7.7 Proof patterns · Matrix Theory

The chapter asked one question about an operator $𝑇$ on a finite-dimensional space: in which basis is its matrix as simple as possible, and how simple can it be made? Diagonal when there is an eigenbasis, triangular when the characteristic polynomial splits, and in every case the answer is controlled by polynomials in $𝑇$ . Most of the chapter’s proofs use a few moves to turn a hypothesis about $𝑇$ into information about vectors: apply a polynomial in $𝑇$ to kill terms, follow a vector along its chain $𝐯, 𝑇 𝐯, 𝑇^{2} 𝐯, \dots$ , look for invariant subspaces among kernels of operators that commute with $𝑇$ , test an identity on eigenvectors, or triangularize and read off the diagonal. This section sorts the chapter’s arguments by the question they answer, solves three problems in several ways to compare the moves, and lists the errors that the chapter’s exercises warn against. The exercises at the end practice one move at a time, and the problem set (Problems for Chapter 7) mixes them with earlier chapters.

7.7.1Polynomials in $𝑇$ kill terms

On an eigenvector, every polynomial in $𝑇$ acts as a scalar: $𝑝 (𝑇) 𝐯 = 𝑝 (𝜆) 𝐯$ (Proposition 7.2.28). So a polynomial that vanishes at some eigenvalues and not at others removes exactly the terms it vanishes on, and this is the most common move of the chapter.

Kill all terms but one (Technique 4.1.25). To show that eigenvectors $𝐯_{1}, \dots, 𝐯_{𝑚}$ for distinct eigenvalues are independent, apply $\prod_{𝑗 \neq 𝑖} (𝑇 - 𝜆_{𝑗} id)$ to a relation; only the $𝑖$ th term survives, multiplied by the nonzero scalar $\prod_{𝑗 \neq 𝑖} (𝜆_{𝑖} - 𝜆_{𝑗})$ (Theorem 7.2.21). With the normalized product, the Lagrange polynomial $𝐿_{𝑖}$ , the surviving term comes out with coefficient $1$ , and $𝐿_{𝑖} (𝑇)$ is the projection onto $𝐸_{𝜆_{𝑖}}$ (Technique 7.6.13).
Read eigenvalues off an annihilating polynomial (Technique 7.6.17). If $𝑝 (𝑇) = 0$ , every eigenvalue is a root of $𝑝$ (Proposition 7.2.28). If $𝑝$ is a product of distinct linear factors, $𝑇$ is diagonalizable (Proposition 7.6.12). If $𝑝 (0) \neq 0$ , then $𝑇$ is invertible and $𝑇^{- 1}$ is a polynomial in $𝑇$ , read off from $𝑝$ (Corollary 7.6.9). Note the direction: a root of $𝑝$ need not be an eigenvalue, since $𝑇 = 0$ on $𝑉 \neq {𝟎}$ is killed by $𝑥 (𝑥 - 1)$ and $1$ is not an eigenvalue of it.
Bézout in $𝑇$ . If $𝑢 𝑝 + 𝑣 𝑞 = 1$ in $𝐹 [𝑥]$ , then $𝑢 (𝑇) 𝑝 (𝑇) +$ $𝑣 (𝑇) 𝑞 (𝑇) =$ ${id}_{𝑉}$ , so a vector killed by both $𝑝 (𝑇)$ and $𝑞 (𝑇)$ is $𝟎$ , and when $𝑝 (𝑇) 𝑞 (𝑇) = 0$ the space splits as $ker 𝑝 (𝑇) \oplus$ $ker 𝑞 (𝑇)$ (Corollary 7.1.48, Exercise 7.2.55). Lagrange interpolation is the case of linear factors.

Behind all three is the fact that polynomials in $𝑇$ commute with each other and with $𝑇$ (Proposition 7.2.27), so they can be applied to a relation in any order.

7.7.2Chains

The chain move of Section 4.8.1 returns with $𝑇$ replaced by $𝑁 = 𝑇 - 𝜆 id$ . If $𝑁^{𝑘} 𝐯 \neq 𝟎$ and $𝑁^{𝑘 + 1} 𝐯 = 𝟎$ , then $𝐯, 𝑁 𝐯, \dots, 𝑁^{𝑘} 𝐯$ is independent (Proposition 7.2.37), its span is the Krylov subspace $𝐾_{𝑇} (𝐯)$ , the smallest $𝑇$ -invariant subspace containing $𝐯$ , and the last nonzero vector $𝑁^{𝑘} 𝐯$ is an eigenvector with eigenvalue $𝜆$ (Corollary 7.2.38). Three things make this the most useful single fact of the chapter.

It turns “killed by a power” into “an eigenvector exists.” Whenever some ${(𝑇 - 𝜆 id)}^{𝑘}$ kills a nonzero $𝐯$ , walk along the chain until the next step gives $𝟎$ ; the last vector is an eigenvector. This is how Technique 7.3.28 produces eigenvectors: the first dependence among $𝐯, 𝑇 𝐯, 𝑇^{2} 𝐯, \dots$ is a polynomial $𝑝$ with $𝑝 (𝑇) 𝐯 = 𝟎$ (Proposition 7.2.35), and peeling off one factor $𝑥 - 𝜆$ of $𝑝$ leaves a nonzero vector killed by $𝑇 - 𝜆 id$ . Over $ℂ$ , $𝑝$ always has such a factor, which is the determinant-free proof that eigenvalues exist (Theorem 7.3.27).
It counts. A chain is independent, so it has at most $dim 𝑉$ vectors, and a chain of length $dim 𝑉$ is a basis in which $𝑇$ is $𝜆$ on the diagonal and $1$ just above it (Corollary 7.2.38(c)). Combined with the kernel count $dim ker 𝑁^{𝑗 + 1} \leq$ $dim ker 𝑁^{𝑗} +$ $dim ker 𝑁$ , it decides when one chain fills the space (Problem 7.3).
It works one vector at a time, with no hypothesis on $𝑇$ beyond the two conditions on $𝐯$ , and in any dimension and over any field. That is why it decides problems where nothing is known about $𝑇$ globally, such as whether a matrix has a square root (Exercise 7.7.9).

Example 7.7.1 (Two eigenvalues cannot share a generalized eigenvector, three ways). Let $𝑇 \in L (𝑉)$ , let $𝜆 \neq 𝜇$ in $𝐹$ , let $𝑘, 𝑚 \geq 1$ , and let $𝐯 \in 𝑉$ satisfy

{(𝑇 - 𝜆 id)}^{𝑘} 𝐯 = 𝟎 and {(𝑇 - 𝜇 id)}^{𝑚} 𝐯 = 𝟎 .

We show $𝐯 = 𝟎$ . Here $𝑉$ is any vector space over any field. Write $𝐴 = 𝑇 - 𝜆 id$ and $𝐵 = 𝑇 - 𝜇 id$ ; these commute, and $𝐵 - 𝐴 = (𝜆 - 𝜇) id$ .

Kill every term. By the binomial theorem for commuting operators (Exercise A.2.8), with $𝑟 = 𝑘 + 𝑚 - 1$ ,

{(𝜆 - 𝜇)}^{𝑟} 𝐯 = {(𝐵 - 𝐴)}^{𝑟} 𝐯 = \sum_{𝑗 = 0}^{𝑟} (\binom{𝑟}{𝑗}) {(- 1)}^{𝑟 - 𝑗} 𝐴^{𝑟 - 𝑗} 𝐵^{𝑗} 𝐯 .

In each term, either $𝑗 \geq 𝑚$ , and then $𝐵^{𝑗} 𝐯 = 𝟎$ , or $𝑗 \leq 𝑚 - 1$ , and then $𝑟 - 𝑗 \geq 𝑘$ and $𝐴^{𝑟 - 𝑗} 𝐵^{𝑗} 𝐯 =$ $𝐵^{𝑗} 𝐴^{𝑟 - 𝑗} 𝐯 =$ $𝟎$ . So ${(𝜆 - 𝜇)}^{𝑟} 𝐯 = 𝟎$ , and since $𝜆 - 𝜇 \neq 0$ , $𝐯 = 𝟎$ .

A chain. Suppose $𝐯 \neq 𝟎$ . Since $𝐴^{𝑘} 𝐯 = 𝟎$ , there is a largest $𝑗$ with $0 \leq 𝑗 < 𝑘$ and $𝐴^{𝑗} 𝐯 \neq 𝟎$ . Then $𝐰 = 𝐴^{𝑗} 𝐯$ satisfies $𝐴 𝐰 = 𝟎$ , so it is an eigenvector of $𝑇$ with eigenvalue $𝜆$ (Corollary 7.2.38(a)). By Proposition 7.2.28, $𝐵^{𝑚} 𝐰 =$ ${(𝜆 - 𝜇)}^{𝑚} 𝐰 \neq$ $𝟎$ . But $𝐵^{𝑚} 𝐰 =$ $𝐴^{𝑗} 𝐵^{𝑚} 𝐯 =$ $𝟎$ , a contradiction.

Restrict, then invert. Let $𝑈 =$ $ker 𝐴^{𝑘} \cap$ $ker 𝐵^{𝑚}$ . Both kernels are $𝑇$ -invariant (Proposition 7.2.30(a)), so $𝑈$ is, and on $𝑈$ the restriction $𝑁 = 𝐴 ↾_{𝑈}$ satisfies $𝑁^{𝑘} = 0$ . Now $𝐵 ↾_{𝑈} =$ $𝑁 +$ $(𝜆 - 𝜇) {id}_{𝑈} =$ $(𝜆 - 𝜇) ({id}_{𝑈} + 𝑐 𝑁)$ with $𝑐 = {(𝜆 - 𝜇)}^{- 1}$ , and ${id}_{𝑈} + 𝑐 𝑁$ is invertible, with inverse ${id}_{𝑈} -$ $𝑐 𝑁 +$ $𝑐^{2} 𝑁^{2} -$ $\dots +$ ${(- 𝑐)}^{𝑘 - 1} 𝑁^{𝑘 - 1}$ (the computation of Exercise 4.1.44). So $𝐵 ↾_{𝑈}$ is invertible, and so is its $𝑚$ th power. But ${𝐵 ↾_{𝑈}}^{𝑚} = 0$ . An invertible zero operator lives only on ${𝟎}$ , so $𝑈 = {𝟎}$ and $𝐯 = 𝟎$ .

The first proof is Bézout’s identity for ${(𝑥 - 𝜆)}^{𝑘}$ and ${(𝑥 - 𝜇)}^{𝑚}$ , written out by hand, and it is the one that generalizes: for any coprime $𝑝, 𝑞$ , a vector killed by $𝑝 (𝑇)$ and $𝑞 (𝑇)$ is $𝟎$ . The chain proof reduces the question to eigenvectors, where polynomials act as scalars, and it is the shortest once the chain is in your hands. The third proof sees the hypothesis as a statement about the operator on $𝑈$ : there $𝑇$ is “ $𝜆$ plus nilpotent,” and a nilpotent perturbation of the nonzero scalar $𝜆 - 𝜇$ cannot be $0$ . All three are the key step in splitting $𝑉$ into generalized eigenspaces (Section 10.2).

Practice: Exercise 7.7.9, Problem 7.3, Problem 7.12, and Problem 7.9.

