9.10 Spectral Mapping, Products in Both Orders, Left Eigenvectors

Once the eigenvalues of $T$ are known, what can we say about the eigenvalues of $T^2$ , of $T^2 - 3T + I$ , or of any polynomial in $T$ ? And when two matrices are multiplied in the two possible orders, how are the eigenvalues of $\mathbf{A} \mathbf{B}$ and $\mathbf{B} \mathbf{A}$ related? This section answers both questions using the triangular forms of Section 7 and the block matrices of Chapter 8. Along the way it settles two debts: Chapter 7 showed $p_{\mathbf{A} \mathbf{B}} = p_{\mathbf{B} \mathbf{A}}$ only when $\mathbf{A}$ is invertible, and Chapter 8 proved that vanishing traces of all powers force a matrix to be nilpotent by a long computation, with no explanation of why. We end with left eigenvectors, the eigenvectors of the transpose, which pair with ordinary eigenvectors in a useful way and prepare the Markov chains of Section 11.

9.10.1. Eigenvalues of a polynomial in $T$

One half of the answer needs no theory. If $Tv = \lambdav$ with $v \ne 0$ , then $T^kv = \lambda^kv$ for every $k$ , and so $q(T)v = q(\lambda)v$ for every $q \in F[x]$ (Exercise (B3: Eigenvalues of powers and inverses)). So $q$ maps eigenvalues of $T$ to eigenvalues of $q(T)$ , over every field and in every dimension.

The other half, that every eigenvalue of $q(T)$ arises this way, can fail. Part (c) of Exercise (B3: Eigenvalues of powers and inverses) already met the rotation $\mathbf{R} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ on $\mathbb{R}^2$ : it has no real eigenvalue, while $\mathbf{R}^2 = -\mathbf{I}$ has the eigenvalue $-1$ . The eigenvalues of $\mathbf{R}$ that would map to $-1$ , namely $\pm i$ , are missing from $\mathbb{R}$ . So we expect the full statement to need a characteristic polynomial that splits. With that hypothesis we even get the multiplicities, because a triangular form shows all the eigenvalues at once on its diagonal.

The eigenvalues of $q(T)$ are the values of $q$ at the eigenvalues of $T$ , counted with multiplicity.

Theorem 9.10.1 (Spectral Mapping Theorem)

Let $V$ be a finite-dimensional vector space over $F$ with $\dim V = n \ge 1$ , and let $T \in \mathcal{L}(V)$ . Suppose $p_T$ splits over $F$ , say $p_T(x) = (x - \lambda_1)(x - \lambda_2)\cdots(x - \lambda_n),$ where each eigenvalue appears as often as its algebraic multiplicity. Then for every $q \in F[x]$ , $p_{q(T)}(x) = \big(x - q(\lambda_1)\big)\big(x - q(\lambda_2)\big)\cdots\big(x - q(\lambda_n)\big),$ and consequently $\operatorname{spec}\big(q(T)\big) = q\big(\operatorname{spec}(T)\big) = \{ q(\lambda) : \lambda \in \operatorname{spec}(T) \} .$ In particular this holds for every operator on a nonzero finite-dimensional complex space, and for every $\mathbf{A} \in M_n(\mathbb{C})$ .

Idea. Choose a basis $\mathcal{B}$ in which $T$ is upper triangular (Section 7). Its diagonal is $\lambda_1, \dots, \lambda_n$ . Products and sums of upper triangular matrices are upper triangular, and their diagonals are computed position by position, so $q([T]_{\mathcal{B}}^{\mathcal{B}})$ is upper triangular with diagonal $q(\lambda_1), \dots, q(\lambda_n)$ . The characteristic polynomial of a triangular matrix is read off its diagonal.

Proof. By Theorem 9.7.2 there is a basis $\mathcal{B}$ of $V$ such that $\mathbf{U} = [T]_{\mathcal{B}}^{\mathcal{B}}$ is upper triangular. By Theorem 9.7.4, its diagonal entries are $\lambda_1, \dots, \lambda_n$ in some order; renumbering the $\lambda_i$ , we may assume $u_{ii} = \lambda_i$ .

Claim. If $\mathbf{X}, \mathbf{Y} \in M_n(F)$ are upper triangular, then so are $\mathbf{X} + \mathbf{Y}$ , $c\mathbf{X}$ for $c \in F$ , and $\mathbf{X} \mathbf{Y}$ , with diagonal entries $x_{ii} + y_{ii}$ , $cx_{ii}$ , and $x_{ii}y_{ii}$ .

Proof. The statements for $\mathbf{X} + \mathbf{Y}$ and $c\mathbf{X}$ are entrywise. For the product, $(\mathbf{X} \mathbf{Y})_{ij} = \sum_{k=1}^{n} x_{ik}y_{kj}$ (Definition 0.9.9). If $i > j$ , every $k$ has $k < i$ or $k \ge i > j$ , so $x_{ik} = 0$ or $y_{kj} = 0$ , and the sum is $0$ . If $i = j$ , the only term that can be nonzero has $k \ge i$ and $k \le i$ , so $(\mathbf{X} \mathbf{Y})_{ii} = x_{ii}y_{ii}$ .

By the claim and induction on $k$ , $\mathbf{U}^k$ is upper triangular with diagonal entries $\lambda_1^k, \dots, \lambda_n^k$ (for $k = 0$ , $\mathbf{U}^0 = \mathbf{I}$ ). Writing $q = \sum_k a_kx^k$ , the claim again shows that $q(\mathbf{U}) = \sum_k a_k\mathbf{U}^k$ is upper triangular with diagonal entries $\sum_k a_k\lambda_i^k = q(\lambda_i)$ . By Corollary 4.1.3, $[q(T)]_{\mathcal{B}}^{\mathcal{B}} = q(\mathbf{U})$ , so by Definition 7.10.2 and the examples after Definition 7.10.1, $p_{q(T)} = p_{q(\mathbf{U})} = \prod_{i=1}^{n} \big(x - q(\lambda_i)\big).$ For the spectra: by Theorem 9.2.1, $\mu \in F$ is an eigenvalue of $q(T)$ if and only if $p_{q(T)}(\mu) = \prod_i (\mu - q(\lambda_i)) = 0$ , that is, if and only if $\mu = q(\lambda_i)$ for some $i$ , since a product of nonzero field elements is nonzero. The $\lambda_i$ are exactly the eigenvalues of $T$ , each listed at least once (Theorem 9.2.1). Hence $\operatorname{spec}(q(T)) = \{ q(\lambda) : \lambda \in \operatorname{spec}(T) \}$ . Over $\mathbb{C}$ , $p_T$ always splits by Corollary 6.5.5.

The formula for $p_{q(T)}$ carries more than the spectrum. The algebraic multiplicity of $\mu$ as an eigenvalue of $q(T)$ is the number of indices $i$ with $q(\lambda_i) = \mu$ (Lemma 6.4.1), so several eigenvalues of $T$ can merge into one eigenvalue of $q(T)$ , and their multiplicities add. Combined with Theorem 9.3.3, applied to $q(T)$ , whose characteristic polynomial splits by the theorem, we get the most used consequence.

Corollary 9.10.2 (Trace and Determinant of a Polynomial in

T

)

Under the hypotheses of Theorem 9.10.1, for every $q \in F[x]$ , $\operatorname{tr} q(T) = q(\lambda_1) + \dots + q(\lambda_n), \qquad \det q(T) = q(\lambda_1)\cdots q(\lambda_n).$ In particular $\operatorname{tr} T^k = \lambda_1^k + \dots + \lambda_n^k$ for every $k \in \mathbb{N}$ .

Proof. By Theorem 9.10.1, $p_{q(T)} = \prod_i (x - q(\lambda_i))$ splits. Group equal values: if $\mu_1, \dots, \mu_s$ are the distinct values among the $q(\lambda_i)$ , then $\mu_j$ occurs exactly $a_{q(T)}(\mu_j)$ times in the list (Lemma 6.4.1). By Theorem 9.3.3 applied to $q(T)$ , $\operatorname{tr} q(T) = \sum_j a_{q(T)}(\mu_j)\,\mu_j = \sum_i q(\lambda_i)$ and $\det q(T) = \prod_j \mu_j^{\,a_{q(T)}(\mu_j)} = \prod_i q(\lambda_i)$ . Take $q = x^k$ for the last statement.

Example (A Polynomial in a Non-Diagonalizable Matrix)

Let $\mathbf{A} = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 0 & 1 \\ -3 & 3 & 2 \end{pmatrix} \in M_3(\mathbb{R})$ , for which Exercise (B1: Diagonalize or show impossible) found $p_{\mathbf{A}} = (x + 1)(x - 2)^2$ and showed that $\mathbf{A}$ is not diagonalizable. Without computing any eigenvectors, find $p_{\mathbf{A}^2}$ , $\operatorname{tr} \mathbf{A}^2$ , and $\det \mathbf{A}^2$ . Then find $p_{q(\mathbf{A})}$ for $q = x^2 - x - 2$ , and decide whether $q(\mathbf{A}) = \mathbf{O}$ .

