)(v) + (\varphi \circ T)(v) \\ &= \big(S^{\prime}(\varphi)\big)(v) + \big(T^{\prime}(\varphi)\big)(v) \\ &= \big(S^{\prime}(\varphi) + T^{\prime}(\varphi)\big)(v). \end{aligned} \end{equation*}

Here the first and sixth equalities use the definition of the dual map, the third uses the definition of the sum \(S+T\) of linear maps (3.5), the fourth uses the additivity of the linear functional \(\varphi\), and the last uses the definition of addition in \(V^{\prime}\) (pointwise addition of functions).

Since this holds for every \(v \in V\), we conclude \((S+T)^{\prime}(\varphi) = S^{\prime}(\varphi) + T^{\prime}(\varphi)\). By the definition of addition in \(\mathcal{L}(W^{\prime}, V^{\prime})\), the right side equals \((S^{\prime} + T^{\prime})(\varphi)\). Since \(\varphi \in W^{\prime}\) was arbitrary, \((S+T)^{\prime} = S^{\prime} + T^{\prime}\).

(b) Let \(\lambda \in \mathbf{F}\), let \(\varphi \in W^{\prime}\), and let \(v \in V\). Then

\begin{equation*} \begin{aligned} \big((\lambda T)^{\prime}(\varphi)\big)(v) &= \big(\varphi \circ (\lambda T)\big)(v) \\ &= \varphi\big((\lambda T)v\big) \\ &= \varphi(\lambda\, Tv) \\ &= \lambda\, \varphi(Tv) \\ &= \lambda\, \big(T^{\prime}(\varphi)\big)(v) \\ &= \big(\lambda\, T^{\prime}(\varphi)\big)(v). \end{aligned} \end{equation*}

The first and fifth equalities use the definition of the dual map, the third uses the definition of the scalar multiple \(\lambda T\) of a linear map (3.5), the fourth uses the homogeneity of the linear functional \(\varphi\), and the last uses the definition of scalar multiplication in \(V^{\prime}\) (pointwise).

Since this holds for every \(v \in V\), we get \((\lambda T)^{\prime}(\varphi) = \lambda\, T^{\prime}(\varphi)\), and the right side equals \((\lambda T^{\prime})(\varphi)\) by the definition of scalar multiplication in \(\mathcal{L}(W^{\prime}, V^{\prime})\). Since \(\varphi \in W^{\prime}\) was arbitrary, \((\lambda T)^{\prime} = \lambda T^{\prime}\).

Together, (a) and (b) say that the map \(T \mapsto T^{\prime}\) from \(\mathcal{L}(V, W)\) to \(\mathcal{L}(W^{\prime}, V^{\prime})\) is itself linear, as remarked in the text just before 3.120.

Let \(I\) denote the identity operator on \(V\), so \(I \in \mathcal{L}(V, V)\) and \(Iv = v\) for every \(v \in V\). Its dual map \(I^{\prime}\) is then an element of \(\mathcal{L}(V^{\prime}, V^{\prime})\), so it is an operator on \(V^{\prime}\); we must show it is the identity operator on \(V^{\prime}\).

Let \(\varphi \in V^{\prime}\). By the definition of the dual map (3.118),

\begin{equation*} I^{\prime}(\varphi) = \varphi \circ I . \end{equation*}

To see that \(\varphi \circ I = \varphi\), let \(v \in V\); then

\begin{equation*} (\varphi \circ I)(v) = \varphi(Iv) = \varphi(v). \end{equation*}

Since \(\varphi \circ I\) and \(\varphi\) are functions on \(V\) taking the same value at every \(v \in V\), they are equal. Hence \(I^{\prime}(\varphi) = \varphi\).

Because \(\varphi \in V^{\prime}\) was arbitrary, \(I^{\prime}\) sends every element of \(V^{\prime}\) to itself. Thus \(I^{\prime}\) is the identity operator on \(V^{\prime}\), as desired.

Remark: combined with 3.120(c), which states \((ST)^{\prime} = T^{\prime}S^{\prime}\), this gives a quick proof of one direction of Exercise 17 of this section. Suppose \(T \in \mathcal{L}(V, W)\) is invertible, with inverse \(T^{-1} \in \mathcal{L}(W, V)\). Taking duals in \(T^{-1}T = I_V\) and using 3.120(c) gives \(T^{\prime}(T^{-1})^{\prime} = (I_V)^{\prime} = I_{V^{\prime}}\); taking duals in \(TT^{-1} = I_W\) gives \((T^{-1})^{\prime}T^{\prime} = (I_W)^{\prime} = I_{W^{\prime}}\). Since \((T^{-1})^{\prime} \in \mathcal{L}(V^{\prime}, W^{\prime})\) and \(T^{\prime} \in \mathcal{L}(W^{\prime}, V^{\prime})\), these two equations say exactly that \(T^{\prime}\) is invertible with \((T^{\prime})^{-1} = (T^{-1})^{\prime}\). (The converse direction of Exercise 17 needs a separate argument, for instance via 3.129 and 3.131.)

By 3.113, the dual basis of the standard basis consists of the coordinate-selection functionals. Thus

\begin{equation*} \varphi_1(a, b) = a, \qquad \varphi_2(a, b) = b \qquad \text{for } (a,b) \in \mathbf{R}^2, \end{equation*}

and

\begin{equation*} \psi_1(x, y, z) = x, \qquad \psi_2(x, y, z) = y, \qquad \psi_3(x, y, z) = z \qquad \text{for } (x,y,z) \in \mathbf{R}^3 . \end{equation*}

Note that \(T^{\prime} \in \mathcal{L}\big((\mathbf{R}^2)^{\prime}, (\mathbf{R}^3)^{\prime}\big)\), so \(T^{\prime}(\varphi_1)\) and \(T^{\prime}(\varphi_2)\) are linear functionals on \(\mathbf{R}^3\).

(a) By the definition of the dual map (3.118), \(T^{\prime}(\varphi_j) = \varphi_j \circ T\). Hence for \((x,y,z) \in \mathbf{R}^3\),

\begin{equation*} \big(T^{\prime}(\varphi_1)\big)(x, y, z) = \varphi_1\big(T(x,y,z)\big) = \varphi_1(4x + 5y + 6z,\; 7x + 8y + 9z) = 4x + 5y + 6z, \end{equation*}

since \(\varphi_1\) selects the first coordinate. Similarly, since \(\varphi_2\) selects the second coordinate,

\begin{equation*} \big(T^{\prime}(\varphi_2)\big)(x, y, z) = \varphi_2\big(T(x,y,z)\big) = 7x + 8y + 9z . \end{equation*}

So \(T^{\prime}(\varphi_1)\) is the linear functional on \(\mathbf{R}^3\) taking \((x,y,z)\) to \(4x + 5y + 6z\), and \(T^{\prime}(\varphi_2)\) is the linear functional on \(\mathbf{R}^3\) taking \((x,y,z)\) to \(7x + 8y + 9z\). In words: \(T^{\prime}(\varphi_j)\) reads off the \(j^{\text{th}}\) coordinate of the output of \(T\), which is the corresponding row of the formula defining \(T\).

(b) Using the formulas for \(\psi_1, \psi_2, \psi_3\) above, for every \((x,y,z) \in \mathbf{R}^3\) we have

\begin{equation*} (4\psi_1 + 5\psi_2 + 6\psi_3)(x,y,z) = 4x + 5y + 6z = \big(T^{\prime}(\varphi_1)\big)(x,y,z), \end{equation*}

and

\begin{equation*} (7\psi_1 + 8\psi_2 + 9\psi_3)(x,y,z) = 7x + 8y + 9z = \big(T^{\prime}(\varphi_2)\big)(x,y,z). \end{equation*}

Since these are equalities of functions at every point of \(\mathbf{R}^3\),

\begin{equation*} T^{\prime}(\varphi_1) = 4\psi_1 + 5\psi_2 + 6\psi_3 \qquad \text{and} \qquad T^{\prime}(\varphi_2) = 7\psi_1 + 8\psi_2 + 9\psi_3 . \end{equation*}

(Alternatively, part (b) follows from Exercise 11 of this section: the coefficient of \(\psi_k\) in the expansion of a functional \(\eta\) on \(\mathbf{R}^3\) is \(\eta(e_k)\), where \(e_1, e_2, e_3\) is the standard basis. Here \(\big(T^{\prime}(\varphi_1)\big)(e_1) = 4\), \(\big(T^{\prime}(\varphi_1)\big)(e_2) = 5\), \(\big(T^{\prime}(\varphi_1)\big)(e_3) = 6\), and similarly \(7, 8, 9\) for \(T^{\prime}(\varphi_2)\).)

This illustrates 3.132. With respect to the standard bases,

\begin{equation*} \mathcal{M}(T) = \begin{pmatrix} 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}, \end{equation*}

so 3.132 predicts that the matrix of \(T^{\prime}\) with respect to the dual bases is

\begin{equation*} \mathcal{M}(T^{\prime}) = \big(\mathcal{M}(T)\big)^{\mathrm{t}} = \begin{pmatrix} 4 & 7 \\ 5 & 8 \\ 6 & 9 \end{pmatrix}. \end{equation*}

The columns of this matrix are the coefficient lists \((4,5,6)\) and \((7,8,9)\) found above, exactly as the computation in (b) shows.

Recall the definition of the dual map (3.118): for \(T \in \mathcal{L}(V,W)\) and \(\varphi \in W^{\prime}\), the linear functional \(T^{\prime}(\varphi) \in V^{\prime}\) is \(\varphi \circ T\), so that \((T^{\prime}(\varphi))(v) = \varphi(Tv)\) for every \(v \in V\). Here \(V = W = \mathcal{P}(\mathbb{R})\), so \(T^{\prime}(\varphi)\) is again a linear functional on \(\mathcal{P}(\mathbb{R})\).

(a) Let \(p \in \mathcal{P}(\mathbb{R})\). By the product rule,

\begin{equation*} (Tp)^{\prime}(x) = \frac{d}{dx}\bigl(x^2 p(x) + p^{\prime\prime}(x)\bigr) = 2x\,p(x) + x^2 p^{\prime}(x) + p^{\prime\prime\prime}(x). \end{equation*}

Hence

\begin{equation*} (T^{\prime}(\varphi))(p) = \varphi(Tp) = (Tp)^{\prime}(4) = 8p(4) + 16\,p^{\prime}(4) + p^{\prime\prime\prime}(4). \end{equation*}

Thus \(T^{\prime}(\varphi)\) is the linear functional on \(\mathcal{P}(\mathbb{R})\) that takes \(p\) to

\begin{equation*} 8p(4) + 16\,p^{\prime}(4) + p^{\prime\prime\prime}(4). \end{equation*}

(b) Here \(x^3\) denotes the polynomial \(p\) with \(p(x) = x^3\). Then \(p^{\prime\prime}(x) = 6x\), so

\begin{equation*} (Tp)(x) = x^2 \cdot x^3 + 6x = x^5 + 6x. \end{equation*}

Therefore

\begin{equation*} (T^{\prime}(\varphi))(x^3) = \varphi(Tp) = \int_0^1 (x^5 + 6x)\,dx = \left[\frac{x^6}{6} + 3x^2\right]_0^1 = \frac{1}{6} + 3 = \frac{19}{6}. \end{equation*}

First suppose \(T = 0\). If \(\varphi \in W^{\prime}\), then for every \(v \in V\) we have

\begin{equation*} \bigl(T^{\prime}(\varphi)\bigr)(v) = \varphi(Tv) = \varphi(0) = 0 . \end{equation*}

Thus \(T^{\prime}(\varphi) = 0\) for every \(\varphi \in W^{\prime}\), which means \(T^{\prime} = 0\). Note that this direction does not use the hypothesis that \(W\) is finite-dimensional.

Now suppose \(T \neq 0\); we show \(T^{\prime} \neq 0\). Because \(T \neq 0\), there exists \(v \in V\) with \(Tv \neq 0\). The list consisting of the single vector \(Tv\) is linearly independent in \(W\), so because \(W\) is finite-dimensional this list can be extended to a basis

\begin{equation*} w_1 = Tv,\ w_2,\ \ldots,\ w_m \end{equation*}

of \(W\) (by 2.32). Let \(\psi_1, \ldots, \psi_m \in W^{\prime}\) be the dual basis of \(w_1, \ldots, w_m\) (see 3.112), so that \(\psi_1(w_1) = 1\). Then

\begin{equation*} \bigl(T^{\prime}(\psi_1)\bigr)(v) = \psi_1(Tv) = \psi_1(w_1) = 1 \neq 0 . \end{equation*}

Hence \(T^{\prime}(\psi_1) \neq 0\), and therefore \(T^{\prime} \neq 0\). Taking the contrapositive, \(T^{\prime} = 0\) implies \(T = 0\).

Combining the two paragraphs gives \(T^{\prime} = 0 \iff T = 0\).

Alternatively, the second direction follows quickly from the results in this section. Suppose \(T^{\prime} = 0\). Then \(\operatorname{null} T^{\prime} = W^{\prime}\). By 3.128(a) (whose proof, as noted in the book, does not use finite-dimensionality), \(\operatorname{null} T^{\prime} = (\operatorname{range} T)^0\). Thus \((\operatorname{range} T)^0 = W^{\prime}\), and since \(W\) is finite-dimensional and \(\operatorname{range} T\) is a subspace of \(W\), part (b) of 3.127 gives \(\operatorname{range} T = \{0\}\). Hence \(Tv = 0\) for every \(v \in V\), so \(T = 0\).

By 3.63, a linear map is invertible if and only if it is both injective and surjective. Because \(V\) and \(W\) are finite-dimensional, so are \(V^{\prime}\) and \(W^{\prime}\) (indeed \(\dim V^{\prime} = \dim V\) and \(\dim W^{\prime} = \dim W\), by 3.111), so 3.63 applies to \(T^{\prime}\) as well. Now

\begin{equation*} \begin{aligned} T \text{ is invertible} \ &\iff\ T \text{ is injective and } T \text{ is surjective}\\ &\iff\ T^{\prime} \text{ is surjective and } T^{\prime} \text{ is injective}\\ &\iff\ T^{\prime} \text{ is invertible}, \end{aligned} \end{equation*}

where the first and third equivalences come from 3.63 and the second equivalence comes from 3.131 (which says \(T\) is injective \(\iff\) \(T^{\prime}\) is surjective) together with 3.129 (which says \(T\) is surjective \(\iff\) \(T^{\prime}\) is injective).

It is worth recording the inverse explicitly. Suppose \(T\) is invertible, and let \(S = T^{-1} \in \mathcal{L}(W, V)\), so that \(ST = I_V\) and \(TS = I_W\). Applying 3.120(c) and Exercise 13 in this section (which says that the dual of an identity operator is the identity operator on the dual space), we get

\begin{equation*} T^{\prime} S^{\prime} = (ST)^{\prime} = (I_V)^{\prime} = I_{V^{\prime}} \qquad\text{and}\qquad S^{\prime} T^{\prime} = (TS)^{\prime} = (I_W)^{\prime} = I_{W^{\prime}} . \end{equation*}

Thus \(T^{\prime}\) is invertible with \((T^{\prime})^{-1} = (T^{-1})^{\prime}\). Note that this half of the argument uses neither 3.129, 3.131, nor finite-dimensionality.

Define

\begin{equation*} \Lambda \colon \mathcal{L}(V, W) \to \mathcal{L}(W^{\prime}, V^{\prime}) \qquad\text{by}\qquad \Lambda T = T^{\prime} . \end{equation*}

This makes sense: if \(T \in \mathcal{L}(V, W)\) then \(T^{\prime}\) is indeed a linear map from \(W^{\prime}\) to \(V^{\prime}\) (see 3.118). We must show that \(\Lambda\) is linear, injective, and surjective; by definition 3.69 that makes \(\Lambda\) an isomorphism onto \(\mathcal{L}(W^{\prime}, V^{\prime})\).

Step 1: \(\Lambda\) is linear. If \(S, T \in \mathcal{L}(V, W)\) and \(\lambda \in \mathbb{F}\), then 3.120(a) and 3.120(b) give

\begin{equation*} \Lambda(S + T) = (S + T)^{\prime} = S^{\prime} + T^{\prime} = \Lambda S + \Lambda T, \qquad \Lambda(\lambda T) = (\lambda T)^{\prime} = \lambda T^{\prime} = \lambda\,\Lambda T . \end{equation*}

Thus \(\Lambda \in \mathcal{L}\bigl(\mathcal{L}(V, W), \mathcal{L}(W^{\prime}, V^{\prime})\bigr)\).

Step 2: \(\Lambda\) is injective. Because \(\Lambda\) is linear, it suffices to show that \(\operatorname{null} \Lambda = \{0\}\) (by 3.15). So suppose \(T \in \mathcal{L}(V, W)\) and \(\Lambda T = 0\), that is, \(T^{\prime} = 0\). Since \(W\) is finite-dimensional, Exercise 16 in this section gives \(T = 0\). (Explicitly: if \(T \neq 0\), pick \(v \in V\) with \(Tv \neq 0\), extend the linearly independent list \(Tv\) to a basis \(Tv, w_2, \ldots, w_m\) of \(W\) using 2.32, and let \(\psi_1, \ldots, \psi_m\) be the dual basis; then \((T^{\prime}(\psi_1))(v) = \psi_1(Tv) = 1 \neq 0\), so \(T^{\prime} \neq 0\).) Hence \(\operatorname{null} \Lambda = \{0\}\) and \(\Lambda\) is injective.

Step 3: the two spaces have the same dimension. By 3.72,

\begin{equation*} \dim \mathcal{L}(V, W) = (\dim V)(\dim W). \end{equation*}

Also \(V^{\prime}\) and \(W^{\prime}\) are finite-dimensional with \(\dim V^{\prime} = \dim V\) and \(\dim W^{\prime} = \dim W\), by 3.111. Hence, using 3.72 again,

\begin{equation*} \dim \mathcal{L}(W^{\prime}, V^{\prime}) = (\dim W^{\prime})(\dim V^{\prime}) = (\dim W)(\dim V) = \dim \mathcal{L}(V, W). \end{equation*}

Step 4: \(\Lambda\) is surjective. By Steps 2 and 3, \(\Lambda\) is an injective linear map between finite-dimensional vector spaces of equal dimension, so 3.65 implies that \(\Lambda\) is surjective (and indeed invertible).

Therefore \(\Lambda\) is an invertible linear map from \(\mathcal{L}(V, W)\) onto \(\mathcal{L}(W^{\prime}, V^{\prime})\), i.e., an isomorphism, as desired. In particular \(\mathcal{L}(V, W)\) and \(\mathcal{L}(W^{\prime}, V^{\prime})\) are isomorphic vector spaces.

Both sides are subsets of \(V^{\prime}\), so it suffices to check that a linear functional \(\varphi \in V^{\prime}\) belongs to one side exactly when it belongs to the other.

By the definition of the annihilator (3.121),

\begin{equation*} U^0 = \{\varphi \in V^{\prime} : \varphi(u) = 0 \text{ for all } u \in U\}, \end{equation*}

and by the definition of null space (3.11),

\begin{equation*} \operatorname{null} \varphi = \{v \in V : \varphi(v) = 0\}. \end{equation*}

Now fix \(\varphi \in V^{\prime}\). Then

\begin{equation*} \begin{aligned} \varphi \in U^0 \ &\iff\ \varphi(u) = 0 \text{ for every } u \in U\\ &\iff\ u \in \operatorname{null}\varphi \text{ for every } u \in U\\ &\iff\ U \subseteq \operatorname{null} \varphi , \end{aligned} \end{equation*}

where the second equivalence is just the definition of \(\operatorname{null}\varphi\) (each \(u \in U\) is in particular an element of \(V\), since \(U \subseteq V\)), and the third equivalence is the definition of set inclusion.

Because \(\varphi \in V^{\prime}\) was arbitrary, the two sets have exactly the same elements, so

\begin{equation*} U^0 = \{\varphi \in V^{\prime} : U \subseteq \operatorname{null} \varphi\}. \end{equation*}

Note that nothing here required \(U\) to be a subspace of \(V\); the statement holds for an arbitrary subset \(U\) of \(V\).

Write

\begin{equation*} X = \{v \in V : \varphi(v) = 0 \text{ for every } \varphi \in U^0\}. \end{equation*}

We prove \(U \subseteq X\) and \(X \subseteq U\).

The inclusion \(U \subseteq X\). Suppose \(u \in U\). If \(\varphi \in U^0\), then \(\varphi(u) = 0\) by the definition of the annihilator (3.121). Since this holds for every \(\varphi \in U^0\), we have \(u \in X\). Hence \(U \subseteq X\). (This direction does not use finite-dimensionality.)

The inclusion \(X \subseteq U\). We prove the contrapositive statement: if \(v \in V\) and \(v \notin U\), then there exists \(\varphi \in U^0\) with \(\varphi(v) \neq 0\), so that \(v \notin X\).

So suppose \(v \in V\) with \(v \notin U\). Because \(V\) is finite-dimensional, so is its subspace \(U\) (by 2.25); let \(u_1, \ldots, u_m\) be a basis of \(U\) (take the empty list if \(U = \{0\}\)). We claim the list \(u_1, \ldots, u_m, v\) is linearly independent. Indeed, suppose \(a_1, \ldots, a_m, b \in \mathbb{F}\) satisfy

\begin{equation*} a_1 u_1 + \cdots + a_m u_m + b v = 0 . \end{equation*}

If \(b \neq 0\), then \(v = -\tfrac{1}{b}(a_1 u_1 + \cdots + a_m u_m) \in U\), contradicting \(v \notin U\). Hence \(b = 0\), and then \(a_1 = \cdots = a_m = 0\) because \(u_1, \ldots, u_m\) is linearly independent. This proves the claim.