7.7.3Invariant subspaces from commuting operators

To find subspaces that $𝑇$ maps into themselves, look at kernels, images, and eigenspaces of operators that commute with $𝑇$ (Technique 7.2.31, Proposition 7.2.30). Polynomials in $𝑇$ always qualify, so $ker {(𝑇 - 𝜆 id)}^{𝑘}$ and $im 𝑝 (𝑇)$ are always available, and Problem 7.4 finds every invariant subspace of a $3 \times 3$ matrix this way. When two given operators $𝑆, 𝑇$ commute, each eigenspace of $𝑆$ is a smaller space on which $𝑇$ still acts, and the problem can be continued there. This is the whole proof that commuting operators have a common eigenvector (Theorem 7.6.31): $𝑇$ has an eigenvalue on $𝐸_{𝜆} (𝑆)$ when the field is $ℂ$ . Two cautions. The restriction of $𝑇$ to $𝐸_{𝜆} (𝑆)$ inherits diagonalizability (Corollary 7.6.16), but nothing forces it to be diagonalizable when $𝑇$ is not, and the eigenspaces of $𝑆$ need not fill $𝑉$ ; so commuting alone gives a common eigenvector and a common triangular form (Theorem 7.6.32), and a common eigenbasis only for diagonalizable operators (Theorem 7.6.34, Exercise 7.7.6). The kernel move does not need eigenvalues at all, which makes it work over any field: two commuting nilpotent operators have a common kernel vector, because $ker 𝑆$ is $𝑇$ -invariant and $𝑇$ is still nilpotent there (Exercise 7.7.13).

Practice: Exercise 7.7.6, Exercise 7.7.13, Exercise 7.7.12, and Problem 7.4.

7.7.4Proving and refuting diagonalizability

There are four standard ways to show that $𝑇$ is diagonalizable, in roughly increasing order of generality:

$𝑛$ distinct eigenvalues (Corollary 7.4.5);
an annihilating polynomial that is a product of distinct linear factors (Theorem 7.6.15), which is the route for any identity such as $𝑇^{2} = 𝑇$ , $𝑇^{3} = 𝑇$ , or $𝑇^{𝑘} = id$ over $ℂ$ ;
a direct sum decomposition of $𝑉$ into eigenspaces, often produced by idempotents (Technique 4.4.28) or by Lagrange polynomials in $𝑇$ ;
counting: $𝑝_{𝑇}$ splits and $gm (𝜆) = am (𝜆)$ for each eigenvalue (Theorem 7.4.16), with $gm (𝜆) =$ $𝑛 -$ $rank (𝐴 - 𝜆 𝐼_{𝑛})$ computed as a rank (Lemma 7.4.13).

To refute it, find one of the obstructions of Obstructions to diagonalizability: $𝑝_{𝑇}$ does not split over $𝐹$ (the rotation, Example 7.4.9), or some eigenvalue has $gm < am$ (the shear, Example 7.4.8). Two shortcuts catch most examples. A diagonalizable operator with a single eigenvalue $𝜆$ is $𝜆 id$ (Proposition 7.4.7), so a nonzero nilpotent operator, or $𝐼 + 𝑁$ with $𝑁 \neq 0$ nilpotent, is never diagonalizable. And over a field where the roots of the relevant polynomial collide, the polynomial route fails: $𝑥^{2} - 1 = {(𝑥 - 1)}^{2}$ over $𝔽_{2}$ , and $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ satisfies $𝐴^{2} = 𝐼_{2}$ there without being diagonalizable.

Example 7.7.2 (An idempotent is diagonalizable, three ways). Let $dim 𝑉 = 𝑛$ and let $𝑃 \in L (𝑉)$ satisfy $𝑃^{2} = 𝑃$ . We show that $𝑃$ is diagonalizable, with $𝑃$ similar to $diag (1, \dots, 1, 0, \dots, 0)$ , where the number of ones is $rank 𝑃$ .

An annihilating polynomial. The polynomial $𝑥^{2} - 𝑥 = 𝑥 (𝑥 - 1)$ kills $𝑃$ , and its factors are distinct because $0 \neq 1$ in every field. By Proposition 7.6.12, $𝑉 = 𝐸_{0} \oplus 𝐸_{1}$ and $𝑃$ is diagonalizable.

Image and kernel. By Theorem 4.4.27, $𝑉 = im 𝑃 \oplus ker 𝑃$ and $im 𝑃 = {𝐯 | 𝑃 𝐯 = 𝐯}$ . So $im 𝑃 = 𝐸_{1}$ and $ker 𝑃 = 𝐸_{0}$ , and a basis of $im 𝑃$ followed by a basis of $ker 𝑃$ is a basis of $𝑉$ (Theorem 4.4.13) consisting of eigenvectors. There are $rank 𝑃$ vectors with eigenvalue $1$ .

Eigenvalues and rank. Every eigenvalue is a root of $𝑥^{2} - 𝑥$ (Proposition 7.2.28), so it is $0$ or $1$ . By Lemma 7.4.13, $gm (0) = 𝑛 - rank 𝑃$ and $gm (1) =$ $𝑛 -$ $rank (𝑃 - id)$ . Since $𝑃 + (id - 𝑃) = id$ , subadditivity gives $rank 𝑃 +$ $rank (id - 𝑃) \geq$ $𝑛$ (Proposition 4.3.20); since $𝑃 (id - 𝑃) = 0$ , Sylvester’s inequality gives $0 \geq$ $rank 𝑃 +$ $rank (id - 𝑃) -$ $𝑛$ (Exercise 4.3.30). So $gm (0) +$ $gm (1) =$ $2 𝑛 -$ $(rank 𝑃 + rank (id - 𝑃)) =$ $𝑛$ , and Theorem 7.4.16(III) applies.

The polynomial proof is the shortest and generalizes at once: if $(𝑇 - 𝑎 id) (𝑇 - 𝑏 id) =$ $0$ with $𝑎 \neq 𝑏$ , then $𝑇$ is diagonalizable. It is also the one that breaks first when the field changes, as the involution over $𝔽_{2}$ above shows. The image-and-kernel proof produces the eigenspaces themselves, and it explains the polynomial proof: for $(𝑇 - 𝑎 id) (𝑇 - 𝑏 id) =$ $0$ , the Lagrange polynomial $(𝑇 - 𝑏 id) / (𝑎 - 𝑏)$ is an idempotent with image $𝐸_{𝑎}$ . The rank proof is the computational one, and run backwards it shows that $rank 𝑃 +$ $rank (id - 𝑃) =$ $𝑛$ characterizes idempotents (Exercise 4.3.42).

Practice: Exercise 7.7.4, Exercise 7.7.11, Problem 7.2, and Problem 7.6.

7.7.5Proving an identity for every operator

To prove an identity such as $tr 𝑇 = \sum 𝜆_{𝑖}$ , $det 𝑝 (𝑇) =$ $\prod 𝑝 (𝜆_{𝑖})$ , or $𝑝_{𝑇} (𝑇) = 0$ , there are four routes, and they apply to fewer operators the more concrete they are.

Test on eigenvectors (Technique 7.4.25). For diagonalizable $𝑇$ , two polynomials in $𝑇$ agree as soon as they agree on an eigenbasis, which reduces the identity to scalars. For other operators the eigenvectors do not span, and the test proves nothing (Exercise 7.7.7).
Diagonalize, solve, conjugate back (Technique 7.4.22). For computations with a diagonalizable matrix: powers, square roots, recurrences, matrix equations. The equation $𝐴 𝑋 = 𝑋 𝐵$ becomes $𝐷 𝑌 = 𝑌 𝐸$ with diagonal $𝐷, 𝐸$ , which is solved entry by entry (Exercise 7.7.11).
Triangularize, then read the diagonal (Technique 7.5.23). Over $ℂ$ every operator is triangular in some basis (Theorem 7.5.13), and products and polynomials of triangular matrices have the expected diagonals. This proves anything that depends only on the diagonal: characteristic polynomials, traces, determinants, eigenvalues with multiplicity (Theorem 7.5.22, Theorem 7.6.28).
A polynomial identity, or perturbation (Technique 5.5.24, Technique 7.6.23). If both sides are polynomials in the entries, it is enough to prove the identity for $𝐴 - 𝑡 𝐼_{𝑛}$ (or another path through $𝐴$ ) at infinitely many values of $𝑡$ , for instance at those $𝑡$ for which it is invertible or has distinct eigenvalues; both sides are then polynomials in $𝑡$ that agree infinitely often. This route works over every infinite field and needs no eigenvalues of the original matrix.

Example 7.7.3 (A unipotent matrix has determinant $1$ , four ways). Let $𝑁 \in 𝐹^{𝑛 \times 𝑛}$ satisfy $𝑁^{𝑘} = 0$ for some $𝑘 \geq 1$ . We show $det (𝐼_{𝑛} + 𝑁) = 1$ .

Triangularize (over $ℂ$ ). The only eigenvalue of $𝑁$ is $0$ , since every eigenvalue is a root of $𝑥^{𝑘}$ (Proposition 7.2.28). By Corollary 7.5.14, $𝑁 = 𝑆 𝑅 𝑆^{- 1}$ with $𝑅$ upper triangular, and the diagonal entries of $𝑅$ are eigenvalues of $𝑁$ (Proposition 7.5.8), so they are $0$ . Then $𝐼_{𝑛} +$ $𝑁 =$ $𝑆 (𝐼_{𝑛} + 𝑅) 𝑆^{- 1}$ , and $𝐼_{𝑛} + 𝑅$ is upper triangular with ones on the diagonal, so $det (𝐼_{𝑛} + 𝑁) =$ $det (𝐼_{𝑛} + 𝑅) =$ $1$ (Proposition 5.4.17).

The characteristic polynomial (any field). By Exercise 7.3.48, $𝑝_{𝑁} = 𝑥^{𝑛}$ . Evaluating $𝑝_{𝑁} =$ $det (𝑥 𝐼_{𝑛} - 𝑁)$ at $𝑥 = - 1$ is allowed, since evaluation respects sums and products (Proposition 7.1.10, Proposition 7.3.2(g)), and gives $det (- 𝐼_{𝑛} - 𝑁) =$ ${(- 1)}^{𝑛}$ . So $det (𝐼_{𝑛} + 𝑁) =$ ${(- 1)}^{𝑛} det (- 𝐼_{𝑛} - 𝑁) =$ $1$ .

A flag of kernels (any field). The kernels $ker 𝑁 \subseteq$ $ker 𝑁^{2} \subseteq$ $\dots \subseteq$ $ker 𝑁^{𝑘} =$ $𝐹^{𝑛}$ are nested. Choose a basis of $ker 𝑁$ , extend it to a basis of $ker 𝑁^{2}$ , and so on up to $𝐹^{𝑛}$ . If a basis vector $𝐛$ was added at stage $𝑗 + 1$ , then $𝑁 𝐛 \in ker 𝑁^{𝑗}$ , which is spanned by basis vectors added strictly before $𝐛$ . So in this basis $𝑁$ is strictly upper triangular, and $𝐼_{𝑛} + 𝑁$ is similar to an upper triangular matrix with ones on the diagonal (Technique 7.5.4).

