7.6 Minimal polynomials, Cayley–Hamilton, and commuting operators

By Proposition 4.1.31, every operator on an $𝑛$ -dimensional space is a root of some nonzero polynomial of degree at most $𝑛^{2}$ . Polynomial identities of this kind have already done a lot of work. A polynomial that kills $𝑇$ confines the eigenvalues of $𝑇$ to its roots (Proposition 7.2.28), and a diagonalizable operator is killed by the product of the factors $𝑥 - 𝜆$ over its eigenvalues (Proposition 7.4.26). This section studies the best such polynomial, the one of least degree. Its roots are exactly the eigenvalues, and whether it has a repeated root decides whether $𝑇$ is diagonalizable, a test that needs no eigenvectors at all. The Cayley–Hamilton theorem says that the characteristic polynomial is always among the polynomials that kill $𝑇$ , which brings the degree bound down from $𝑛^{2}$ to $𝑛$ ; we prove it four ways. The last two topics also run on polynomials in an operator: commuting operators, which over $ℂ$ can be triangularized by a single basis (and, if each is diagonalizable, diagonalized by one over any field), and Gershgorin’s theorem, which locates eigenvalues without computing them.

Throughout, $𝑉$ is a finite-dimensional vector space over a field $𝐹$ with $dim 𝑉 = 𝑛 \geq 1$ , and $𝑇 \in L (𝑉)$ .

7.6.1Annihilating polynomials and the minimal polynomial

Definition 7.6.1 (Annihilating polynomial). A polynomial

𝑝 \in 𝐹 [𝑥]

annihilates

𝑇

𝑝 (𝑇) = 0

. Similarly,

𝑝

annihilates

𝐴 \in 𝐹^{𝑛 \times 𝑛}

𝑝 (𝐴) = 0

The zero polynomial annihilates every operator, so the question is which nonzero polynomials do. We have met several. A projection ( $𝑃^{2} = 𝑃$ ) is annihilated by $𝑥^{2} - 𝑥$ , an operator with $𝑁^{𝑘} = 0$ by $𝑥^{𝑘}$ , the cyclic permutation $𝑍$ of Example 7.2.36 by $𝑥^{3} - 1$ , and a diagonalizable operator with distinct eigenvalues $𝜆_{1}, \dots, 𝜆_{𝑚}$ by $(𝑥 - 𝜆_{1}) \dots (𝑥 - 𝜆_{𝑚})$ . Annihilating polynomials never come alone: if $𝑝 (𝑇) = 0$ , then $(𝑞 𝑝) (𝑇) =$ $𝑞 (𝑇) 𝑝 (𝑇) =$ $0$ for every $𝑞 \in 𝐹 [𝑥]$ (Proposition 7.2.27(a)). So the natural candidate for “the” polynomial of $𝑇$ is one of least degree.

Theorem 7.6.2 (Existence and uniqueness of the minimal polynomial). There is exactly one monic polynomial $𝑚 \in 𝐹 [𝑥]$ such that $𝑚 (𝑇) = 0$ and $deg 𝑚 \leq deg 𝑝$ for every nonzero $𝑝 \in 𝐹 [𝑥]$ with $𝑝 (𝑇) = 0$ . Moreover, for every $𝑝 \in 𝐹 [𝑥]$ ,

𝑝 (𝑇) = 0 ⟺ 𝑚 ∣ 𝑝 .

Proof.standard

Uses. Proposition 4.1.31 (Every square matrix satisfies a polynomial), Proposition 7.2.27 (Evaluating polynomials at an operator), Theorem 7.1.14 (Division with remainder), Definition 7.1.5 (Monic polynomial), Definition 7.1.16 (Divisibility)

Techniques. Extremal choice and minimal counterexamples (Technique 1.5.9), Uniqueness via two candidates (Technique 1.1.8), Compare degrees and leading coefficients (Technique 7.1.6)

Hint 1. Among the nonzero annihilating polynomials, take one of smallest degree. Why is there one, and how do you make it monic?