Because \(V\) is finite-dimensional, the linearly independent list \(u_1, \ldots, u_m, v\) extends to a basis

\begin{equation*} u_1, \ldots, u_m,\ v,\ w_1, \ldots, w_k \end{equation*}

of \(V\) (by 2.32). Let

\begin{equation*} \varphi_1, \ldots, \varphi_m,\ \varphi,\ \psi_1, \ldots, \psi_k \end{equation*}

be the corresponding dual basis of \(V^{\prime}\) (3.112), where \(\varphi\) denotes the functional dual to \(v\). By the defining property of a dual basis,

\begin{equation*} \varphi(u_j) = 0 \text{ for } j = 1, \ldots, m, \qquad \varphi(v) = 1 . \end{equation*}

Since \(\varphi\) vanishes on the basis \(u_1, \ldots, u_m\) of \(U\), linearity gives \(\varphi(u) = 0\) for every \(u \in U\); that is, \(\varphi \in U^0\). But \(\varphi(v) = 1 \neq 0\), so \(v \notin X\). This completes the proof of the contrapositive, hence \(X \subseteq U\).

Combining the two inclusions gives \(U = X\), as desired.

Remark (a dimension-counting alternative for the second inclusion). The set \(X\) is a subspace of \(V\): it is the intersection of the subspaces \(\operatorname{null}\varphi\) over \(\varphi \in U^0\), or directly, \(0 \in X\) and \(X\) is closed under addition and scalar multiplication because each condition \(\varphi(v) = 0\) is linear in \(v\). Under the canonical identification of \(V\) with \(V^{\prime\prime}\) that sends \(v\) to the functional \(\varphi \mapsto \varphi(v)\) (this map is injective by the construction above, applied with \(U = \{0\}\), and \(\dim V^{\prime\prime} = \dim V^{\prime}= \dim V\) by 3.111, so it is an isomorphism by 3.65), the subspace \(X\) corresponds exactly to \((U^0)^0\). Hence, by 3.125 applied twice,

\begin{equation*} \dim X = \dim V^{\prime} - \dim U^0 = \dim V - (\dim V - \dim U) = \dim U . \end{equation*}

Since \(U \subseteq X\) and \(\dim U = \dim X\), we get \(U = X\) by 2.39.

(a) First suppose \(U \subseteq W\); we show \(W^0 \subseteq U^0\). Let \(\varphi \in W^0\). Then \(\varphi \in V^{\prime}\) and \(\varphi(w) = 0\) for every \(w \in W\), by 3.121. If \(u \in U\), then \(u \in W\) (because \(U \subseteq W\)), and hence \(\varphi(u) = 0\). Thus \(\varphi\) vanishes on all of \(U\), so \(\varphi \in U^0\). This shows \(W^0 \subseteq U^0\). Note that this implication holds for arbitrary subsets \(U \subseteq W\) of \(V\) and does not use finite-dimensionality: annihilation reverses inclusions.

Conversely, suppose \(W^0 \subseteq U^0\); we show \(U \subseteq W\). Let \(u \in U\). Let \(\varphi \in W^0\) be arbitrary. Because \(W^0 \subseteq U^0\), we have \(\varphi \in U^0\), and hence \(\varphi(u) = 0\) since \(u \in U\). Thus

\begin{equation*} \varphi(u) = 0 \quad\text{for every } \varphi \in W^0 . \end{equation*}

Because \(V\) is finite-dimensional and \(W\) is a subspace of \(V\), Exercise 20 in this section gives

\begin{equation*} W = \{v \in V : \varphi(v) = 0 \text{ for every } \varphi \in W^0\}. \end{equation*}

The displayed condition therefore says exactly that \(u \in W\). Since \(u \in U\) was arbitrary, \(U \subseteq W\).

This completes the proof of (a): \(W^0 \subseteq U^0 \iff U \subseteq W\).

(b) First suppose \(U = W\). Then \(U^0\) and \(W^0\) are annihilators of the same subspace, so trivially \(W^0 = U^0\).

Conversely, suppose \(W^0 = U^0\). Then in particular \(W^0 \subseteq U^0\), so part (a) gives \(U \subseteq W\). Also \(U^0 \subseteq W^0\), so part (a) with the roles of \(U\) and \(W\) interchanged gives \(W \subseteq U\). Combining the two inclusions yields \(U = W\).

This completes the proof of (b): \(W^0 = U^0 \iff U = W\).

Remark. Part (b) says that on the collection of subspaces of a finite-dimensional vector space \(V\) the map \(U \mapsto U^0\) is injective, and part (a) says that it is an inclusion-reversing correspondence. In fact \(U \mapsto U^0\) is an inclusion-reversing bijection from the subspaces of \(V\) onto the subspaces of \(V^{\prime}\). Only surjectivity remains to be checked: given a subspace \(Y\) of \(V^{\prime}\), put

\begin{equation*} U = \{v \in V : \varphi(v) = 0 \text{ for every } \varphi \in Y\}, \end{equation*}

which is a subspace of \(V\) (each condition \(\varphi(v) = 0\) is linear in \(v\)). If \(\varphi \in Y\) and \(u \in U\), then \(\varphi(u) = 0\) by the definition of \(U\), so \(Y \subseteq U^0\). Under the canonical identification of \(V\) with \(V^{\prime\prime}\) the subspace \(U\) is exactly \(Y^0\), so 3.125 applied to the subspace \(Y\) of \(V^{\prime}\) gives \(\dim U = \dim V^{\prime} - \dim Y = \dim V - \dim Y\), using 3.111. Hence 3.125 applied to \(U\) gives

\begin{equation*} \dim U^0 = \dim V - \dim U = \dim Y . \end{equation*}

Since \(Y \subseteq U^0\) and these subspaces of \(V^{\prime}\) have the same dimension, 2.39 gives \(Y = U^0\).

Throughout, recall the definition 3.121 of the annihilator: for \(X \subseteq V\),

\begin{equation*} X^0 = \{\varphi \in V^{\prime} : \varphi(x) = 0 \text{ for every } x \in X\}, \end{equation*}

and recall 3.124, which says that \(X^0\) is a subspace of \(V^{\prime}\).

Part (a). This part does not use the hypothesis that \(V\) is finite-dimensional.

First suppose \(\varphi \in (U + W)^0\). Because \(0 \in W\), every \(u \in U\) can be written \(u = u + 0 \in U + W\); thus \(U \subseteq U + W\), and similarly \(W \subseteq U + W\). Hence \(\varphi\) vanishes on \(U\) and on \(W\), which means \(\varphi \in U^0\) and \(\varphi \in W^0\). Therefore \((U + W)^0 \subseteq U^0 \cap W^0\).

Conversely, suppose \(\varphi \in U^0 \cap W^0\). A typical element of \(U + W\) has the form \(u + w\) with \(u \in U\) and \(w \in W\), and

\begin{equation*} \varphi(u + w) = \varphi(u) + \varphi(w) = 0 + 0 = 0 . \end{equation*}

Thus \(\varphi \in (U + W)^0\), giving \(U^0 \cap W^0 \subseteq (U + W)^0\).

The two inclusions give \((U + W)^0 = U^0 \cap W^0\).

Part (b). First we prove the inclusion \(U^0 + W^0 \subseteq (U \cap W)^0\). If \(\varphi \in U^0\), then \(\varphi\) vanishes on \(U\) and hence on the smaller set \(U \cap W\); thus \(U^0 \subseteq (U \cap W)^0\). Similarly \(W^0 \subseteq (U \cap W)^0\). Because \((U \cap W)^0\) is a subspace of \(V^{\prime}\) (by 3.124), it is closed under addition, and therefore

\begin{equation*} U^0 + W^0 \subseteq (U \cap W)^0 . \end{equation*}

Now we show that these two subspaces of \(V^{\prime}\) have the same dimension. Let \(n = \dim V\). Because \(V\) is finite-dimensional, so is \(V^{\prime}\), with \(\dim V^{\prime} = \dim V = n\) (by 3.111); hence 2.43 may legitimately be applied to the subspaces \(U^0\) and \(W^0\) of \(V^{\prime}\), and every dimension written below is finite. Using the formula for the dimension of a sum of two subspaces (2.43) applied inside \(V^{\prime}\), then part (a), then the formula 3.125 for the dimension of an annihilator, we get

\begin{equation*} \begin{aligned} \dim(U^0 + W^0) &= \dim U^0 + \dim W^0 - \dim(U^0 \cap W^0) \\ &= \dim U^0 + \dim W^0 - \dim\bigl((U + W)^0\bigr) \\ &= (n - \dim U) + (n - \dim W) - \bigl(n - \dim(U + W)\bigr) \\ &= n - \dim U - \dim W + \dim(U + W). \end{aligned} \end{equation*}

Applying 2.43 once more, this time to \(U\) and \(W\) inside \(V\), we have \(\dim(U + W) = \dim U + \dim W - \dim(U \cap W)\). Substituting this into the last line above gives

\begin{equation*} \dim(U^0 + W^0) = n - \dim(U \cap W) = \dim\bigl((U \cap W)^0\bigr), \end{equation*}

where the last equality is 3.125 applied to the subspace \(U \cap W\) of \(V\).

Thus \(U^0 + W^0\) is a subspace of the finite-dimensional vector space \((U \cap W)^0\) with \(\dim(U^0 + W^0) = \dim (U \cap W)^0\). By 2.39, this forces

\begin{equation*} (U \cap W)^0 = U^0 + W^0, \end{equation*}

as desired.

Write

\begin{equation*} W = (\operatorname{null} \varphi_1) \cap \cdots \cap (\operatorname{null} \varphi_m), \end{equation*}

which is a subspace of \(V\), being an intersection of subspaces.

Step 1: the sets in (b) and (c) are equal. For \(\varphi \in V^{\prime}\) we have

\begin{equation*} \varphi \in W^0 \iff \varphi(v) = 0 \text{ for every } v \in W \iff W \subseteq \operatorname{null} \varphi , \end{equation*}

where the first equivalence is the definition 3.121 of the annihilator and the second is the definition of null space. Thus \(W^0\) is exactly the set in (c). (This is Exercise 19 in this section.)

Step 2: for a single linear functional \(\psi \in V^{\prime}\) we have \((\operatorname{null} \psi)^0 = \operatorname{span}(\psi)\).

If \(\psi = 0\), then \(\operatorname{null} \psi = V\) and \(V^0 = \{0\} = \operatorname{span}(0) = \operatorname{span}(\psi)\); here \(V^0 = \{0\}\) by 3.127(a) (or directly: a linear functional vanishing on all of \(V\) is the zero functional).

Now suppose \(\psi \neq 0\). Then \(\operatorname{range} \psi\) is a nonzero subspace of the one-dimensional space \(\mathbf{F}\), so \(\operatorname{range} \psi = \mathbf{F}\) and \(\dim \operatorname{range} \psi = 1\). The fundamental theorem of linear maps (3.21) gives

\begin{equation*} \dim \operatorname{null} \psi = \dim V - 1 . \end{equation*}

Hence by 3.125,

\begin{equation*} \dim (\operatorname{null} \psi)^0 = \dim V - \dim \operatorname{null} \psi = 1 . \end{equation*}

Also \(\psi\) itself vanishes on \(\operatorname{null} \psi\), so \(\psi \in (\operatorname{null} \psi)^0\), and \(\psi \neq 0\). Thus \(\operatorname{span}(\psi)\) is a one-dimensional subspace of the one-dimensional space \((\operatorname{null} \psi)^0\), and 2.39 gives \((\operatorname{null} \psi)^0 = \operatorname{span}(\psi)\).

Step 3: the annihilator of an intersection is the sum of the annihilators. We claim that if \(U_1, \ldots, U_m\) are subspaces of \(V\), then

\begin{equation*} (U_1 \cap \cdots \cap U_m)^0 = U_1^{\,0} + \cdots + U_m^{\,0} . \end{equation*}

This holds trivially for \(m = 1\). Suppose \(m > 1\) and the claim holds for \(m - 1\) subspaces. Since \(U_1 \cap \cdots \cap U_{m-1}\) is a subspace of \(V\), Exercise 22(b) applied to the two subspaces \(U_1 \cap \cdots \cap U_{m-1}\) and \(U_m\) gives

\begin{equation*} (U_1 \cap \cdots \cap U_m)^0 = \bigl((U_1 \cap \cdots \cap U_{m-1}) \cap U_m\bigr)^0 = (U_1 \cap \cdots \cap U_{m-1})^0 + U_m^{\,0}, \end{equation*}

and the induction hypothesis turns the first summand into \(U_1^{\,0} + \cdots + U_{m-1}^{\,0}\). This completes the induction.

Step 4: conclusion. Applying Step 3 with \(U_k = \operatorname{null} \varphi_k\) and then Step 2 to each term, we get

\begin{equation*} W^0 = (\operatorname{null} \varphi_1)^0 + \cdots + (\operatorname{null} \varphi_m)^0 = \operatorname{span}(\varphi_1) + \cdots + \operatorname{span}(\varphi_m) = \operatorname{span}(\varphi_1, \ldots, \varphi_m), \end{equation*}

where the last equality holds because a sum of spans of single vectors consists precisely of all sums \(a_1 \varphi_1 + \cdots + a_m \varphi_m\) with \(a_1, \ldots, a_m \in \mathbf{F}\), which is the span of the list.

Thus the set in (a) equals the set in (b), and by Step 1 the sets in (b) and (c) are equal. Hence all three sets coincide.

Let \(U = \operatorname{span}(v_1, \ldots, v_m)\), a subspace of \(V\).

The key computation, used in both parts, is

\begin{equation*} \operatorname{null} \Gamma = U^0 . \end{equation*}

To see this, first suppose \(\varphi \in \operatorname{null} \Gamma\). Then \((\varphi(v_1), \ldots, \varphi(v_m)) = (0, \ldots, 0)\), so \(\varphi(v_k) = 0\) for each \(k\). If \(u \in U\), then \(u = a_1 v_1 + \cdots + a_m v_m\) for some \(a_1, \ldots, a_m \in \mathbf{F}\), and linearity of \(\varphi\) gives

\begin{equation*} \varphi(u) = a_1 \varphi(v_1) + \cdots + a_m \varphi(v_m) = 0 . \end{equation*}

Hence \(\varphi \in U^0\). Conversely, if \(\varphi \in U^0\), then \(\varphi\) vanishes on \(U\) and in particular \(\varphi(v_k) = 0\) for each \(k\) (because \(v_k \in U\)), so \(\Gamma(\varphi) = 0\). This proves the displayed equality.

Part (a). Using 3.15 (a linear map is injective if and only if its null space is \(\{0\}\)), the computation above, and 3.127(a), we have

\begin{equation*} \Gamma \text{ is injective} \iff \operatorname{null} \Gamma = \{0\} \iff U^0 = \{0\} \iff U = V . \end{equation*}

Since \(U = V\) says exactly that \(v_1, \ldots, v_m\) spans \(V\), part (a) follows.

Part (b). We compute \(\dim \operatorname{range} \Gamma\). By the fundamental theorem of linear maps (3.21) applied to \(\Gamma \in \mathcal{L}(V^{\prime}, \mathbf{F}^m)\), together with \(\dim V^{\prime} = \dim V\) (by 3.111), the equality \(\operatorname{null} \Gamma = U^0\), and 3.125,

\begin{equation*} \begin{aligned} \dim \operatorname{range} \Gamma &= \dim V^{\prime} - \dim \operatorname{null} \Gamma \\ &= \dim V - \dim U^0 \\ &= \dim V - (\dim V - \dim U) \\ &= \dim U . \end{aligned} \end{equation*}

Because \(\operatorname{range} \Gamma\) is a subspace of \(\mathbf{F}^m\), which has dimension \(m\), the map \(\Gamma\) is surjective if and only if \(\dim \operatorname{range} \Gamma = m\) (by 2.39). By the computation above, this happens if and only if \(\dim U = m\).

It remains to note that \(\dim U = m\) if and only if \(v_1, \ldots, v_m\) is linearly independent. Indeed, \(v_1, \ldots, v_m\) is by construction a list of length \(m\) spanning \(U\). If the list is linearly independent, then it is a basis of \(U\), so \(\dim U = m\). Conversely, if \(\dim U = m\), then \(v_1, \ldots, v_m\) is a spanning list of \(U\) whose length equals \(\dim U\), so it is a basis of \(U\) by 2.42, and in particular it is linearly independent.

Combining the last two paragraphs: \(\Gamma\) is surjective if and only if \(v_1, \ldots, v_m\) is linearly independent.

Let

\begin{equation*} W = (\operatorname{null} \varphi_1) \cap \cdots \cap (\operatorname{null} \varphi_m), \end{equation*}

a subspace of \(V\).

The key observation, used in both parts, is

\begin{equation*} \operatorname{null} \Gamma = W . \end{equation*}

Indeed, for \(v \in V\) we have \(\Gamma(v) = 0\) if and only if \(\varphi_k(v) = 0\) for every \(k \in \{1, \ldots, m\}\), which says exactly that \(v \in \operatorname{null} \varphi_k\) for every \(k\), that is, \(v \in W\).

We also use Exercise 23 of this section, which shows that

\begin{equation*} W^0 = \operatorname{span}(\varphi_1, \ldots, \varphi_m). \end{equation*}

Part (a). By 3.15, the equality \(\operatorname{null} \Gamma = W\), and 3.127(b) applied to the subspace \(W\) of \(V\), we have

\begin{equation*} \Gamma \text{ is injective} \iff \operatorname{null} \Gamma = \{0\} \iff W = \{0\} \iff W^0 = V^{\prime} . \end{equation*}

By the displayed consequence of Exercise 23, the condition \(W^0 = V^{\prime}\) says precisely that \(\operatorname{span}(\varphi_1, \ldots, \varphi_m) = V^{\prime}\), that is, that \(\varphi_1, \ldots, \varphi_m\) spans \(V^{\prime}\). This proves (a).

Part (b). The fundamental theorem of linear maps (3.21) applied to \(\Gamma\), together with \(\operatorname{null} \Gamma = W\) and 3.125, gives

\begin{equation*} \dim \operatorname{range} \Gamma = \dim V - \dim \operatorname{null} \Gamma = \dim V - \dim W = \dim W^0 = \dim \operatorname{span}(\varphi_1, \ldots, \varphi_m), \end{equation*}

where the last equality again uses Exercise 23.

Because \(\operatorname{range} \Gamma\) is a subspace of the \(m\)-dimensional space \(\mathbf{F}^m\), the map \(\Gamma\) is surjective if and only if \(\dim \operatorname{range} \Gamma = m\) (by 2.39). By the display above, this happens if and only if \(\dim \operatorname{span}(\varphi_1, \ldots, \varphi_m) = m\).

Finally, \(\dim \operatorname{span}(\varphi_1, \ldots, \varphi_m) = m\) if and only if \(\varphi_1, \ldots, \varphi_m\) is linearly independent. Indeed, the list \(\varphi_1, \ldots, \varphi_m\) has length \(m\) and spans the subspace \(\operatorname{span}(\varphi_1, \ldots, \varphi_m)\) of \(V^{\prime}\). If it is linearly independent, it is a basis of its span, so that span has dimension \(m\). Conversely, if the span has dimension \(m\), then \(\varphi_1, \ldots, \varphi_m\) is a spanning list of that span with length equal to its dimension, hence a basis of it by 2.42, and in particular linearly independent.

Combining the last two paragraphs proves (b).

Set

\begin{equation*} W = \{v \in V : \varphi(v) = 0 \text{ for every } \varphi \in \Omega\} . \end{equation*}

We must show that \(W^0 = \Omega\). Note first that \(W\) is indeed a subspace of \(V\): it is the intersection of the subspaces \(\operatorname{null} \varphi\) over all \(\varphi \in \Omega\) (and \(W = V\) if \(\Omega = \{0\}\), since then the only condition imposed is \(0(v) = 0\)).

Case 1: \(\Omega = \{0\}\). Then every \(v \in V\) satisfies the defining condition of \(W\), so \(W = V\). Hence \(W^0 = V^0 = \{0\} = \Omega\), where \(V^0 = \{0\}\) by 3.127(a).

Case 2: \(\Omega \neq \{0\}\). Since \(V\) is finite-dimensional, so is \(V^{\prime}\) (indeed \(\dim V^{\prime} = \dim V\) by 3.111), hence so is its subspace \(\Omega\) (by 2.25 and 2.37). Let \(\varphi_1, \ldots, \varphi_m\) be a basis of \(\Omega\), where \(m = \dim \Omega \geq 1\).

We claim that

\begin{equation*} W = (\operatorname{null} \varphi_1) \cap \cdots \cap (\operatorname{null} \varphi_m). \end{equation*}

For one inclusion: if \(v \in W\), then \(\varphi(v) = 0\) for every \(\varphi \in \Omega\), and in particular \(\varphi_k(v) = 0\) for each \(k\), since each \(\varphi_k \in \Omega\); thus \(v\) lies in the intersection. For the other inclusion: suppose \(\varphi_k(v) = 0\) for every \(k\), and let \(\varphi \in \Omega\) be arbitrary. Because \(\varphi_1, \ldots, \varphi_m\) is a basis of \(\Omega\), there exist \(a_1, \ldots, a_m \in \mathbf{F}\) with \(\varphi = a_1 \varphi_1 + \cdots + a_m \varphi_m\), and therefore

\begin{equation*} \varphi(v) = a_1 \varphi_1(v) + \cdots + a_m \varphi_m(v) = 0 . \end{equation*}

Hence \(v \in W\), proving the claim.