A polynomial identity in $𝑡$ (any field). In ${𝐹 [𝑡]}^{𝑛 \times 𝑛}$ , the matrices $𝐼_{𝑛} + 𝑡 𝑁$ and $𝐺 =$ $\sum_{𝑗 = 0}^{𝑘 - 1} {(- 𝑡)}^{𝑗} 𝑁^{𝑗}$ satisfy $(𝐼_{𝑛} + 𝑡 𝑁) 𝐺 =$ $𝐼_{𝑛} -$ ${(- 𝑡 𝑁)}^{𝑘} =$ $𝐼_{𝑛}$ , by telescoping. Let $𝑓 (𝑡) =$ $det (𝐼_{𝑛} + 𝑡 𝑁) \in$ $𝐹 [𝑡]$ . By multiplicativity over $𝐹 [𝑡]$ (Proposition 7.3.2(c)), $𝑓 (𝑡) det 𝐺 = 1$ , so $𝑓$ is a unit of $𝐹 [𝑡]$ , hence a nonzero constant (Corollary 7.1.7(c)). Its value at $𝑡 = 0$ is $det 𝐼_{𝑛} = 1$ , so $𝑓 = 1$ , and at $𝑡 = 1$ this is $det (𝐼_{𝑛} + 𝑁) = 1$ .

The triangular proof is the quickest, but it needs the eigenvalues of $𝑁$ in the field, which is why it is stated over $ℂ$ ; the characteristic polynomial proof removes that need but uses Exercise 7.3.48. The flag proof is the most informative: it shows that over any field a nilpotent matrix is similar to a strictly upper triangular one, so $𝐼_{𝑛} + 𝑁$ has trace $𝑛$ as well. The last proof is the strangest and the most portable. At $𝑡 = 1$ alone, $(𝐼_{𝑛} + 𝑁) 𝐺 = 𝐼_{𝑛}$ shows only that $det (𝐼_{𝑛} + 𝑁) \neq 0$ ; the parameter $𝑡$ is what pins the value down, because a polynomial with a polynomial inverse cannot vary. That is the shape of every perturbation argument in Technique 7.6.23.

Practice: Exercise 7.7.7, Exercise 7.7.14, Problem 7.10, and Problem 7.13.

7.7.6Induction on dimension

The existence theorems of Section 7.5 and Section 7.6.6 all have the same shape (Technique 7.5.12): find an invariant subspace $𝑈$ with ${𝟎} \neq 𝑈 \neq 𝑉$ , apply the induction hypothesis to a smaller operator, and glue with a block triangular matrix. The smaller operator can be the restriction to $𝑈$ , the compression to a complement $𝑊$ (Lemma 7.5.11), or the quotient operator on $𝑉 / 𝑈$ (Section 4.6). The point to check in every such proof is that the smaller operator satisfies the hypotheses again. A restriction of a complex operator is a complex operator, so triangularization goes through; the compressions of two commuting operators to a complement of a common invariant subspace commute, so simultaneous triangularization goes through (Exercise 7.7.13). A compression is not a restriction, and treating it as one is a real error (Problem 7.14). For Schur’s theorem one compresses to $𝑈^{⟂}$ , which keeps orthonormality (Theorem 7.5.18), and the same move proves the spectral theorems in Section 8.2.

7.7.7Real matrices Beyond 21-242

Complexification is not covered in lecture.

A real matrix need not have a real eigenvalue, and the arguments of the chapter that start from an eigenvector then have nothing to start from. Three moves get around this. In odd dimension there is always a real eigenvalue, by the intermediate value theorem applied to $𝑝_{𝐴}$ (Proposition 7.3.30). Otherwise, regard $𝐴$ as a complex matrix, use complex eigenvalues and eigenvectors, and come back by taking real and imaginary parts (Technique 7.3.35): a complex eigenvector $𝐮 + 𝑖 𝐰$ gives the real invariant plane $span (𝐮, 𝐰)$ (Proposition 7.3.34), and the nonreal eigenvalues come in conjugate pairs with equal multiplicities (Proposition 7.1.38), so $det 𝐴$ is a product of real eigenvalues and factors ${| 𝜆 |}^{2} > 0$ . That last fact settles sign questions such as $det (𝐴^{2} + 𝐵^{2}) \geq 0$ for commuting $𝐴, 𝐵$ (Exercise 7.6.52). Finally, a real identity can sometimes be factored over $ℂ$ as $𝑍 𝑍$ , whose determinant is ${| det 𝑍 |}^{2} \geq 0$ (Problem 7.18).

Practice: Exercise 7.7.8, Exercise 7.7.12, Problem 7.11, and Problem 7.17.

7.7.8Reading the chapter’s proofs

Which field? Each statement of the chapter depends on the field in one of three ways. Some hold over every field: eigenvectors for distinct eigenvalues are independent, chains are independent, $gm \leq am$ , and the characteristic polynomial has $tr$ and $det$ as coefficients. Some hold whenever the relevant polynomial splits: triangularization (Theorem 7.5.17), and $tr$ and $det$ as sum and product of eigenvalues. Over $ℂ$ that is automatic by the Fundamental Theorem of Algebra, and over a finite field it can fail for every size $𝑛 \geq 2$ (Exercise 7.3.44). And some depend on the characteristic: “an involution is diagonalizable” needs $1 + 1 \neq 0$ , and “ $tr 𝐴^{𝑘} = 0$ for all $𝑘$ forces $𝐴$ nilpotent” needs characteristic $0$ . When a proof divides by an integer, or uses that roots such as $\pm 1$ are distinct, that is where the field enters.

Where finite dimension enters. Eigenvectors for distinct eigenvalues are independent in any vector space, and so are chains. Existence of eigenvalues is where finite dimension is needed: multiplication by $𝑥$ on $𝐹 [𝑥]$ has none (Remark 7.2.10), because the Krylov sequence $𝐯, 𝑇 𝐯, \dots$ never becomes dependent.

Two polynomials. The characteristic polynomial records multiplicities: $am$ , trace, determinant. The minimal polynomial records structure: the eigenvalues again, and whether $𝑇$ is diagonalizable. Neither determines the other (Remark 7.6.11), and by Cayley–Hamilton the minimal polynomial divides the characteristic one (Theorem 7.6.20). For a vector $𝐯$ , the first dependence in its Krylov sequence is a third polynomial, and the minimal polynomial is assembled from these one basis vector at a time (Problem 7.9).

Similarity invariants. Characteristic polynomial, minimal polynomial, geometric multiplicities, and the ranks of ${(𝐴 - 𝜆 𝐼_{𝑛})}^{𝑘}$ are all invariant under similarity, and each of the first three fails to determine the similarity class on its own (Exercise 7.3.38, Exercise 7.6.43). When $𝑝_{𝐴}$ splits, as over $ℂ$ , the ranks of the powers do determine it; that is the Jordan form (Section 10.3).

7.7.9Common errors

Most of these appear in the chapter as find-the-error exercises.

Assuming the eigenvectors span. An argument that checks an identity on eigenvectors and concludes it everywhere has assumed that $𝑇$ is diagonalizable (Exercise 7.7.7, Exercise 7.6.44). The same gap hides in “choose an eigenbasis” and in “the eigenspaces of $𝐴$ add up to $𝑉$ ” (Exercise 7.7.6).
Algebraic and geometric multiplicity confused. $dim ker 𝐴 = gm (0)$ , while “the eigenvalue $0$ counted with multiplicity” means $am (0)$ , which can be larger (Exercise 7.7.5). The Fundamental Theorem of Algebra gives $am$ ’s that add up to $𝑛$ , not $gm$ ’s (Exercise 7.4.39).
Assuming eigenvalues exist. A real matrix need not have a real eigenvalue, and a matrix over a finite field need not have any (Example 7.4.9, Exercise 7.3.44). “The only real eigenvalue is $1$ ” does not make $1$ the only eigenvalue over $ℂ$ (Exercise 7.7.8).
A one-way implication used both ways. “ $det 𝐴 = 0$ if and only if $0$ is an eigenvalue” is correct (Theorem 7.2.9). But $𝑝 (𝑇) = 0$ makes every eigenvalue a root of $𝑝$ , not every root an eigenvalue; $𝐴$ diagonalizable makes $𝐴^{2}$ diagonalizable, not conversely (Exercise 7.4.38); and equal characteristic polynomials follow from similarity without implying it (Exercise 7.3.38).
Eigenvalues of sums and products. For noncommuting $𝐴, 𝐵$ the eigenvalues of $𝐴 + 𝐵$ and $𝐴 𝐵$ have nothing to do with sums and products of eigenvalues of $𝐴$ and $𝐵$ (Exercise 7.2.43, Exercise 7.5.37); commuting is what makes them pair up (Corollary 7.6.33).
Substituting a matrix for $𝑥$ inside a determinant, as in the false proof of Cayley–Hamilton (Exercise 7.3.47). The entries of $𝑥 𝐼_{𝑛} - 𝐴$ are polynomials; $𝐴 𝐼_{𝑛} - 𝐴$ is a matrix with matrix entries, and its “determinant” is not $𝑝_{𝐴} (𝐴)$ .
Eigenvectors are not closed under addition, and an eigenvector of $𝑇^{2}$ need not be one of $𝑇$ (Exercise 7.2.25, Exercise 7.2.53).
Restriction and compression confused. Only an invariant subspace carries a restriction of $𝑇$ ; a complement usually is not invariant (Problem 7.14).

Exercises

Exercise 7.7.4.routine For each statement, decide whether it is true, and name the fact that decides it.

(a)Every $𝐴 \in ℚ^{3 \times 3}$ with $𝐴^{3} = 𝐴$ is diagonalizable over $ℚ$ . Does the same hold over $𝔽_{2}$ ?
(b)Every $𝐴 \in ℂ^{4 \times 4}$ with $𝑝_{𝐴} = 𝑥^{2} {(𝑥 - 1)}^{2}$ , $rank 𝐴 = 2$ , and $rank (𝐴 - 𝐼_{4}) = 3$ is diagonalizable.
(c)Every $𝐴 \in ℝ^{5 \times 5}$ has a real eigenvalue.
(d)If $𝐴 \in ℂ^{3 \times 3}$ has eigenvalues $1, 2, 3$ , then $rank (𝐴^{2} - 3 𝐴 + 2 𝐼_{3}) =$ $1$ .
(e)Every $𝐴 \in ℝ^{2 \times 2}$ with $tr 𝐴 = 0$ and $det 𝐴 < 0$ is diagonalizable over $ℝ$ .

Solution.

Uses. Theorem 7.6.15 (Diagonalizability and the minimal polynomial), Lemma 7.4.13 (Geometric multiplicity through a rank), Theorem 7.4.16 (Criteria for diagonalizability), Proposition 7.3.30 (Odd dimension), Proposition 7.4.23 (Powers and polynomials of a diagonalizable matrix), Corollary 7.4.5 (Distinct eigenvalues give a diagonalization)

Techniques. Use an annihilating polynomial (Technique 7.6.17), Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. (a) Factor

𝑥^{3} - 𝑥

over

ℚ

and over

𝔽_{2}

. (b) Compute

gm (1)

from a rank. (d) Diagonalize and apply the polynomial to the diagonal.