Solution. The square. $p_{\mathbf{A}}$ splits over $\mathbb{R}$ with the list $\lambda_1 = -1$ , $\lambda_2 = \lambda_3 = 2$ . By Theorem 9.10.1 with $q = x^2$ , the list for $\mathbf{A}^2$ is $1, 4, 4$ , so $p_{\mathbf{A}^2} = (x - 1)(x - 4)^2, \qquad \operatorname{tr} \mathbf{A}^2 = 1 + 4 + 4 = 9, \qquad \det \mathbf{A}^2 = 1 \cdot 4 \cdot 4 = 16$ by Corollary 9.10.2. Check. Multiplying out, $\mathbf{A}^2 = \begin{pmatrix} 0 & 4 & 4 \\ -1 & 5 & 4 \\ -3 & 3 & 4 \end{pmatrix},$ whose trace is $0 + 5 + 4 = 9$ ; and $\det \mathbf{A} = (-1)\cdot 2 \cdot 2 = -4$ by Theorem 9.3.3, so $\det \mathbf{A}^2 = (\det \mathbf{A})^2 = 16$ by Theorem 7.5.4.

The polynomial $q$ . Here $q = (x - 2)(x + 1)$ , so $q(-1) = 0$ and $q(2) = 0$ . The list for $q(\mathbf{A})$ is $0, 0, 0$ , and $p_{q(\mathbf{A})} = x^3$ : the only eigenvalue of $q(\mathbf{A})$ is $0$ . Yet $q(\mathbf{A}) \ne \mathbf{O}$ . Indeed $q(\mathbf{A}) = \mathbf{A}^2 - \mathbf{A} - 2\mathbf{I} = \begin{pmatrix} -3 & 3 & 3 \\ -3 & 3 & 3 \\ 0 & 0 & 0 \end{pmatrix},$ and one checks $q(\mathbf{A})^2 = \mathbf{O}$ . The spectral mapping theorem sees only eigenvalues, and a nonzero matrix can have $0$ as its only eigenvalue. That $q(\mathbf{A}) \ne \mathbf{O}$ here is no accident: if $q(\mathbf{A}) = \mathbf{O}$ , the minimal polynomial of $\mathbf{A}$ would divide (Theorem 9.5.1) $(x - 2)(x + 1)$ , which has distinct roots, and $\mathbf{A}$ would be diagonalizable by Theorem 9.8.2.

Warning. Without splitting, spectral mapping fails. For the real rotation $\mathbf{R}$ above, $\operatorname{spec}(\mathbf{R}) = \varnothing$ in $\mathbb{R}$ , so $q(\operatorname{spec}(\mathbf{R})) = \varnothing$ for $q = x^2$ , while $\operatorname{spec}(\mathbf{R}^2) = \operatorname{spec}(-\mathbf{I}) = \{-1\}$ . Over $\mathbb{C}$ the theorem is restored: $\operatorname{spec}(\mathbf{R}) = \{i, -i\}$ , both square to $-1$ , and $p_{\mathbf{R}^2} = (x + 1)^2$ . Also, the theorem is about eigenvalues and algebraic multiplicities only: $\mathbf{A} = \operatorname{diag}(1, -1)$ has two one-dimensional eigenspaces, while $\mathbf{A}^2 = \mathbf{I}$ has the single eigenspace $\mathbb{R}^2$ .

Quick check. Let $\mathbf{A} \in M_2(\mathbb{C})$ have eigenvalues $i$ and $-i$ . Find $p_{\mathbf{A}^2}$ and $\det(\mathbf{A}^2 + \mathbf{I})$ .

Solution. The list for $\mathbf{A}$ is $i, -i$ (two distinct roots of the degree- $2$ polynomial $p_{\mathbf{A}}$ ). By Theorem 9.10.1 with $q = x^2$ , the list for $\mathbf{A}^2$ is $-1, -1$ , so $p_{\mathbf{A}^2} = (x + 1)^2$ . With $q = x^2 + 1$ , Corollary 9.10.2 gives $\det(\mathbf{A}^2 + \mathbf{I}) = q(i)\,q(-i) = 0 \cdot 0 = 0$ .

9.10.2. Traces of powers detect nilpotency

Chapter 8 proved, in Exercise (C2: When $\mathbf{A}$ commutes with $[\mathbf{A}, \mathbf{B}]$ ), that a matrix $\mathbf{C}$ over a field of characteristic $0$ with $\operatorname{tr} \mathbf{C}^k = 0$ for every $k \ge 1$ is nilpotent, by a careful argument with annihilating polynomials that gave no reason for the result. Here is the view through eigenvalues, for complex matrices. The traces of the powers are the power sums $\lambda_1^k + \dots + \lambda_n^k$ of the eigenvalues (Corollary 9.10.2), and a nilpotent matrix is one whose eigenvalues are all $0$ . So the question becomes: if all power sums of $n$ complex numbers vanish, must the numbers vanish? A Vandermonde determinant says yes.

Theorem 9.10.3 (Nilpotent Matrices via Eigenvalues and Traces)

Let $n \ge 1$ and $\mathbf{A} \in M_n(\mathbb{C})$ . The following are equivalent:

(a) $\mathbf{A}$ is nilpotent, that is, $\mathbf{A}^s = \mathbf{O}$ for some $s \ge 1$ ;
(b) $\operatorname{spec}(\mathbf{A}) = \{0\}$ ;
(c) $p_{\mathbf{A}} = x^n$ ;
(d) $\mathbf{A}^n = \mathbf{O}$ ;
(e) $\operatorname{tr} \mathbf{A}^k = 0$ for $k = 1, 2, \dots, n$ .

The same holds for $\mathbf{A} \in M_n(F)$ when $F$ is a subfield of $\mathbb{C}$ , such as $\mathbb{Q}$ or $\mathbb{R}$ , with $\operatorname{spec}(\mathbf{A})$ in (b) taken over $\mathbb{C}$ .

Idea. Plan: (a) $\Rightarrow$ (b) $\Rightarrow$ (c) $\Rightarrow$ (d) $\Rightarrow$ (a), and (c) $\Rightarrow$ (e) $\Rightarrow$ (b). All but one arrow are short. For (e) $\Rightarrow$ (b), suppose some eigenvalue is nonzero. Group the nonzero eigenvalues into distinct values $\mu_1, \dots, \mu_r$ with multiplicities $m_1, \dots, m_r \ge 1$ . The vanishing traces say $\sum_j m_j\mu_j^k = 0$ for $k = 1, \dots, r$ : a square linear system for the vector $(m_1, \dots, m_r)$ , whose matrix is a Vandermonde matrix times the invertible $\operatorname{diag}(\mu_j)$ . So the $m_j$ would be $0$ , but they are positive integers, and positive integers are nonzero in $\mathbb{C}$ .

Proof.