Now Exercise 23 of this section says that

\begin{equation*} \bigl((\operatorname{null} \varphi_1) \cap \cdots \cap (\operatorname{null} \varphi_m)\bigr)^0 = \operatorname{span}(\varphi_1, \ldots, \varphi_m). \end{equation*}

Combining this with the claim, and using that \(\varphi_1, \ldots, \varphi_m\) is a basis of \(\Omega\) (so its span is \(\Omega\)), we conclude

\begin{equation*} W^0 = \operatorname{span}(\varphi_1, \ldots, \varphi_m) = \Omega , \end{equation*}

as desired.

Write \(V = \mathcal{P}_5(\mathbf{R})\), so \(\dim V = 6\) (the list \(1, x, x^2, x^3, x^4, x^5\) is a basis of \(V\)). The map \(\varphi : V \to \mathbf{R}\) given by \(\varphi(p) = p(8)\) is a linear functional on \(V\), and the set in question is exactly \(\operatorname{null} \varphi\):

\begin{equation*} \{p \in \mathcal{P}_5(\mathbf{R}) : p(8) = 0\} = \operatorname{null} \varphi . \end{equation*}

First observe that \(\varphi \neq 0\), because \(\varphi(1) = 1 \neq 0\) for the constant polynomial \(1\). Hence \(\operatorname{range} \varphi\) is a nonzero subspace of the one-dimensional vector space \(\mathbf{R}\), so \(\operatorname{range} \varphi = \mathbf{R}\) and \(\dim \operatorname{range} \varphi = 1\). The fundamental theorem of linear maps (3.21) now gives

\begin{equation*} \dim \operatorname{null} \varphi = \dim V - \dim \operatorname{range} \varphi = 6 - 1 = 5 . \end{equation*}

Also, since \(\varphi \neq 0\), the list consisting of the single vector \(\varphi\) is linearly independent and hence a basis of \(\operatorname{span}(\varphi)\), so \(\dim \operatorname{span}(\varphi) = 1\).

Next we show that \(\operatorname{range} T \subseteq \operatorname{null} \varphi\). By 3.128(a) applied to \(T \in \mathcal{L}(V, V)\),

\begin{equation*} (\operatorname{range} T)^0 = \operatorname{null} T^{\prime} = \operatorname{span}(\varphi) . \end{equation*}

In particular \(\varphi \in (\operatorname{range} T)^0\), which by the definition 3.121 of the annihilator means \(\varphi(w) = 0\) for every \(w \in \operatorname{range} T\). Hence every \(w \in \operatorname{range} T\) lies in \(\operatorname{null} \varphi\), that is, \(\operatorname{range} T \subseteq \operatorname{null} \varphi\).

Finally we compare dimensions. Since \(\operatorname{range} T\) is a subspace of the finite-dimensional space \(V\), 3.125 gives

\begin{equation*} \dim \operatorname{range} T = \dim V - \dim (\operatorname{range} T)^0 = 6 - \dim \operatorname{span}(\varphi) = 6 - 1 = 5 . \end{equation*}

Thus \(\operatorname{range} T\) is a subspace of \(\operatorname{null} \varphi\) and

\begin{equation*} \dim \operatorname{range} T = 5 = \dim \operatorname{null} \varphi . \end{equation*}

By 2.39 (a subspace of full dimension equals the whole space, applied inside the vector space \(\operatorname{null} \varphi\)), we conclude

\begin{equation*} \operatorname{range} T = \operatorname{null} \varphi = \{p \in \mathcal{P}_5(\mathbf{R}) : p(8) = 0\}, \end{equation*}

as desired.

Let

\begin{equation*} W = (\operatorname{null} \varphi_1) \cap \cdots \cap (\operatorname{null} \varphi_m), \end{equation*}

a subspace of \(V\), being an intersection of subspaces.

Define \(\Gamma : V \to \mathbf{F}^m\) by

\begin{equation*} \Gamma(v) = (\varphi_1(v), \ldots, \varphi_m(v)), \end{equation*}

as in Exercise 25 of this section. This map is linear: for \(u, v \in V\) and \(\lambda \in \mathbf{F}\), linearity of each \(\varphi_k\) gives \(\Gamma(u + v) = \Gamma(u) + \Gamma(v)\) and \(\Gamma(\lambda v) = \lambda \Gamma(v)\).

As computed in the solution to Exercise 25,

\begin{equation*} \operatorname{null} \Gamma = W , \end{equation*}

because \(\Gamma(v) = 0\) holds if and only if \(\varphi_k(v) = 0\) for every \(k \in \{1, \ldots, m\}\), which says exactly that \(v\) belongs to each \(\operatorname{null} \varphi_k\).

Because \(\varphi_1, \ldots, \varphi_m\) is linearly independent, Exercise 25(b) shows that \(\Gamma\) is surjective, so

\begin{equation*} \operatorname{range} \Gamma = \mathbf{F}^m \quad\text{and}\quad \dim \operatorname{range} \Gamma = m . \end{equation*}

Now the fundamental theorem of linear maps (3.21) applied to \(\Gamma \in \mathcal{L}(V, \mathbf{F}^m)\) gives

\begin{equation*} \dim V = \dim \operatorname{null} \Gamma + \dim \operatorname{range} \Gamma = \dim W + m . \end{equation*}

Therefore

\begin{equation*} \dim\bigl((\operatorname{null} \varphi_1) \cap \cdots \cap (\operatorname{null} \varphi_m)\bigr) = \dim W = (\dim V) - m , \end{equation*}

as desired.

An alternative route avoids Exercise 25: by Exercise 23 of this section, \(W^0 = \operatorname{span}(\varphi_1, \ldots, \varphi_m)\), and this span has dimension \(m\) because \(\varphi_1, \ldots, \varphi_m\) is linearly independent and hence a basis of its span. Then 3.125 gives

\begin{equation*} \dim W = \dim V - \dim W^0 = (\dim V) - m . \end{equation*}

Note in passing that the hypothesis of linear independence is essential: for instance, if \(\varphi_1 = \varphi_2 \neq 0\), then \(W = \operatorname{null} \varphi_1\) has dimension \((\dim V) - 1\), not \((\dim V) - 2\).

We use repeatedly the following elementary observation: if \(X\) is a finite-dimensional vector space and \(\alpha \in X^{\prime}\) is nonzero, then \(\operatorname{range} \alpha\) is a nonzero subspace of \(\mathbf{F}\), hence equals \(\mathbf{F}\), so the fundamental theorem of linear maps (3.21) gives

\begin{equation*} \dim \operatorname{null} \alpha = \dim X - \dim \operatorname{range} \alpha = \dim X - 1 . \end{equation*}

Part (a). By 3.128(a) we have \(\operatorname{null} T^{\prime} = (\operatorname{range} T)^0\), so the hypothesis says

\begin{equation*} (\operatorname{range} T)^0 = \operatorname{span}(\varphi) . \end{equation*}

First suppose \(\varphi = 0\). Then \((\operatorname{range} T)^0 = \{0\}\), so \(\operatorname{range} T = W\) by 3.127(a). Also \(\operatorname{null} \varphi = \operatorname{null} 0 = W\). Hence \(\operatorname{range} T = W = \operatorname{null} \varphi\).

Now suppose \(\varphi \neq 0\). Because \(\varphi \in \operatorname{span}(\varphi) = (\operatorname{range} T)^0\), the definition of the annihilator (3.121) gives \(\varphi(w) = 0\) for every \(w \in \operatorname{range} T\); in other words,

\begin{equation*} \operatorname{range} T \subseteq \operatorname{null} \varphi . \end{equation*}

Because \(\varphi \neq 0\), the list \(\varphi\) is linearly independent, so \(\dim \operatorname{span}(\varphi) = 1\). Thus \(\dim (\operatorname{range} T)^0 = 1\), and 3.125 (applied to the subspace \(\operatorname{range} T\) of \(W\)) gives

\begin{equation*} 1 = \dim (\operatorname{range} T)^0 = \dim W - \dim \operatorname{range} T , \end{equation*}

so \(\dim \operatorname{range} T = \dim W - 1\). By the observation at the start of this solution, \(\dim \operatorname{null} \varphi = \dim W - 1\) as well. Hence \(\operatorname{range} T\) is a subspace of \(\operatorname{null} \varphi\) of the same (finite) dimension, so 2.39 gives \(\operatorname{range} T = \operatorname{null} \varphi\).

Part (b). By 3.130(b) we have \(\operatorname{range} T^{\prime} = (\operatorname{null} T)^0\), so the hypothesis says

\begin{equation*} (\operatorname{null} T)^0 = \operatorname{span}(\psi) . \end{equation*}

First suppose \(\psi = 0\). Then \((\operatorname{null} T)^0 = \{0\}\), so \(\operatorname{null} T = V\) by 3.127(a). Also \(\operatorname{null} \psi = \operatorname{null} 0 = V\). Hence \(\operatorname{null} T = V = \operatorname{null} \psi\).

Now suppose \(\psi \neq 0\). Because \(\psi \in \operatorname{span}(\psi) = (\operatorname{null} T)^0\), we have \(\psi(v) = 0\) for every \(v \in \operatorname{null} T\); that is,

\begin{equation*} \operatorname{null} T \subseteq \operatorname{null} \psi . \end{equation*}

As in (a), \(\dim \operatorname{span}(\psi) = 1\), so 3.125 (applied to the subspace \(\operatorname{null} T\) of \(V\)) gives

\begin{equation*} 1 = \dim (\operatorname{null} T)^0 = \dim V - \dim \operatorname{null} T , \end{equation*}

hence \(\dim \operatorname{null} T = \dim V - 1\). By the observation at the start, \(\dim \operatorname{null} \psi = \dim V - 1\) too. Thus \(\operatorname{null} T\) is a subspace of \(\operatorname{null} \psi\) of the same dimension, and 2.39 gives \(\operatorname{null} T = \operatorname{null} \psi\).

By 3.111 we have \(\dim V = \dim V^{\prime} = n\).

We first record a standard fact: if \(v \in V\) and \(v \neq 0\), then there exists \(\varphi \in V^{\prime}\) with \(\varphi(v) \neq 0\). Indeed, the list \(v\) is linearly independent, so it extends to a basis \(v, u_2, \ldots, u_n\) of \(V\) (by 2.32), and the linear map lemma (3.4) produces a linear map \(\varphi \colon V \to \mathbf{F}\) with \(\varphi(v) = 1\) and \(\varphi(u_k) = 0\) for \(k = 2, \ldots, n\). Then \(\varphi \in V^{\prime}\) and \(\varphi(v) = 1 \neq 0\).

Now define \(\Gamma \colon V \to \mathbf{F}^n\) by

\begin{equation*} \Gamma(v) = \bigl( \varphi_1(v), \ldots, \varphi_n(v) \bigr) . \end{equation*}

Each coordinate of \(\Gamma\) is a linear functional, so \(\Gamma\) is linear.

\(\Gamma\) is injective. Suppose \(v \in \operatorname{null} \Gamma\), so \(\varphi_1(v) = \cdots = \varphi_n(v) = 0\). If \(\varphi \in V^{\prime}\) is arbitrary, then, because \(\varphi_1, \ldots, \varphi_n\) spans \(V^{\prime}\), there exist \(a_1, \ldots, a_n \in \mathbf{F}\) with \(\varphi = a_1 \varphi_1 + \cdots + a_n \varphi_n\), and hence

\begin{equation*} \varphi(v) = a_1 \varphi_1(v) + \cdots + a_n \varphi_n(v) = 0 . \end{equation*}

Thus \(\varphi(v) = 0\) for every \(\varphi \in V^{\prime}\). By the standard fact above, this forces \(v = 0\). Hence \(\operatorname{null} \Gamma = \{0\}\) and \(\Gamma\) is injective by 3.15.

Because \(\dim V = n = \dim \mathbf{F}^n\) and \(\Gamma\) is injective, 3.65 shows that \(\Gamma\) is also surjective, so \(\Gamma\) is invertible (3.63).

Let \(e_1, \ldots, e_n\) denote the standard basis of \(\mathbf{F}^n\) and set

\begin{equation*} v_j = \Gamma^{-1}(e_j) \qquad \text{for } j = 1, \ldots, n . \end{equation*}

The list \(v_1, \ldots, v_n\) is a basis of \(V\). To see that it is linearly independent, suppose \(a_1, \ldots, a_n \in \mathbf{F}\) satisfy \(a_1 v_1 + \cdots + a_n v_n = 0\). Applying \(\Gamma\) and using linearity gives \(a_1 e_1 + \cdots + a_n e_n = 0\), and thus \(a_1 = \cdots = a_n = 0\) because \(e_1, \ldots, e_n\) is linearly independent. So \(v_1, \ldots, v_n\) is a linearly independent list in \(V\) of length \(n = \dim V\), and therefore is a basis of \(V\) by 2.38.

Finally, \(\Gamma(v_j) = e_j\) means precisely that

\begin{equation*} \bigl( \varphi_1(v_j), \ldots, \varphi_n(v_j) \bigr) = e_j , \end{equation*}

that is, \(\varphi_k(v_j) = 1\) if \(k = j\) and \(\varphi_k(v_j) = 0\) if \(k \neq j\). By the definition of the dual basis (3.112), this says exactly that \(\varphi_1, \ldots, \varphi_n\) is the dual basis of \(v_1, \ldots, v_n\), as desired.

Throughout, note that for \(\varphi \in V^{\prime}\) the definition of the dual map (3.118) gives \(i^{\prime}(\varphi) = \varphi \circ i\), and for \(u \in U\) we have \((\varphi \circ i)(u) = \varphi(i(u)) = \varphi(u)\). Thus

\begin{equation*} i^{\prime}(\varphi) = \varphi|_U , \end{equation*}

the restriction of \(\varphi\) to \(U\).

Part (a). Let \(\varphi \in V^{\prime}\). Then \(\varphi \in \operatorname{null} i^{\prime}\) if and only if \(i^{\prime}(\varphi)\) is the zero linear functional on \(U\), which by the displayed formula holds if and only if \(\varphi(u) = 0\) for every \(u \in U\). By the definition of the annihilator (3.121), that last condition says exactly \(\varphi \in U^0\). Hence \(\operatorname{null} i^{\prime} = U^0\). (No finite-dimensionality is used here.)

Part (b). Certainly \(\operatorname{range} i^{\prime} \subseteq U^{\prime}\), since \(i^{\prime} \in \mathcal{L}(V^{\prime}, U^{\prime})\). For the reverse inclusion, suppose \(\varphi \in U^{\prime}\). Because \(V\) is finite-dimensional, so is its subspace \(U\) (by 2.25); let \(u_1, \ldots, u_m\) be a basis of \(U\) and extend it to a basis \(u_1, \ldots, u_m, u_{m+1}, \ldots, u_n\) of \(V\) (by 2.32). The linear map lemma (3.4) gives a linear map \(\psi \colon V \to \mathbf{F}\) with

\begin{equation*} \psi(u_j) = \begin{cases} \varphi(u_j) & \text{if } 1 \le j \le m, \\ 0 & \text{if } m < j \le n. \end{cases} \end{equation*}

Then \(\psi \in V^{\prime}\), and the linear functionals \(\psi|_U\) and \(\varphi\) on \(U\) agree on the basis \(u_1, \ldots, u_m\) of \(U\), hence agree on all of \(U\). Thus \(i^{\prime}(\psi) = \psi|_U = \varphi\), so \(\varphi \in \operatorname{range} i^{\prime}\). Therefore \(\operatorname{range} i^{\prime} = U^{\prime}\).

Part (c). By the notation of 3.106 applied to the linear map \(i^{\prime} \in \mathcal{L}(V^{\prime}, U^{\prime})\), the map \(\widetilde{i^{\prime}}\) is the linear map

\begin{equation*} \widetilde{i^{\prime}} \colon V^{\prime}/(\operatorname{null} i^{\prime}) \to U^{\prime} , \qquad \widetilde{i^{\prime}}\bigl( \varphi + \operatorname{null} i^{\prime} \bigr) = i^{\prime}(\varphi) . \end{equation*}

By part (a), \(\operatorname{null} i^{\prime} = U^0\), so the domain of \(\widetilde{i^{\prime}}\) is \(V^{\prime}/U^0\).

By 3.107(b), \(\widetilde{i^{\prime}}\) is injective. By 3.107(c), \(\operatorname{range} \widetilde{i^{\prime}} = \operatorname{range} i^{\prime}\), which equals \(U^{\prime}\) by part (b); hence \(\widetilde{i^{\prime}}\) is surjective. A linear map that is both injective and surjective is invertible (3.63), so \(\widetilde{i^{\prime}}\) is an isomorphism from \(V^{\prime}/U^0\) onto \(U^{\prime}\).

Part (a). First we check that \(\Lambda v\) really is an element of \(V^{\prime\prime} = (V^{\prime})^{\prime}\), that is, that \(\Lambda v\) is a linear functional on \(V^{\prime}\). If \(\varphi, \psi \in V^{\prime}\), then

\begin{equation*} (\Lambda v)(\varphi + \psi) = (\varphi + \psi)(v) = \varphi(v) + \psi(v) = (\Lambda v)(\varphi) + (\Lambda v)(\psi) , \end{equation*}

where the middle equality is the definition of addition in \(V^{\prime} = \mathcal{L}(V, \mathbf{F})\). If \(\lambda \in \mathbf{F}\) and \(\varphi \in V^{\prime}\), then

\begin{equation*} (\Lambda v)(\lambda \varphi) = (\lambda \varphi)(v) = \lambda \bigl( \varphi(v) \bigr) = \lambda (\Lambda v)(\varphi) , \end{equation*}

using the definition of scalar multiplication in \(V^{\prime}\). Thus \(\Lambda v \colon V^{\prime} \to \mathbf{F}\) is linear, so \(\Lambda v \in V^{\prime\prime}\).

Now we check that \(\Lambda\) is linear. Let \(u, v \in V\). For every \(\varphi \in V^{\prime}\),

\begin{equation*} \bigl( \Lambda(u + v) \bigr)(\varphi) = \varphi(u + v) = \varphi(u) + \varphi(v) = (\Lambda u)(\varphi) + (\Lambda v)(\varphi) = (\Lambda u + \Lambda v)(\varphi) . \end{equation*}

Because two elements of \(V^{\prime\prime}\) that agree at every \(\varphi \in V^{\prime}\) are equal, \(\Lambda(u + v) = \Lambda u + \Lambda v\). Similarly, if \(\lambda \in \mathbf{F}\) and \(v \in V\), then for every \(\varphi \in V^{\prime}\),

\begin{equation*} \bigl( \Lambda(\lambda v) \bigr)(\varphi) = \varphi(\lambda v) = \lambda \varphi(v) = \lambda (\Lambda v)(\varphi) = \bigl( \lambda (\Lambda v) \bigr)(\varphi) , \end{equation*}

so \(\Lambda(\lambda v) = \lambda \Lambda v\). Hence \(\Lambda \in \mathcal{L}(V, V^{\prime\prime})\).

Part (b). Suppose \(T \in \mathcal{L}(V)\). Then \(T^{\prime} \in \mathcal{L}(V^{\prime}, V^{\prime})\) and \(T^{\prime\prime} = (T^{\prime})^{\prime} \in \mathcal{L}(V^{\prime\prime}, V^{\prime\prime})\). Both \(T^{\prime\prime} \circ \Lambda\) and \(\Lambda \circ T\) are maps from \(V\) to \(V^{\prime\prime}\), so it suffices to show they agree at each \(v \in V\), and for that it suffices to show that the two elements of \(V^{\prime\prime}\) agree at each \(\varphi \in V^{\prime}\).

Fix \(v \in V\) and \(\varphi \in V^{\prime}\). Using the definition of the dual map (3.118) twice,

\begin{equation*} \begin{aligned} \bigl( (T^{\prime\prime} \circ \Lambda)(v) \bigr)(\varphi) &= \bigl( T^{\prime\prime}(\Lambda v) \bigr)(\varphi) \\ &= \bigl( (\Lambda v) \circ T^{\prime} \bigr)(\varphi) \\ &= (\Lambda v)\bigl( T^{\prime}(\varphi) \bigr) \\ &= (\Lambda v)(\varphi \circ T) \\ &= (\varphi \circ T)(v) \\ &= \varphi(Tv) \\ &= \bigl( \Lambda(Tv) \bigr)(\varphi) \\ &= \bigl( (\Lambda \circ T)(v) \bigr)(\varphi) , \end{aligned} \end{equation*}

where the second equality uses \(T^{\prime\prime}(\alpha) = \alpha \circ T^{\prime}\) for \(\alpha \in V^{\prime\prime}\), the fourth uses \(T^{\prime}(\varphi) = \varphi \circ T\), and the fifth and seventh use the definition of \(\Lambda\). Since \(\varphi \in V^{\prime}\) was arbitrary, \((T^{\prime\prime} \circ \Lambda)(v) = (\Lambda \circ T)(v)\); since \(v \in V\) was arbitrary, \(T^{\prime\prime} \circ \Lambda = \Lambda \circ T\).