Hint 2. For the divisibility statement, divide

𝑝

𝑚

with remainder and evaluate the identity at

𝑇

. The remainder annihilates

𝑇

too. What does the choice of

𝑚

say about it?

Hint 3. For uniqueness, if

𝑚

and

𝑚^{'}

both qualify, they have the same degree. Look at

𝑚 - 𝑚^{'}

Sketch. Nonzero annihilating polynomials exist by Proposition 4.1.31(b); take one of least degree

𝑑

and divide it by its leading coefficient. If

𝑝 (𝑇) = 0

, write

𝑝 = 𝑞 𝑚 + 𝑟

with

𝑟 = 𝟎

deg 𝑟 < 𝑑

; then

𝑟 (𝑇) =

𝑝 (𝑇) -

𝑞 (𝑇) 𝑚 (𝑇) =

0

, so

𝑟 = 𝟎

by minimality. Multiples of

𝑚

annihilate

𝑇

. A second candidate

𝑚^{'}

also has degree

𝑑

, and

𝑚 - 𝑚^{'}

annihilates

𝑇

with degree less than

𝑑

(the leading terms cancel), so

𝑚 = 𝑚^{'}

Existence. By Proposition 4.1.31(b), some nonzero polynomial annihilates $𝑇$ . So the degrees of the nonzero annihilating polynomials form a nonempty set of nonnegative integers, which has a least element $𝑑$ . Let $𝑝_{0}$ be a nonzero annihilating polynomial of degree $𝑑$ , with leading coefficient $𝑎 \neq 0$ , and put $𝑚 = 𝑎^{- 1} 𝑝_{0}$ . Then $𝑚$ is monic of degree $𝑑$ , and $𝑚 (𝑇) =$ $𝑎^{- 1} 𝑝_{0} (𝑇) =$ $0$ by Proposition 7.2.27(a). By the choice of $𝑑$ , $deg 𝑚 \leq deg 𝑝$ for every nonzero $𝑝$ with $𝑝 (𝑇) = 0$ .

Divisibility. If $𝑚 ∣ 𝑝$ , say $𝑝 = 𝑞 𝑚$ , then $𝑝 (𝑇) =$ $𝑞 (𝑇) 𝑚 (𝑇) =$ $0$ by Proposition 7.2.27(a). Conversely, suppose $𝑝 (𝑇) = 0$ . By Theorem 7.1.14 there are $𝑞, 𝑟 \in 𝐹 [𝑥]$ with $𝑝 = 𝑞 𝑚 + 𝑟$ , where $𝑟 = 𝟎$ or $deg 𝑟 < 𝑑$ . Then

𝑟 (𝑇) = 𝑝 (𝑇) - 𝑞 (𝑇) 𝑚 (𝑇) = 0 - 0 = 0 .

If $𝑟$ were nonzero, it would be a nonzero annihilating polynomial of degree less than $𝑑$ , contradicting the choice of $𝑑$ . So $𝑟 = 𝟎$ , and $𝑚 ∣ 𝑝$ .

Uniqueness. Suppose $𝑚^{'}$ is also monic, $𝑚^{'} (𝑇) = 0$ , and $deg 𝑚^{'} \leq deg 𝑝$ for every nonzero annihilating $𝑝$ . Applying this to $𝑝 = 𝑚$ , and the same property of $𝑚$ to $𝑝 = 𝑚^{'}$ , gives $deg 𝑚^{'} = deg 𝑚 = 𝑑$ . Both are monic of degree $𝑑$ , so the coefficients of $𝑥^{𝑑}$ cancel in $𝑚 - 𝑚^{'}$ , which is therefore $𝟎$ or of degree less than $𝑑$ . It annihilates $𝑇$ , so $𝑚 - 𝑚^{'} = 𝟎$ by the minimality of $𝑑$ . □