Hint 2. (a) Over

𝔽_{2}

𝑥^{3} - 𝑥 = 𝑥 {(𝑥 + 1)}^{2}

; try

(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})

in a block. (e)

𝑝_{𝐴} = 𝑥^{2} + det 𝐴

Sketch. (a) True over

ℚ

𝑥 (𝑥 - 1) (𝑥 + 1)

has distinct roots. False over

𝔽_{2}

𝐴 = (\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix})

has

𝐴^{3} = 𝐴

and

gm (1) =

1 <

2 =

am (1)

. (b) False:

gm (1) = 4 - 3 = 1 < 2

. (c) True: odd dimension. (d) True:

𝐴 =

𝑆 diag (1, 2, 3) 𝑆^{- 1}

, so

𝐴^{2} -

3 𝐴 +

2 𝐼_{3} =

𝑆 diag (0, 0, 2) 𝑆^{- 1}

. (e) True:

𝑝_{𝐴} = 𝑥^{2} + det 𝐴

has the distinct roots

\pm \sqrt{- det 𝐴}

(a) True over $ℚ$ . The polynomial $𝑥^{3} -$ $𝑥 =$ $𝑥 (𝑥 - 1) (𝑥 + 1)$ annihilates $𝐴$ and has the distinct roots $0, 1, - 1$ , so $𝐴$ is diagonalizable by Theorem 7.6.15. Over $𝔽_{2}$ it is false, because $- 1 = 1$ there and $𝑥^{3} - 𝑥 = 𝑥 {(𝑥 + 1)}^{2}$ has a repeated root. For

𝐴 = (\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}) \in 𝔽_{2}^{3 \times 3},

the block $𝐽 = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ has $𝐽^{2} = (\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}) = 𝐼_{2}$ over $𝔽_{2}$ , so $𝐽^{3} = 𝐽$ and $𝐴^{3} = 𝐴$ . But $𝑝_{𝐴} = 𝑥 {(𝑥 - 1)}^{2}$ and $𝐴 -$ $𝐼_{3} =$ $(\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix})$ has rank $2$ , so $gm (1) =$ $1 <$ $2 =$ $am (1)$ (Lemma 7.4.13), and $𝐴$ is not diagonalizable (Theorem 7.4.16).

(b) False. By Lemma 7.4.13, $gm (1) =$ $4 -$ $rank (𝐴 - 𝐼_{4}) =$ $1$ , while $am (1) = 2$ , so $𝐴$ is not diagonalizable. Such matrices exist: the block diagonal matrix with blocks $0_{2 \times 2}$ and $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ has $𝑝_{𝐴} = 𝑥^{2} {(𝑥 - 1)}^{2}$ , rank $2$ , and $rank (𝐴 - 𝐼_{4}) =$ $2 +$ $1 =$ $3$ .

(c) True by Proposition 7.3.30: $𝑝_{𝐴}$ is a real polynomial of odd degree $5$ , so it has a real root.

(d) True. Three distinct eigenvalues make $𝐴$ diagonalizable (Corollary 7.4.5): $𝐴 = 𝑆 𝐷 𝑆^{- 1}$ with $𝐷 = diag (1, 2, 3)$ . With $𝑞 =$ $𝑥^{2} -$ $3 𝑥 +$ $2 =$ $(𝑥 - 1) (𝑥 - 2)$ , Proposition 7.4.23 gives $𝑞 (𝐴) =$ $𝑆 𝑞 (𝐷) 𝑆^{- 1} =$ $𝑆 diag (0, 0, 2) 𝑆^{- 1}$ , which has rank $1$ .

(e) True. By Proposition 7.3.18, $𝑝_{𝐴} =$ $𝑥^{2} -$ $(tr 𝐴) 𝑥 +$ $det 𝐴 =$ $𝑥^{2} -$ $𝑐$ with $𝑐 = - det 𝐴 > 0$ . Its roots $\pm \sqrt{𝑐}$ are real and distinct, so $𝐴$ is diagonalizable over $ℝ$ (Corollary 7.4.5).

Exercise 7.7.5.routine Find the error in the following argument, decide whether the claim is true, and prove a correct version.

Claim. If $𝐴 \in ℂ^{𝑛 \times 𝑛}$ has rank $𝑟$ , then exactly $𝑟$ of the eigenvalues of $𝐴$ , counted with multiplicity, are nonzero.

“Proof.” By rank–nullity, $dim ker 𝐴 = 𝑛 - 𝑟$ . Since $ker 𝐴 = 𝐸_{0}$ , the eigenvalue $0$ has multiplicity $𝑛 - 𝑟$ . There are $𝑛$ eigenvalues counted with multiplicity, so the other $𝑟$ are nonzero.

Solution.

Uses. Definition 7.4.12 (Geometric multiplicity), Definition 7.4.11 (Algebraic multiplicity), Theorem 7.4.15 (Geometric multiplicity is at most algebraic multiplicity), Proposition 7.1.34 (Roots of a split polynomial)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. Which multiplicity is

dim 𝐸_{0}

, and which one does “counted with multiplicity” refer to?

Hint 2. Try the shift

𝑁

with

𝑁 𝐞_{1} = 𝟎

and

𝑁 𝐞_{𝑗} = 𝐞_{𝑗 - 1}

. For the correct version, use

gm \leq am

Sketch.

dim 𝐸_{0} = gm (0)

, but “counted with multiplicity” means

am (0)

. The claim is false: the

𝑛 \times 𝑛

shift (

𝑛 \geq 2

) has rank

𝑛 - 1

and no nonzero eigenvalue. Correct: the number of nonzero eigenvalues with multiplicity is

𝑛 -

am (0) \leq

𝑛 -

gm (0) =

𝑟

, with equality when

𝐴

is diagonalizable.

The error. $dim ker 𝐴 = dim 𝐸_{0}$ is the geometric multiplicity $gm (0)$ . The eigenvalues “counted with multiplicity” are the roots of $𝑝_{𝐴}$ counted with algebraic multiplicity (Proposition 7.1.34), and $am (0)$ can be larger than $gm (0)$ . The claim is false: for $𝑛 \geq 2$ , the shift $𝑁 \in ℂ^{𝑛 \times 𝑛}$ with $𝑁 𝐞_{1} = 𝟎$ and $𝑁 𝐞_{𝑗} = 𝐞_{𝑗 - 1}$ for $𝑗 \geq 2$ has rank $𝑛 - 1$ , but $𝑁^{𝑛} = 0$ , so its only eigenvalue is $0$ (Proposition 7.2.28) and no eigenvalue is nonzero.

A correct version. At most $𝑟$ eigenvalues, counted with multiplicity, are nonzero. Indeed $𝑝_{𝐴}$ splits over $ℂ$ , its roots with multiplicity number $𝑛$ , and $0$ occurs $am (0)$ times, so the number of nonzero ones is $𝑛 -$ $am (0) \leq$ $𝑛 -$ $gm (0) =$ $𝑛 -$ $(𝑛 - 𝑟) =$ $𝑟$ by Theorem 7.4.15. Equality holds exactly when $am (0) = gm (0)$ , for instance when $𝐴$ is diagonalizable.

Exercise 7.7.6.routine Find the errors in the following argument, and give a pair of matrices showing that the claim is false.

Claim. If $𝐴, 𝐵 \in ℂ^{𝑛 \times 𝑛}$ commute, then $ℂ^{𝑛}$ has a basis of common eigenvectors of $𝐴$ and $𝐵$ .

“Proof.” Let $𝜆$ be an eigenvalue of $𝐴$ . For $𝐯 \in 𝐸_{𝜆} (𝐴)$ , $𝐴 (𝐵 𝐯) =$ $𝐵 𝐴 𝐯 =$ $𝜆 𝐵 𝐯$ , so $𝐸_{𝜆} (𝐴)$ is $𝐵$ -invariant. Choose a basis of $𝐸_{𝜆} (𝐴)$ consisting of eigenvectors of $𝐵$ . Doing this for each eigenvalue of $𝐴$ and putting the bases together gives a basis of $ℂ^{𝑛}$ consisting of common eigenvectors.

Solution.

Uses. Proposition 7.2.30 (Kernels and images of commuting operators are invariant), Theorem 7.6.34 (Simultaneous diagonalization), Theorem 7.6.31 (Commuting operators have a common eigenvector)

Techniques. Look for invariant subspaces among kernels and images of commuting operators (Technique 7.2.31), Finding counterexamples and minimal examples (Technique 1.4.18)

Hint 1. Two steps need a hypothesis that is not given. Which operator must be diagonalizable for each of them?

Hint 2. Try

𝐴 = 𝐼_{2}

with

𝐵

a nonzero nilpotent matrix, and the other way round.

Sketch. The basis of

𝐸_{𝜆} (𝐴)

made of eigenvectors of

𝐵

exists only if

𝐵 ↾_{𝐸_{𝜆} (𝐴)}

is diagonalizable, and the eigenspaces of

𝐴

add up to

ℂ^{𝑛}

only if

𝐴

is diagonalizable.

𝐴 = 𝐼_{2}

𝐵 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})

breaks the first step;

𝐴 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})

𝐵 = 𝐼_{2}

breaks the second. With both diagonalizable the argument is correct (Theorem 7.6.34).

The first step is right: $𝐸_{𝜆} (𝐴)$ is $𝐵$ -invariant (Proposition 7.2.30(b)). There are two errors after it.

First, $𝐸_{𝜆} (𝐴)$ need not have a basis of eigenvectors of $𝐵$ : that requires the restriction of $𝐵$ to $𝐸_{𝜆} (𝐴)$ to be diagonalizable, which follows if $𝐵$ is (Corollary 7.6.16) and can fail otherwise. With $𝐴 = 𝐼_{2}$ and $𝐵 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ , which commute, $𝐸_{1} (𝐴) = ℂ^{2}$ and the only eigenvectors of $𝐵$ are the nonzero multiples of $𝐞_{1}$ . So no basis of common eigenvectors exists, and the claim is false.

Second, the eigenspaces of $𝐴$ need not add up to $ℂ^{𝑛}$ ; that is the statement that $𝐴$ is diagonalizable. With $𝐴 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ and $𝐵 = 𝐼_{2}$ , the bases chosen span only $𝐸_{0} (𝐴) = span (𝐞_{1})$ .

When $𝐴$ and $𝐵$ are both diagonalizable, both gaps close and the argument is the proof of Theorem 7.6.34. Without that, commuting still gives one common eigenvector (Theorem 7.6.31), and a common triangular form.

Exercise 7.7.7.standard Find the error in the following argument, show that the claim is false, and then prove the two correct statements below.

Claim. If $𝐴 \in ℂ^{𝑛 \times 𝑛}$ satisfies $𝐴^{3} = 𝐴^{2}$ , then $𝐴^{2} = 𝐴$ .

“Proof.” Let $𝐴 𝐯 = 𝜆 𝐯$ with $𝐯 \neq 𝟎$ . Then $𝜆^{3} 𝐯 =$ $𝐴^{3} 𝐯 =$ $𝐴^{2} 𝐯 =$ $𝜆^{2} 𝐯$ , so $𝜆^{2} (𝜆 - 1) = 0$ and $𝜆 \in {0, 1}$ . Hence $𝜆^{2} = 𝜆$ , and $𝐴^{2} 𝐯 =$ $𝜆^{2} 𝐯 =$ $𝜆 𝐯 =$ $𝐴 𝐯$ . So $𝐴^{2}$ and $𝐴$ agree on every eigenvector of $𝐴$ , and therefore $𝐴^{2} = 𝐴$ .