(a) $\Rightarrow$ (b). Let $\mathbf{A}^{s} = \mathbf{O}$ with $s \ge 1$ . If $\mathbf{A}\mathbf{v} = \lambda\mathbf{v}$ with $\mathbf{v} \ne \mathbf{0}$ , then $\mathbf{A}^{s}\mathbf{v} = \lambda^{s}\mathbf{v}$ by induction on $s$ , so $\lambda^{s}\mathbf{v} = \mathbf{0}$ and $\lambda^{s} = 0$ by Theorem 1.1.9; hence $\lambda = 0$ , since a product of nonzero scalars is nonzero (Theorem 0.7.1 (f)). So $\operatorname{spec}(\mathbf{A}) \subseteq \{0\}$ . Conversely $\mathbf{A}$ is not injective: if it were, its $s$ -fold composite $\mathbf{A}^{s}$ would be injective too (Theorem 0.4.3 (a)), whereas $\mathbf{A}^{s} = \mathbf{O}$ is not injective on $\mathbb{C}^{n} \ne \{\mathbf{0}\}$ . Hence $0 \in \operatorname{spec}(\mathbf{A})$ by Theorem 9.2.1 ((b) $\Rightarrow$ (a)), and $\operatorname{spec}(\mathbf{A}) = \{0\}$ .
(b) $\Rightarrow$ (c). By Corollary 6.5.5, $p_{\mathbf{A}} = (x - z_1)\cdots(x - z_n)$ with $z_i \in \mathbb{C}$ . Each $z_i$ is an eigenvalue (Theorem 9.2.1), hence $0$ , so $p_{\mathbf{A}} = x^n$ .
(c) $\Rightarrow$ (d). By Corollary 9.7.3, $\mathbf{A} = \mathbf{P} \mathbf{U} \mathbf{P}^{-1}$ with $\mathbf{P}$ invertible and $\mathbf{U}$ upper triangular, and by Theorem 9.7.4 the diagonal entries of $\mathbf{U}$ are the roots of $p_{\mathbf{A}} = x^n$ , all $0$ . Then $\mathbf{U}\mathbf{e}_1 = \mathbf{0}$ and $\mathbf{U}\mathbf{e}_j \in \operatorname{span}(\mathbf{e}_1, \dots, \mathbf{e}_{j-1})$ for $j \ge 2$ , since column $j$ of $\mathbf{U}$ has zeros from row $j$ down. By induction on $j$ , $\mathbf{U}^j\mathbf{e}_j = \mathbf{0}$ , hence $\mathbf{U}^n\mathbf{e}_j = \mathbf{U}^{n-j}\mathbf{U}^j\mathbf{e}_j = \mathbf{0}$ for every $j$ , and $\mathbf{U}^n = \mathbf{O}$ . By Proposition 4.2.6 (c), $\mathbf{A}^n = \mathbf{P} \mathbf{U}^n\mathbf{P}^{-1} = \mathbf{O}$ .
(d) $\Rightarrow$ (a). Take $s = n$ .
(e) $\Rightarrow$ (e). The list of eigenvalues of $\mathbf{A}$ with multiplicity is $0, \dots, 0$ , so by Corollary 9.10.2, $\operatorname{tr} \mathbf{A}^k = 0^k + \dots + 0^k = 0$ for every $k \ge 1$ .
(f) $\Rightarrow$ (b). By Corollary 6.5.5, $p_{\mathbf{A}}$ splits; let $\lambda_1, \dots, \lambda_n$ be the list of its roots with multiplicity. Since $n \ge 1$ , $\operatorname{spec}(\mathbf{A})$ is nonempty, so it suffices to show that every $\lambda_i = 0$ . Suppose not, let $\mu_1, \dots, \mu_r$ ( $1 \le r \le n$ ) be the distinct nonzero values among the $\lambda_i$ , and let $m_j \ge 1$ be the number of indices $i$ with $\lambda_i = \mu_j$ . For $1 \le k \le r$ , the zero eigenvalues contribute $0^k = 0$ , so by Corollary 9.10.2 and (e), $\sum_{j=1}^{r} m_j\,\mu_j^{\,k} = \operatorname{tr} \mathbf{A}^k = 0 .$ In matrix form, $\mathbf{W}\mathbf{m} = \mathbf{0}$ , where $\mathbf{m} = (m_1, \dots, m_r) \in \mathbb{C}^r$ and $\mathbf{W} \in M_r(\mathbb{C})$ has $(k, j)$ -entry $\mu_j^{\,k} = \mu_j^{\,k-1}\mu_j$ . Hence $\mathbf{W} = V(\mu_1, \dots, \mu_r)^{\top}\,\operatorname{diag}(\mu_1, \dots, \mu_r)$ , where $V(\mu_1, \dots, \mu_r)$ is the Vandermonde matrix with $(j, k)$ -entry $\mu_j^{\,k-1}$ (Theorem 7.8.1). The $\mu_j$ are distinct, so $V$ is invertible (Corollary 7.8.2), and so is $V^{\top}$ , with inverse $(V^{-1})^{\top}$ (Theorem 0.9.5). The $\mu_j$ are nonzero, so $\operatorname{diag}(\mu_1, \dots, \mu_r)$ is invertible. Therefore $\mathbf{W}$ is invertible, and $\mathbf{m} = \mathbf{W}^{-1}\mathbf{0} = \mathbf{0}$ . But $m_1$ is a positive integer, and $m_1 \ne 0$ in $\mathbb{C}$ . This contradiction shows that every $\lambda_i = 0$ , so $\operatorname{spec}(\mathbf{A}) = \{0\}$ .

For a subfield $F \subseteq \mathbb{C}$ , regard $\mathbf{A}$ as a complex matrix. Its powers, their traces, its characteristic polynomial, and the condition $\mathbf{A}^s = \mathbf{O}$ are computed by the same arithmetic in $F$ and in $\mathbb{C}$ , so (a), (c), (d), and (e) mean the same thing over $F$ and over $\mathbb{C}$ , and the equivalences carry over.

In the setting of Exercise (C2: When $\mathbf{A}$ commutes with $[\mathbf{A}, \mathbf{B}]$ ), where $\mathbf{C} = [\mathbf{A}, \mathbf{B}]$ commutes with $\mathbf{A}$ , part (a) of that exercise gives $\operatorname{tr} \mathbf{C}^k = 0$ for all $k \ge 1$ , and for complex matrices the theorem now shows at once that $\mathbf{C}^n = \mathbf{O}$ . That exercise remains the route for a general field of characteristic $0$ .

Warning. The characteristic matters. Over $\mathbb{F}_p$ , the identity $\mathbf{I}_p \in M_p(\mathbb{F}_p)$ has $\operatorname{tr} \mathbf{I}_p^{\,k} = p \cdot 1 = 0$ for every $k$ , yet $\mathbf{I}_p^{\,k} = \mathbf{I}_p \ne \mathbf{O}$ . The proof above breaks exactly where it used that the positive integer $m_1$ is nonzero in the field: here the multiplicity $m_1 = p$ of the eigenvalue $1$ is $0$ in $\mathbb{F}_p$ .

9.10.3. $\mathbf{A} \mathbf{B}$ and $\mathbf{B} \mathbf{A}$

Chapter 7, in Exercise (B3: $\mathbf{A}\mathbf{B}$ and $\mathbf{B}\mathbf{A}$ ), showed that $p_{\mathbf{A} \mathbf{B}} = p_{\mathbf{B} \mathbf{A}}$ for square $\mathbf{A}$ and $\mathbf{B}$ when $\mathbf{A}$ is invertible, because then $\mathbf{B} \mathbf{A} = \mathbf{A}^{-1}(\mathbf{A} \mathbf{B})\mathbf{A}$ is similar to $\mathbf{A} \mathbf{B}$ . Without invertibility the two products need not be similar, and for rectangular $\mathbf{A}$ and $\mathbf{B}$ they do not even have the same size: if $\mathbf{A}$ is $m \times n$ and $\mathbf{B}$ is $n \times m$ , then $\mathbf{A} \mathbf{B}$ is $m \times m$ and $\mathbf{B} \mathbf{A}$ is $n \times n$ . Their characteristic polynomials have degrees $m$ and $n$ . Multiplying by powers of $x$ repairs the degrees, and the identity then holds in general.

Theorem 9.10.4 (

\mathbf{A} \mathbf{B}

and

\mathbf{B} \mathbf{A}

Have the Same Nonzero Eigenvalues)

Let $m, n \ge 1$ , $\mathbf{A} \in M_{m \times n}(F)$ , and $\mathbf{B} \in M_{n \times m}(F)$ . Then $x^n\,p_{\mathbf{A} \mathbf{B}}(x) = x^m\,p_{\mathbf{B} \mathbf{A}}(x).$ Consequently:

(a) for every $\lambda \ne 0$ in $F$ , $a_{\mathbf{A} \mathbf{B}}(\lambda) = a_{\mathbf{B} \mathbf{A}}(\lambda)$ ; in particular $\mathbf{A} \mathbf{B}$ and $\mathbf{B} \mathbf{A}$ have the same nonzero eigenvalues;
(b) if $m = n$ , then $p_{\mathbf{A} \mathbf{B}} = p_{\mathbf{B} \mathbf{A}}$ ; and if $m \ge n$ , then $p_{\mathbf{A} \mathbf{B}} = x^{m-n}\,p_{\mathbf{B} \mathbf{A}}$ .

Idea. We want one big matrix that “contains” $\mathbf{A} \mathbf{B}$ and another that “contains” $\mathbf{B} \mathbf{A}$ , and that are similar to each other. Put them in $(m + n) \times (m + n)$ block form: $\mathbf{M} = \begin{pmatrix} \mathbf{A} \mathbf{B} & \mathbf{O} \\ \mathbf{B} & \mathbf{O} \end{pmatrix}, \qquad \mathbf{N} = \begin{pmatrix} \mathbf{O} & \mathbf{O} \\ \mathbf{B} & \mathbf{B} \mathbf{A} \end{pmatrix}.$ Both are block triangular, so their characteristic polynomials are $p_{\mathbf{A} \mathbf{B}} \cdot x^n$ and $x^m \cdot p_{\mathbf{B} \mathbf{A}}$ . The invertible block matrix $\mathbf{S} = \begin{pmatrix} \mathbf{I}_m & \mathbf{A} \\ \mathbf{O} & \mathbf{I}_n \end{pmatrix}$ turns one into the other: $\mathbf{M} \mathbf{S}$ and $\mathbf{S} \mathbf{N}$ are both $\begin{pmatrix} \mathbf{A} \mathbf{B} & \mathbf{A} \mathbf{B} \mathbf{A} \\ \mathbf{B} & \mathbf{B} \mathbf{A} \end{pmatrix}$ .