Part (c). Suppose \(V\) is finite-dimensional. Applying 3.111 twice gives

\begin{equation*} \dim V^{\prime\prime} = \dim V^{\prime} = \dim V , \end{equation*}

so \(V\) and \(V^{\prime\prime}\) are finite-dimensional of the same dimension.

We show \(\Lambda\) is injective. Suppose \(v \in \operatorname{null} \Lambda\), so \(\Lambda v = 0\), which means \(\varphi(v) = 0\) for every \(\varphi \in V^{\prime}\). If \(v \neq 0\), then the list \(v\) is linearly independent and so extends to a basis \(v, u_2, \ldots, u_n\) of \(V\) (by 2.32); the linear map lemma (3.4) then provides \(\varphi \in V^{\prime}\) with \(\varphi(v) = 1\) and \(\varphi(u_k) = 0\) for \(k \ge 2\), contradicting \(\varphi(v) = 0\). Hence \(v = 0\), so \(\operatorname{null} \Lambda = \{0\}\) and \(\Lambda\) is injective by 3.15.

Because \(\Lambda \in \mathcal{L}(V, V^{\prime\prime})\) is injective and \(\dim V = \dim V^{\prime\prime} < \infty\), 3.65 shows that \(\Lambda\) is surjective as well. An injective and surjective linear map is invertible (3.63), so \(\Lambda\) is an isomorphism from \(V\) onto \(V^{\prime\prime}\).

Recall that the quotient map is defined by \(\pi(v) = v + U\) for \(v \in V\), and that for \(\varphi \in (V/U)^{\prime}\) the dual map is given by \(\pi^{\prime}(\varphi) = \varphi \circ \pi\) (3.118), so

\begin{equation*} \bigl( \pi^{\prime}(\varphi) \bigr)(v) = \varphi(v + U) \qquad \text{for all } v \in V . \end{equation*}

No finite-dimensionality is assumed anywhere below.

Part (a). Suppose \(\varphi \in \operatorname{null} \pi^{\prime}\), so \(\varphi \circ \pi = 0\), that is, \(\varphi(v + U) = 0\) for every \(v \in V\). Every element of \(V/U\) has the form \(v + U\) for some \(v \in V\) (equivalently, \(\pi\) is surjective), so \(\varphi\) vanishes on all of \(V/U\); hence \(\varphi = 0\). Thus \(\operatorname{null} \pi^{\prime} = \{0\}\), and \(\pi^{\prime}\) is injective by 3.15.

Part (b). First we show \(\operatorname{range} \pi^{\prime} \subseteq U^0\). Suppose \(\varphi \in (V/U)^{\prime}\) and \(u \in U\). Since \(u \in U\), we have \(u + U = 0 + U\) by 3.101, and therefore

\begin{equation*} \bigl( \pi^{\prime}(\varphi) \bigr)(u) = \varphi(u + U) = \varphi(0 + U) = 0 , \end{equation*}

the last equality because \(0 + U\) is the additive identity of \(V/U\) and \(\varphi\) is linear (3.10). As \(u \in U\) was arbitrary, \(\pi^{\prime}(\varphi) \in U^0\) by the definition of the annihilator (3.121).

Now we show \(U^0 \subseteq \operatorname{range} \pi^{\prime}\). Suppose \(\psi \in U^0\); thus \(\psi \in V^{\prime}\) and \(\psi(u) = 0\) for all \(u \in U\). Define

\begin{equation*} \varphi \colon V/U \to \mathbf{F} \qquad \text{by} \qquad \varphi(v + U) = \psi(v) . \end{equation*}

This is well defined: if \(v + U = w + U\), then \(v - w \in U\) (by 3.101), so \(\psi(v) - \psi(w) = \psi(v - w) = 0\), giving \(\psi(v) = \psi(w)\). It is linear: for \(v, w \in V\) and \(\lambda \in \mathbf{F}\),

\begin{equation*} \begin{aligned} \varphi\bigl( (v + U) + (w + U) \bigr) &= \varphi\bigl( (v + w) + U \bigr) = \psi(v + w) = \psi(v) + \psi(w) = \varphi(v + U) + \varphi(w + U) , \\ \varphi\bigl( \lambda(v + U) \bigr) &= \varphi(\lambda v + U) = \psi(\lambda v) = \lambda \psi(v) = \lambda \varphi(v + U) , \end{aligned} \end{equation*}

using the definitions of addition and scalar multiplication on \(V/U\). Hence \(\varphi \in (V/U)^{\prime}\). Finally, for every \(v \in V\),

\begin{equation*} \bigl( \pi^{\prime}(\varphi) \bigr)(v) = \varphi(v + U) = \psi(v) , \end{equation*}

so \(\pi^{\prime}(\varphi) = \psi\) and therefore \(\psi \in \operatorname{range} \pi^{\prime}\).

Combining the two inclusions gives \(\operatorname{range} \pi^{\prime} = U^0\).

Part (c). By 3.124, \(U^0\) is a subspace of \(V^{\prime}\), so we may regard \(\pi^{\prime}\) as a linear map from \((V/U)^{\prime}\) into \(U^0\); call this map \(S\), so \(S(\varphi) = \pi^{\prime}(\varphi)\) for \(\varphi \in (V/U)^{\prime}\), which is legitimate by part (b). Then \(S\) is injective, since \(\operatorname{null} S = \operatorname{null} \pi^{\prime} = \{0\}\) by part (a), and \(S\) is surjective onto \(U^0\), since \(\operatorname{range} S = \operatorname{range} \pi^{\prime} = U^0\) by part (b). A linear map that is injective and surjective is invertible (3.63). Hence \(\pi^{\prime}\) is an isomorphism from \((V/U)^{\prime}\) onto \(U^0\).

Throughout, write

\begin{equation*} z = a + bi, \qquad w = c + di, \end{equation*}

where \(a = \operatorname{Re} z\), \(b = \operatorname{Im} z\), \(c = \operatorname{Re} w\), \(d = \operatorname{Im} w\) are real numbers. By the definitions in 4.2 we then have \(\bar{z} = a - bi\), \(\bar{w} = c - di\), and

\begin{equation*} |z| = \sqrt{a^2 + b^2}, \qquad |w| = \sqrt{c^2 + d^2}. \end{equation*}

(a) \(z + \bar{z} = (a + bi) + (a - bi) = 2a = 2\operatorname{Re} z\).

(b) \(z - \bar{z} = (a + bi) - (a - bi) = 2bi = 2(\operatorname{Im} z)i\).

(c) Using \(i^2 = -1\),

\begin{equation*} z\bar{z} = (a + bi)(a - bi) = a^2 - abi + abi - b^2i^2 = a^2 + b^2 = |z|^2 . \end{equation*}

(d) First, \(w + z = (c + a) + (d + b)i\), and \(c + a\) and \(d + b\) are real, so

\begin{equation*} \overline{w + z} = (c + a) - (d + b)i = (c - di) + (a - bi) = \bar{w} + \bar{z}. \end{equation*}

Second, \(wz = (ca - db) + (cb + da)i\), and \(ca - db\) and \(cb + da\) are real, so

\begin{equation*} \overline{wz} = (ca - db) - (cb + da)i . \end{equation*}

On the other hand,

\begin{equation*} \bar{w}\,\bar{z} = (c - di)(a - bi) = ca - cbi - dai + db\,i^2 = (ca - db) - (cb + da)i . \end{equation*}

The two right sides agree, so \(\overline{wz} = \bar{w}\,\bar{z}\).

(e) Because \(\bar{z} = a + (-b)i\) with \(a\) and \(-b\) real, we have \(\operatorname{Re}\bar{z} = a\) and \(\operatorname{Im}\bar{z} = -b\). Hence

\begin{equation*} \bar{\bar{z}} = a - (-b)i = a + bi = z . \end{equation*}

(f) Because \(b^2 \ge 0\) we have \(a^2 \le a^2 + b^2\), and the square root function is increasing on \([0, \infty)\), so

\begin{equation*} |\operatorname{Re} z| = |a| = \sqrt{a^2} \le \sqrt{a^2 + b^2} = |z| . \end{equation*}

Similarly \(a^2 \ge 0\) gives \(b^2 \le a^2 + b^2\) and therefore

\begin{equation*} |\operatorname{Im} z| = |b| = \sqrt{b^2} \le \sqrt{a^2 + b^2} = |z| . \end{equation*}

(g) As noted in (e), \(\operatorname{Re}\bar{z} = a\) and \(\operatorname{Im}\bar{z} = -b\). Hence

\begin{equation*} |\bar{z}| = \sqrt{a^2 + (-b)^2} = \sqrt{a^2 + b^2} = |z| . \end{equation*}

(h) Using (c) three times and (d) once, along with commutativity of complex multiplication,

\begin{equation*} |wz|^2 = (wz)\overline{(wz)} = wz\,\bar{w}\,\bar{z} = (w\bar{w})(z\bar{z}) = |w|^2 |z|^2 = \bigl(|w|\,|z|\bigr)^2 . \end{equation*}

Both \(|wz|\) and \(|w|\,|z|\) are nonnegative real numbers with equal squares, so taking square roots gives \(|wz| = |w|\,|z|\).

First note that if \(u \in \mathbb{C}\), then by the multiplicativity of absolute value (4.4, proved in Exercise 4.1(h)),

\begin{equation*} |-u| = |(-1)u| = |-1|\,|u| = |u| , \end{equation*}

because \(|-1| = \sqrt{(-1)^2 + 0^2} = 1\). In particular \(|z - w| = |-(w - z)| = |w - z|\).

Now apply the triangle inequality (4.4) to the decomposition \(w = (w - z) + z\):

\begin{equation*} |w| = \bigl|(w - z) + z\bigr| \le |w - z| + |z| , \end{equation*}

and hence

\begin{equation*} |w| - |z| \le |w - z| . \end{equation*}

Interchanging the roles of \(w\) and \(z\) in the display above gives

\begin{equation*} |z| - |w| \le |z - w| = |w - z| . \end{equation*}

The real number \(\bigl| \, |w| - |z| \, \bigr|\) equals either \(|w| - |z|\) or \(|z| - |w|\), and both of those quantities have just been shown to be at most \(|w - z|\). Therefore

\begin{equation*} \bigl| \, |w| - |z| \, \bigr| \le |w - z| , \end{equation*}

as desired.

Recall that \(V^{\prime}\) denotes the dual space of \(V\), so \(\varphi\) is a linear functional from \(V\) to \(\mathbb{C}\); in particular \(\varphi(\lambda v) = \lambda\varphi(v)\) for all \(\lambda \in \mathbb{C}\) and all \(v \in V\), because \(V\) is a complex vector space.

Fix \(v \in V\) and write

\begin{equation*} \varphi(v) = \alpha + \beta i , \end{equation*}

where \(\alpha = \operatorname{Re}\varphi(v)\) and \(\beta = \operatorname{Im}\varphi(v)\) are real numbers. Then \(\sigma(v) = \alpha\).

By the homogeneity of \(\varphi\) with respect to the complex scalar \(i\),

\begin{equation*} \varphi(iv) = i\varphi(v) = i(\alpha + \beta i) = -\beta + \alpha i . \end{equation*}

Because \(-\beta\) and \(\alpha\) are real, the last expression exhibits the real and imaginary parts of \(\varphi(iv)\); thus

\begin{equation*} \sigma(iv) = \operatorname{Re}\varphi(iv) = -\beta . \end{equation*}

Therefore

\begin{equation*} \sigma(v) - i\sigma(iv) = \alpha - i(-\beta) = \alpha + \beta i = \varphi(v) . \end{equation*}

Because \(v \in V\) was arbitrary, the desired equation holds for all \(v \in V\).

Note that the point of this exercise is that a complex linear functional is completely determined by its real part: if \(\varphi_1, \varphi_2 \in V^{\prime}\) satisfy \(\operatorname{Re}\varphi_1(v) = \operatorname{Re}\varphi_2(v)\) for all \(v \in V\), then the formula just proved forces \(\varphi_1 = \varphi_2\).

No. Call the set above \(U\). Then \(U\) is not closed under addition, so \(U\) is not a subspace of \(\mathcal{P}(\mathbb{F})\).

To see this, define \(p, q \in \mathcal{P}(\mathbb{F})\) by

\begin{equation*} p(z) = z^m \qquad \text{and} \qquad q(z) = 1 - z^m \end{equation*}

for \(z \in \mathbb{F}\). The coefficient of \(z^m\) in \(p\) is \(1 \ne 0\) and the coefficient of \(z^m\) in \(q\) is \(-1 \ne 0\), and no higher powers appear in either polynomial; hence \(\deg p = \deg q = m\) and so \(p, q \in U\). However

\begin{equation*} (p + q)(z) = z^m + (1 - z^m) = 1 \end{equation*}

for all \(z \in \mathbb{F}\). Thus \(p + q\) is the constant polynomial \(1\), which is not the zero polynomial and has degree \(0\). Because \(m\) is a positive integer, \(0 \ne m\), and therefore \(p + q \notin U\).

Two remarks. First, the other two requirements for a subspace do hold: \(0 \in U\) by the definition of \(U\), and \(U\) is closed under scalar multiplication, because if \(\lambda \in \mathbb{F}\) and \(p \in U\), then either \(\lambda = 0\) or \(p = 0\) (in which case \(\lambda p = 0 \in U\)), or else \(\lambda \ne 0\) and \(\deg p = m\), in which case multiplying each coefficient of \(p\) by the nonzero scalar \(\lambda\) leaves the coefficient of \(z^m\) nonzero, so \(\deg(\lambda p) = m\) and \(\lambda p \in U\). Only additive closure fails.

Second, the hypothesis that \(m\) is positive is needed for the counterexample as phrased, but in fact it is needed for the conclusion as well: if \(m = 0\), then \(\{0\} \cup \{p : \deg p = 0\}\) is the set of constant polynomials, which is a subspace of \(\mathcal{P}(\mathbb{F})\).

No. Call the set above \(U\). As in Exercise 4.4, the failure is closure under addition.

Define \(p, q \in \mathcal{P}(\mathbb{F})\) by

\begin{equation*} p(z) = z^2 + z \qquad \text{and} \qquad q(z) = -z^2 \end{equation*}

for \(z \in \mathbb{F}\). Both \(p\) and \(q\) have degree \(2\), which is even, so \(p, q \in U\). But

\begin{equation*} (p + q)(z) = (z^2 + z) + (-z^2) = z \end{equation*}

for all \(z \in \mathbb{F}\). Thus \(p + q\) is the polynomial \(z \mapsto z\), which is not the zero polynomial and has degree \(1\), an odd number. Hence \(p + q \notin U\), and so \(U\) is not a subspace of \(\mathcal{P}(\mathbb{F})\).

As in Exercise 4.4, note that the other two subspace requirements are satisfied: \(0 \in U\) by definition, and \(U\) is closed under scalar multiplication, because \(\lambda p = 0 \in U\) when \(\lambda = 0\), while for \(\lambda \ne 0\) and \(p \ne 0\) we have \(\deg(\lambda p) = \deg p\), which is even when \(p \in U\).

The point worth noticing is that cancellation of leading terms can change the degree by an odd amount: adding two polynomials of even degree can produce a polynomial of odd degree. Choosing the two polynomials to have leading coefficients that cancel while the degree-one terms do not, as above, exhibits exactly that.

Define \(p \in \mathcal{P}(\mathbb{F})\) by

\begin{equation*} p(z) = (z - \lambda_1)^{\,n - m + 1}(z - \lambda_2)(z - \lambda_3)\cdots(z - \lambda_m) \end{equation*}

for \(z \in \mathbb{F}\); if \(m = 1\) this means simply \(p(z) = (z - \lambda_1)^{\,n}\). Note that \(n - m + 1 \ge 1\) because \(m \le n\), so the exponent is a positive integer and \(p\) is indeed a product of polynomials.

We verify the three required properties.

First, \(\deg p = n\). Each factor \(z - \lambda_k\) is a monic polynomial of degree \(1\), meaning its coefficient of \(z^1\) equals \(1\). A product of monic polynomials is monic of degree equal to the sum of the degrees: indeed, if \(r\) and \(s\) are monic of degrees \(j\) and \(k\), then multiplying out \(rs\) produces the term \(z^{j+k}\) with coefficient \(1\) and no higher power of \(z\), so \(rs\) is monic of degree \(j + k\). Applying this repeatedly to the \((n - m + 1) + (m - 1) = n\) degree-one factors in the definition of \(p\) shows that \(p\) is monic of degree

\begin{equation*} (n - m + 1) + \underbrace{1 + \cdots + 1}_{m - 1 \text{ times}} = (n - m + 1) + (m - 1) = n . \end{equation*}

In particular \(p \ne 0\).

Second, \(p(\lambda_k) = 0\) for each \(k \in \{1, \ldots, m\}\). If \(k = 1\), then the factor \((\lambda_1 - \lambda_1)^{\,n - m + 1} = 0\) makes the whole product \(0\). If \(k \ge 2\), then the factor \(\lambda_k - \lambda_k = 0\) occurs in the product, again making the product \(0\).

Third, \(p\) has no zeros other than \(\lambda_1, \ldots, \lambda_m\). Suppose \(\lambda \in \mathbb{F}\) and \(p(\lambda) = 0\), so

\begin{equation*} (\lambda - \lambda_1)^{\,n - m + 1}(\lambda - \lambda_2)\cdots(\lambda - \lambda_m) = 0 . \end{equation*}

Because \(\mathbb{F}\) (which is \(\mathbb{R}\) or \(\mathbb{C}\)) has no zero divisors, a product of elements of \(\mathbb{F}\) equals \(0\) only if one of the factors equals \(0\). Hence \(\lambda - \lambda_k = 0\) for some \(k \in \{1, \ldots, m\}\), that is, \(\lambda \in \{\lambda_1, \ldots, \lambda_m\}\).

Thus \(p\) has degree \(n\) and its set of zeros is exactly \(\{\lambda_1, \ldots, \lambda_m\}\), as required.

Two comments. The numbers \(\lambda_1, \ldots, \lambda_m\) are not assumed to be distinct, and nothing above used distinctness; if some of them coincide, the set of zeros is just the (possibly smaller) set \(\{\lambda_1, \ldots, \lambda_m\}\), and the conclusion reads the same. Also, the trick of piling the extra \(n - m\) degrees onto the single factor \(z - \lambda_1\) avoids any need to find a polynomial of degree \(n - m\) with no zeros in \(\mathbb{F}\); over \(\mathbb{R}\) no such polynomial exists when \(n - m\) is odd (see Exercise 4.9), and over \(\mathbb{C}\) no such polynomial exists when \(n - m \ge 1\), by the fundamental theorem of algebra 4.12.

Define \(T \colon \mathcal{P}_m(\mathbb{F}) \to \mathbb{F}^{m+1}\) by

\begin{equation*} Tp = \bigl(p(z_1), \ldots, p(z_{m+1})\bigr) . \end{equation*}

\(T\) is linear. Indeed, if \(p, q \in \mathcal{P}_m(\mathbb{F})\) and \(\lambda \in \mathbb{F}\), then for each \(k\) we have \((p + q)(z_k) = p(z_k) + q(z_k)\) and \((\lambda p)(z_k) = \lambda\, p(z_k)\), by the definition of addition and scalar multiplication of functions; hence \(T(p + q) = Tp + Tq\) and \(T(\lambda p) = \lambda Tp\).

\(T\) is injective. Because \(T\) is linear, it suffices to show that \(\operatorname{null} T = \{0\}\) (by 3.15). So suppose \(p \in \mathcal{P}_m(\mathbb{F})\) and \(Tp = 0\), meaning

\begin{equation*} p(z_1) = \cdots = p(z_{m+1}) = 0 . \end{equation*}

Because \(z_1, \ldots, z_{m+1}\) are distinct, this says that \(p\) has at least \(m + 1\) distinct zeros in \(\mathbb{F}\). Suppose \(p \ne 0\) and let \(k = \deg p\), so \(0 \le k \le m\). If \(k = 0\), then \(p\) is a nonzero constant polynomial and thus has no zeros at all, contradicting the display above (there is at least one \(z_k\), since \(m + 1 \ge 1\)). If \(k \ge 1\), then 4.8 tells us that \(p\) has at most \(k \le m\) zeros in \(\mathbb{F}\), again contradicting the fact that \(p\) has at least \(m + 1\) zeros. Hence \(p = 0\), and so \(\operatorname{null} T = \{0\}\), proving that \(T\) is injective.

Now note that

\begin{equation*} \dim \mathcal{P}_m(\mathbb{F}) = m + 1 = \dim \mathbb{F}^{m+1}, \end{equation*}

where the first equality holds because \(1, z, \ldots, z^m\) is a basis of \(\mathcal{P}_m(\mathbb{F})\) (see 2.36). Both spaces are finite-dimensional and have the same dimension, so by 3.65 the injectivity of \(T\) implies that \(T\) is invertible; in particular \(T\) is surjective as well as injective.

Surjectivity of \(T\) applied to the vector \((w_1, \ldots, w_{m+1}) \in \mathbb{F}^{m+1}\) produces \(p \in \mathcal{P}_m(\mathbb{F})\) with

\begin{equation*} \bigl(p(z_1), \ldots, p(z_{m+1})\bigr) = (w_1, \ldots, w_{m+1}) , \end{equation*}

that is, \(p(z_k) = w_k\) for each \(k = 1, \ldots, m + 1\). This proves existence.