Definition 7.6.3 (Minimal polynomial). The polynomial

𝑚

of Theorem 7.6.2 is the minimal polynomial of

𝑇

, written

𝑚_{𝑇}

. For

𝐴 \in 𝐹^{𝑛 \times 𝑛}

, the minimal polynomial

𝑚_{𝐴}

𝐴

is the minimal polynomial of

𝑇_{𝐴} \in L (𝐹^{𝑛})

. Since

𝑝 (𝑇_{𝐴}) = 𝑇_{𝑝 (𝐴)}

(Proposition 7.2.27(c)),

𝑚_{𝐴}

is the monic polynomial of least degree with

𝑚_{𝐴} (𝐴) = 0

So $𝑚_{𝑇}$ is minimal in two senses: no nonzero annihilating polynomial has smaller degree, and every annihilating polynomial is a multiple of it. Since $𝑉 \neq {𝟎}$ , a nonzero constant $𝑐$ gives $𝑐 {id}_{𝑉} \neq 0$ , so $deg 𝑚_{𝑇} \geq 1$ . Moreover $deg 𝑚_{𝑇} = 1$ exactly when $𝑇 - 𝜆 {id}_{𝑉} = 0$ for some $𝜆 \in 𝐹$ , that is, when $𝑇 = 𝜆 {id}_{𝑉}$ is a scalar multiple of the identity; then $𝑚_{𝑇} = 𝑥 - 𝜆$ .

Aside (Ideals). The set

𝐼_{𝑇} =

{𝑝 \in 𝐹 [𝑥] | 𝑝 (𝑇) = 0}

is closed under addition and under multiplication by arbitrary polynomials. Such a subset of a commutative ring is called an ideal. The proof of Theorem 7.6.2 used nothing about

𝐼_{𝑇}

except these two closure properties and

𝐼_{𝑇} \neq {𝟎}

, so it shows that every nonzero ideal of

𝐹 [𝑥]

consists of the multiples of one monic polynomial. This makes

𝐹 [𝑥]

a principal ideal domain, the setting of Section 14.4. The greatest common divisor (Theorem 7.1.45) is the same phenomenon for the ideal

{𝑢 𝑝 + 𝑣 𝑠 | 𝑢, 𝑣 \in 𝐹 [𝑥]}

The least degree can be found by looking for the first linear dependence among the powers of $𝑇$ , exactly as the Krylov polynomial of a vector was found among $𝐯, 𝑇 𝐯, 𝑇^{2} 𝐯, \dots$ in Proposition 7.2.35.

Proposition 7.6.4 (The minimal polynomial is the first dependence among the powers). Let $𝑑 = deg 𝑚_{𝑇}$ . Then ${id}_{𝑉}, 𝑇, \dots, 𝑇^{𝑑 - 1}$ are linearly independent in $L (𝑉)$ , and $𝑑$ is the smallest integer $𝑘 \geq 1$ with $𝑇^{𝑘} \in$ $span ({id}_{𝑉}, 𝑇, \dots, 𝑇^{𝑘 - 1})$ . If

𝑇^{𝑑} = - (𝑐_{0} {id}_{𝑉} + 𝑐_{1} 𝑇 + \dots + 𝑐_{𝑑 - 1} 𝑇^{𝑑 - 1}),

then $𝑚_{𝑇} =$ $𝑥^{𝑑} +$ $𝑐_{𝑑 - 1} 𝑥^{𝑑 - 1} +$ $\dots +$ $𝑐_{1} 𝑥 +$ $𝑐_{0}$ . Moreover, ${id}_{𝑉}, 𝑇, \dots, 𝑇^{𝑑 - 1}$ is a basis of the subspace $𝐹 [𝑇] =$ ${𝑝 (𝑇) | 𝑝 \in 𝐹 [𝑥]}$ of $L (𝑉)$ , so $dim 𝐹 [𝑇] = deg 𝑚_{𝑇}$ .