Proof. Let $\mathbf{M}, \mathbf{N}, \mathbf{S} \in M_{m+n}(F)$ be the block matrices displayed in the Idea, partitioned by $m, n$ in rows and columns. By Theorem 8.1.1, $\begin{aligned} \mathbf{M} \mathbf{S} &= \begin{pmatrix} \mathbf{A} \mathbf{B}\cdot \mathbf{I}_m + \mathbf{O} & \mathbf{A} \mathbf{B}\cdot \mathbf{A} + \mathbf{O} \\ \mathbf{B}\cdot \mathbf{I}_m + \mathbf{O} & \mathbf{B}\cdot \mathbf{A} + \mathbf{O} \end{pmatrix} = \begin{pmatrix} \mathbf{A} \mathbf{B} & \mathbf{A} \mathbf{B} \mathbf{A} \\ \mathbf{B} & \mathbf{B} \mathbf{A} \end{pmatrix}, \\ \mathbf{S} \mathbf{N} &= \begin{pmatrix} \mathbf{O} + \mathbf{A} \mathbf{B} & \mathbf{O} + \mathbf{A}\cdot \mathbf{B} \mathbf{A} \\ \mathbf{O} + \mathbf{B} & \mathbf{O} + \mathbf{B} \mathbf{A} \end{pmatrix} = \begin{pmatrix} \mathbf{A} \mathbf{B} & \mathbf{A} \mathbf{B} \mathbf{A} \\ \mathbf{B} & \mathbf{B} \mathbf{A} \end{pmatrix}, \end{aligned}$ using associativity (Theorem 0.9.4). So $\mathbf{M} \mathbf{S} = \mathbf{S} \mathbf{N}$ . The matrix $\mathbf{S}$ is invertible: by Theorem 8.1.1, $\begin{pmatrix} \mathbf{I}_m & \mathbf{A} \\ \mathbf{O} & \mathbf{I}_n \end{pmatrix}\begin{pmatrix} \mathbf{I}_m & -\mathbf{A} \\ \mathbf{O} & \mathbf{I}_n \end{pmatrix} = \begin{pmatrix} \mathbf{I}_m & -\mathbf{A} + \mathbf{A} \\ \mathbf{O} & \mathbf{I}_n \end{pmatrix} = \mathbf{I}_{m+n}$ , and Theorem 3.4.5 applies. Hence $\mathbf{S}^{-1}\mathbf{M} \mathbf{S} = \mathbf{N}$ , and $p_{\mathbf{M}} = p_{\mathbf{N}}$ by Theorem 7.10.3.

Now compute both sides. The matrix $x\mathbf{I}_{m+n} - \mathbf{M} = \begin{pmatrix} x\mathbf{I}_m - \mathbf{A} \mathbf{B} & \mathbf{O} \\ -\mathbf{B} & x\mathbf{I}_n \end{pmatrix}$ is lower block triangular with square diagonal blocks, so by Theorem 7.8.4, which holds over the commutative ring $F[x]$ (the remarks before Example (A $3 \times 3$ Characteristic Polynomial)), and by Theorem 7.4.6 for $x\mathbf{I}_n$ , $p_{\mathbf{M}}(x) = \det(x\mathbf{I}_m - \mathbf{A} \mathbf{B})\,\det(x\mathbf{I}_n) = p_{\mathbf{A} \mathbf{B}}(x)\,x^n .$ Likewise $x\mathbf{I}_{m+n} - \mathbf{N} = \begin{pmatrix} x\mathbf{I}_m & \mathbf{O} \\ -\mathbf{B} & x\mathbf{I}_n - \mathbf{B} \mathbf{A} \end{pmatrix}$ gives $p_{\mathbf{N}}(x) = x^m\,p_{\mathbf{B} \mathbf{A}}(x)$ . Since $p_{\mathbf{M}} = p_{\mathbf{N}}$ , this proves $x^np_{\mathbf{A} \mathbf{B}} = x^mp_{\mathbf{B} \mathbf{A}}$ .

(a) Let $\lambda \ne 0$ . Then $\lambda$ is not a root of $x^n$ , so $\operatorname{mult}_\lambda(x^n) = 0$ , and similarly for $x^m$ . By Theorem 6.4.2, $a_{\mathbf{A} \mathbf{B}}(\lambda) = \operatorname{mult}_\lambda(x^np_{\mathbf{A} \mathbf{B}}) = \operatorname{mult}_\lambda(x^mp_{\mathbf{B} \mathbf{A}}) = a_{\mathbf{B} \mathbf{A}}(\lambda).$ A scalar is an eigenvalue exactly when its algebraic multiplicity is positive (Theorem 9.2.1, Definition 6.4.1), so the nonzero eigenvalues agree.
(b) If $m \ge n$ , then $x^np_{\mathbf{A} \mathbf{B}} = x^n\big(x^{m-n}p_{\mathbf{B} \mathbf{A}}\big)$ , and canceling the nonzero polynomial $x^n$ (Corollary 0.8.4) gives $p_{\mathbf{A} \mathbf{B}} = x^{m-n}p_{\mathbf{B} \mathbf{A}}$ . For $m = n$ this is $p_{\mathbf{A} \mathbf{B}} = p_{\mathbf{B} \mathbf{A}}$ .

This pays off the promise of Exercise (B3: $\mathbf{A}\mathbf{B}$ and $\mathbf{B}\mathbf{A}$ ): $p_{\mathbf{A} \mathbf{B}} = p_{\mathbf{B} \mathbf{A}}$ for all square $\mathbf{A}$ and $\mathbf{B}$ , invertible or not. Comparing coefficients (Theorem 7.10.2) recovers $\operatorname{tr}(\mathbf{A} \mathbf{B}) = \operatorname{tr}(\mathbf{B} \mathbf{A})$ and $\det(\mathbf{A} \mathbf{B}) = \det(\mathbf{B} \mathbf{A})$ once more.

The eigenvectors travel between the two products as well, and by hand. If $\mathbf{A} \mathbf{B}\mathbf{w} = \lambda\mathbf{w}$ with $\lambda \ne 0$ , then $\mathbf{B} \mathbf{A}(\mathbf{B}\mathbf{w}) = \mathbf{B}(\mathbf{A} \mathbf{B}\mathbf{w}) = \lambda\,\mathbf{B}\mathbf{w} ,$ and $\mathbf{B}\mathbf{w} \ne \mathbf{0}$ , since $\mathbf{A}(\mathbf{B}\mathbf{w}) = \lambda\mathbf{w} \ne \mathbf{0}$ . So $\mathbf{B}\mathbf{w}$ is an eigenvector of $\mathbf{B} \mathbf{A}$ for the same $\lambda$ : multiplying an eigenvector of $\mathbf{A} \mathbf{B}$ by $\mathbf{B}$ produces one of $\mathbf{B} \mathbf{A}$ . For $\lambda = 0$ the recipe can fail, because $\mathbf{B}\mathbf{w}$ may be $\mathbf{0}$ — which is the warning below in another guise.

A rank-one example. Let $\mathbf{u}, \mathbf{v} \in F^n$ with $n \ge 2$ . Take $\mathbf{A} = \mathbf{u} \in M_{n \times 1}(F)$ and $\mathbf{B} = \mathbf{v}^{\top} \in M_{1 \times n}(F)$ . Then $\mathbf{A} \mathbf{B} = \mathbf{u}\mathbf{v}^{\top}$ is $n \times n$ , and $\mathbf{B} \mathbf{A} = \mathbf{v}^{\top}\mathbf{u}$ is the $1 \times 1$ matrix whose entry is the number $c = \mathbf{v}^{\top}\mathbf{u}$ , so $p_{\mathbf{B} \mathbf{A}} = x - c$ . Applying the theorem with its $m$ equal to our $n$ , and its $n$ equal to $1$ , gives $x\,p_{\mathbf{u}\mathbf{v}^{\top}} = x^n(x - c)$ , and canceling $x$ , $p_{\mathbf{u}\mathbf{v}^{\top}}(x) = x^{n-1}\,(x - \mathbf{v}^{\top}\mathbf{u}).$ For instance, the all-ones matrix $\mathbf{1}\mathbf{1}^{\top} \in M_n(\mathbb{R})$ , with $\mathbf{1} = (1, \dots, 1)$ , has $p = x^{n-1}(x - n)$ : eigenvalue $0$ with multiplicity $n - 1$ , and eigenvalue $n$ . Evaluating at $-1$ recovers Theorem 7.8.5: $\begin{aligned} \det(\mathbf{I} + \mathbf{u}\mathbf{v}^{\top}) &= (-1)^n\det(-\mathbf{I} - \mathbf{u}\mathbf{v}^{\top}) = (-1)^n\,p_{\mathbf{u}\mathbf{v}^{\top}}(-1) \\ &= (-1)^n(-1)^{n-1}(-1 - \mathbf{v}^{\top}\mathbf{u}) = 1 + \mathbf{v}^{\top}\mathbf{u} . \end{aligned}$

Warning. The zero eigenvalue behaves differently, and $\mathbf{A} \mathbf{B}$ need not be similar to $\mathbf{B} \mathbf{A}$ . Take $\mathbf{A} = \mathbf{E}_{12}$ and $\mathbf{B} = \mathbf{E}_{11}$ in $M_2(F)$ . Then $\mathbf{A} \mathbf{B} = \mathbf{E}_{12}\mathbf{E}_{11} = \mathbf{O}$ and $\mathbf{B} \mathbf{A} = \mathbf{E}_{11}\mathbf{E}_{12} = \mathbf{E}_{12}$ . Both have $p = x^2$ , as the theorem demands. But $g_{\mathbf{A} \mathbf{B}}(0) = 2$ while $g_{\mathbf{B} \mathbf{A}}(0) = 1$ , and $\mathbf{A} \mathbf{B} = \mathbf{O}$ is similar only to $\mathbf{O}$ , so $\mathbf{A} \mathbf{B} \not\sim \mathbf{B} \mathbf{A}$ . For rectangular matrices even the algebraic multiplicities of $0$ differ, by $|m - n|$ .