Injectivity of \(T\) proves uniqueness: if \(p, \tilde{p} \in \mathcal{P}_m(\mathbb{F})\) both satisfy \(p(z_k) = w_k = \tilde{p}(z_k)\) for each \(k\), then \(Tp = T\tilde{p}\), and hence \(p = \tilde{p}\).

Thus there is exactly one \(p \in \mathcal{P}_m(\mathbb{F})\) with \(p(z_k) = w_k\) for \(k = 1, \ldots, m + 1\).

For the record, the polynomial can be written down explicitly (the Lagrange interpolation formula):

\begin{equation*} p(z) = \sum_{k=1}^{m+1} w_k \prod_{\substack{j = 1 \\ j \ne k}}^{m+1} \frac{z - z_j}{z_k - z_j}, \end{equation*}

where each product is a well-defined polynomial of degree \(m\) because the \(z_j\) are distinct, so no denominator vanishes; evaluating at \(z = z_k\) makes the \(k\)-th product equal \(1\) and every other product equal \(0\), giving \(p(z_k) = w_k\). The linear algebra proof above has the advantage of delivering uniqueness for free, from injectivity, rather than by a separate argument.

Throughout, the derivative of \(p(z) = a_0 + a_1 z + \cdots + a_m z^m\) is the polynomial

\begin{equation*} p^{\prime}(z) = a_1 + 2a_2 z + \cdots + m a_m z^{m-1}. \end{equation*}

The map \(p \mapsto p^{\prime}\) is linear, and the product rule

\begin{equation*} (fg)^{\prime} = f^{\prime}g + fg^{\prime} \end{equation*}

holds for all \(f, g \in \mathcal{P}(\mathbb{C})\): both sides are bilinear in the pair \((f,g)\), so it suffices to check it for monomials \(f(z) = z^j\), \(g(z) = z^k\), where it reads \((j+k)z^{j+k-1} = j z^{j-1} z^k + z^j k z^{k-1}\), which is true.

First dispose of the case \(m = 0\). Then \(p\) is a nonzero constant, so \(p\) has no zeros at all; thus \(p\) has \(m = 0\) distinct zeros. Here \(p^{\prime}\) is the zero polynomial, every complex number being one of its zeros, but since \(p\) has no zeros the two polynomials still have no zeros in common. So both sides of the asserted equivalence hold, and the statement is true. Assume from now on that \(m \geq 1\).

By the second version of the fundamental theorem of algebra (4.13) there are \(c, \lambda_1, \ldots, \lambda_m \in \mathbb{C}\) with \(c \neq 0\) such that

\begin{equation*} p(z) = c(z - \lambda_1) \cdots (z - \lambda_m) \end{equation*}

for all \(z \in \mathbb{C}\). The zeros of \(p\) are precisely \(\lambda_1, \ldots, \lambda_m\), so \(p\) has \(m\) distinct zeros if and only if \(\lambda_1, \ldots, \lambda_m\) are distinct.

Suppose first that \(\lambda_1, \ldots, \lambda_m\) are distinct. Repeated application of the product rule to the factorization above gives

\begin{equation*} p^{\prime}(z) = c \sum_{k=1}^{m} \; \prod_{j \neq k} (z - \lambda_j). \end{equation*}

Fix \(k \in \{1, \ldots, m\}\) and evaluate at \(z = \lambda_k\). For every index \(i \neq k\), the product \(\prod_{j \neq i}(\lambda_k - \lambda_j)\) contains the factor \(\lambda_k - \lambda_k = 0\) and hence vanishes. Only the \(k^{\text{th}}\) summand survives, giving

\begin{equation*} p^{\prime}(\lambda_k) = c \prod_{j \neq k} (\lambda_k - \lambda_j), \end{equation*}

which is a product of nonzero numbers (here we use that the \(\lambda\)’s are distinct, so \(\lambda_k - \lambda_j \neq 0\) for \(j \neq k\)) times \(c \neq 0\). Hence \(p^{\prime}(\lambda_k) \neq 0\). Since \(\lambda_1, \ldots, \lambda_m\) are all the zeros of \(p\), no zero of \(p\) is a zero of \(p^{\prime}\); that is, \(p\) and \(p^{\prime}\) have no zeros in common.

Conversely, suppose \(p\) does not have \(m\) distinct zeros. Then in the factorization above two of the \(\lambda\)’s coincide; after relabeling, say \(\lambda_1 = \lambda_2 =: \lambda\). Put \(g(z) = c(z - \lambda_3)\cdots(z - \lambda_m)\), so that

\begin{equation*} p(z) = (z - \lambda)^2 g(z). \end{equation*}

By the product rule,

\begin{equation*} p^{\prime}(z) = 2(z - \lambda)g(z) + (z - \lambda)^2 g^{\prime}(z), \end{equation*}

and therefore \(p^{\prime}(\lambda) = 0\). Since also \(p(\lambda) = 0\), the number \(\lambda\) is a zero common to \(p\) and \(p^{\prime}\). Taking the contrapositive: if \(p\) and \(p^{\prime}\) have no zeros in common, then \(p\) has \(m\) distinct zeros.

Combining the two implications gives the desired equivalence.

Suppose \(p \in \mathcal{P}(\mathbb{R})\) has odd degree \(n\). In particular \(n \geq 1\), so \(p\) is nonconstant and the factorization theorem over \(\mathbb{R}\) (4.16) applies: there exist \(c, \lambda_1, \ldots, \lambda_m, b_1, \ldots, b_M, c_1, \ldots, c_M \in \mathbb{R}\) with \(b_k^2 < 4c_k\) for each \(k\), such that

\begin{equation*} p(x) = c(x - \lambda_1) \cdots (x - \lambda_m)(x^2 + b_1 x + c_1) \cdots (x^2 + b_M x + c_M) \end{equation*}

for all \(x \in \mathbb{R}\). Here \(c \neq 0\): if \(c\) were \(0\), the right side would be the zero polynomial, contradicting \(p \neq 0\). Thus \(c\) is the coefficient of \(x^n\) in \(p\).

Comparing degrees on the two sides of this equation gives

\begin{equation*} n = m + 2M. \end{equation*}

Because \(n\) is odd and \(2M\) is even, \(m\) is odd. In particular \(m \neq 0\), so \(m \geq 1\) and the real number \(\lambda_1\) occurs in the factorization. Evaluating at \(x = \lambda_1\) makes the factor \(x - \lambda_1\) equal to \(0\), so

\begin{equation*} p(\lambda_1) = 0. \end{equation*}

Thus \(\lambda_1 \in \mathbb{R}\) is a real zero of \(p\), as desired.

A remark on the structure of the argument: the quadratic factors are exactly the ones that contribute no real zeros, since \(b_k^2 < 4c_k\) forces \(x^2 + b_k x + c_k = \left(x + \frac{b_k}{2}\right)^2 + \left(c_k - \frac{b_k^2}{4}\right) > 0\) for all \(x \in \mathbb{R}\) (this completing-the-square computation is the one used in the proof of 4.15). Each such factor uses up two degrees. Hence a polynomial with no real zeros must have even degree, which is the contrapositive of what was to be proved.

Fix \(p \in \mathcal{P}(\mathbb{R})\) and let \(\tilde{p}\) denote the polynomial \(\tilde{p}(x) = p(x) - p(3)\). Then \(\tilde{p}(3) = 0\).

If \(p\) is constant, then \(\tilde{p}\) is the zero polynomial and \(p^{\prime} = 0\); hence \((Tp)(x) = 0\) for \(x \neq 3\) and \((Tp)(3) = p^{\prime}(3) = 0\). So \(Tp\) is the zero polynomial, which lies in \(\mathcal{P}(\mathbb{R})\).

Now suppose \(\deg p = m \geq 1\). Subtracting the constant \(p(3)\) changes only the constant coefficient, so \(\deg \tilde{p} = m \geq 1\) as well, and \(3\) is a zero of \(\tilde{p}\). By 4.6 there is a polynomial \(q \in \mathcal{P}(\mathbb{R})\) of degree \(m - 1\) such that

\begin{equation*} p(x) - p(3) = \tilde{p}(x) = (x - 3)q(x) \end{equation*}

for every \(x \in \mathbb{R}\). Dividing by \(x - 3\) shows that

\begin{equation*} (Tp)(x) = q(x) \qquad \text{for all } x \neq 3 . \end{equation*}

It remains to check that \(Tp\) and \(q\) also agree at \(x = 3\). Differentiating the identity \(p(x) - p(3) = (x-3)q(x)\) (the right side by the product rule) gives

\begin{equation*} p^{\prime}(x) = q(x) + (x - 3)q^{\prime}(x) \end{equation*}

for all \(x \in \mathbb{R}\), and setting \(x = 3\) yields \(p^{\prime}(3) = q(3)\). Since \((Tp)(3) = p^{\prime}(3)\) by definition, we get \((Tp)(3) = q(3)\).

Therefore \(Tp = q\) as functions on \(\mathbb{R}\), so \(Tp\) is a polynomial with real coefficients; that is, \(Tp \in \mathcal{P}(\mathbb{R})\). (Its degree is \(\deg p - 1\) when \(p\) is nonconstant.) This shows \(T\) is a well-defined map from \(\mathcal{P}(\mathbb{R})\) to \(\mathcal{P}(\mathbb{R})\).

Now we verify linearity. Let \(p_1, p_2 \in \mathcal{P}(\mathbb{R})\) and \(\alpha \in \mathbb{R}\). Two polynomials are equal precisely when they are equal as functions on \(\mathbb{R}\), so it suffices to compare values at each \(x \in \mathbb{R}\).

For \(x \neq 3\),

\begin{equation*} \bigl(T(p_1 + p_2)\bigr)(x) = \frac{(p_1 + p_2)(x) - (p_1 + p_2)(3)}{x - 3} = \frac{p_1(x) - p_1(3)}{x - 3} + \frac{p_2(x) - p_2(3)}{x - 3} = (Tp_1)(x) + (Tp_2)(x). \end{equation*}

For \(x = 3\), using that differentiation is additive,

\begin{equation*} \bigl(T(p_1 + p_2)\bigr)(3) = (p_1 + p_2)^{\prime}(3) = p_1^{\prime}(3) + p_2^{\prime}(3) = (Tp_1)(3) + (Tp_2)(3). \end{equation*}

Hence \(T(p_1 + p_2) = Tp_1 + Tp_2\).

Similarly, for \(x \neq 3\),

\begin{equation*} \bigl(T(\alpha p_1)\bigr)(x) = \frac{\alpha p_1(x) - \alpha p_1(3)}{x - 3} = \alpha \cdot \frac{p_1(x) - p_1(3)}{x - 3} = \alpha (Tp_1)(x), \end{equation*}

and at \(x = 3\), since \((\alpha p_1)^{\prime} = \alpha p_1^{\prime}\),

\begin{equation*} \bigl(T(\alpha p_1)\bigr)(3) = (\alpha p_1)^{\prime}(3) = \alpha p_1^{\prime}(3) = \alpha (Tp_1)(3). \end{equation*}

Hence \(T(\alpha p_1) = \alpha \, T p_1\).

Thus \(T\) is additive and homogeneous, so \(T \colon \mathcal{P}(\mathbb{R}) \to \mathcal{P}(\mathbb{R})\) is a linear map.

Write

\begin{equation*} p(z) = a_0 + a_1 z + \cdots + a_m z^m \end{equation*}

with \(a_0, \ldots, a_m \in \mathbb{C}\), and define a second polynomial \(\bar{p} \in \mathcal{P}(\mathbb{C})\) by

\begin{equation*} \bar{p}(z) = \overline{a_0} + \overline{a_1} z + \cdots + \overline{a_m} z^m . \end{equation*}

The first step is to identify \(\overline{p(\bar{z})}\) with \(\bar{p}(z)\). Using that conjugation is additive and multiplicative and that \(\overline{\bar{z}} = z\) (see 4.4), we have, for every \(z \in \mathbb{C}\),

\begin{equation*} \overline{p(\bar{z})} = \overline{\sum_{k=0}^{m} a_k (\bar{z})^k} = \sum_{k=0}^{m} \overline{a_k} \, \overline{(\bar{z})^{k}} = \sum_{k=0}^{m} \overline{a_k} \, \bigl(\overline{\bar{z}}\bigr)^{k} = \sum_{k=0}^{m} \overline{a_k} z^k = \bar{p}(z). \end{equation*}

Therefore

\begin{equation*} q(z) = p(z) \, \bar{p}(z) \qquad \text{for all } z \in \mathbb{C}, \end{equation*}

so \(q\) is a product of two polynomials and hence is itself a polynomial: \(q \in \mathcal{P}(\mathbb{C})\).

It remains to prove that the coefficients of \(q\) are real. Multiplying out the product and collecting powers of \(z\) gives

\begin{equation*} q(z) = \sum_{n=0}^{2m} c_n z^n, \qquad \text{where } \; c_n = \sum_{\substack{j + k = n \\ 0 \leq j,\, k \leq m}} a_j \overline{a_k}. \end{equation*}

(These are the coefficients of \(q\), and by the uniqueness of the coefficients of a polynomial — a consequence of 4.8 — they are the only ones.)

Fix \(n\) and conjugate:

\begin{equation*} \overline{c_n} = \overline{\sum_{j+k=n} a_j \overline{a_k}} = \sum_{j+k=n} \overline{a_j} \, \overline{\overline{a_k}} = \sum_{j+k=n} \overline{a_j} a_k = \sum_{j+k=n} a_k \overline{a_j}. \end{equation*}

The index set \(\{(j,k) : j + k = n, \; 0 \leq j, k \leq m\}\) is carried to itself by the swap \((j,k) \mapsto (k,j)\), so relabeling the summation in the last expression by interchanging the names of \(j\) and \(k\) turns it back into \(\sum_{j+k=n} a_j \overline{a_k} = c_n\). Hence

\begin{equation*} \overline{c_n} = c_n, \end{equation*}

which by 4.4 means \(c_n \in \mathbb{R}\) for every \(n \in \{0, 1, \ldots, 2m\}\).

Thus \(q\) is a polynomial all of whose coefficients are real, as desired.

(An alternative way to see that each \(c_n\) is real: \(\overline{q(\bar z)} = \overline{p(\bar z)}\,\overline{\bar p (\bar z)} = \bar p(z) p(z) = q(z)\) for all \(z\), since \(\overline{\bar p(\bar z)} = p(z)\) by the computation above applied to \(\bar p\). Taking \(z = x\) real gives \(\overline{q(x)} = q(x)\), so \(q\) takes real values on \(\mathbb{R}\); then the imaginary parts of the coefficients form a polynomial vanishing identically on \(\mathbb{R}\), hence vanish by 4.8.)

Write

\begin{equation*} p(z) = a_0 + a_1 z + \cdots + a_m z^m \end{equation*}

with \(a_0, \ldots, a_m \in \mathbb{C}\) (some of these coefficients may be \(0\), since \(p\) is only assumed to have degree at most \(m\)). Define

\begin{equation*} u(x) = (\operatorname{Im} a_0) + (\operatorname{Im} a_1)x + \cdots + (\operatorname{Im} a_m)x^m , \end{equation*}

which is a polynomial with real coefficients of degree at most \(m\).

Suppose \(x \in \mathbb{R}\). Then \(x^k \in \mathbb{R}\) for each \(k\), and since \(\operatorname{Im}\) is additive and satisfies \(\operatorname{Im}(a t) = t \operatorname{Im} a\) for \(a \in \mathbb{C}\) and \(t \in \mathbb{R}\), we get

\begin{equation*} \operatorname{Im} p(x) = \sum_{k=0}^{m} \operatorname{Im}\bigl(a_k x^k\bigr) = \sum_{k=0}^{m} (\operatorname{Im} a_k) x^k = u(x). \end{equation*}

By hypothesis \(p(x_k) \in \mathbb{R}\), so \(\operatorname{Im} p(x_k) = 0\), for each \(k = 0, 1, \ldots, m\). Hence

\begin{equation*} u(x_0) = u(x_1) = \cdots = u(x_m) = 0, \end{equation*}

so the polynomial \(u\) has at least \(m + 1\) distinct zeros in \(\mathbb{R}\) (the numbers \(x_0, \ldots, x_m\) are distinct by assumption).

Now suppose, for the sake of contradiction, that \(u\) is not the zero polynomial, and let \(d = \deg u\), so \(0 \leq d \leq m\). If \(d = 0\), then \(u\) is a nonzero constant and has no zeros at all, contradicting the display above (there is at least one \(x_k\), since \(m \geq 0\)). If \(d \geq 1\), then by 4.8 the polynomial \(u\) has at most \(d \leq m\) zeros, again contradicting the fact that it has at least \(m + 1\) distinct zeros. Therefore \(u = 0\).

Since the coefficients of a polynomial are uniquely determined (a consequence of 4.8), \(u = 0\) forces

\begin{equation*} \operatorname{Im} a_k = 0 \qquad \text{for each } k = 0, 1, \ldots, m, \end{equation*}

that is, \(a_k \in \mathbb{R}\) for every \(k\). Thus all coefficients of \(p\) are real, as desired.

The hypothesis that there are \(m + 1\) such points cannot be weakened to \(m\) points: for instance \(p(z) = iz\) lies in \(\mathcal{P}_1(\mathbb{C})\) and satisfies \(p(0) = 0 \in \mathbb{R}\), yet its coefficients are not all real.

Alternative argument via Exercise 4.7. Set \(w_k = p(x_k) \in \mathbb{R}\) for \(k = 0, \ldots, m\). Applying Exercise 4.7 over the field \(\mathbb{R}\) produces a polynomial \(\tilde{p} \in \mathcal{P}_m(\mathbb{R})\) with \(\tilde{p}(x_k) = w_k\) for each \(k\). Regarding \(\tilde{p}\) as an element of \(\mathcal{P}_m(\mathbb{C})\), both \(p\) and \(\tilde{p}\) lie in \(\mathcal{P}_m(\mathbb{C})\) and take the value \(w_k\) at \(x_k\) for each of the \(m+1\) distinct points \(x_0, \ldots, x_m\). The uniqueness assertion of Exercise 4.7 over \(\mathbb{C}\) then gives \(p = \tilde{p}\), so all coefficients of \(p\) are real.

Let \(m = \deg p\); since \(p \neq 0\) we have \(m \geq 0\).

First note that \(U\) is a subspace of \(\mathcal{P}(\mathbb{F})\): it contains \(0 = p \cdot 0\), and it is closed under addition and scalar multiplication because \(pq_1 + pq_2 = p(q_1 + q_2)\) and \(\lambda(pq) = p(\lambda q)\) for all \(q_1, q_2, q \in \mathcal{P}(\mathbb{F})\) and \(\lambda \in \mathbb{F}\). Hence the quotient space \(\mathcal{P}(\mathbb{F})/U\) makes sense.

We prove (a) and (b) together by exhibiting a basis with \(m\) elements. Let \(z^k\) denote the polynomial \(z \mapsto z^k\), and consider the list of cosets

\begin{equation*} 1 + U, \; z + U, \; z^2 + U, \; \ldots, \; z^{m-1} + U \end{equation*}

in \(\mathcal{P}(\mathbb{F})/U\). (When \(m = 0\) this is the empty list.)

Spanning. Let \(r \in \mathcal{P}(\mathbb{F})\) be arbitrary. Because \(p \neq 0\), the division algorithm for polynomials (4.9) provides \(q, s \in \mathcal{P}(\mathbb{F})\) with

\begin{equation*} r = pq + s \quad \text{and} \quad \deg s < \deg p = m. \end{equation*}

Since \(pq \in U\), we have

\begin{equation*} r + U = s + U . \end{equation*}

The condition \(\deg s < m\) says exactly that \(s \in \mathcal{P}_{m-1}(\mathbb{F})\), so \(s = a_0 + a_1 z + \cdots + a_{m-1}z^{m-1}\) for some \(a_0, \ldots, a_{m-1} \in \mathbb{F}\). Because the quotient map \(r \mapsto r + U\) is linear (3.104), this gives

\begin{equation*} r + U = s + U = a_0(1 + U) + a_1(z + U) + \cdots + a_{m-1}\bigl(z^{m-1} + U\bigr). \end{equation*}

Hence the list above spans \(\mathcal{P}(\mathbb{F})/U\). (When \(m = 0\) the polynomial \(s\) must be \(0\), so every coset equals \(U\) and the empty list spans the zero space, consistent with the fact that a nonzero constant \(p\) gives \(U = \mathcal{P}(\mathbb{F})\).)

Linear independence. Suppose \(a_0, \ldots, a_{m-1} \in \mathbb{F}\) satisfy

\begin{equation*} a_0(1 + U) + a_1(z + U) + \cdots + a_{m-1}\bigl(z^{m-1} + U\bigr) = 0 + U . \end{equation*}

Setting \(s = a_0 + a_1 z + \cdots + a_{m-1}z^{m-1}\), this says \(s + U = 0 + U\), which by 3.101 means \(s \in U\). So \(s = pq\) for some \(q \in \mathcal{P}(\mathbb{F})\). If \(q \neq 0\), then

\begin{equation*} \deg s = \deg p + \deg q = m + \deg q \geq m, \end{equation*}

contradicting \(\deg s \leq m - 1\). Hence \(q = 0\) and therefore \(s = 0\). Since the coefficients of a polynomial are uniquely determined (a consequence of 4.8), \(a_0 = a_1 = \cdots = a_{m-1} = 0\). Thus the list is linearly independent.