(a)If $𝐴^{3} = 𝐴^{2}$ , then $𝐴^{2}$ is idempotent.
(b)If $𝐴^{3} = 𝐴^{2}$ , then $𝐴^{2} = 𝐴$ if and only if $𝐴$ is diagonalizable.

Solution.

Uses. Proposition 7.2.28 (Polynomials act on eigenvectors by their values), Theorem 7.6.15 (Diagonalizability and the minimal polynomial), Example 7.7.2 (An idempotent is diagonalizable, three ways)

Techniques. Test an identity on eigenvectors (Technique 7.4.25), Use an annihilating polynomial (Technique 7.6.17)

Hint 1. The last step needs a basis of eigenvectors. Look for a counterexample with only one eigenvector line.

Hint 2. (a) Compute

𝐴^{4}

. (b) If

𝐴

is diagonalizable, the argument above is correct. Conversely,

𝐴^{2} = 𝐴

is an annihilating polynomial with distinct roots.

Sketch. Agreement on eigenvectors gives agreement everywhere only when the eigenvectors span.

𝑁 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})

has

𝑁^{3} = 𝑁^{2} = 0

and

𝑁^{2} \neq 𝑁

. (a)

𝐴^{4} =

𝐴 𝐴^{3} =

𝐴 𝐴^{2} =

𝐴^{3} =

𝐴^{2}

. (b) If

𝐴

is diagonalizable, the argument works on an eigenbasis; if

𝐴^{2} = 𝐴

, then

𝑥 (𝑥 - 1)

annihilates

𝐴

The error. Two linear maps that agree on a spanning list are equal, but the eigenvectors of $𝐴$ need not span $ℂ^{𝑛}$ ; the last step assumes that $𝐴$ is diagonalizable. The claim is false: $𝑁 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ has $𝑁^{2} = 0$ , so $𝑁^{3} = 𝑁^{2}$ , but $𝑁^{2} = 0 \neq 𝑁$ . Its eigenvectors are the nonzero multiples of $𝐞_{1}$ , on which $𝑁^{2}$ and $𝑁$ do agree.

(a) $𝐴^{4} =$ $𝐴 𝐴^{3} =$ $𝐴 𝐴^{2} =$ $𝐴^{3} =$ $𝐴^{2}$ , so ${(𝐴^{2})}^{2} = 𝐴^{2}$ .

(b) If $𝐴$ is diagonalizable, it has an eigenbasis, and the argument above shows that $𝐴^{2}$ and $𝐴$ agree on it. So $𝐴^{2} = 𝐴$ (Technique 7.4.25). Conversely, if $𝐴^{2} = 𝐴$ , then $𝑥 (𝑥 - 1)$ annihilates $𝐴$ , and $𝐴$ is diagonalizable (Theorem 7.6.15, or Example 7.7.2).

In terms of polynomials: $𝑥^{2} (𝑥 - 1)$ annihilates $𝐴$ , so $𝑚_{𝐴}$ divides it, and $𝐴$ is diagonalizable exactly when $𝑚_{𝐴}$ divides $𝑥 (𝑥 - 1)$ . The “proof” tested the identity on eigenvectors, which sees only the roots of $𝑚_{𝐴}$ , never their multiplicities.

Exercise 7.7.8.standard Find the error in the following argument, give a counterexample with integer entries, and prove the correct statements below.

Claim. If $𝐴 \in ℝ^{3 \times 3}$ satisfies $𝐴^{3} = 𝐼_{3}$ , then $𝐴 = 𝐼_{3}$ .

“Proof.” If $𝜆$ is an eigenvalue of $𝐴$ , then $𝜆^{3} = 1$ by Proposition 7.2.28, and the only real cube root of $1$ is $1$ . So $1$ is the only eigenvalue of $𝐴$ . The minimal polynomial divides $𝑥^{3} - 1$ and its roots are the eigenvalues of $𝐴$ , so $𝑚_{𝐴} = 𝑥 - 1$ , that is, $𝐴 = 𝐼_{3}$ .

(a) $1$ is an eigenvalue of $𝐴$ .
(b) $𝐴 = 𝐼_{3}$ if and only if $𝐴$ is diagonalizable over $ℝ$ .

Solution.

Uses. Theorem 7.6.7 (The roots of the minimal polynomial are the eigenvalues), Proposition 7.3.30 (Odd dimension), Proposition 7.4.7 (One eigenvalue), Theorem 7.1.39 (Factorization over $ℝ$ )

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18), Use an annihilating polynomial (Technique 7.6.17)

Hint 1. The roots of

𝑚_{𝐴}

ℝ

are the real eigenvalues. Can

𝑚_{𝐴}

have a factor without real roots?

Hint 2. Over

ℝ

𝑥^{3} -

1 =

(𝑥 - 1) (𝑥^{2} + 𝑥 + 1)

. Look for a

2 \times 2

integer matrix with

𝑝 = 𝑥^{2} + 𝑥 + 1

, and add a block

1

Sketch. Knowing the real roots of

𝑚_{𝐴}

does not determine

𝑚_{𝐴}

: it can be

(𝑥 - 1) (𝑥^{2} + 𝑥 + 1)

. Counterexample:

𝐴

with diagonal blocks

1

and

𝐶 = (\begin{matrix} 0 & - 1 \\ 1 & - 1 \end{matrix})

, where

𝐶^{3} = 𝐼_{2}

. (a)

𝑝_{𝐴}

has odd degree, so there is a real eigenvalue

𝜆

, and

𝜆^{3} = 1

forces

𝜆 = 1

. (b) If

𝐴

is diagonalizable over

ℝ

with only eigenvalue

1

, then

𝐴 = 𝐼_{3}

by Proposition 7.4.7.

The error. By Theorem 7.6.7, the roots of $𝑚_{𝐴}$ in $ℝ$ are exactly the real eigenvalues of $𝐴$ , so $𝑚_{𝐴}$ has $1$ as its only real root. That does not make $𝑚_{𝐴} = 𝑥 - 1$ : over $ℝ$ , $𝑥^{3} -$ $1 =$ $(𝑥 - 1) (𝑥^{2} + 𝑥 + 1)$ , and $𝑥^{2} + 𝑥 + 1$ has no real root, so $𝑚_{𝐴}$ could be $(𝑥 - 1) (𝑥^{2} + 𝑥 + 1)$ . The step “its roots are the eigenvalues, so it is $𝑥 - 1$ ” assumed that $𝑚_{𝐴}$ splits over $ℝ$ .

A counterexample. Let $𝐶 = (\begin{matrix} 0 & - 1 \\ 1 & - 1 \end{matrix})$ . Then $𝐶^{2} = (\begin{matrix} - 1 & 1 \\ - 1 & 0 \end{matrix})$ and $𝐶^{3} = 𝐶 𝐶^{2} = 𝐼_{2}$ . The block diagonal matrix $𝐴$ with blocks $(1)$ and $𝐶$ has $𝐴^{3} = 𝐼_{3}$ and $𝐴 \neq 𝐼_{3}$ . Over $ℂ$ its eigenvalues are $1$ and the two primitive cube roots of unity, the roots of $𝑝_{𝐶} = 𝑥^{2} + 𝑥 + 1$ ; geometrically, $𝐶$ is similar to a rotation of the plane by $2 𝜋 / 3$ .

(a) $𝐴$ has a real eigenvalue $𝜆$ by Proposition 7.3.30, and $𝜆^{3} = 1$ with $𝜆 \in ℝ$ forces $𝜆 = 1$ .

(b) If $𝐴 = 𝐼_{3}$ , it is diagonal. Conversely, if $𝐴$ is diagonalizable over $ℝ$ , its eigenvalues are real roots of $𝑥^{3} - 1$ , so $1$ is its only eigenvalue, and $𝐴 = 𝐼_{3}$ by Proposition 7.4.7. Over $ℂ$ the situation is different: $𝑥^{3} - 1$ has three distinct complex roots, so every $𝐴$ with $𝐴^{3} = 𝐼_{3}$ is diagonalizable over $ℂ$ (Theorem 7.6.15), and $𝐴 \neq 𝐼_{3}$ exactly when the eigenvalues are $1, 𝜔, 𝜔$ with $𝜔 = 𝑒^{2 𝜋 𝑖 / 3}$ .

Exercise 7.7.9.standard Let

𝑛 \geq 2

, and let

𝑁 \in 𝐹^{𝑛 \times 𝑛}

satisfy

𝑁^{𝑛} = 0

and

𝑁^{𝑛 - 1} \neq 0

. Show that there is no

𝐵 \in 𝐹^{𝑛 \times 𝑛}

with

𝐵^{𝑘} = 𝑁

for some

𝑘 \geq 2

. Give two proofs: one using powers of

𝐵

, and one using kernels.

Solution.

Uses. Corollary 4.1.27 (A nilpotent operator dies within $dim 𝑉$ steps), Proposition 4.1.26 (Nilpotent chains are independent), Theorem 4.2.11 (Rank–nullity), Exercise 4.2.51

Techniques. Apply an operator to a relation to kill terms (Technique 4.1.25), Dimension counting (Technique 4.2.24)

Hint 1. If

𝐵^{𝑘} = 𝑁

, then

𝐵

is nilpotent. How fast must a nilpotent

𝑛 \times 𝑛

matrix die?

Hint 2. Powers:

𝑁^{𝑗} = 𝐵^{𝑘 𝑗}

, and

𝑘 𝑗 \geq 𝑛

already for

𝑗 = ⌈ 𝑛 / 𝑘 ⌉

. Kernels:

dim ker 𝑁 = 1

, and

ker 𝐵 \subseteq

ker 𝐵^{2} \subseteq

ker 𝑁

Sketch. Powers:

𝐵^{𝑘 𝑛} = 𝑁^{𝑛} = 0

, so

𝐵^{𝑛} = 0

(Corollary 4.1.27). With

𝑗 = ⌈ 𝑛 / 𝑘 ⌉ \leq 𝑛 - 1

𝑁^{𝑗} = 𝐵^{𝑘 𝑗} = 0

, so

𝑁^{𝑛 - 1} = 0

, a contradiction. Kernels: a chain of length

𝑛

shows

dim ker 𝑁 = 1

𝐵

is nilpotent, so

{𝟎} \neq

ker 𝐵 \subseteq

ker 𝐵^{2} \subseteq

ker 𝐵^{𝑘} =

ker 𝑁

, forcing

ker 𝐵 = ker 𝐵^{2}

; then all

ker 𝐵^{𝑗}

equal

ker 𝐵

, which is not

𝐹^{𝑛}

, but

𝐵^{𝑘 𝑛} = 0

Suppose $𝐵^{𝑘} = 𝑁$ with $𝑘 \geq 2$ . Then $𝐵^{𝑘 𝑛} = 𝑁^{𝑛} = 0$ , so $𝐵$ is nilpotent.

Powers. By Corollary 4.1.27, $𝐵^{𝑛} = 0$ . Let $𝑗 = ⌈ 𝑛 / 𝑘 ⌉$ , so that $𝑘 𝑗 \geq 𝑛$ and $𝑁^{𝑗} =$ $𝐵^{𝑘 𝑗} =$ $𝐵^{𝑘 𝑗 - 𝑛} 𝐵^{𝑛} =$ $0$ . Since $𝑘 \geq 2$ and $𝑛 \geq 2$ , $𝑗 \leq ⌈ 𝑛 / 2 ⌉ \leq 𝑛 - 1$ , so $𝑁^{𝑛 - 1} =$ $𝑁^{𝑛 - 1 - 𝑗} 𝑁^{𝑗} =$ $0$ , a contradiction.