Quick check. Let $\mathbf{A} \in M_{2 \times 5}(\mathbb{R})$ and $\mathbf{B} \in M_{5 \times 2}(\mathbb{R})$ with $p_{\mathbf{A} \mathbf{B}} = x^2 - 3x + 2$ . What is $p_{\mathbf{B} \mathbf{A}}$ , and what are $\operatorname{tr}(\mathbf{B} \mathbf{A})$ and $\det(\mathbf{B} \mathbf{A})$ ?

Solution. By Theorem 9.10.4 with $m = 2$ , $n = 5$ , $x^5p_{\mathbf{A} \mathbf{B}} = x^2p_{\mathbf{B} \mathbf{A}}$ , so $p_{\mathbf{B} \mathbf{A}} = x^3(x^2 - 3x + 2) = x^5 - 3x^4 + 2x^3$ . By Theorem 7.10.2, $\operatorname{tr}(\mathbf{B} \mathbf{A}) = 3$ (equal to $\operatorname{tr}(\mathbf{A} \mathbf{B})$ ), and $\det(\mathbf{B} \mathbf{A}) = (-1)^5 \cdot 0 = 0$ , as it must be, since $\mathbf{B} \mathbf{A}$ has rank at most $2 < 5$ .

9.10.4. Left eigenvectors

An eigenvector multiplies $\mathbf{A}$ from the right: $\mathbf{A}\mathbf{v} = \lambda\mathbf{v}$ . Row vectors can multiply from the left, and in applications, especially Markov chains, the row version is often the one with a meaning.

A left eigenvector is a row that $\mathbf{A}$ only stretches when it acts from the right of the row.

Definition 9.10.1 (Left Eigenvector)

Let $\mathbf{A} \in M_n(F)$ and $\lambda \in F$ . A left eigenvector of $\mathbf{A}$ for $\lambda$ is a nonzero $\mathbf{y} \in F^n$ with $\mathbf{y}^{\top} \mathbf{A} = \lambda\,\mathbf{y}^{\top} .$ To distinguish them, the eigenvectors of Definition 9.2.1 are also called right eigenvectors.

In words: $\mathbf{y}^{\top}$ is a $1 \times n$ row, $\mathbf{y}^{\top} \mathbf{A}$ is again a row, and we ask that it be a multiple of $\mathbf{y}^{\top}$ . For example, $\mathbf{A} = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}$ and $\mathbf{y} = (1, 1)$ give $\mathbf{y}^{\top} \mathbf{A} = \begin{pmatrix} 4 & 4 \end{pmatrix} = 4\mathbf{y}^{\top}$ , so $(1, 1)$ is a left eigenvector for $4$ , whereas the right eigenvectors for $4$ are the multiples of $(2, 3)$ (Example (Eigenvalues and Eigenspaces of a $2 \times 2$ Matrix)). For a diagonal matrix, $\mathbf{e}_i$ is both a left and a right eigenvector for the $i$ -th diagonal entry. And $\mathbf{y} = \mathbf{0}$ always satisfies the equation, which is why it is excluded, exactly as for right eigenvectors.

Nonexample by minimal change. For the same $\mathbf{A}$ , change $\mathbf{y} = (1, 1)$ to $\mathbf{y} = (1, -1)$ . Then $\mathbf{y}^{\top} \mathbf{A} = \begin{pmatrix} 1 - 3 & 2 - 2 \end{pmatrix} = \begin{pmatrix} -2 & 0 \end{pmatrix}$ , which is not a multiple of $\begin{pmatrix} 1 & -1 \end{pmatrix}$ : the row is nonzero, but the defining equation $\mathbf{y}^{\top} \mathbf{A} = \lambda\mathbf{y}^{\top}$ has no solution $\lambda$ . Note that $(1, -1)$ is a right eigenvector of $\mathbf{A}$ , for $-1$ ; the two notions do not agree.

Transposing turns the definition into one we know.

Proposition 9.10.5 (Left Eigenvectors Are Eigenvectors of the Transpose)

Let $\mathbf{A} \in M_n(F)$ and $\lambda \in F$ .

(a) $\mathbf{y}$ is a left eigenvector of $\mathbf{A}$ for $\lambda$ if and only if $\mathbf{y}$ is an eigenvector of $\mathbf{A}^{\top}$ for $\lambda$ .
(b) $\mathbf{A}$ has a left eigenvector for $\lambda$ if and only if $\lambda \in \operatorname{spec}(\mathbf{A})$ . Moreover $p_{\mathbf{A}^{\top}} = p_{\mathbf{A}}$ , and $a_{\mathbf{A}^{\top}}(\lambda) = a_{\mathbf{A}}(\lambda)$ , $g_{\mathbf{A}^{\top}}(\lambda) = g_{\mathbf{A}}(\lambda)$ .

Proof.

(a) By Theorem 0.9.5, $(\mathbf{y}^{\top} \mathbf{A})^{\top} = \mathbf{A}^{\top}\mathbf{y}$ and $(\lambda\mathbf{y}^{\top})^{\top} = \lambda\mathbf{y}$ . Since transposing twice gives back the original, $\mathbf{y}^{\top} \mathbf{A} = \lambda\mathbf{y}^{\top}$ if and only if $\mathbf{A}^{\top}\mathbf{y} = \lambda\mathbf{y}$ .
(b) First, $p_{\mathbf{A}^{\top}} = p_{\mathbf{A}}$ : by Theorem 0.9.5, $(x\mathbf{I} - \mathbf{A})^{\top} = x\mathbf{I} - \mathbf{A}^{\top}$ , and transposing does not change a determinant (Theorem 7.4.5, which holds over the commutative ring $F[x]$ by the remark after it), so $p_{\mathbf{A}^{\top}} = \det(x\mathbf{I} - \mathbf{A}^{\top}) = \det\big((x\mathbf{I} - \mathbf{A})^{\top}\big) = \det(x\mathbf{I} - \mathbf{A}) = p_{\mathbf{A}} .$ Hence $a_{\mathbf{A}^{\top}}(\lambda) = a_{\mathbf{A}}(\lambda)$ , and $\operatorname{spec}(\mathbf{A}^{\top}) = \operatorname{spec}(\mathbf{A})$ by Theorem 9.2.1. With (a), $\mathbf{A}$ has a left eigenvector for $\lambda$ if and only if $\lambda \in \operatorname{spec}(\mathbf{A}^{\top}) = \operatorname{spec}(\mathbf{A})$ . Finally, $\mathbf{A}^{\top} - \lambda \mathbf{I} = (\mathbf{A} - \lambda \mathbf{I})^{\top}$ , which has the same rank as $\mathbf{A} - \lambda \mathbf{I}$ (Theorem 3.5.5), so by Theorem 3.5.7, $g_{\mathbf{A}^{\top}}(\lambda) = n - \operatorname{rank}(\mathbf{A} - \lambda \mathbf{I})^{\top} = n - \operatorname{rank}(\mathbf{A} - \lambda \mathbf{I}) = g_{\mathbf{A}}(\lambda)$ .

Remark. For an operator $T$ on $V$ , left eigenvectors correspond to eigenvectors of the dual map $T'$ on $V^{*}$ (Chapter 5), whose matrix in the dual bases is the transpose (Theorem 5.4.2): an eigenvector of $T'$ is a functional $\varphi \ne 0$ with $\varphi \circ T = \lambda\varphi$ .

Left and right eigenvectors for the same eigenvalue are usually unrelated, as $(1, 1)$ and $(2, 3)$ show. For different eigenvalues, though, they are always orthogonal in the sense of the plain product $\mathbf{y}^{\top}\mathbf{v}$ .

Theorem 9.10.6 (Left and Right Eigenvectors Are Biorthogonal)

Let $\mathbf{A} \in M_n(F)$ , let $\mathbf{y}$ be a left eigenvector of $\mathbf{A}$ for $\mu$ , and let $\mathbf{v}$ be a right eigenvector of $\mathbf{A}$ for $\lambda$ . If $\lambda \ne \mu$ , then $\mathbf{y}^{\top}\mathbf{v} = 0$ .

Proof. Compute the $1 \times 1$ matrix $\mathbf{y}^{\top} \mathbf{A}\mathbf{v}$ in two ways, using associativity (Theorem 0.9.4): $\mu\,\mathbf{y}^{\top}\mathbf{v} = (\mathbf{y}^{\top} \mathbf{A})\mathbf{v} = \mathbf{y}^{\top} \mathbf{A}\mathbf{v} = \mathbf{y}^{\top}(\mathbf{A}\mathbf{v}) = \lambda\,\mathbf{y}^{\top}\mathbf{v} .$ Hence $(\lambda - \mu)\,\mathbf{y}^{\top}\mathbf{v} = 0$ . Since $\lambda - \mu \ne 0$ , this forces $\mathbf{y}^{\top}\mathbf{v} = 0$ (Theorem 0.7.1).

For the example, $(1, 1)$ is a left eigenvector for $4$ and $(1, -1)$ a right eigenvector for $-1$ , and indeed $1 - 1 = 0$ .

For a diagonalizable matrix the left eigenvectors come for free with the diagonalization, and they give a formula for all powers that we use in Section 11.