Conclusion for (b). The list

\begin{equation*} 1 + U, \; z + U, \; \ldots, \; z^{m-1} + U \end{equation*}

is a basis of \(\mathcal{P}(\mathbb{F})/U\), where \(m = \deg p\). Equivalently: the cosets \(z^k + U\) for \(k = 0, 1, \ldots, (\deg p) - 1\) form a basis.

Conclusion for (a). This basis has length \(m\), so

\begin{equation*} \dim \mathcal{P}(\mathbb{F})/U = m = \deg p, \end{equation*}

as claimed. Note in particular that although \(\mathcal{P}(\mathbb{F})\) is infinite-dimensional, the quotient \(\mathcal{P}(\mathbb{F})/U\) is finite-dimensional.

Since \(p\) and \(q\) are nonconstant, \(m \geq 1\) and \(n \geq 1\).

Preliminaries. Recall that \(\deg(uv) = \deg u + \deg v\) for all \(u, v \in \mathcal{P}(\mathbb{C})\): if \(u, v \neq 0\) the leading coefficient of \(uv\) is the product of the leading coefficients of \(u\) and \(v\), which is nonzero because \(\mathbb{C}\) has no zero divisors; and if one of them is \(0\) then both sides equal \(-\infty\). Two consequences we shall use: (i) \(uv = 0\) implies \(u = 0\) or \(v = 0\); (ii) cancellation holds, i.e. if \(u \neq 0\) and \(uv = uw\) then \(u(v-w) = 0\) and hence \(v = w\).

\(T\) is well defined and linear. If \(\deg r \leq n - 1\) and \(\deg s \leq m - 1\), then \(\deg(rp) \leq (n-1) + m = m + n - 1\) and \(\deg(sq) \leq (m-1) + n = m + n - 1\), so \(rp + sq \in \mathcal{P}_{m+n-1}(\mathbb{C})\). Linearity is immediate from the distributive law:

\begin{equation*} T\bigl((r_1,s_1) + (r_2,s_2)\bigr) = (r_1+r_2)p + (s_1+s_2)q = T(r_1,s_1) + T(r_2,s_2), \end{equation*}

and \(T(\lambda r, \lambda s) = \lambda r p + \lambda s q = \lambda\, T(r,s)\) for \(\lambda \in \mathbb{C}\).

(a) Injectivity. By 3.15 it suffices to show \(\operatorname{null} T = \{(0,0)\}\). So suppose \((r,s) \in \mathcal{P}_{n-1}(\mathbb{C}) \times \mathcal{P}_{m-1}(\mathbb{C})\) satisfies

\begin{equation*} rp + sq = 0, \qquad \text{i.e.} \qquad rp = -sq . \end{equation*}

Because \(q\) is nonconstant, the second version of the fundamental theorem of algebra (4.13) gives \(c \in \mathbb{C}\) with \(c \neq 0\) and \(\lambda_1, \ldots, \lambda_n \in \mathbb{C}\) (not necessarily distinct) such that

\begin{equation*} q(z) = c(z - \lambda_1) \cdots (z - \lambda_n). \end{equation*}

Each \(\lambda_k\) is a zero of \(q\), hence is not a zero of \(p\); that is,

\begin{equation*} p(\lambda_k) \neq 0 \qquad \text{for } k = 1, \ldots, n . \end{equation*}

Assume, seeking a contradiction, that \(r \neq 0\). We claim that for each \(j \in \{0, 1, \ldots, n\}\) there is a polynomial \(r_j \in \mathcal{P}(\mathbb{C})\) with \(r_j \neq 0\) such that

\begin{equation*} r(z) = (z - \lambda_1)\cdots(z - \lambda_j)\, r_j(z) . \end{equation*}

We prove the claim by induction on \(j\). For \(j = 0\) take \(r_0 = r\) (the empty product is \(1\)), which is nonzero by assumption. Suppose the claim holds for some \(j\) with \(0 \leq j < n\). Substituting \(r = (z-\lambda_1)\cdots(z-\lambda_j) r_j\) into \(rp = -sq\) and using the factorization of \(q\) gives

\begin{equation*} (z-\lambda_1)\cdots(z-\lambda_j)\, r_j(z) p(z) = -s(z)\, c\, (z-\lambda_1)\cdots(z-\lambda_j)(z - \lambda_{j+1})\cdots(z-\lambda_n). \end{equation*}

The polynomial \((z-\lambda_1)\cdots(z-\lambda_j)\) is nonzero, so cancellation (consequence (ii) above) yields

\begin{equation*} r_j(z) p(z) = -c\, s(z)\, (z - \lambda_{j+1}) \cdots (z - \lambda_n) \end{equation*}

for all \(z \in \mathbb{C}\). Evaluating at \(z = \lambda_{j+1}\) makes the right side \(0\), so

\begin{equation*} r_j(\lambda_{j+1})\, p(\lambda_{j+1}) = 0 . \end{equation*}

Since \(p(\lambda_{j+1}) \neq 0\), we get \(r_j(\lambda_{j+1}) = 0\). Now \(r_j \neq 0\) and \(r_j\) has a zero, so \(r_j\) is not a nonzero constant; hence \(\deg r_j \geq 1\) and 4.6 provides a polynomial \(r_{j+1}\) with \(\deg r_{j+1} = \deg r_j - 1\) (in particular \(r_{j+1} \neq 0\)) such that \(r_j(z) = (z - \lambda_{j+1}) r_{j+1}(z)\). Substituting this into the formula for \(r\) completes the induction.

Taking \(j = n\) in the claim gives \(r = (z-\lambda_1)\cdots(z-\lambda_n) r_n\) with \(r_n \neq 0\), so

\begin{equation*} \deg r = n + \deg r_n \geq n, \end{equation*}

contradicting \(\deg r \leq n - 1\). Therefore \(r = 0\).

With \(r = 0\) the equation \(rp = -sq\) becomes \(sq = 0\); since \(q \neq 0\), consequence (i) gives \(s = 0\). Hence \(\operatorname{null} T = \{(0,0)\}\) and \(T\) is injective.

(b) Surjectivity. Because \(1, z, \ldots, z^k\) is a basis of \(\mathcal{P}_k(\mathbb{C})\), we have (see 2.36) \(\dim \mathcal{P}_k(\mathbb{C}) = k+1\) for every nonnegative integer \(k\). Hence, using the formula for the dimension of a product (3.92),

\begin{equation*} \dim \bigl(\mathcal{P}_{n-1}(\mathbb{C}) \times \mathcal{P}_{m-1}(\mathbb{C})\bigr) = \dim \mathcal{P}_{n-1}(\mathbb{C}) + \dim \mathcal{P}_{m-1}(\mathbb{C}) = n + m, \end{equation*}

while

\begin{equation*} \dim \mathcal{P}_{m+n-1}(\mathbb{C}) = m + n . \end{equation*}

So \(T\) is a linear map between finite-dimensional vector spaces of the same dimension. By 3.65, injectivity is equivalent to surjectivity for such a map, and \(T\) is injective by (a). Therefore \(T\) is surjective.

(c) Conclusion. Since \(m + n - 1 \geq 1 \geq 0\), the constant polynomial \(1\) belongs to \(\mathcal{P}_{m+n-1}(\mathbb{C})\). By (b) there exists a pair

\begin{equation*} (r,s) \in \mathcal{P}_{n-1}(\mathbb{C}) \times \mathcal{P}_{m-1}(\mathbb{C}) \end{equation*}

with \(T(r,s) = 1\), that is,

\begin{equation*} rp + sq = 1, \end{equation*}

which is exactly what was to be proved. (Injectivity of \(T\) shows moreover that this pair \((r,s)\) is unique.)

Recall the definition 5.2: a subspace \(U\) of \(V\) is invariant under \(T\) precisely when \(Tu \in U\) for every \(u \in U\). So in each part we must take an arbitrary \(u \in U\) and show \(Tu\) lands back in \(U\).

(a) Suppose \(U \subseteq \operatorname{null} T\). Let \(u \in U\). Then \(u \in \operatorname{null} T\), so

\begin{equation*} Tu = 0 . \end{equation*}

Since \(U\) is a subspace of \(V\), it contains \(0\). Hence \(Tu = 0 \in U\). As \(u \in U\) was arbitrary, \(U\) is invariant under \(T\).

(b) Suppose \(\operatorname{range} T \subseteq U\). Let \(u \in U\). Then \(u \in V\), so \(Tu \in \operatorname{range} T\) by the definition of the range. Therefore

\begin{equation*} Tu \in \operatorname{range} T \subseteq U , \end{equation*}

so \(Tu \in U\). As \(u \in U\) was arbitrary, \(U\) is invariant under \(T\).

Note that the two hypotheses are extreme cases of a general principle: in (a) the operator crushes all of \(U\) to the single point \(0\), which every subspace contains, and in (b) the operator cannot escape \(U\) no matter where it starts, since everything in the image of \(T\) already lies in \(U\). In particular, taking \(U = \operatorname{null} T\) in (a) and \(U = \operatorname{range} T\) in (b) recovers the observation of 5.4 that \(\operatorname{null} T\) and \(\operatorname{range} T\) are invariant under \(T\).

First recall that a sum of subspaces of \(V\) is again a subspace of \(V\) (1.40), so \(V_1 + \dots + V_m\) is a subspace of \(V\) and it makes sense to ask whether it is invariant under \(T\).

Let \(v \in V_1 + \dots + V_m\). By the definition of the sum of subspaces, there exist vectors

\begin{equation*} v_1 \in V_1, \quad \dots, \quad v_m \in V_m \end{equation*}

such that

\begin{equation*} v = v_1 + \dots + v_m . \end{equation*}

Applying \(T\) and using the additivity of \(T\),

\begin{equation*} Tv = T(v_1 + \dots + v_m) = Tv_1 + \dots + Tv_m . \end{equation*}

For each \(k \in \{1, \dots, m\}\) we have \(v_k \in V_k\) and \(V_k\) is invariant under \(T\), so \(Tv_k \in V_k\). Hence \(Tv\) is a sum of a vector in \(V_1\), a vector in \(V_2\), and so on through a vector in \(V_m\), which means precisely that

\begin{equation*} Tv \in V_1 + \dots + V_m . \end{equation*}

Since \(v \in V_1 + \dots + V_m\) was arbitrary, \(V_1 + \dots + V_m\) is invariant under \(T\).

Let \(\{U_a\}_{a \in A}\) be a collection of subspaces of \(V\), each invariant under \(T\), indexed by a set \(A\), and set

\begin{equation*} U = \bigcap_{a \in A} U_a . \end{equation*}

(If \(A\) is empty the intersection is interpreted as all of \(V\), which is invariant under \(T\) because \(T\) maps \(V\) into \(V\); so assume \(A\) is nonempty.)

First, \(U\) is a subspace of \(V\). Indeed, \(0 \in U_a\) for every \(a \in A\) because each \(U_a\) is a subspace, so \(0 \in U\). If \(u, w \in U\) and \(\lambda \in \mathbf{F}\), then for each \(a \in A\) we have \(u, w \in U_a\), and since \(U_a\) is closed under addition and scalar multiplication we get \(u + w \in U_a\) and \(\lambda u \in U_a\). As this holds for every \(a \in A\), we conclude \(u + w \in U\) and \(\lambda u \in U\). Thus \(U\) contains \(0\) and is closed under addition and scalar multiplication, so \(U\) is a subspace of \(V\) (1.34).

Now we show \(U\) is invariant under \(T\). Let \(u \in U\). Fix any \(a \in A\). Then \(u \in U_a\), and \(U_a\) is invariant under \(T\), so

\begin{equation*} Tu \in U_a . \end{equation*}

Since \(a \in A\) was arbitrary, \(Tu\) belongs to \(U_a\) for every \(a \in A\), and therefore

\begin{equation*} Tu \in \bigcap_{a \in A} U_a = U . \end{equation*}

As \(u \in U\) was arbitrary, \(U\) is invariant under \(T\).

The statement is true; we prove it.

Suppose \(V\) is finite-dimensional and \(U\) is a subspace of \(V\) invariant under every operator on \(V\). If \(U = \{0\}\) we are done, so assume \(U \neq \{0\}\). We show \(U = V\).

Because \(U \neq \{0\}\), we may choose \(u \in U\) with \(u \neq 0\). The list consisting of the single vector \(u\) is linearly independent (a one-vector list is linearly independent exactly when the vector is nonzero). Since \(V\) is finite-dimensional, 2.32 lets us extend this list to a basis

\begin{equation*} u, e_2, \dots, e_n \end{equation*}

of \(V\), where \(n = \dim V\).

Now let \(v \in V\) be arbitrary. By the linear map lemma 3.4 there is a (unique) linear map \(T \colon V \to V\) with

\begin{equation*} Tu = v, \qquad Te_2 = 0, \quad \dots, \quad Te_n = 0 , \end{equation*}

since a linear map may be defined by prescribing its values arbitrarily on a basis. This \(T\) is an operator on \(V\), so by hypothesis \(U\) is invariant under \(T\). Since \(u \in U\), invariance gives

\begin{equation*} v = Tu \in U . \end{equation*}

As \(v \in V\) was arbitrary, \(V \subseteq U\). Combined with \(U \subseteq V\), this gives \(U = V\).

Hence \(U = \{0\}\) or \(U = V\), as claimed.

Two remarks. First, the hypothesis that \(U\) is invariant under every operator is a very strong assumption, and we used only a small part of it: for each target vector \(v\) we needed just the single operator \(T\) constructed above. Second, finite-dimensionality entered only through 2.32, the extension of a linearly independent list to a basis. If one is willing to use the existence of a basis of an arbitrary vector space (a fact whose proof requires the axiom of choice and which is not proved in this book), the same argument gives the same conclusion without the finite-dimensionality hypothesis.

The answer: \(T\) has no eigenvalues.

By definition 5.5, a number \(\lambda \in \mathbf{R}\) is an eigenvalue of \(T\) if and only if there is a nonzero \((x, y) \in \mathbf{R}^2\) with \(T(x, y) = \lambda(x, y)\). Written out, this equation says

\begin{equation*} (-3y,\, x) = (\lambda x,\, \lambda y) , \end{equation*}

which is equivalent to the simultaneous equations

\begin{equation*} -3y = \lambda x, \qquad x = \lambda y . \end{equation*}

Substituting the second equation into the first gives

\begin{equation*} -3y = \lambda(\lambda y) = \lambda^2 y , \end{equation*}

that is,

\begin{equation*} (\lambda^2 + 3) y = 0 . \end{equation*}

Suppose \(\lambda \in \mathbf{R}\) is an eigenvalue, with eigenvector \((x, y) \neq (0, 0)\). If \(y = 0\), then the equation \(x = \lambda y\) forces \(x = 0\), so \((x, y) = (0, 0)\), a contradiction. Hence \(y \neq 0\), and dividing the displayed equation by \(y\) gives

\begin{equation*} \lambda^2 = -3 . \end{equation*}

But \(\lambda^2 \geq 0\) for every real number \(\lambda\), so no real \(\lambda\) satisfies this. Therefore \(T\) has no eigenvalues (and hence no eigenvectors).

Geometrically this is unsurprising. Writing \(S(x, y) = (x, 3y)\) for the stretch by a factor of \(3\) in the second coordinate and \(R(x, y) = (-y, x)\) for the counterclockwise rotation by \(90\) degrees about the origin, we have

\begin{equation*} R\bigl( S(x,y) \bigr) = R(x, 3y) = (-3y,\, x) = T(x, y) , \end{equation*}

so \(T = RS\). A stretch along an axis followed by a quarter turn moves every nonzero vector off the line through it, which is exactly the computation carried out above. Equivalently, by the comments following 5.5, \(T\) has no one-dimensional invariant subspace.

For contrast, if the same formula defined an operator on \(\mathbf{C}^2\), then \(\lambda^2 = -3\) would have the two solutions \(\lambda = \sqrt{3}\, i\) and \(\lambda = -\sqrt{3}\, i\), and both would be eigenvalues. This illustrates the point made in 5.9 that the field matters.

The answer: the eigenvalues of \(T\) are \(1\) and \(-1\). The eigenvectors corresponding to \(1\) are the vectors \((w, w)\) with \(w \in \mathbf{F}\) and \(w \neq 0\); the eigenvectors corresponding to \(-1\) are the vectors \((w, -w)\) with \(w \in \mathbf{F}\) and \(w \neq 0\).

A scalar \(\lambda \in \mathbf{F}\) is an eigenvalue of \(T\) with eigenvector \((w, z)\) exactly when \((w, z) \neq (0, 0)\) and \(T(w, z) = \lambda(w, z)\), that is, when

\begin{equation*} (z,\, w) = (\lambda w,\, \lambda z) . \end{equation*}

This is equivalent to the pair of equations

\begin{equation*} z = \lambda w, \qquad w = \lambda z . \end{equation*}

Substituting the first equation into the second gives

\begin{equation*} w = \lambda(\lambda w) = \lambda^2 w , \end{equation*}

so

\begin{equation*} (\lambda^2 - 1) w = 0 . \end{equation*}

If \(w = 0\), then \(z = \lambda w = 0\) and \((w, z)\) is the zero vector, which is not an eigenvector. Hence \(w \neq 0\), and dividing by \(w\) gives \(\lambda^2 = 1\), so

\begin{equation*} \lambda = 1 \quad \text{or} \quad \lambda = -1 . \end{equation*}

Thus \(1\) and \(-1\) are the only candidates for eigenvalues. (This is consistent with 5.12: an operator on the two-dimensional space \(\mathbf{F}^2\) has at most \(2\) distinct eigenvalues.) Note that these two candidates are distinct scalars because \(\mathbf{F}\) is \(\mathbf{R}\) or \(\mathbf{C}\).

Now we check each candidate and determine its eigenvectors.

Case \(\lambda = 1\). The equations become \(z = w\) and \(w = z\), which together say exactly \(z = w\). So the vectors satisfying \(T(w, z) = (w, z)\) are precisely those of the form \((w, w)\) with \(w \in \mathbf{F}\). Indeed \(T(w, w) = (w, w)\). Taking \(w = 1\) shows \((1, 1)\) is a nonzero such vector, so \(1\) is an eigenvalue, and the eigenvectors corresponding to \(1\) are the vectors \((w, w)\) with \(w \neq 0\). Equivalently,

\begin{equation*} \operatorname{null}(T - I) = \{ (w, w) : w \in \mathbf{F} \} = \operatorname{span}\bigl( (1,1) \bigr) . \end{equation*}

Case \(\lambda = -1\). The equations become \(z = -w\) and \(w = -z\), which are the same condition \(z = -w\). So the solutions are the vectors \((w, -w)\) with \(w \in \mathbf{F}\), and indeed

\begin{equation*} T(w, -w) = (-w,\, w) = -(w, -w) . \end{equation*}

Taking \(w = 1\) shows \((1, -1)\) is nonzero, so \(-1\) is an eigenvalue, and the eigenvectors corresponding to \(-1\) are the vectors \((w, -w)\) with \(w \neq 0\). Equivalently,

\begin{equation*} \operatorname{null}(T + I) = \{ (w, -w) : w \in \mathbf{F} \} = \operatorname{span}\bigl( (1,-1) \bigr) . \end{equation*}

The answer is the same for \(\mathbf{F} = \mathbf{R}\) and \(\mathbf{F} = \mathbf{C}\), in contrast with the operator \((w, z) \mapsto (-z, w)\) of 5.9. Note also that \((1,1)\) and \((1,-1)\) are eigenvectors corresponding to the distinct eigenvalues \(1\) and \(-1\), so the list \((1,1), (1,-1)\) is linearly independent by 5.11; since its length equals \(\dim \mathbf{F}^2 = 2\), it is a basis of \(\mathbf{F}^2\) by 2.38. Thus \(\mathbf{F}^2\) has a basis consisting of eigenvectors of \(T\), and the matrix of \(T\) with respect to that basis is the diagonal matrix with diagonal entries \(1\) and \(-1\).

The answer: the eigenvalues of \(T\) are \(0\) and \(5\). The eigenvectors corresponding to \(0\) are the vectors \((z_1, 0, 0)\) with \(z_1 \in \mathbf{F}\) and \(z_1 \neq 0\); the eigenvectors corresponding to \(5\) are the vectors \((0, 0, z_3)\) with \(z_3 \in \mathbf{F}\) and \(z_3 \neq 0\).

A scalar \(\lambda \in \mathbf{F}\) is an eigenvalue with eigenvector \((z_1, z_2, z_3) \neq (0,0,0)\) exactly when

\begin{equation*} (2z_2,\, 0,\, 5z_3) = (\lambda z_1,\, \lambda z_2,\, \lambda z_3) , \end{equation*}

which is equivalent to the three equations

\begin{equation*} 2z_2 = \lambda z_1, \qquad 0 = \lambda z_2, \qquad 5z_3 = \lambda z_3 . \end{equation*}

We analyze these by cases on \(\lambda\).