Proof.routine

Uses. Theorem 7.6.2 (Existence and uniqueness of the minimal polynomial), Theorem 7.1.14 (Division with remainder), Theorem 1.5.5 (Independence is decided at zero), Definition 7.2.26 (Polynomials in an operator)

Techniques. Too many vectors force a dependence (Technique 4.1.30)

Hint 1. A linear relation

𝑏_{0} {id}_{𝑉} +

𝑏_{1} 𝑇 +

\dots +

𝑏_{𝑘 - 1} 𝑇^{𝑘 - 1} =

0

says that the polynomial

𝑏_{0} +

𝑏_{1} 𝑥 +

\dots +

𝑏_{𝑘 - 1} 𝑥^{𝑘 - 1}

annihilates

𝑇

Hint 2. For the basis of

𝐹 [𝑇]

, divide an arbitrary

𝑝

𝑚_{𝑇}

with remainder.

Sketch. A nontrivial relation among

{id}_{𝑉}, \dots, 𝑇^{𝑑 - 1}

would be a nonzero annihilating polynomial of degree less than

𝑑

. The relation

𝑚_{𝑇} (𝑇) = 0

expresses

𝑇^{𝑑}

through lower powers, and since those are independent, the coefficients are unique, so they are those of

𝑚_{𝑇}

. If

𝑇^{𝑘}

were a combination of lower powers for some

𝑘 < 𝑑

, there would be a monic annihilating polynomial of degree

𝑘 < 𝑑

. Finally,

𝑝 = 𝑞 𝑚_{𝑇} + 𝑟

gives

𝑝 (𝑇) =

𝑟 (𝑇) \in

span ({id}_{𝑉}, \dots, 𝑇^{𝑑 - 1})

For scalars $𝑏_{0}, \dots, 𝑏_{𝑘 - 1}$ , the relation $\sum_{𝑖 < 𝑘} 𝑏_{𝑖} 𝑇^{𝑖} =$ $0$ says exactly that $𝑞 =$ $\sum_{𝑖 < 𝑘} 𝑏_{𝑖} 𝑥^{𝑖}$ annihilates $𝑇$ (Definition 7.2.26). If ${id}_{𝑉}, 𝑇, \dots, 𝑇^{𝑑 - 1}$ satisfied a nontrivial relation, $𝑞$ would be a nonzero annihilating polynomial of degree less than $𝑑$ , which Theorem 7.6.2 rules out. So the list is independent.

Writing $𝑚_{𝑇} =$ $𝑥^{𝑑} +$ $𝑎_{𝑑 - 1} 𝑥^{𝑑 - 1} +$ $\dots +$ $𝑎_{0}$ , the equation $𝑚_{𝑇} (𝑇) = 0$ says $𝑇^{𝑑} =$ $- (𝑎_{0} {id}_{𝑉} + \dots + 𝑎_{𝑑 - 1} 𝑇^{𝑑 - 1})$ . So $𝑇^{𝑑}$ lies in the span of the lower powers, and since these are independent, its coefficients are unique (Theorem 1.5.5). Hence $𝑐_{𝑖} = 𝑎_{𝑖}$ for every $𝑖$ . If $𝑇^{𝑘} =$ $\sum_{𝑖 < 𝑘} 𝑏_{𝑖} 𝑇^{𝑖}$ for some $1 \leq 𝑘 < 𝑑$ , then $𝑥^{𝑘} -$ $\sum_{𝑖 < 𝑘} 𝑏_{𝑖} 𝑥^{𝑖}$ would be a monic annihilating polynomial of degree $𝑘 < 𝑑$ , which is impossible. So $𝑑$ is the smallest such $𝑘$ .