Proposition 9.10.7 (Expansion in Left and Right Eigenvectors)

Let $\mathbf{A} = \mathbf{P} \mathbf{D} \mathbf{P}^{-1}$ with $\mathbf{P} \in M_n(F)$ invertible and $\mathbf{D} = \operatorname{diag}(\lambda_1, \dots, \lambda_n)$ . Let $\mathbf{v}_1, \dots, \mathbf{v}_n$ be the columns of $\mathbf{P}$ , and let $\mathbf{y}_1^{\top}, \dots, \mathbf{y}_n^{\top}$ be the rows of $\mathbf{P}^{-1}$ . Then:

(a) $\mathbf{v}_i$ is a right eigenvector and $\mathbf{y}_i$ a left eigenvector of $\mathbf{A}$ for $\lambda_i$ ;
(b) $\mathbf{y}_i^{\top}\mathbf{v}_j = 1$ if $i = j$ and $0$ if $i \ne j$ ;
(c) for every $q \in F[x]$ , $q(\mathbf{A}) = \sum_{i=1}^{n} q(\lambda_i)\,\mathbf{v}_i\mathbf{y}_i^{\top}, \qquad\text{in particular}\qquad \mathbf{A}^k = \sum_{i=1}^{n} \lambda_i^{\,k}\,\mathbf{v}_i\mathbf{y}_i^{\top} \quad (k \in \mathbb{N}).$

Proof.

(a) From $\mathbf{A} \mathbf{P} = \mathbf{P} \mathbf{D}$ , column $i$ reads $\mathbf{A}\mathbf{v}_i = \lambda_i\mathbf{v}_i$ (Section 4), and $\mathbf{v}_i \ne \mathbf{0}$ because $\mathbf{P}$ is invertible. From $\mathbf{P}^{-1}\mathbf{A} = \mathbf{D} \mathbf{P}^{-1}$ , row $i$ reads $\mathbf{y}_i^{\top} \mathbf{A} = \lambda_i\mathbf{y}_i^{\top}$ (Theorem 0.9.2), and $\mathbf{y}_i \ne \mathbf{0}$ because $\mathbf{P}^{-1}$ is invertible.
(b) The $(i, j)$ -entry of $\mathbf{P}^{-1}\mathbf{P} = \mathbf{I}$ is row $i$ of $\mathbf{P}^{-1}$ times column $j$ of $\mathbf{P}$ , that is, $\mathbf{y}_i^{\top}\mathbf{v}_j$ .
(c) By Theorem 9.4.3, $q(\mathbf{A}) = \mathbf{P}\,q(\mathbf{D})\,\mathbf{P}^{-1}$ . The columns of $\mathbf{P}\,q(\mathbf{D})$ are $q(\lambda_i)\mathbf{v}_i$ , and the rows of $\mathbf{P}^{-1}$ are $\mathbf{y}_i^{\top}$ , so Corollary 8.1.3 gives $q(\mathbf{A}) = \sum_i q(\lambda_i)\mathbf{v}_i\mathbf{y}_i^{\top}$ . Take $q = x^k$ .

Part (b) is biorthogonality with a normalization: $\mathbf{y}_i^{\top}\mathbf{v}_i = 1$ . Part (c) splits $\mathbf{A}$ into rank-one pieces $\mathbf{v}_i\mathbf{y}_i^{\top}$ , one per eigenvalue, and each piece is simply multiplied by $\lambda_i^k$ when $\mathbf{A}$ is raised to the power $k$ . In the running example, $\mathbf{P} = \begin{pmatrix} 2 & 1 \\ 3 & -1 \end{pmatrix}$ and $\mathbf{D} = \operatorname{diag}(4, -1)$ give $\mathbf{P}^{-1} = \frac15\begin{pmatrix} 1 & 1 \\ 3 & -2 \end{pmatrix}$ , so $\mathbf{y}_1 = \frac15(1, 1)$ and $\mathbf{y}_2 = \frac15(3, -2)$ , and $\mathbf{A}^k = \frac{4^k}{5}\begin{pmatrix} 2 & 2 \\ 3 & 3 \end{pmatrix} + \frac{(-1)^k}{5}\begin{pmatrix} 3 & -2 \\ -3 & 2 \end{pmatrix}.$ For $k = 1$ this is $\frac15\begin{pmatrix} 8 - 3 & 8 + 2 \\ 12 + 3 & 12 - 2 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}$ , as it should be.

9.10.5. Exercises

A. Check your understanding

Exercise (A1)

(a) State the Spectral Mapping Theorem, including its hypothesis on $p_T$ .
(b) True or false: for every $\mathbf{A} \in M_n(\mathbb{R})$ , every real eigenvalue of $\mathbf{A}^2$ is the square of a real eigenvalue of $\mathbf{A}$ . Justify your answer.
(c) True or false: if $\mathbf{A} \in M_{2 \times 3}(F)$ and $\mathbf{B} \in M_{3 \times 2}(F)$ , then $p_{\mathbf{A} \mathbf{B}} = p_{\mathbf{B} \mathbf{A}}$ . Justify your answer.
(d) True or false: for all $\mathbf{A}, \mathbf{B} \in M_n(F)$ , $\mathbf{A} \mathbf{B}$ is similar to $\mathbf{B} \mathbf{A}$ . Justify your answer.
(e) Define a left eigenvector of $\mathbf{A} \in M_n(F)$ , and say how left eigenvectors of $\mathbf{A}$ are related to $\mathbf{A}^{\top}$ .
(f) True or false: a matrix $\mathbf{A} \in M_3(\mathbb{C})$ with $\operatorname{tr} \mathbf{A} = \operatorname{tr} \mathbf{A}^2 = \operatorname{tr} \mathbf{A}^3 = 0$ satisfies $\mathbf{A}^3 = \mathbf{O}$ . Justify your answer.

Solution.

(a) If $V$ is finite-dimensional with $\dim V = n \ge 1$ , $T \in \mathcal{L}(V)$ , and $p_T = \prod_{i=1}^{n}(x - \lambda_i)$ splits over $F$ , then for every $q \in F[x]$ , $p_{q(T)} = \prod_i (x - q(\lambda_i))$ and $\operatorname{spec}(q(T)) = q(\operatorname{spec}(T))$ (Theorem 9.10.1).
(b) False. The rotation $\mathbf{R} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ has no real eigenvalue, while $\mathbf{R}^2 = -\mathbf{I}$ has the real eigenvalue $-1$ .
(c) False. They have different degrees, $2$ and $3$ . The correct identity is $x^3p_{\mathbf{A} \mathbf{B}} = x^2p_{\mathbf{B} \mathbf{A}}$ , that is, $p_{\mathbf{B} \mathbf{A}} = x\,p_{\mathbf{A} \mathbf{B}}$ (Theorem 9.10.4).
(d) False. $\mathbf{A} = \mathbf{E}_{12}$ , $\mathbf{B} = \mathbf{E}_{11}$ give $\mathbf{A} \mathbf{B} = \mathbf{O}$ and $\mathbf{B} \mathbf{A} = \mathbf{E}_{12} \ne \mathbf{O}$ , and the only matrix similar to $\mathbf{O}$ is $\mathbf{O}$ .
(e) A nonzero $\mathbf{y} \in F^n$ with $\mathbf{y}^{\top} \mathbf{A} = \lambda\mathbf{y}^{\top}$ for some $\lambda \in F$ . Equivalently, $\mathbf{y}$ is an eigenvector of $\mathbf{A}^{\top}$ for $\lambda$ (Proposition 9.10.5).
(f) True. By Theorem 9.10.3 with $n = 3$ , (e) $\Rightarrow$ (d).

B. Practice

Exercise (B1: Eigenvalues of a polynomial in

\mathbf{A}

)

Let $\mathbf{A} \in M_3(\mathbb{C})$ have $p_{\mathbf{A}} = (x - 1)^2(x + 2)$ .

(a) Find $p_{\mathbf{A}^2 + \mathbf{A} + \mathbf{I}}$ , $\operatorname{tr}(\mathbf{A}^2 + \mathbf{A} + \mathbf{I})$ , and $\det(\mathbf{A}^2 + \mathbf{A} + \mathbf{I})$ . Is $\mathbf{A}^2 + \mathbf{A} + \mathbf{I}$ invertible?
(b) Find $p_{\mathbf{A}^2 + \mathbf{A} - 2\mathbf{I}}$ . Hence show that $\mathbf{A}^2 + \mathbf{A} - 2\mathbf{I}$ is nilpotent.
(c) Find $\operatorname{tr} \mathbf{A}^{5}$ and $\det(\mathbf{A}^{-1})$ .

Solution. The list of eigenvalues of $\mathbf{A}$ with multiplicity is $1, 1, -2$ , since $p_{\mathbf{A}}$ splits.