Kernels. Choose $𝐯$ with $𝑁^{𝑛 - 1} 𝐯 \neq 𝟎$ . By Proposition 4.1.26, $𝐯,$ $𝑁 𝐯,$ $\dots,$ $𝑁^{𝑛 - 1} 𝐯$ is independent, so it is a basis of $𝐹^{𝑛}$ , and its image under $𝑁$ spans $span (𝑁 𝐯, \dots, 𝑁^{𝑛 - 1} 𝐯)$ , of dimension $𝑛 - 1$ . By Theorem 4.2.11, $dim ker 𝑁 = 1$ . Since $𝐵$ is nilpotent and $𝑛 \geq 1$ , $𝐵$ is not injective, so $ker 𝐵 \neq {𝟎}$ . Also $ker 𝐵 \subseteq$ $ker 𝐵^{2} \subseteq$ $ker 𝐵^{𝑘} =$ $ker 𝑁$ , because $𝑘 \geq 2$ . A nonzero subspace of a line is the line, so $ker 𝐵 =$ $ker 𝐵^{2} =$ $ker 𝑁$ . By Exercise 4.2.51(b), $ker 𝐵^{𝑚} = ker 𝐵$ for every $𝑚 \geq 1$ . But $𝐵^{𝑘 𝑛} = 0$ , so $ker 𝐵^{𝑘 𝑛} = 𝐹^{𝑛}$ , while $ker 𝐵$ is a line and $𝑛 \geq 2$ . This is a contradiction.

The first proof needs only the bound $𝐵^{𝑛} = 0$ ; the second shows where a root would have to fail: $ker 𝐵^{2}$ would have to lie in the line $ker 𝑁$ and also strictly contain the nonzero $ker 𝐵$ , because $𝐵$ is nilpotent and $𝑛 \geq 2$ . By contrast $𝐼_{𝑛} + 𝑁$ does have a square root over $ℚ$ , $ℝ$ , or $ℂ$ , namely the truncated binomial series $\sum_{𝑗 = 0}^{𝑛 - 1} (\binom{1 / 2}{𝑗}) 𝑁^{𝑗}$ , since the series for ${(1 + 𝑡)}^{1 / 2}$ squares to $1 + 𝑡$ and $𝑁^{𝑛} = 0$ .

Exercise 7.7.10.standardBeyond 21-242 Let $𝐴 \in 𝐹^{𝑚 \times 𝑛}$ and $𝐵 \in 𝐹^{𝑛 \times 𝑚}$ with $𝑚, 𝑛 \geq 1$ .

(a)Show that $𝐵 𝑓 (𝐴 𝐵) 𝐴 =$ $(𝑥 𝑓) (𝐵 𝐴)$ for every $𝑓 \in 𝐹 [𝑥]$ , and deduce that $𝑚_{𝐵 𝐴}$ divides $𝑥 𝑚_{𝐴 𝐵}$ .
(b)Deduce that $𝐴 𝐵$ and $𝐵 𝐴$ have the same nonzero eigenvalues, and give an example with $𝑚_{𝐴 𝐵} \neq 𝑚_{𝐵 𝐴}$ .
(c)Show that ${gm}_{𝐴 𝐵} (𝜆) =$ ${gm}_{𝐵 𝐴} (𝜆)$ for every $𝜆 \neq 0$ .
(d)Suppose $𝐴 𝐵$ is diagonalizable and invertible. Show that $𝐵 𝐴$ is diagonalizable, and show that “invertible” cannot be dropped.

Solution.

Uses. Theorem 7.6.2 (Existence and uniqueness of the minimal polynomial), Theorem 7.6.7 (The roots of the minimal polynomial are the eigenvalues), Theorem 7.6.15 (Diagonalizability and the minimal polynomial), Exercise 7.3.40, Exercise 7.4.40

Techniques. Compare $𝐴 𝐵$ and $𝐵 𝐴$ (Technique 7.3.21), Use an annihilating polynomial (Technique 7.6.17)

Hint 1.

𝐵 {(𝐴 𝐵)}^{𝑗} 𝐴 =

{(𝐵 𝐴)}^{𝑗 + 1}

. Apply this with

𝑓 = 𝑚_{𝐴 𝐵}

Hint 2. (c) Map

𝐸_{𝜆} (𝐴 𝐵)

𝐸_{𝜆} (𝐵 𝐴)

𝐯 \mapsto 𝐵 𝐯

. (d) If

0

is not a root of

𝑚_{𝐴 𝐵}

, then

𝑥 𝑚_{𝐴 𝐵}

still has distinct roots.

Sketch. (a)

𝐵 {(𝐴 𝐵)}^{𝑗} 𝐴 =

{(𝐵 𝐴)}^{𝑗 + 1}

, sum. With

𝑓 = 𝑚_{𝐴 𝐵}

the left side is

0

. (b) A nonzero root of

𝑚_{𝐵 𝐴}

is a root of

𝑚_{𝐴 𝐵}

, and symmetrically.

𝐴 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})

𝐵 = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})

𝐴 𝐵 = 0

𝐵 𝐴 = 𝐴

. (c)

𝐯 \mapsto 𝐵 𝐯

is injective on

𝐸_{𝜆} (𝐴 𝐵)

and lands in

𝐸_{𝜆} (𝐵 𝐴)

; symmetric. (d)

𝑥 𝑚_{𝐴 𝐵}

is a product of distinct linear factors and kills

𝐵 𝐴

. The example in (b) has

𝐴 𝐵 = 0

diagonalizable and

𝐵 𝐴

not.

(a) By associativity, $𝐵 {(𝐴 𝐵)}^{𝑗} 𝐴 =$ ${(𝐵 𝐴)}^{𝑗 + 1}$ for every $𝑗 \geq 0$ . If $𝑓 = \sum_{𝑗} 𝑐_{𝑗} 𝑥^{𝑗}$ , summing gives $𝐵 𝑓 (𝐴 𝐵) 𝐴 =$ $\sum_{𝑗} 𝑐_{𝑗} {(𝐵 𝐴)}^{𝑗 + 1} =$ $(𝑥 𝑓) (𝐵 𝐴)$ . With $𝑓 = 𝑚_{𝐴 𝐵}$ the left side is $𝐵 0 𝐴 = 0$ , so $𝑥 𝑚_{𝐴 𝐵}$ annihilates $𝐵 𝐴$ , and $𝑚_{𝐵 𝐴}$ divides it (Theorem 7.6.2).

(b) By symmetry, $𝑚_{𝐴 𝐵}$ also divides $𝑥 𝑚_{𝐵 𝐴}$ . If $𝜆 \neq 0$ is an eigenvalue of $𝐵 𝐴$ , it is a root of $𝑚_{𝐵 𝐴}$ (Theorem 7.6.7), hence of $𝑥 𝑚_{𝐴 𝐵}$ , hence of $𝑚_{𝐴 𝐵}$ , so it is an eigenvalue of $𝐴 𝐵$ . The converse is symmetric. For $𝐴 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ and $𝐵 = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ , $𝐴 𝐵 = 0$ and $𝐵 𝐴 = 𝐴$ , so $𝑚_{𝐴 𝐵} = 𝑥$ while $𝑚_{𝐵 𝐴} = 𝑥^{2}$ .

(c) Let $𝜆 \neq 0$ and $𝐯 \in 𝐸_{𝜆} (𝐴 𝐵)$ . Then $𝐵 𝐴 (𝐵 𝐯) =$ $𝐵 (𝐴 𝐵 𝐯) =$ $𝜆 𝐵 𝐯$ , so $𝐵 𝐯 \in 𝐸_{𝜆} (𝐵 𝐴)$ . If $𝐵 𝐯 = 𝟎$ , then $𝜆 𝐯 = 𝐴 𝐵 𝐯 = 𝟎$ and $𝐯 = 𝟎$ . So $𝐯 \mapsto 𝐵 𝐯$ is an injective linear map $𝐸_{𝜆} (𝐴 𝐵) \to$ $𝐸_{𝜆} (𝐵 𝐴)$ , and ${gm}_{𝐴 𝐵} (𝜆) \leq$ ${gm}_{𝐵 𝐴} (𝜆)$ . Exchanging the roles of $𝐴$ and $𝐵$ gives the reverse inequality.

(d) By Theorem 7.6.15, $𝑚_{𝐴 𝐵} =$ $(𝑥 - 𝜆_{1}) \dots (𝑥 - 𝜆_{𝑟})$ with distinct $𝜆_{𝑖}$ , and none of them is $0$ since $𝐴 𝐵$ is invertible. So $𝑥 𝑚_{𝐴 𝐵}$ is a product of distinct linear factors, and by (a) it annihilates $𝐵 𝐴$ . Hence $𝐵 𝐴$ is diagonalizable. In the example of (b), $𝐴 𝐵 = 0$ is diagonalizable but not invertible, and $𝐵 𝐴 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ is not diagonalizable.

For $𝑚 = 1$ , write $𝐴 = 𝐯^{𝖳}$ and $𝐵 = 𝐮$ . Then $𝐴 𝐵 = 𝐯^{𝖳} 𝐮$ is a $1 \times 1$ matrix and $𝐵 𝐴 = 𝐮 𝐯^{𝖳}$ , and (d) says that $𝐮 𝐯^{𝖳}$ is diagonalizable when $𝐯^{𝖳} 𝐮 \neq 0$ , one half of Exercise 7.4.40. Compare the three proofs of “same nonzero eigenvalues” in Exercise 7.3.40: characteristic polynomials see multiplicities, eigenvectors see geometric multiplicities as in (c), and minimal polynomials see diagonalizability.

Exercise 7.7.11.standard Let $𝐴 \in 𝐹^{𝑚 \times 𝑚}$ and $𝐵 \in 𝐹^{𝑛 \times 𝑛}$ be diagonalizable, and let $𝑊 =$ ${𝑋 \in 𝐹^{𝑚 \times 𝑛} | 𝐴 𝑋 = 𝑋 𝐵}$ .

(a)Show that $dim 𝑊 =$ $\sum_{𝜆} {gm}_{𝐴} (𝜆) {gm}_{𝐵} (𝜆)$ , the sum over the common eigenvalues of $𝐴$ and $𝐵$ .
(b)Deduce that $𝐴 𝑋 = 𝑋 𝐵$ has only the solution $𝑋 = 0$ if and only if $𝐴$ and $𝐵$ have no common eigenvalue.
(c)Deduce that the matrices commuting with $𝐴$ form a subspace of dimension at least $𝑚$ , with equality if and only if $𝐴$ has $𝑚$ distinct eigenvalues.

Solution.

Uses. Theorem 7.4.16 (Criteria for diagonalizability), Theorem 7.4.2 (Diagonalizable means an eigenbasis), Proposition 7.4.23 (Powers and polynomials of a diagonalizable matrix)

Techniques. Diagonalize, solve, conjugate back (Technique 7.4.22), Study a linear map on a space of matrices (Technique 4.2.29)

Hint 1. Write

𝐴 = 𝑆 𝐷 𝑆^{- 1}

and

𝐵 = 𝑅 𝐸 𝑅^{- 1}

, and substitute

𝑋 = 𝑆 𝑌 𝑅^{- 1}

Hint 2.