Finally, let $𝑝 \in 𝐹 [𝑥]$ . By Theorem 7.1.14, $𝑝 = 𝑞 𝑚_{𝑇} + 𝑟$ with $𝑟 = 𝟎$ or $deg 𝑟 < 𝑑$ , so $𝑝 (𝑇) =$ $𝑞 (𝑇) 𝑚_{𝑇} (𝑇) +$ $𝑟 (𝑇) =$ $𝑟 (𝑇)$ , a combination of ${id}_{𝑉}, \dots, 𝑇^{𝑑 - 1}$ . Thus this independent list spans $𝐹 [𝑇]$ , and it is a basis. □

In the language of Proposition 7.2.35: the Krylov sequence of the “vector” ${id}_{𝑉} \in L (𝑉)$ under the operator $𝑆 \mapsto 𝑇 𝑆$ on $L (𝑉)$ is ${id}_{𝑉}, 𝑇, 𝑇^{2}, \dots$ , and its Krylov polynomial is $𝑚_{𝑇}$ .

Example 7.6.5 (First minimal polynomials).

(a)Projections. Let $𝑃 \in L (𝑉)$ with $𝑃^{2} = 𝑃$ and $𝑃 \neq 0, {id}_{𝑉}$ . Then $𝑥^{2} - 𝑥$ annihilates $𝑃$ . If $𝑃 = 𝑐 {id}_{𝑉}$ , then $𝑐^{2} = 𝑐$ , so $𝑐 \in {0, 1}$ ; this is excluded, so no polynomial of degree $1$ annihilates $𝑃$ , and $𝑚_{𝑃} =$ $𝑥^{2} -$ $𝑥 =$ $𝑥 (𝑥 - 1)$ .
(b)The shift. Let $𝑁 \in L (𝐹^{𝑛})$ with $𝑁 𝐞_{1} = 𝟎$ and $𝑁 𝐞_{𝑗} = 𝐞_{𝑗 - 1}$ for $𝑗 \geq 2$ (Example 7.2.20). Then $𝑁^{𝑛} = 0$ . For scalars $𝑏_{0}, \dots, 𝑏_{𝑛 - 1}$ ,
$(𝑏_{0} id + 𝑏_{1} 𝑁 + \dots + 𝑏_{𝑛 - 1} 𝑁^{𝑛 - 1}) 𝐞_{𝑛} = 𝑏_{0} 𝐞_{𝑛} + 𝑏_{1} 𝐞_{𝑛 - 1} + \dots + 𝑏_{𝑛 - 1} 𝐞_{1},$
which is $𝟎$ only if every $𝑏_{𝑖} = 0$ . So $id, 𝑁, \dots, 𝑁^{𝑛 - 1}$ are independent, and Proposition 7.6.4 gives $deg 𝑚_{𝑁} \geq 𝑛$ . Since $𝑥^{𝑛}$ annihilates $𝑁$ , $𝑚_{𝑁} ∣ 𝑥^{𝑛}$ , so $deg 𝑚_{𝑁} = 𝑛$ , and a monic divisor of $𝑥^{𝑛}$ of degree $𝑛$ is $𝑥^{𝑛}$ itself. Hence $𝑚_{𝑁} = 𝑥^{𝑛}$ .
(c)A rotation. Let $𝑅 = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ . Then $𝑅^{2} = - 𝐼_{2}$ , so $𝑥^{2} + 1$ annihilates $𝑅$ , and $𝑅$ is not a scalar matrix, so $𝑚_{𝑅} = 𝑥^{2} + 1$ , over $ℝ$ and over $ℂ$ alike. Over $ℝ$ this has no roots, and $𝑅$ has no eigenvalues; over $ℂ$ its roots are $\pm 𝑖$ , which are the eigenvalues of $𝑅$ (Non-example 7.5.6).
(d)The cyclic permutation. Let $𝑍 (𝑥, 𝑦, 𝑧) =$ $(𝑧, 𝑥, 𝑦)$ on $𝐹^{3}$ , so $𝑍^{3} = id$ . The matrices of $id$ , $𝑍$ , $𝑍^{2}$ have their $1$ ′s in disjoint positions,
$𝐼_{3} = (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}), [𝑍] = (\begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}), {[𝑍]}^{2} = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}),$
so they are independent, and $𝑚_{𝑍} = 𝑥^{3} - 1$ over every field.