(a) With $q = x^2 + x + 1$ , $q(1) = 3$ and $q(-2) = 4 - 2 + 1 = 3$ . By Theorem 9.10.1, $p_{q(\mathbf{A})} = (x - 3)^3$ , and by Corollary 9.10.2, $\operatorname{tr} q(\mathbf{A}) = 9$ and $\det q(\mathbf{A}) = 27 \ne 0$ . So $\mathbf{A}^2 + \mathbf{A} + \mathbf{I}$ is invertible (Theorem 7.5.3).
(b) With $r = x^2 + x - 2 = (x - 1)(x + 2)$ , $r(1) = r(-2) = 0$ , so $p_{r(\mathbf{A})} = x^3$ by Theorem 9.10.1. By Theorem 9.10.3 ((c) $\Rightarrow$ (d)), $r(\mathbf{A})^3 = \mathbf{O}$ , so $\mathbf{A}^2 + \mathbf{A} - 2\mathbf{I}$ is nilpotent.
(c) By Corollary 9.10.2, $\operatorname{tr} \mathbf{A}^5 = 1 + 1 + (-2)^5 = -30$ . Also $\det \mathbf{A} = 1 \cdot 1 \cdot (-2) = -2 \ne 0$ (Theorem 9.3.3), so $\mathbf{A}$ is invertible and $\det(\mathbf{A}^{-1}) = (\det \mathbf{A})^{-1} = -\frac12$ (Corollary 7.5.6).

Exercise (B2:

\mathbf{A} \mathbf{B}

and

\mathbf{B} \mathbf{A}

for a

2 \times 3

example)

Let $\mathbf{A} = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \end{pmatrix}$ and $\mathbf{B} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ 1 & 1 \end{pmatrix}$ .

(a) Compute $\mathbf{A} \mathbf{B}$ , $\mathbf{B} \mathbf{A}$ , $p_{\mathbf{A} \mathbf{B}}$ , and $p_{\mathbf{B} \mathbf{A}}$ , and verify that $x^3p_{\mathbf{A} \mathbf{B}} = x^2p_{\mathbf{B} \mathbf{A}}$ .
(b) Hence list the eigenvalues of $\mathbf{B} \mathbf{A}$ without further computation, and find an eigenvector of $\mathbf{B} \mathbf{A}$ for each nonzero eigenvalue by applying $\mathbf{B}$ to eigenvectors of $\mathbf{A} \mathbf{B}$ .

Solution.

(a) By row-times-column, $\mathbf{A} \mathbf{B} = \begin{pmatrix} 2 & 1 \\ 0 & -2 \end{pmatrix}, \qquad \mathbf{B} \mathbf{A} = \begin{pmatrix} 1 & 0 & 1 \\ 0 & -2 & 0 \\ 1 & 2 & 1 \end{pmatrix}.$ $\mathbf{A} \mathbf{B}$ is upper triangular, so $p_{\mathbf{A} \mathbf{B}} = (x - 2)(x + 2) = x^2 - 4$ . Expanding $\det(x\mathbf{I} - \mathbf{B} \mathbf{A})$ along the second row, whose only nonzero entry is $x + 2$ in position $(2, 2)$ , $\begin{aligned} p_{\mathbf{B} \mathbf{A}} &= (x + 2)\det\begin{pmatrix} x - 1 & -1 \\ -1 & x - 1 \end{pmatrix} \\ &= (x + 2)\big((x - 1)^2 - 1\big) = (x + 2)\,x\,(x - 2) = x^3 - 4x . \end{aligned}$ Then $x^3p_{\mathbf{A} \mathbf{B}} = x^5 - 4x^3 = x^2(x^3 - 4x) = x^2p_{\mathbf{B} \mathbf{A}}$ .
(b) The eigenvalues of $\mathbf{B} \mathbf{A}$ are the roots of $x(x - 2)(x + 2)$ : $0, 2, -2$ , matching Theorem 9.10.4 (the nonzero ones are those of $\mathbf{A} \mathbf{B}$ ). If $\mathbf{A} \mathbf{B}\mathbf{w} = \lambda\mathbf{w}$ , then $\mathbf{B} \mathbf{A}(\mathbf{B}\mathbf{w}) = \mathbf{B}(\mathbf{A} \mathbf{B}\mathbf{w}) = \lambda \mathbf{B}\mathbf{w}$ , so $\mathbf{B}\mathbf{w}$ is an eigenvector of $\mathbf{B} \mathbf{A}$ as long as $\mathbf{B}\mathbf{w} \ne \mathbf{0}$ . For $\lambda = 2$ : $\mathbf{w} = (1, 0)$ , $\mathbf{B}\mathbf{w} = (1, 0, 1)$ , and $\mathbf{B} \mathbf{A}(1, 0, 1) = (2, 0, 2)$ . For $\lambda = -2$ : $\mathbf{A} \mathbf{B} + 2\mathbf{I} = \begin{pmatrix} 4 & 1 \\ 0 & 0 \end{pmatrix}$ gives $\mathbf{w} = (1, -4)$ , $\mathbf{B}\mathbf{w} = (1, 4, -3)$ , and $\mathbf{B} \mathbf{A}(1, 4, -3) = (1 - 3, -8, 1 + 8 - 3) = (-2, -8, 6) = -2(1, 4, -3)$ .

Exercise (B3: Left eigenvectors of a

2 \times 2

matrix)

Let $\mathbf{A} = \begin{pmatrix} 0 & 1 \\ -6 & 5 \end{pmatrix} \in M_2(\mathbb{R})$ , whose right eigenvectors $(1, 2)$ for $2$ and $(1, 3)$ for $3$ were found in Exercise (B1: Diagonalize or show impossible).

(a) Find a left eigenvector of $\mathbf{A}$ for each eigenvalue.
(b) Verify Theorem 9.10.6 for these vectors, and rescale the left eigenvectors so that $\mathbf{y}_i^{\top}\mathbf{v}_i = 1$ .
(c) Hence write $\mathbf{A}^k$ as a combination of two rank-one matrices, and check the formula for $k = 0$ and $k = 1$ .

Solution.

(a) By Proposition 9.10.5, we need eigenvectors of $\mathbf{A}^{\top} = \begin{pmatrix} 0 & -6 \\ 1 & 5 \end{pmatrix}$ . For $2$ : $\mathbf{A}^{\top} - 2\mathbf{I} = \begin{pmatrix} -2 & -6 \\ 1 & 3 \end{pmatrix}$ gives $y_1 = -3y_2$ , so $(3, -1)$ . For $3$ : $\mathbf{A}^{\top} - 3\mathbf{I} = \begin{pmatrix} -3 & -6 \\ 1 & 2 \end{pmatrix}$ gives $y_1 = -2y_2$ , so $(2, -1)$ . Check: $\begin{pmatrix} 3 & -1 \end{pmatrix}\mathbf{A} = \begin{pmatrix} 6 & -2 \end{pmatrix}$ and $\begin{pmatrix} 2 & -1 \end{pmatrix}\mathbf{A} = \begin{pmatrix} 6 & -3 \end{pmatrix}$ .
(b) $(3, -1)\cdot(1, 3) = 0$ and $(2, -1)\cdot(1, 2) = 0$ , as the theorem predicts for different eigenvalues. Also $(3, -1)\cdot(1, 2) = 1$ and $(2, -1)\cdot(1, 3) = -1$ , so take $\mathbf{y}_1 = (3, -1)$ and $\mathbf{y}_2 = (-2, 1)$ .
(c) With $\mathbf{P} = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}$ , the rows of $\mathbf{P}^{-1} = \begin{pmatrix} 3 & -1 \\ -2 & 1 \end{pmatrix}$ are exactly $\mathbf{y}_1^{\top}, \mathbf{y}_2^{\top}$ , and by Proposition 9.10.7, $\mathbf{A}^k = 2^k\begin{pmatrix} 1 \\ 2 \end{pmatrix}\begin{pmatrix} 3 & -1 \end{pmatrix} + 3^k\begin{pmatrix} 1 \\ 3 \end{pmatrix}\begin{pmatrix} -2 & 1 \end{pmatrix} = 2^k\begin{pmatrix} 3 & -1 \\ 6 & -2 \end{pmatrix} + 3^k\begin{pmatrix} -2 & 1 \\ -6 & 3 \end{pmatrix}.$ For $k = 0$ : $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ . For $k = 1$ : $\begin{pmatrix} 6 - 6 & -2 + 3 \\ 12 - 18 & -4 + 9 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -6 & 5 \end{pmatrix} = \mathbf{A}$ .

C. Going deeper

Exercise (C1: Traces of powers of

\mathbf{A} \mathbf{B}

and

\mathbf{B} \mathbf{A}

)

Let $\mathbf{A} \in M_{m \times n}(F)$ and $\mathbf{B} \in M_{n \times m}(F)$ .

(a) Prove that $\mathbf{B}(\mathbf{A} \mathbf{B})^{k-1}\mathbf{A} = (\mathbf{B} \mathbf{A})^k$ for every $k \ge 1$ , and deduce that $\operatorname{tr}\big((\mathbf{A} \mathbf{B})^k\big) = \operatorname{tr}\big((\mathbf{B} \mathbf{A})^k\big)$ .
(b) Now let $F = \mathbb{C}$ . Give a second proof of the trace identity from Theorem 9.10.4 and Corollary 9.10.2.

Solution.