𝐷 𝑌 = 𝑌 𝐸

says

(𝑑_{𝑖} - 𝑒_{𝑗}) 𝑦_{𝑖 𝑗} =

0

for all

𝑖, 𝑗

. Count the pairs with

𝑑_{𝑖} = 𝑒_{𝑗}

Sketch.

𝑋 \mapsto 𝑆^{- 1} 𝑋 𝑅

is an isomorphism of

𝐹^{𝑚 \times 𝑛}

taking

𝑊

{𝑌 | 𝐷 𝑌 = 𝑌 𝐸}

, whose elements have

𝑦_{𝑖 𝑗} = 0

unless

𝑑_{𝑖} = 𝑒_{𝑗}

. The number of such pairs is

\sum_{𝜆} {gm}_{𝐴} (𝜆) {gm}_{𝐵} (𝜆)

. (b) is the case

dim 𝑊 = 0

. (c) With

𝐵 = 𝐴

\sum {gm (𝜆)}^{2} \geq

\sum gm (𝜆) =

𝑚

, with equality iff every

gm (𝜆) \leq 1

(a) Write $𝐴 = 𝑆 𝐷 𝑆^{- 1}$ and $𝐵 = 𝑅 𝐸 𝑅^{- 1}$ with $𝐷 =$ $diag (𝑑_{1}, \dots, 𝑑_{𝑚})$ and $𝐸 =$ $diag (𝑒_{1}, \dots, 𝑒_{𝑛})$ . The map $𝑋 \mapsto 𝑆^{- 1} 𝑋 𝑅$ is linear on $𝐹^{𝑚 \times 𝑛}$ with inverse $𝑌 \mapsto 𝑆 𝑌 𝑅^{- 1}$ , so it is an isomorphism. For $𝑌 = 𝑆^{- 1} 𝑋 𝑅$ ,

𝐴 𝑋 = 𝑋 𝐵 ⟺ 𝑆 𝐷 𝑆^{- 1} 𝑋 = 𝑋 𝑅 𝐸 𝑅^{- 1} ⟺ 𝐷 𝑌 = 𝑌 𝐸,

multiplying on the left by $𝑆^{- 1}$ and on the right by $𝑅$ . The $(𝑖, 𝑗)$ entry of $𝐷 𝑌 - 𝑌 𝐸$ is $(𝑑_{𝑖} - 𝑒_{𝑗}) 𝑦_{𝑖 𝑗}$ . So $𝑊$ corresponds to the space of $𝑌$ with $𝑦_{𝑖 𝑗} = 0$ whenever $𝑑_{𝑖} \neq 𝑒_{𝑗}$ , which has as a basis the matrix units at the positions with $𝑑_{𝑖} = 𝑒_{𝑗}$ . By Theorem 7.4.16, each $𝜆$ occurs exactly ${gm}_{𝐴} (𝜆)$ times among the $𝑑_{𝑖}$ and ${gm}_{𝐵} (𝜆)$ times among the $𝑒_{𝑗}$ , so the number of positions is $\sum_{𝜆} {gm}_{𝐴} (𝜆) {gm}_{𝐵} (𝜆)$ .

(b) $𝑊 = {0}$ exactly when every term of the sum vanishes, that is, when no $𝜆$ is an eigenvalue of both.

(c) Take $𝐵 = 𝐴$ . Then $dim 𝑊 = \sum_{𝜆} {gm (𝜆)}^{2}$ , where $\sum_{𝜆} gm (𝜆) = 𝑚$ since $𝐴$ is diagonalizable. As each $gm (𝜆)$ is a nonnegative integer, ${gm (𝜆)}^{2} \geq gm (𝜆)$ , with equality iff $gm (𝜆) \in {0, 1}$ . So $dim 𝑊 \geq 𝑚$ , with equality iff every eigenvalue has $gm = 1$ , that is, iff there are $𝑚$ distinct eigenvalues.

The “if” direction of (b) holds for all $𝐴 \in ℂ^{𝑚 \times 𝑚}$ and $𝐵 \in ℂ^{𝑛 \times 𝑛}$ , diagonalizable or not. If $𝐴 𝑋 = 𝑋 𝐵$ , then $𝐴^{𝑗} 𝑋 = 𝑋 𝐵^{𝑗}$ for all $𝑗$ , so $𝑝_{𝐵} (𝐴) 𝑋 =$ $𝑋 𝑝_{𝐵} (𝐵) =$ $0$ by Cayley–Hamilton (Theorem 7.6.20). The eigenvalues of $𝑝_{𝐵} (𝐴)$ are the numbers $𝑝_{𝐵} (𝜆)$ for eigenvalues $𝜆$ of $𝐴$ (Theorem 7.6.28), and these are nonzero when no eigenvalue is shared, so $𝑝_{𝐵} (𝐴)$ is invertible and $𝑋 = 0$ .

Exercise 7.7.12.standardBeyond 21-242 Let $𝐴, 𝐵 \in ℝ^{𝑛 \times 𝑛}$ with $𝑛 \geq 1$ and $𝐴 𝐵 = 𝐵 𝐴$ .

(a)Show that there is a subspace $𝑈 \subseteq ℝ^{𝑛}$ of dimension $1$ or $2$ with $𝐴 𝑈 \subseteq 𝑈$ and $𝐵 𝑈 \subseteq 𝑈$ .
(b)Show by an example that dimension $1$ cannot always be achieved.

Solution.

Uses. Theorem 7.6.31 (Commuting operators have a common eigenvector), Proposition 7.3.34 (Eigenvalues of a complexification), Example 7.2.18 (Rotations)

Techniques. Complexify, then take real and imaginary parts (Technique 7.3.35), Look for invariant subspaces among kernels and images of commuting operators (Technique 7.2.31)

Hint 1. Regard

𝐴, 𝐵

as commuting complex matrices and take a common eigenvector

𝐳 = 𝐮 + 𝑖 𝐰

with

𝐮, 𝐰 \in ℝ^{𝑛}

Hint 2. Compare real and imaginary parts of

𝐴 𝐳 = 𝜆 𝐳

. For (b), a rotation has no invariant line.

Sketch. By Theorem 7.6.31 there is

𝐳 = 𝐮 + 𝑖 𝐰 \neq 𝟎

with

𝐴 𝐳 = 𝜆 𝐳

and

𝐵 𝐳 = 𝜇 𝐳

. With

𝜆 = 𝑎 + 𝑏 𝑖

, real and imaginary parts give

𝐴 𝐮 = 𝑎 𝐮 - 𝑏 𝐰

and

𝐴 𝐰 = 𝑏 𝐮 + 𝑎 𝐰

; the same for

𝐵

. So

𝑈 = span (𝐮, 𝐰)

works. (b)

𝐴 = 𝐵 = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})

(a) As complex matrices, $𝐴$ and $𝐵$ commute, so by Theorem 7.6.31 there is a nonzero $𝐳 \in ℂ^{𝑛}$ with $𝐴 𝐳 = 𝜆 𝐳$ and $𝐵 𝐳 = 𝜇 𝐳$ . Write $𝐳 = 𝐮 + 𝑖 𝐰$ with $𝐮, 𝐰 \in ℝ^{𝑛}$ , and $𝜆 = 𝑎 + 𝑏 𝑖$ , $𝜇 = 𝑐 + 𝑑 𝑖$ with $𝑎, 𝑏, 𝑐, 𝑑 \in ℝ$ . Since $𝐴$ has real entries,

𝐴 𝐮 + 𝑖 𝐴 𝐰 = 𝐴 𝐳 = (𝑎 + 𝑏 𝑖) (𝐮 + 𝑖 𝐰) = (𝑎 𝐮 - 𝑏 𝐰) + 𝑖 (𝑏 𝐮 + 𝑎 𝐰),

and comparing real and imaginary parts, $𝐴 𝐮 = 𝑎 𝐮 - 𝑏 𝐰$ and $𝐴 𝐰 = 𝑏 𝐮 + 𝑎 𝐰$ . In the same way $𝐵 𝐮 = 𝑐 𝐮 - 𝑑 𝐰$ and $𝐵 𝐰 = 𝑑 𝐮 + 𝑐 𝐰$ . So $𝑈 = span (𝐮, 𝐰)$ is invariant under $𝐴$ and $𝐵$ (Proposition 7.2.4). It is nonzero because $𝐳 \neq 𝟎$ , and it has dimension at most $2$ .

(b) Let $𝐴 = 𝐵 = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ , the rotation by $𝜋 / 2$ . An invariant line of $𝐴$ would be spanned by a real eigenvector (Proposition 7.2.7), and $𝐴$ has no real eigenvalue (Example 7.2.18). So the only common invariant subspace of dimension $1$ or $2$ is $ℝ^{2}$ .

This is the real counterpart of Theorem 7.6.31, and the proof is Technique 7.3.35: solve over $ℂ$ , then take real and imaginary parts. The same move with one matrix is Theorem 7.3.31.

Exercise 7.7.13.hard Let $𝐴, 𝐵 \in 𝐹^{𝑛 \times 𝑛}$ , $𝑛 \geq 1$ , be nilpotent with $𝐴 𝐵 = 𝐵 𝐴$ , over any field $𝐹$ .

(a)Show that there is a nonzero $𝐯$ with $𝐴 𝐯 = 𝐵 𝐯 = 𝟎$ .
(b)Show that there is an invertible $𝑆$ such that $𝑆^{- 1} 𝐴 𝑆$ and $𝑆^{- 1} 𝐵 𝑆$ are both strictly upper triangular.
(c)Deduce that $𝐴 + 𝐵$ and $𝐴 𝐵$ are nilpotent.

Solution.

Uses. Proposition 7.2.30 (Kernels and images of commuting operators are invariant), Lemma 7.5.11 (Compressing an operator to a complement), Lemma 7.6.21 (A triangular operator is killed by its diagonal)

Techniques. Look for invariant subspaces among kernels and images of commuting operators (Technique 7.2.31), Induction on dimension via an invariant subspace (Technique 7.5.12)

Hint 1. (a)

ker 𝐴

is nonzero and

𝐵

-invariant. What is

𝐵

on it?

Hint 2. (b) Induct on

𝑛

. Take

𝑈 = span (𝐯)

from (a) and compress

𝐴

and

𝐵

to a complement

𝑊

. Check that

𝑇 𝐰 - 𝑇^{'} 𝐰 \in 𝑈

(Lemma 7.5.11(a)) gives

{(𝐴 𝐵)}^{'} = 𝐴^{'} 𝐵^{'}

and

{(𝐴^{𝑘})}^{'} = {(𝐴^{'})}^{𝑘}

Sketch. (a)

ker 𝐴 \neq {𝟎}

𝐵

-invariant and

𝐵 ↾_{ker 𝐴}

is nilpotent, hence not injective. (b) Induct on

𝑛

. Compress to a complement

𝑊

𝑈 = span (𝐯)

; since

𝑈

is invariant under both, compressions respect products, so

𝐴^{'}, 𝐵^{'}

commute and are nilpotent. By induction they are strictly upper triangular in a basis of

𝑊

; prepend

𝐯

. (c) Sums and products of strictly upper triangular matrices are strictly upper triangular, hence nilpotent.