Case \(\lambda = 0\). The equations reduce to \(2z_2 = 0\), \(0 = 0\), and \(5z_3 = 0\), that is, \(z_2 = 0\) and \(z_3 = 0\), with \(z_1\) unrestricted. So the solutions are exactly the vectors \((z_1, 0, 0)\) with \(z_1 \in \mathbf{F}\), and indeed

\begin{equation*} T(z_1, 0, 0) = (0, 0, 0) = 0 \cdot (z_1, 0, 0) . \end{equation*}

Taking \(z_1 = 1\) gives the nonzero vector \((1,0,0)\), so \(0\) is an eigenvalue of \(T\), and the corresponding eigenvectors are the vectors \((z_1, 0, 0)\) with \(z_1 \neq 0\). In other words,

\begin{equation*} \operatorname{null} T = \{ (z_1, 0, 0) : z_1 \in \mathbf{F} \} = \operatorname{span}\bigl( (1,0,0) \bigr) . \end{equation*}

(That \(0\) is an eigenvalue is equivalent to \(T\) not being injective, by 5.7 with \(\lambda = 0\); and indeed \(T(1,0,0) = 0\).)

Case \(\lambda \neq 0\). From the second equation \(0 = \lambda z_2\) and \(\lambda \neq 0\) we get \(z_2 = 0\). Substituting into the first equation gives \(\lambda z_1 = 2 z_2 = 0\), and again since \(\lambda \neq 0\) we get \(z_1 = 0\). So any eigenvector for a nonzero \(\lambda\) has the form \((0, 0, z_3)\), and since an eigenvector is nonzero we must have \(z_3 \neq 0\). The third equation \(5 z_3 = \lambda z_3\) then gives, after dividing by \(z_3 \neq 0\),

\begin{equation*} \lambda = 5 . \end{equation*}

So \(\lambda = 5\) is the only possible nonzero eigenvalue, and it really is one: for \(z_3 \in \mathbf{F}\),

\begin{equation*} T(0, 0, z_3) = (0, 0, 5 z_3) = 5 \cdot (0, 0, z_3) , \end{equation*}

so \((0,0,1)\) is an eigenvector corresponding to \(5\). The eigenvectors corresponding to \(5\) are exactly the vectors \((0, 0, z_3)\) with \(z_3 \neq 0\), that is,

\begin{equation*} \operatorname{null}(T - 5I) = \operatorname{span}\bigl( (0,0,1) \bigr) . \end{equation*}

Hence the complete list of eigenvalues is \(0, 5\), with eigenvectors as described. Notice that \(T\) has only \(2\) distinct eigenvalues even though \(\dim \mathbf{F}^3 = 3\); 5.12 gives an upper bound of \(3\), not an exact count. Indeed \(\mathbf{F}^3\) has no basis consisting of eigenvectors of \(T\): every eigenvector of \(T\) lies in \(\operatorname{span}\bigl( (1,0,0), (0,0,1) \bigr)\), which is a two-dimensional subspace, and a list of three vectors lying in a two-dimensional subspace cannot be linearly independent (by 2.22 applied inside that subspace). The obstruction is visible in the matrix of \(T\) with respect to the standard basis,

\begin{equation*} \begin{pmatrix} 0 & 2 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 5 \end{pmatrix}, \end{equation*}

whose upper-left \(2 \times 2\) block is nilpotent but nonzero.

Suppose \(\lambda \in \mathbf{F}\) is an eigenvalue of \(P\). By the definition of eigenvalue (5.5) there exists \(v \in V\) with \(v \neq 0\) and

\begin{equation*} Pv = \lambda v . \end{equation*}

Apply \(P\) to both sides of this equation. On the left we get \(P(Pv) = P^2 v = Pv = \lambda v\), using the hypothesis \(P^2 = P\) and then the eigenvalue equation. On the right, linearity of \(P\) gives \(P(\lambda v) = \lambda Pv = \lambda(\lambda v) = \lambda^2 v\). Hence

\begin{equation*} \lambda v = \lambda^2 v , \end{equation*}

which we rewrite as

\begin{equation*} (\lambda^2 - \lambda) v = 0 . \end{equation*}

Because \(v \neq 0\), the scalar \(\lambda^2 - \lambda\) must equal \(0\) (if a scalar \(c\) satisfies \(cv = 0\) with \(v \neq 0\), then \(c = 0\), since otherwise multiplying by \(c^{-1}\) would give \(v = 0\)). Thus

\begin{equation*} \lambda(\lambda - 1) = 0 . \end{equation*}

Because \(\mathbf{F}\) (which is \(\mathbf{R}\) or \(\mathbf{C}\)) has no zero divisors, this forces \(\lambda = 0\) or \(\lambda = 1\), as desired.

The only eigenvalue of \(T\) is \(0\), and the eigenvectors corresponding to \(0\) are exactly the nonzero constant polynomials.

First we verify that \(0\) is an eigenvalue. Let \(p\) be the constant polynomial \(p(x) = 1\). Then \(p \neq 0\) and \(Tp = p^{\prime} = 0 = 0 \cdot p\), so \(0\) is an eigenvalue of \(T\) and \(p\) is a corresponding eigenvector.

Now suppose \(\lambda \in \mathbf{R}\) is an eigenvalue of \(T\) with \(\lambda \neq 0\); we derive a contradiction. There exists \(p \in \mathcal{P}(\mathbf{R})\) with \(p \neq 0\) and \(p^{\prime} = \lambda p\). Because \(p \neq 0\), it has a well-defined degree \(m \geq 0\); write

\begin{equation*} p(x) = a_0 + a_1 x + \cdots + a_m x^m , \qquad a_m \neq 0 . \end{equation*}

If \(m = 0\), then \(p^{\prime} = 0\) while \(\lambda p\) is the nonzero constant \(\lambda a_0\) (nonzero because \(\lambda \neq 0\) and \(a_0 = a_m \neq 0\)), so \(p^{\prime} \neq \lambda p\), a contradiction. If \(m \geq 1\), then

\begin{equation*} p^{\prime}(x) = a_1 + 2 a_2 x + \cdots + m a_m x^{m-1} , \end{equation*}

whose coefficient of \(x^{m-1}\) is \(m a_m \neq 0\) (here we use that \(\mathbf{R}\) has characteristic \(0\), so \(m a_m \neq 0\)). Thus \(p^{\prime}\) has degree \(m - 1\), whereas \(\lambda p\) has degree \(m\) because \(\lambda a_m \neq 0\). Two polynomials of different degrees cannot be equal, so again \(p^{\prime} \neq \lambda p\), a contradiction.

Hence no nonzero scalar is an eigenvalue of \(T\), so \(0\) is the only eigenvalue.

Finally we identify the eigenvectors corresponding to \(0\). A vector \(p \neq 0\) is such an eigenvector if and only if \(Tp = 0 \cdot p\), that is, \(p^{\prime} = 0\). Writing \(p(x) = a_0 + a_1 x + \cdots + a_m x^m\), the condition \(p^{\prime} = 0\) says \(a_1 = 2a_2 = \cdots = m a_m = 0\), hence \(a_1 = \cdots = a_m = 0\), i.e. \(p\) is constant. So the eigenvectors of \(T\) are precisely the nonzero constant polynomials.

The eigenvalues of \(T\) are exactly \(0, 1, 2, 3, 4\). For each \(j \in \{0,1,2,3,4\}\), the eigenvectors of \(T\) corresponding to \(j\) are exactly the nonzero scalar multiples of the polynomial \(x^j\).

Work with the standard basis \(1, x, x^2, x^3, x^4\) of \(\mathcal{P}_4(\mathbf{R})\). For each \(j \in \{0,1,2,3,4\}\), if \(p(x) = x^j\) then \(p^{\prime}(x) = j x^{j-1}\) (interpreting this as \(0\) when \(j = 0\)), so

\begin{equation*} (Tp)(x) = x \cdot j x^{j-1} = j x^j = j \, p(x) . \end{equation*}

Thus \(T x^j = j x^j\), and since \(x^j \neq 0\), each of \(0, 1, 2, 3, 4\) is an eigenvalue of \(T\) with \(x^j\) a corresponding eigenvector.

Now suppose \(\lambda \in \mathbf{R}\) and \(p \in \mathcal{P}_4(\mathbf{R})\) satisfy \(Tp = \lambda p\). Write

\begin{equation*} p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 \end{equation*}

with \(a_0, \dots, a_4 \in \mathbf{R}\). Then

\begin{equation*} (Tp)(x) = x\bigl(a_1 + 2a_2 x + 3a_3 x^2 + 4a_4 x^3\bigr) = a_1 x + 2a_2 x^2 + 3a_3 x^3 + 4a_4 x^4 , \end{equation*}

so \(Tp = \sum_{j=0}^{4} j\, a_j x^j\). The equation \(Tp = \lambda p\) therefore says

\begin{equation*} \sum_{j=0}^{4} j\, a_j x^j = \sum_{j=0}^{4} \lambda\, a_j x^j . \end{equation*}

Because \(1, x, x^2, x^3, x^4\) is a basis of \(\mathcal{P}_4(\mathbf{R})\), coefficients may be compared, giving

\begin{equation*} (j - \lambda) a_j = 0 \qquad \text{for } j = 0, 1, 2, 3, 4 . \end{equation*}

If \(\lambda \notin \{0,1,2,3,4\}\), then \(j - \lambda \neq 0\) for every such \(j\), forcing \(a_0 = a_1 = a_2 = a_3 = a_4 = 0\) and hence \(p = 0\). Thus no such \(\lambda\) is an eigenvalue, and the list \(0,1,2,3,4\) found above is the complete list of eigenvalues. (Consistently with 5.12, there are \(5 = \dim \mathcal{P}_4(\mathbf{R})\) of them, the maximum possible.)

If \(\lambda = k\) for some \(k \in \{0,1,2,3,4\}\), then \((j-k) a_j = 0\) gives \(a_j = 0\) for every \(j \neq k\), so \(p(x) = a_k x^k\). Conversely every such \(p\) satisfies \(Tp = k p\). Hence the eigenvectors corresponding to the eigenvalue \(k\) are exactly the polynomials \(a_k x^k\) with \(a_k \neq 0\).

The point is that \(T\) has only finitely many eigenvalues, so the eigenvalues other than \(\alpha\) stay a positive distance away from \(\alpha\).

By 5.12, the operator \(T\) has at most \(\dim V\) distinct eigenvalues. In particular the set

\begin{equation*} E = \{\mu \in \mathbf{F} : \mu \text{ is an eigenvalue of } T \text{ and } \mu \neq \alpha\} \end{equation*}

is finite.

Case 1: \(E = \emptyset\). Take \(\delta = 1\).

Case 2: \(E \neq \emptyset\). Write \(E = \{\lambda_1, \dots, \lambda_m\}\) and set

\begin{equation*} \delta = \min\{|\alpha - \lambda_1|, \dots, |\alpha - \lambda_m|\} . \end{equation*}

This is a minimum of finitely many numbers, each of which is strictly positive because \(\lambda_j \neq \alpha\) for each \(j\). Hence \(\delta > 0\).

In either case, suppose \(\lambda \in \mathbf{F}\) satisfies \(0 < |\alpha - \lambda| < \delta\). We claim \(\lambda\) is not an eigenvalue of \(T\). Indeed, \(|\alpha - \lambda| > 0\) gives \(\lambda \neq \alpha\), so if \(\lambda\) were an eigenvalue of \(T\) then \(\lambda \in E\). In Case 1 this is impossible since \(E\) is empty. In Case 2 it would give \(\lambda = \lambda_j\) for some \(j\), whence

\begin{equation*} |\alpha - \lambda| = |\alpha - \lambda_j| \geq \delta , \end{equation*}

contradicting \(|\alpha - \lambda| < \delta\). So \(\lambda\) is not an eigenvalue of \(T\).

Because \(V\) is finite-dimensional, 5.7 tells us that \(\lambda\) is an eigenvalue of \(T\) if and only if \(T - \lambda I\) is not invertible. Since \(\lambda\) is not an eigenvalue of \(T\), the operator \(T - \lambda I\) is invertible. This holds for every \(\lambda \in \mathbf{F}\) with \(0 < |\alpha - \lambda| < \delta\), completing the proof.

(In the language of analysis: the set of eigenvalues of \(T\) is finite, hence every point of \(\mathbf{F}\) has a punctured neighborhood containing no eigenvalues. Note that the hypothesis says nothing about \(\alpha\) itself; \(\alpha\) may well be an eigenvalue of \(T\), which is exactly why the punctured condition \(0 < |\alpha - \lambda|\) appears.)

The eigenvalues of \(P\) are exactly \(0\) and \(1\). The eigenvectors corresponding to \(1\) are the nonzero vectors in \(U\); the eigenvectors corresponding to \(0\) are the nonzero vectors in \(W\).

First note that \(P\) is well defined: because \(V = U \oplus W\), every \(v \in V\) has a unique representation \(v = u + w\) with \(u \in U\) and \(w \in W\) (this uniqueness is exactly the definition of direct sum, 1.41), so the formula assigns exactly one value to each \(v\).

Only \(0\) and \(1\) can be eigenvalues. For \(v = u + w\) as above we have \(Pv = u\), and since \(u = u + 0\) is the decomposition of \(u\), we get \(P(Pv) = Pu = u = Pv\). Thus \(P^2 = P\), so by Exercise 5A.8 every eigenvalue of \(P\) equals \(0\) or \(1\).

Both \(0\) and \(1\) are eigenvalues. Because \(U \neq \{0\}\) there exists \(u \in U\) with \(u \neq 0\), and \(Pu = u = 1 \cdot u\), so \(1\) is an eigenvalue of \(P\). Because \(W \neq \{0\}\) there exists \(w \in W\) with \(w \neq 0\), and \(Pw = 0 = 0 \cdot w\) (here \(w = 0 + w\) is the decomposition of \(w\)), so \(0\) is an eigenvalue of \(P\).

Eigenvectors corresponding to \(1\). Suppose \(v \in V\) with \(v \neq 0\) and \(Pv = v\). Write \(v = u + w\) with \(u \in U\), \(w \in W\). Then \(u = Pv = v = u + w\), so \(w = 0\) and hence \(v = u \in U\). Conversely, if \(v \in U\) and \(v \neq 0\), then \(Pv = v\) as computed above. So the eigenvectors of \(P\) corresponding to \(1\) are precisely the nonzero elements of \(U\); equivalently \(\operatorname{null}(P - I) = U\).

Eigenvectors corresponding to \(0\). Suppose \(v \in V\) with \(v \neq 0\) and \(Pv = 0\). Write \(v = u + w\) with \(u \in U\), \(w \in W\). Then \(u = Pv = 0\), so \(v = w \in W\). Conversely, if \(v \in W\) and \(v \neq 0\), then \(Pv = 0\). So the eigenvectors of \(P\) corresponding to \(0\) are precisely the nonzero elements of \(W\); equivalently \(\operatorname{null} P = W\).

(a) Suppose \(\lambda \in \mathbf{F}\) is an eigenvalue of \(T\). Then there exists \(v \in V\) with \(v \neq 0\) and \(Tv = \lambda v\). Put

\begin{equation*} w = S^{-1} v . \end{equation*}

Then \(w \neq 0\): if \(w = 0\) then applying \(S\) gives \(v = S w = 0\), contradicting \(v \neq 0\). Moreover, using \(S S^{-1} = I\),

\begin{equation*} (S^{-1} T S) w = S^{-1} T S S^{-1} v = S^{-1} T v = S^{-1} (\lambda v) = \lambda S^{-1} v = \lambda w , \end{equation*}

where the second-to-last equality uses linearity of \(S^{-1}\). Hence \(\lambda\) is an eigenvalue of \(S^{-1} T S\), with eigenvector \(w\).

For the reverse inclusion, apply what was just proved with \(T\) replaced by the operator \(S^{-1} T S\) and with \(S\) replaced by the invertible operator \(S^{-1}\) (whose inverse is \(S\)). This shows that every eigenvalue of \(S^{-1} T S\) is an eigenvalue of

\begin{equation*} (S^{-1})^{-1} (S^{-1} T S) S^{-1} = S S^{-1} T S S^{-1} = T . \end{equation*}

Thus \(T\) and \(S^{-1} T S\) have exactly the same eigenvalues.

(Alternative proof, valid in the same generality: for each \(\lambda \in \mathbf{F}\) we have the identity

\begin{equation*} S^{-1} T S - \lambda I = S^{-1} (T - \lambda I) S , \end{equation*}

because \(S^{-1} (\lambda I) S = \lambda S^{-1} S = \lambda I\). Since \(S\) and \(S^{-1}\) are injective and surjective, \(S^{-1}(T - \lambda I)S\) is injective if and only if \(T - \lambda I\) is injective. By 5.5 and the remark following 5.8, \(\lambda\) is an eigenvalue of an operator exactly when that operator minus \(\lambda I\) fails to be injective, so \(\lambda\) is an eigenvalue of \(S^{-1}TS\) if and only if it is an eigenvalue of \(T\).)

(b) The eigenvectors correspond under the invertible map \(S^{-1}\), with the eigenvalue unchanged. Precisely: for \(\lambda \in \mathbf{F}\) and \(w \in V\) with \(w \neq 0\),

\begin{equation*} w \text{ is an eigenvector of } S^{-1} T S \text{ corresponding to } \lambda \iff Sw \text{ is an eigenvector of } T \text{ corresponding to } \lambda . \end{equation*}

To see this, note that \(Sw \neq 0\) if and only if \(w \neq 0\), since \(S\) is injective and \(S 0 = 0\). Now apply \(S\) to the equation \((S^{-1} T S) w = \lambda w\): this equation holds if and only if

\begin{equation*} S (S^{-1} T S) w = S(\lambda w) , \end{equation*}

that is, if and only if \(T (S w) = \lambda (S w)\) (again because \(S\) is injective, applying \(S\) loses no information, and \(S(\lambda w) = \lambda S w\) by linearity). This is exactly the statement that \(Sw\) is an eigenvector of \(T\) corresponding to \(\lambda\).

Equivalently, in terms of null spaces, for every \(\lambda \in \mathbf{F}\),

\begin{equation*} \operatorname{null}\bigl(S^{-1} T S - \lambda I\bigr) = S^{-1}\bigl(\operatorname{null}(T - \lambda I)\bigr) , \end{equation*}

so \(S^{-1}\) maps the set of eigenvectors of \(T\) corresponding to \(\lambda\) bijectively onto the set of eigenvectors of \(S^{-1}TS\) corresponding to \(\lambda\), and \(S\) maps them back.

Define \(T \in \mathcal{L}(\mathbf{R}^4)\) by

\begin{equation*} T(x_1, x_2, x_3, x_4) = (-x_2,\; x_1,\; -x_4,\; x_3) . \end{equation*}

This map is clearly linear (each coordinate of the output is a linear function of the input coordinates). Geometrically, \(T\) is a counterclockwise rotation by \(90^\circ\) in each of the two coordinate planes \(\{(x_1,x_2,0,0)\}\) and \(\{(0,0,x_3,x_4)\}\); compare 5.9, where the same idea is used on \(\mathbf{R}^2\).

We claim \(T\) has no eigenvalues. Suppose \(\lambda \in \mathbf{R}\) and \(v = (x_1, x_2, x_3, x_4) \in \mathbf{R}^4\) satisfy \(Tv = \lambda v\). Comparing coordinates gives the four equations

\begin{equation*} -x_2 = \lambda x_1, \qquad x_1 = \lambda x_2, \qquad -x_4 = \lambda x_3, \qquad x_3 = \lambda x_4 . \end{equation*}

Substituting the second equation into the first gives

\begin{equation*} -x_2 = \lambda(\lambda x_2) = \lambda^2 x_2 , \end{equation*}

hence \((\lambda^2 + 1) x_2 = 0\). Because \(\lambda\) is a real number, \(\lambda^2 + 1 \geq 1 > 0\), so \(x_2 = 0\). Then \(x_1 = \lambda x_2 = 0\).

Substituting the fourth equation into the third gives, in the same way,

\begin{equation*} -x_4 = \lambda^2 x_4 , \end{equation*}

so \((\lambda^2 + 1) x_4 = 0\), forcing \(x_4 = 0\), and then \(x_3 = \lambda x_4 = 0\).

Thus \(v = (0,0,0,0)\). Hence there is no nonzero \(v \in \mathbf{R}^4\) with \(Tv = \lambda v\), for any \(\lambda \in \mathbf{R}\). By the definition of eigenvalue (5.5), \(T\) has no eigenvalues.

Note that the analogous operator on \(\mathbf{C}^4\) does have eigenvalues, namely \(i\) and \(-i\): the obstruction above disappears once \(\lambda^2 + 1 = 0\) has solutions. This is consistent with 5.19, which guarantees that every operator on a nonzero finite-dimensional complex vector space has an eigenvalue.

Throughout, \(I\) denotes the identity operator on \(V\) and \(I_{V^{\prime}}\) denotes the identity operator on the dual space \(V^{\prime}\). Recall (3.118) that for \(S \in \mathcal{L}(V)\) the dual map \(S^{\prime} \in \mathcal{L}(V^{\prime})\) is defined by \(S^{\prime}(\varphi) = \varphi \circ S\) for \(\varphi \in V^{\prime}\).

First observe that the dual of the identity operator is the identity operator on \(V^{\prime}\): for every \(\varphi \in V^{\prime}\) we have

\begin{equation*} I^{\prime}(\varphi) = \varphi \circ I = \varphi , \end{equation*}

so \(I^{\prime} = I_{V^{\prime}}\).

Next, by 3.120(a) and 3.120(b) (the dual map operation is additive and homogeneous),

\begin{equation*} (T - \lambda I)^{\prime} = T^{\prime} + (-\lambda I)^{\prime} = T^{\prime} - \lambda I^{\prime} = T^{\prime} - \lambda I_{V^{\prime}} . \end{equation*}

Because \(V\) is finite-dimensional, so is \(V^{\prime}\); indeed \(\dim V^{\prime} = \dim V\) by 3.111. Hence 5.7 applies both to the operator \(T\) on \(V\) and to the operator \(T^{\prime}\) on \(V^{\prime}\).