(a) By induction on $k$ . For $k = 1$ both sides are $\mathbf{B} \mathbf{A}$ . If $\mathbf{B}(\mathbf{A} \mathbf{B})^{k-1}\mathbf{A} = (\mathbf{B} \mathbf{A})^k$ , then by associativity $\begin{aligned} \mathbf{B}(\mathbf{A} \mathbf{B})^k\mathbf{A} &= \mathbf{B}(\mathbf{A} \mathbf{B})^{k-1}(\mathbf{A} \mathbf{B})\mathbf{A} = \big(\mathbf{B}(\mathbf{A} \mathbf{B})^{k-1}\mathbf{A}\big)(\mathbf{B} \mathbf{A}) \\ &= (\mathbf{B} \mathbf{A})^k(\mathbf{B} \mathbf{A}) = (\mathbf{B} \mathbf{A})^{k+1}. \end{aligned}$ Now $(\mathbf{A} \mathbf{B})^k = \mathbf{A}\,\big(\mathbf{B}(\mathbf{A} \mathbf{B})^{k-1}\big)$ , with $\mathbf{A}$ of size $m \times n$ and $\mathbf{B}(\mathbf{A} \mathbf{B})^{k-1}$ of size $n \times m$ . By Theorem 0.9.6 (3), $\operatorname{tr}\big((\mathbf{A} \mathbf{B})^k\big) = \operatorname{tr}\big(\mathbf{A} \cdot \mathbf{B}(\mathbf{A} \mathbf{B})^{k-1}\big) = \operatorname{tr}\big(\mathbf{B}(\mathbf{A} \mathbf{B})^{k-1} \cdot \mathbf{A}\big) = \operatorname{tr}\big((\mathbf{B} \mathbf{A})^k\big).$
(b) Over $\mathbb{C}$ , $p_{\mathbf{A} \mathbf{B}}$ and $p_{\mathbf{B} \mathbf{A}}$ split. Suppose $m \ge n$ (otherwise swap the roles of $\mathbf{A}$ and $\mathbf{B}$ ). By Theorem 9.10.4, $p_{\mathbf{A} \mathbf{B}} = x^{m-n}p_{\mathbf{B} \mathbf{A}}$ , so if $\lambda_1, \dots, \lambda_n$ is the list of eigenvalues of $\mathbf{B} \mathbf{A}$ with multiplicity, then the list for $\mathbf{A} \mathbf{B}$ is $\lambda_1, \dots, \lambda_n$ together with $m - n$ zeros. For $k \ge 1$ the zeros contribute $0^k = 0$ , so by Corollary 9.10.2, $\operatorname{tr}\big((\mathbf{A} \mathbf{B})^k\big) = \sum_i \lambda_i^k = \operatorname{tr}\big((\mathbf{B} \mathbf{A})^k\big)$ .

Exercise (C2: Adding a commuting nilpotent matrix)

Let $\mathbf{A}, \mathbf{N} \in M_n(\mathbb{C})$ , $n \ge 1$ , with $\mathbf{A} \mathbf{N} = \mathbf{N} \mathbf{A}$ and $\mathbf{N}$ nilpotent.

(a) Prove that $p_{\mathbf{A} + \mathbf{N}} = p_{\mathbf{A}}$ ; in particular $\operatorname{spec}(\mathbf{A} + \mathbf{N}) = \operatorname{spec}(\mathbf{A})$ and $\det(\mathbf{A} + \mathbf{N}) = \det \mathbf{A}$ .
(b) Deduce that if $\mathbf{A}$ is invertible, so is $\mathbf{A} + \mathbf{N}$ .
(c) Show that (a) fails without $\mathbf{A} \mathbf{N} = \mathbf{N} \mathbf{A}$ .

Hint: triangularize $\mathbf{A}$ and $\mathbf{N}$ together.

Solution.

(a) By Theorem 9.9.4 there is an invertible $\mathbf{P}$ with $\mathbf{U} = \mathbf{P}^{-1}\mathbf{A} \mathbf{P}$ and $\mathbf{U}' = \mathbf{P}^{-1}\mathbf{N} \mathbf{P}$ upper triangular. Since $\mathbf{N}$ is nilpotent, $\operatorname{spec}(\mathbf{N}) = \{0\}$ (Exercise (C2: Nilpotent operators)), so every diagonal entry of $\mathbf{U}'$ is $0$ by Theorem 9.7.4. Then $\mathbf{P}^{-1}(\mathbf{A} + \mathbf{N})\mathbf{P} = \mathbf{U} + \mathbf{U}'$ is upper triangular with the same diagonal as $\mathbf{U}$ . By Theorem 7.10.3 and the examples after Definition 7.10.1, $p_{\mathbf{A} + \mathbf{N}} = p_{\mathbf{U} + \mathbf{U}'} = \prod_i (x - u_{ii}) = p_{\mathbf{U}} = p_{\mathbf{A}} .$ Equal characteristic polynomials have the same roots (Theorem 9.2.1) and the same constant term $(-1)^n\det$ (Theorem 7.10.2).
(b) By (a), $\det(\mathbf{A} + \mathbf{N}) = \det \mathbf{A} \ne 0$ , so $\mathbf{A} + \mathbf{N}$ is invertible (Theorem 7.5.3).
(c) Let $\mathbf{A} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ and $\mathbf{N} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ . Both are nilpotent, so $p_{\mathbf{A}} = x^2$ , but $\mathbf{A} + \mathbf{N} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ has $p = x^2 - 1$ and eigenvalues $\pm 1$ . Here $\mathbf{A} \mathbf{N} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \ne \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = \mathbf{N} \mathbf{A}$ .

Exercise (C3: How many traces, and which field?)

(a) Let $\omega = e^{2\pi i/3} \in \mathbb{C}$ and $\mathbf{C} = \operatorname{diag}(1, \omega, \omega^2)$ . Show that $\operatorname{tr} \mathbf{C} = \operatorname{tr} \mathbf{C}^2 = 0$ but $\mathbf{C}$ is not nilpotent. Explain why this does not contradict Theorem 9.10.3.
(b) Let $n \ge 2$ . Give a matrix $\mathbf{C} \in M_n(\mathbb{C})$ with $\operatorname{tr} \mathbf{C}^k = 0$ for $k = 1, \dots, n - 1$ that is not nilpotent.
(c) Let $\mathbf{A}, \mathbf{B} \in M_n(\mathbb{C})$ and suppose $\mathbf{C} = \mathbf{A} \mathbf{B} - \mathbf{B} \mathbf{A}$ commutes with $\mathbf{A}$ . Prove that $\mathbf{C}^n = \mathbf{O}$ .
(d) Over $\mathbb{F}_2$ , find a nonnilpotent $\mathbf{C} \in M_2(\mathbb{F}_2)$ with $\operatorname{tr} \mathbf{C}^k = 0$ for every $k \ge 1$ .

Solution.

(a) $\omega^3 = 1$ and $\omega \ne 1$ , so $0 = \omega^3 - 1 = (\omega - 1)(\omega^2 + \omega + 1)$ gives $1 + \omega + \omega^2 = 0$ . Hence $\operatorname{tr} \mathbf{C} = 1 + \omega + \omega^2 = 0$ , and $\operatorname{tr} \mathbf{C}^2 = 1 + \omega^2 + \omega^4 = 1 + \omega^2 + \omega = 0$ . But $\mathbf{C}^k = \operatorname{diag}(1, \omega^k, \omega^{2k}) \ne \mathbf{O}$ for every $k$ . No contradiction: the theorem, for $n = 3$ , requires $\operatorname{tr} \mathbf{C}^k = 0$ for $k = 1, 2, 3$ , and $\operatorname{tr} \mathbf{C}^3 = 1 + 1 + 1 = 3 \ne 0$ .
(b) Let $\zeta = e^{2\pi i/n}$ and $\mathbf{C} = \operatorname{diag}(1, \zeta, \dots, \zeta^{n-1})$ . For $1 \le k \le n - 1$ , $\zeta^k \ne 1$ and $(\zeta^k)^n = 1$ , so $\operatorname{tr} \mathbf{C}^k = \sum_{j=0}^{n-1} (\zeta^k)^j = \frac{(\zeta^k)^n - 1}{\zeta^k - 1} = 0$ by the geometric sum. But $\mathbf{C}^k$ has the entry $1$ in position $(1, 1)$ for all $k$ , so $\mathbf{C}$ is not nilpotent.
(c) By Exercise (C2: When $\mathbf{A}$ commutes with $[\mathbf{A}, \mathbf{B}]$ ) (a), $\operatorname{tr} \mathbf{C}^k = 0$ for every $k \ge 1$ , in particular for $k = 1, \dots, n$ . By Theorem 9.10.3 ((e) $\Rightarrow$ (d)), $\mathbf{C}^n = \mathbf{O}$ .
(d) $\mathbf{C} = \mathbf{I}_2$ : $\operatorname{tr} \mathbf{C}^k = 1 + 1 = 0$ in $\mathbb{F}_2$ for every $k$ , and $\mathbf{C}^k = \mathbf{I}_2 \ne \mathbf{O}$ .

9.10.1. Eigenvalues of a polynomial in T

9.10.2. Traces of powers detect nilpotency

9.10.3. AB and BA

9.10.4. Left eigenvectors

9.10.5. Exercises

A. Check your understanding

B. Practice

C. Going deeper

9.10.1. Eigenvalues of a polynomial in $T$

9.10.3. $\mathbf{A} \mathbf{B}$ and $\mathbf{B} \mathbf{A}$