(a) If $𝐴$ were injective, so would be every power of $𝐴$ , but $𝐴^{𝑘} = 0$ on $𝐹^{𝑛} \neq {𝟎}$ . So $𝐾 = ker 𝐴 \neq {𝟎}$ . By Proposition 7.2.30, $𝐾$ is $𝐵$ -invariant, and the restriction $𝐵 ↾_{𝐾}$ satisfies ${(𝐵 ↾_{𝐾})}^{𝑘} = 0$ whenever $𝐵^{𝑘} = 0$ , so it is not injective either: some $𝐯 \in 𝐾$ , $𝐯 \neq 𝟎$ , has $𝐵 𝐯 = 𝟎$ . Then $𝐴 𝐯 = 𝐵 𝐯 = 𝟎$ .

(b) We induct on $𝑛$ . For $𝑛 = 1$ , a nilpotent $1 \times 1$ matrix is $0$ . Let $𝑛 \geq 2$ , take $𝐯$ from (a), let $𝑈 = span (𝐯)$ , and choose a complement $𝑊$ of $𝑈$ , with projection $𝑃 = {Pr}_{𝑊, 𝑈}$ . For $𝑇 \in {𝐴, 𝐵}$ , let $𝑇^{'} =$ $𝑃 \circ$ $𝑇 ↾_{𝑊} \in$ $L (𝑊)$ be the compression (Lemma 7.5.11).

We claim that ${(𝑇_{1} 𝑇_{2})}^{'} =$ $𝑇_{1}^{'} 𝑇_{2}^{'}$ for all $𝑇_{1}, 𝑇_{2}$ that map $𝑈$ into $𝑈$ . Let $𝐰 \in 𝑊$ . By Lemma 7.5.11(a), $𝑇_{2} 𝐰 = 𝑇_{2}^{'} 𝐰 + 𝐮$ with $𝐮 \in 𝑈$ , and $𝑇_{1} 𝑇_{2}^{'} 𝐰 =$ $𝑇_{1}^{'} 𝑇_{2}^{'} 𝐰 +$ $𝐮^{'}$ with $𝐮^{'} \in 𝑈$ . So $𝑇_{1} 𝑇_{2} 𝐰 =$ $𝑇_{1}^{'} 𝑇_{2}^{'} 𝐰 +$ $𝐮^{'} +$ $𝑇_{1} 𝐮$ , where $𝐮^{'} + 𝑇_{1} 𝐮 \in 𝑈$ ; applying $𝑃$ gives the claim. Here $𝑈$ is killed by $𝐴$ and $𝐵$ , so the claim applies to all products of $𝐴$ and $𝐵$ . Hence $𝐴^{'} 𝐵^{'} =$ ${(𝐴 𝐵)}^{'} =$ ${(𝐵 𝐴)}^{'} =$ $𝐵^{'} 𝐴^{'}$ , and ${(𝐴^{'})}^{𝑘} = {(𝐴^{𝑘})}^{'} = 0$ , and likewise for $𝐵^{'}$ . By the induction hypothesis, $𝑊$ has a basis $𝐰_{2}, \dots, 𝐰_{𝑛}$ in which $𝐴^{'}$ and $𝐵^{'}$ are strictly upper triangular. In the basis $𝐯, 𝐰_{2}, \dots, 𝐰_{𝑛}$ of $𝐹^{𝑛}$ , Lemma 7.5.11(b) gives

[𝑇_{𝐴}] = (\begin{matrix} 0 & * \\ 0 & [𝐴^{'}] \end{matrix}),

with $0$ in the corner because $𝐴 𝐯 = 𝟎$ , and $[𝐴^{'}]$ strictly upper triangular. So $[𝑇_{𝐴}]$ is strictly upper triangular, and the same holds for $𝐵$ . Let $𝑆$ have columns $𝐯, 𝐰_{2}, \dots, 𝐰_{𝑛}$ .

(c) With $𝑆$ from (b), $𝑆^{- 1} (𝐴 + 𝐵) 𝑆$ and $𝑆^{- 1} 𝐴 𝐵 𝑆 =$ $(𝑆^{- 1} 𝐴 𝑆) (𝑆^{- 1} 𝐵 𝑆)$ are strictly upper triangular. A strictly upper triangular $𝑛 \times 𝑛$ matrix $𝑀$ has $𝑀^{𝑛} = 0$ (Lemma 7.6.21 with all diagonal entries $0$ ), so $𝐴 + 𝐵$ and $𝐴 𝐵$ are nilpotent.

Nothing in this proof uses eigenvalues, which is why it works over every field. Over $ℂ$ , (b) is a special case of Theorem 7.6.32, and (c) also follows from the binomial theorem (Exercise 7.6.51).

Exercise 7.7.14.hardBeyond 21-242 Let $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ with $𝑛 \geq 2$ , and write $𝑝_{𝐴} =$ $𝑥^{𝑛} +$ $𝑐_{𝑛 - 1} 𝑥^{𝑛 - 1} +$ $\dots +$ $𝑐_{1} 𝑥 +$ $𝑐_{0}$ . Show that

adj (𝐴) = {(- 1)}^{𝑛 + 1} (𝐴^{𝑛 - 1} + 𝑐_{𝑛 - 1} 𝐴^{𝑛 - 2} + \dots + 𝑐_{2} 𝐴 + 𝑐_{1} 𝐼_{𝑛}),

first for invertible $𝐴$ , and then for all $𝐴$ when $𝐹$ is infinite. Check the formula for $𝑛 = 2$ .

Solution.

Uses. Theorem 7.6.20 (Cayley–Hamilton), Proposition 7.3.18 (The first and last coefficients), Theorem 5.5.9 (The adjugate and the inverse), Corollary 7.1.23 (Polynomials and their values), Aside (Rational functions) (Rational functions)

Techniques. Perturbation and density (Technique 7.6.23), Polynomial identity arguments (Technique 5.5.24)

Hint 1. For invertible

𝐴

, Cayley–Hamilton gives

𝐴 𝑄 = - 𝑐_{0} 𝐼_{𝑛}

with

𝑄

the bracket. What is

𝑐_{0}

Hint 2. For general

𝐴

, replace

𝐴

𝐴 - 𝑡 𝐼_{𝑛}

. Both sides are matrices of polynomials in

𝑡

, and they agree whenever

𝑡

is not an eigenvalue of

𝐴

Sketch. Let

𝑄

be the bracket. Cayley–Hamilton:

𝐴 𝑄 =

- 𝑐_{0} 𝐼_{𝑛} =

{(- 1)}^{𝑛 + 1} (det 𝐴) 𝐼_{𝑛}

. Multiply by

adj (𝐴)

on the left:

(det 𝐴) 𝑄 =

{(- 1)}^{𝑛 + 1} (det 𝐴) adj (𝐴)

, and cancel

det 𝐴 \neq 0

. For general

𝐴

, both sides for

𝐴 - 𝑡 𝐼_{𝑛}

are polynomial in

𝑡

entrywise (the

𝑐_{𝑗}

𝑝_{𝐴 - 𝑡 𝐼_{𝑛}} (𝑥) =

𝑝_{𝐴} (𝑥 + 𝑡)

are polynomials in

𝑡

), and they agree for all

𝑡

outside the at most

𝑛

eigenvalues of

𝐴

; so they agree at

𝑡 = 0

Write $𝑄 (𝐴) =$ $𝐴^{𝑛 - 1} +$ $𝑐_{𝑛 - 1} 𝐴^{𝑛 - 2} +$ $\dots +$ $𝑐_{1} 𝐼_{𝑛}$ , so that $𝑝_{𝐴} (𝐴) =$ $𝐴 𝑄 (𝐴) +$ $𝑐_{0} 𝐼_{𝑛}$ .

Invertible $𝐴$ . By Cayley–Hamilton (Theorem 7.6.20), $𝐴 𝑄 (𝐴) = - 𝑐_{0} 𝐼_{𝑛}$ , and $𝑐_{0} = {(- 1)}^{𝑛} det 𝐴$ by Proposition 7.3.18, so $𝐴 𝑄 (𝐴) =$ ${(- 1)}^{𝑛 + 1} (det 𝐴) 𝐼_{𝑛}$ . Multiplying on the left by $adj (𝐴)$ and using $adj (𝐴) 𝐴 =$ $(det 𝐴) 𝐼_{𝑛}$ (Theorem 5.5.9),

(det 𝐴) 𝑄 (𝐴) = {(- 1)}^{𝑛 + 1} (det 𝐴) adj (𝐴) .

Since $det 𝐴 \neq 0$ , $adj (𝐴) =$ ${(- 1)}^{𝑛 + 1} 𝑄 (𝐴)$ .

All $𝐴$ , for infinite $𝐹$ . For $𝑡 \in 𝐹$ , let $𝐴_{𝑡} = 𝐴 - 𝑡 𝐼_{𝑛}$ . Its characteristic polynomial is $det (𝑥 𝐼_{𝑛} - 𝐴 + 𝑡 𝐼_{𝑛}) =$ $𝑝_{𝐴} (𝑥 + 𝑡)$ , so each coefficient $𝑐_{𝑗} (𝑡)$ of $𝑝_{𝐴_{𝑡}}$ is a polynomial in $𝑡$ , and every entry of

𝑄_{𝑡} = 𝐴_{𝑡}^{𝑛 - 1} + 𝑐_{𝑛 - 1} (𝑡) 𝐴_{𝑡}^{𝑛 - 2} + \dots + 𝑐_{1} (𝑡) 𝐼_{𝑛}

is a polynomial in $𝑡$ . So is every entry of $adj (𝐴_{𝑡})$ , being a cofactor of $𝐴_{𝑡}$ . By the first part, $adj (𝐴_{𝑡}) =$ ${(- 1)}^{𝑛 + 1} 𝑄_{𝑡}$ whenever $𝐴_{𝑡}$ is invertible, that is, for every $𝑡$ that is not one of the at most $𝑛$ eigenvalues of $𝐴$ . Each entry of the difference is then a polynomial in $𝑡$ with infinitely many roots, hence $𝟎$ (Corollary 7.1.23). At $𝑡 = 0$ this is the claim. (For a finite field, run the same argument over the field of rational functions $𝐹 (𝑡)$ of Aside (Rational functions), where $𝐴 - 𝑡 𝐼_{𝑛}$ is invertible because its determinant ${(- 1)}^{𝑛} 𝑝_{𝐴} (𝑡)$ is nonzero, and then set $𝑡 = 0$ .)

$𝑛 = 2$ . Here $𝑝_{𝐴} =$ $𝑥^{2} -$ $(tr 𝐴) 𝑥 +$ $det 𝐴$ , so $𝑐_{1} = - tr 𝐴$ and the formula reads $adj (𝐴) =$ $- (𝐴 - (tr 𝐴) 𝐼_{2}) =$ $(tr 𝐴) 𝐼_{2} -$ $𝐴$ . For $𝐴 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ this is $(\begin{matrix} 𝑑 & - 𝑏 \\ - 𝑐 & 𝑎 \end{matrix})$ , the adjugate of Example 5.5.7.

In particular $adj (𝐴)$ commutes with every matrix that commutes with $𝐴$ , even when $𝐴$ is singular. The proof is Technique 7.6.23 in its algebraic form: prove the identity where it is easy (invertible matrices), then use that both sides are polynomials.