Now chain the equivalences:

\begin{equation*} \begin{aligned} \lambda \text{ is an eigenvalue of } T &\iff T - \lambda I \text{ is not surjective} \\ &\iff (T - \lambda I)^{\prime} \text{ is not injective} \\ &\iff T^{\prime} - \lambda I_{V^{\prime}} \text{ is not injective} \\ &\iff \lambda \text{ is an eigenvalue of } T^{\prime} . \end{aligned} \end{equation*}

Here the first equivalence is the equivalence of (a) and (c) in 5.7 applied to \(T \in \mathcal{L}(V)\); the second is 3.129 (\(S\) is surjective if and only if \(S^{\prime}\) is injective, valid because \(V\) is finite-dimensional), applied to the operator \(S = T - \lambda I\) and then negated; the third is the displayed identity \((T - \lambda I)^{\prime} = T^{\prime} - \lambda I_{V^{\prime}}\); and the fourth is the equivalence of (a) and (b) in 5.7 applied to \(T^{\prime} \in \mathcal{L}(V^{\prime})\), which is legitimate since \(V^{\prime}\) is finite-dimensional.

This proves that \(\lambda\) is an eigenvalue of \(T\) if and only if \(\lambda\) is an eigenvalue of \(T^{\prime}\).

Two remarks. First, finite-dimensionality is genuinely used: 5.7 and 3.129 both require it. Second, although \(T\) and \(T^{\prime}\) have the same eigenvalues, there is in general no natural identification of their eigenvectors, since \(T^{\prime}\) acts on the dual space rather than on \(V\).

Write \(A = \mathcal{M}(T)\) for the matrix of \(T\) with respect to the basis \(v_1, \dots, v_n\), so that by the definition of the matrix of a linear map (3.31), applied with the basis \(v_1, \dots, v_n\) used for both the domain and the codomain,

\begin{equation*} T v_k = \sum_{j=1}^{n} A_{j,k} \, v_j \qquad \text{for each } k = 1, \dots, n , \end{equation*}

and set

\begin{equation*} C = \max \bigl\{ |A_{j,k}| : 1 \leq j, k \leq n \bigr\} . \end{equation*}

Our goal is to show \(|\lambda| \leq nC\).

Suppose \(\lambda\) is an eigenvalue of \(T\), and let \(v \in V\) be an eigenvector corresponding to \(\lambda\); thus \(v \neq 0\) and \(Tv = \lambda v\). Because \(v_1, \dots, v_n\) is a basis of \(V\), there are unique scalars \(c_1, \dots, c_n \in \mathbf{F}\) with

\begin{equation*} v = \sum_{k=1}^{n} c_k v_k . \end{equation*}

Since \(v \neq 0\), not all the \(c_k\) are \(0\). Choose an index \(m \in \{1, \dots, n\}\) with

\begin{equation*} |c_m| = \max \bigl\{ |c_1|, \dots, |c_n| \bigr\} , \end{equation*}

so in particular \(|c_m| > 0\).

Now compute \(Tv\) two ways. Using linearity of \(T\) and the formula for \(Tv_k\),

\begin{equation*} Tv = \sum_{k=1}^{n} c_k \, T v_k = \sum_{k=1}^{n} c_k \sum_{j=1}^{n} A_{j,k} v_j = \sum_{j=1}^{n} \Bigl( \sum_{k=1}^{n} A_{j,k} c_k \Bigr) v_j . \end{equation*}

On the other hand \(Tv = \lambda v = \sum_{j=1}^{n} \lambda c_j \, v_j\). Because the representation of a vector in terms of a basis is unique, comparing coefficients of \(v_j\) gives

\begin{equation*} \lambda c_j = \sum_{k=1}^{n} A_{j,k} c_k \qquad \text{for each } j = 1, \dots, n . \end{equation*}

Apply this with \(j = m\) and take absolute values. Using the triangle inequality and then the bounds \(|A_{m,k}| \leq C\) and \(|c_k| \leq |c_m|\), we get

\begin{equation*} |\lambda| \, |c_m| = \Bigl| \sum_{k=1}^{n} A_{m,k} c_k \Bigr| \leq \sum_{k=1}^{n} |A_{m,k}| \, |c_k| \leq \sum_{k=1}^{n} C \, |c_m| = n C |c_m| . \end{equation*}

Since \(|c_m| > 0\), we may divide both sides by \(|c_m|\) to conclude

\begin{equation*} |\lambda| \leq nC = n \max \bigl\{ \, \bigl| \mathcal{M}(T)_{j,k} \bigr| : 1 \leq j, k \leq n \, \bigr\} , \end{equation*}

as desired.

Recall the definitions. The complexification \(V_{\mathbf{C}}\) of the real vector space \(V\) is the set \(V \times V\), whose elements are written \(u + iv\) with \(u, v \in V\), with addition computed componentwise and with complex scalar multiplication given by

\begin{equation*} (a + bi)(u + iv) = (au - bv) + i(av + bu) \end{equation*}

for all \(a, b \in \mathbf{R}\) and all \(u, v \in V\) (Exercise 8 in Section 1B). The complexification of \(T \in \mathcal{L}(V)\) is the complex-linear map \(T_{\mathbf{C}} \in \mathcal{L}(V_{\mathbf{C}})\) defined by

\begin{equation*} T_{\mathbf{C}}(u + iv) = Tu + iTv \end{equation*}

for all \(u, v \in V\) (Exercise 33 in Section 3B). Note that \(u + iv\) is the zero vector of \(V_{\mathbf{C}}\) exactly when \(u = 0\) and \(v = 0\), since \(V_{\mathbf{C}}\) is \(V \times V\) as a set.

A preliminary computation will be used in both directions. Since \(\lambda\) is real, \(\lambda = \lambda + 0i\), so for all \(u, v \in V\),

\begin{equation*} \lambda (u + iv) = (\lambda + 0i)(u + iv) = (\lambda u - 0 \cdot v) + i(\lambda v + 0 \cdot u) = \lambda u + i \lambda v . \end{equation*}

First suppose \(\lambda\) is an eigenvalue of \(T\). Then there exists \(v \in V\) with \(v \neq 0\) and \(Tv = \lambda v\). Consider the vector \(v + i0 \in V_{\mathbf{C}}\). It is nonzero, because its first component \(v\) is nonzero. Furthermore

\begin{equation*} T_{\mathbf{C}}(v + i0) = Tv + i T0 = \lambda v + i 0 = \lambda (v + i0), \end{equation*}

where the last equality uses the preliminary computation with \(u = v\) and the second vector equal to \(0\). Thus \(v + i0\) is an eigenvector of \(T_{\mathbf{C}}\) corresponding to \(\lambda\), so \(\lambda\) is an eigenvalue of \(T_{\mathbf{C}}\).

Conversely, suppose \(\lambda\) is an eigenvalue of \(T_{\mathbf{C}}\). Then there exist \(u, v \in V\), not both \(0\), such that

\begin{equation*} T_{\mathbf{C}}(u + iv) = \lambda (u + iv) . \end{equation*}

The left side equals \(Tu + iTv\) and, by the preliminary computation, the right side equals \(\lambda u + i \lambda v\). Two elements of \(V_{\mathbf{C}} = V \times V\) are equal exactly when their two components agree, so

\begin{equation*} Tu = \lambda u \qquad \text{and} \qquad Tv = \lambda v . \end{equation*}

Since \(u\) and \(v\) are not both \(0\), at least one of them is a nonzero vector in \(V\) satisfying the eigenvector equation for \(T\) with the scalar \(\lambda\). Hence \(\lambda\) is an eigenvalue of \(T\).

This completes the proof.

Remark. A slicker route uses Exercise 33(b) in Section 3B, which says that \(S_{\mathbf{C}}\) is injective if and only if \(S\) is injective. Since

\begin{equation*} (T - \lambda I)_{\mathbf{C}}(u + iv) = (T - \lambda I)u + i(T - \lambda I)v = (Tu + iTv) - \lambda(u + iv) = (T_{\mathbf{C}} - \lambda I)(u + iv) \end{equation*}

for all \(u, v \in V\) (using again that \(\lambda\) is real), we have \((T - \lambda I)_{\mathbf{C}} = T_{\mathbf{C}} - \lambda I\). Because a scalar \(\mu\) is an eigenvalue of an operator \(S\) precisely when \(S - \mu I\) fails to be injective (a restatement of the definition, valid without any finite-dimensionality hypothesis), we get

\begin{equation*} \begin{aligned} \lambda \text{ is an eigenvalue of } T &\iff T - \lambda I \text{ is not injective} \\ &\iff T_{\mathbf{C}} - \lambda I \text{ is not injective} \\ &\iff \lambda \text{ is an eigenvalue of } T_{\mathbf{C}} . \end{aligned} \end{equation*}

Note that no finite-dimensionality assumption on \(V\) is needed anywhere.

As in Exercise 17, \(V_{\mathbf{C}} = V \times V\) with elements written \(u + iv\) for \(u, v \in V\), complex scalar multiplication given by

\begin{equation*} (a + bi)(u + iv) = (au - bv) + i(av + bu) \qquad (a, b \in \mathbf{R}, \ u, v \in V), \end{equation*}

and \(T_{\mathbf{C}}(u + iv) = Tu + iTv\).

Define the conjugation map \(\sigma \colon V_{\mathbf{C}} \to V_{\mathbf{C}}\) by

\begin{equation*} \sigma(u + iv) = u - iv = u + i(-v) \qquad \text{for all } u, v \in V . \end{equation*}

This is a well-defined function on \(V_{\mathbf{C}} = V \times V\), namely \((u, v) \mapsto (u, -v)\). We record four properties of \(\sigma\).

First, \(\sigma\) is additive: for \(u_1, v_1, u_2, v_2 \in V\),

\begin{equation*} \sigma\bigl( (u_1 + iv_1) + (u_2 + iv_2) \bigr) = \sigma\bigl( (u_1 + u_2) + i(v_1 + v_2) \bigr) = (u_1 + u_2) + i\bigl(-(v_1 + v_2)\bigr), \end{equation*}

which equals \(\sigma(u_1 + iv_1) + \sigma(u_2 + iv_2)\).

Second, \(\sigma\) is conjugate-homogeneous: if \(\mu = a + bi\) with \(a, b \in \mathbf{R}\) and \(u, v \in V\), then

\begin{equation*} \sigma\bigl( \mu (u + iv) \bigr) = \sigma\bigl( (au - bv) + i(av + bu) \bigr) = (au - bv) + i\bigl(-av - bu\bigr), \end{equation*}

while

\begin{equation*} \overline{\mu} \, \sigma(u + iv) = (a - bi)\bigl( u + i(-v) \bigr) = \bigl( au - (-b)(-v) \bigr) + i\bigl( a(-v) + (-b)u \bigr) = (au - bv) + i(-av - bu) . \end{equation*}

The two right-hand sides agree, so \(\sigma(\mu w) = \overline{\mu} \, \sigma(w)\) for all \(\mu \in \mathbf{C}\) and \(w \in V_{\mathbf{C}}\).

Third, \(\sigma\) is a bijection of \(V_{\mathbf{C}}\) onto itself with \(\sigma \circ \sigma\) equal to the identity, since \(\sigma(\sigma(u + iv)) = \sigma(u + i(-v)) = u + iv\). In particular \(\sigma(w) = 0\) if and only if \(w = 0\).

Fourth, \(\sigma\) commutes with \(T_{\mathbf{C}}\): for all \(u, v \in V\),

\begin{equation*} \sigma\bigl( T_{\mathbf{C}}(u + iv) \bigr) = \sigma(Tu + iTv) = Tu + i(-Tv) = Tu + i\,T(-v) = T_{\mathbf{C}}\bigl( u + i(-v) \bigr) = T_{\mathbf{C}}\bigl( \sigma(u + iv) \bigr), \end{equation*}

where we used the linearity of \(T\) in the form \(T(-v) = -Tv\). Thus \(\sigma \circ T_{\mathbf{C}} = T_{\mathbf{C}} \circ \sigma\).

Now suppose \(\lambda \in \mathbf{C}\) is an eigenvalue of \(T_{\mathbf{C}}\). Choose \(w \in V_{\mathbf{C}}\) with \(w \neq 0\) and \(T_{\mathbf{C}} w = \lambda w\). Then \(\sigma(w) \neq 0\) by the third property, and

\begin{equation*} T_{\mathbf{C}}\bigl( \sigma(w) \bigr) = \sigma\bigl( T_{\mathbf{C}} w \bigr) = \sigma(\lambda w) = \overline{\lambda} \, \sigma(w), \end{equation*}

using the fourth property and then the second. Hence \(\sigma(w)\) is an eigenvector of \(T_{\mathbf{C}}\) corresponding to \(\overline{\lambda}\), so \(\overline{\lambda}\) is an eigenvalue of \(T_{\mathbf{C}}\).

For the other direction, suppose \(\overline{\lambda}\) is an eigenvalue of \(T_{\mathbf{C}}\). Applying what we just proved with \(\overline{\lambda}\) in place of \(\lambda\) shows that \(\overline{\overline{\lambda}}\) is an eigenvalue of \(T_{\mathbf{C}}\); since \(\overline{\overline{\lambda}} = \lambda\), this says \(\lambda\) is an eigenvalue of \(T_{\mathbf{C}}\).

Therefore \(\lambda\) is an eigenvalue of \(T_{\mathbf{C}}\) if and only if \(\overline{\lambda}\) is an eigenvalue of \(T_{\mathbf{C}}\).

Concretely, unwinding the abstraction: if \(\lambda = a + bi\) and \(T_{\mathbf{C}}(u + iv) = \lambda(u + iv)\) with \((u, v) \neq (0, 0)\), then comparing components gives \(Tu = au - bv\) and \(Tv = av + bu\), and one checks directly that these same two equations say exactly that \(T_{\mathbf{C}}(u - iv) = \overline{\lambda}(u - iv)\), with \(u - iv \neq 0\).

Note that the argument uses in an essential way that \(T\) is the complexification of a real operator: this is what makes \(T_{\mathbf{C}}\) commute with \(\sigma\). The statement is false for a general operator on a complex vector space, for example for the operator on \(\mathbf{C}\) given by multiplication by \(i\), whose only eigenvalue is \(i\).

Suppose \(\lambda \in \mathbf{F}\) and \(z = (z_1, z_2, z_3, \dots) \in \mathbf{F}^\infty\) satisfy \(Tz = \lambda z\). We show \(z = 0\); since an eigenvector must be nonzero, this proves that no \(\lambda \in \mathbf{F}\) is an eigenvalue of \(T\).

The equation \(Tz = \lambda z\) reads

\begin{equation*} (0, z_1, z_2, z_3, \dots) = (\lambda z_1, \lambda z_2, \lambda z_3, \dots) . \end{equation*}

Two elements of \(\mathbf{F}^\infty\) are equal exactly when all their coordinates agree, so comparing the first coordinate and then the \((k+1)\)-st coordinate for each \(k \geq 1\) gives

\begin{equation*} \lambda z_1 = 0 \qquad \text{and} \qquad z_k = \lambda z_{k+1} \quad \text{for every positive integer } k . \end{equation*}

Case 1: \(\lambda = 0\). Then the second family of equations reads \(z_k = 0 \cdot z_{k+1} = 0\) for every positive integer \(k\). Hence \(z = 0\).

Case 2: \(\lambda \neq 0\). From \(\lambda z_1 = 0\) and \(\lambda \neq 0\) we get \(z_1 = 0\). Now the equations \(z_k = \lambda z_{k+1}\) can be rewritten as

\begin{equation*} z_{k+1} = \frac{z_k}{\lambda} \qquad \text{for every positive integer } k . \end{equation*}

We prove by induction on \(k\) that \(z_k = 0\) for every positive integer \(k\). The base case \(z_1 = 0\) has just been established. If \(z_k = 0\) for some \(k \geq 1\), then \(z_{k+1} = z_k / \lambda = 0\). Hence \(z_k = 0\) for all \(k\), so \(z = 0\).

In both cases \(z = 0\). Thus for every \(\lambda \in \mathbf{F}\) the only solution of \(Tz = \lambda z\) is \(z = 0\), that is, \(\operatorname{null}(T - \lambda I) = \{0\}\). Therefore \(T\) has no eigenvalues.

This example shows that the equivalence of (b) and (c) in 5.7 genuinely requires finite-dimensionality. Here \(T - 0I = T\) is injective, so \(0\) is not an eigenvalue of \(T\), yet \(T\) is not surjective: no vector in the range of \(T\) has a nonzero first coordinate. Compare Exercise 20, where the backward shift, which is surjective but not injective, has every element of \(\mathbf{F}\) as an eigenvalue. (The equivalence of (a) and (b) in 5.7 does hold in every vector space, since \(Tv = \lambda v\) is the same equation as \((T - \lambda I)v = 0\); it is (c) and (d) that decouple in infinite dimensions.)

(a) Let \(\lambda \in \mathbf{F}\) be arbitrary. Define

\begin{equation*} w_\lambda = (1, \lambda, \lambda^2, \lambda^3, \dots) \in \mathbf{F}^\infty , \end{equation*}

that is, the sequence whose \(k\)-th coordinate is \(\lambda^{k-1}\), with the convention \(\lambda^0 = 1\) (so \(w_0 = (1, 0, 0, \dots)\)). Then \(w_\lambda \neq 0\), because its first coordinate is \(1\).

Applying \(S\) deletes the first coordinate:

\begin{equation*} S w_\lambda = (\lambda, \lambda^2, \lambda^3, \dots) . \end{equation*}

The \(k\)-th coordinate of \(S w_\lambda\) is \(\lambda^{k}\), which equals \(\lambda \cdot \lambda^{k-1}\), the \(k\)-th coordinate of \(\lambda w_\lambda\). Hence

\begin{equation*} S w_\lambda = \lambda w_\lambda , \end{equation*}

so \(w_\lambda\) is an eigenvector of \(S\) corresponding to \(\lambda\). Thus every \(\lambda \in \mathbf{F}\) is an eigenvalue of \(S\). (For \(\lambda = 0\) this reads \(S(1, 0, 0, \dots) = (0, 0, \dots) = 0 \cdot w_0\), which is correct.)

(b) We claim that the eigenvectors of \(S\) are exactly the vectors

\begin{equation*} c \, (1, \lambda, \lambda^2, \lambda^3, \dots) = c \, w_\lambda \qquad \text{with } \lambda \in \mathbf{F} \text{ and } c \in \mathbf{F}, \ c \neq 0 , \end{equation*}

and that \(c \, w_\lambda\) is an eigenvector corresponding to the eigenvalue \(\lambda\).

Each such vector is indeed an eigenvector: it is nonzero because its first coordinate is \(c \neq 0\), and \(S(c w_\lambda) = c \, S w_\lambda = c \lambda w_\lambda = \lambda (c w_\lambda)\) by part (a) and the linearity of \(S\).

Conversely, suppose \(z = (z_1, z_2, z_3, \dots)\) is an eigenvector of \(S\), say \(Sz = \lambda z\) with \(z \neq 0\) and \(\lambda \in \mathbf{F}\). Written out, the equation \(Sz = \lambda z\) is

\begin{equation*} (z_2, z_3, z_4, \dots) = (\lambda z_1, \lambda z_2, \lambda z_3, \dots) , \end{equation*}

and comparing the \(k\)-th coordinates gives

\begin{equation*} z_{k+1} = \lambda z_k \qquad \text{for every positive integer } k . \end{equation*}

By induction on \(k\) this yields

\begin{equation*} z_k = \lambda^{k-1} z_1 \qquad \text{for every positive integer } k : \end{equation*}

the case \(k = 1\) is the trivial identity \(z_1 = \lambda^0 z_1\), and if \(z_k = \lambda^{k-1} z_1\) then \(z_{k+1} = \lambda z_k = \lambda^{k} z_1\).

If \(z_1 = 0\), the formula forces \(z_k = 0\) for every \(k\), so \(z = 0\), contradicting the assumption that \(z\) is an eigenvector. Hence \(z_1 \neq 0\), and setting \(c = z_1\) we have

\begin{equation*} z = (c, c\lambda, c\lambda^2, \dots) = c \, w_\lambda \qquad \text{with } c \neq 0 , \end{equation*}

as claimed.

Two consequences are worth recording. First, for each \(\lambda \in \mathbf{F}\) the eigenspace of \(S\) corresponding to \(\lambda\) is

\begin{equation*} \operatorname{null}(S - \lambda I) = \{ c \, w_\lambda : c \in \mathbf{F} \} = \operatorname{span}(w_\lambda), \end{equation*}

a one-dimensional subspace of \(\mathbf{F}^\infty\). Second, distinct scalars give distinct eigenspaces: if \(\lambda \neq \mu\) then \(w_\lambda\) and \(w_\mu\) differ in their second coordinate, and since both have first coordinate \(1\) neither is a scalar multiple of the other. So the eigenvectors of \(S\) form exactly the union over \(\lambda \in \mathbf{F}\) of the punctured lines \(\operatorname{span}(w_\lambda) \setminus \{0\}\), and these lines are pairwise distinct.

Note that \(S\) has infinitely many distinct eigenvalues when \(\mathbf{F}\) is infinite, which is consistent with 5.12 only because \(\mathbf{F}^\infty\) is infinite-dimensional.