Exercise 7.6.6.routine Let $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ and $𝑐 \in 𝐹$ .

(a)Show that $𝑚_{𝐴^{𝖳}} = 𝑚_{𝐴}$ .
(b)Show that $𝑚_{𝐴 + 𝑐 𝐼_{𝑛}} (𝑥) =$ $𝑚_{𝐴} (𝑥 - 𝑐)$ . Deduce the minimal polynomial of $𝜆 𝐼_{𝑛} + 𝑁$ , where $𝑁$ is the shift of Example 7.6.5(b).

Solution.

Uses. Theorem 7.6.2 (Existence and uniqueness of the minimal polynomial), Proposition 2.4.23 (Transpose of a product), Remark 7.1.11 (Polynomials in a matrix), Definition 7.1.9 (Substitution)

Hint 1. Show that

𝐴

and

𝐴^{𝖳}

have exactly the same annihilating polynomials. What does that say about their minimal polynomials?

Hint 2. For (b), check on

𝑝 = 𝑥^{𝑘}

that the polynomial

𝑞 (𝑥) = 𝑝 (𝑥 - 𝑐)

satisfies

𝑞 (𝐴 + 𝑐 𝐼_{𝑛}) =

𝑝 (𝐴)

, then extend by linearity.

Sketch. (a)

𝑝 (𝐴^{𝖳}) = {𝑝 (𝐴)}^{𝖳}

, so the annihilating polynomials of

𝐴

and

𝐴^{𝖳}

coincide, and so do their monic generators. (b) Substituting

𝑥 - 𝑐

is a bijection of

𝐹 [𝑥]

preserving degrees and monicity, and

𝑞 (𝐴 + 𝑐 𝐼_{𝑛}) =

𝑝 (𝐴)

for

𝑞 (𝑥) = 𝑝 (𝑥 - 𝑐)

; so it maps the annihilating polynomials of

𝐴

onto those of

𝐴 + 𝑐 𝐼_{𝑛}

. Hence

𝑚_{𝜆 𝐼_{𝑛} + 𝑁} =

{(𝑥 - 𝜆)}^{𝑛}

(a) By Proposition 2.4.23, ${(𝐴^{𝑘})}^{𝖳} = {(𝐴^{𝖳})}^{𝑘}$ for every $𝑘 \geq 0$ , and transposition respects sums and scalar multiples, so $𝑝 (𝐴^{𝖳}) = {𝑝 (𝐴)}^{𝖳}$ for every $𝑝 \in 𝐹 [𝑥]$ . A matrix is $0$ if and only if its transpose is, so $𝑝$ annihilates $𝐴^{𝖳}$ if and only if it annihilates $𝐴$ . By Theorem 7.6.2, $𝑚_{𝐴}$ divides every polynomial that annihilates $𝐴^{𝖳}$ , in particular $𝑚_{𝐴^{𝖳}}$ , and likewise $𝑚_{𝐴^{𝖳}} ∣ 𝑚_{𝐴}$ . Two monic polynomials that divide each other have the same degree, so the quotient is a constant, which is $1$ by comparing leading coefficients. Hence $𝑚_{𝐴^{𝖳}} = 𝑚_{𝐴}$ .

(b) For $𝑝 \in 𝐹 [𝑥]$ let $𝑞 (𝑥) = 𝑝 (𝑥 - 𝑐)$ , the result of substituting $𝑥 - 𝑐$ into $𝑝$ (Definition 7.1.9). Both $𝑝 \mapsto 𝑞 (𝐴 + 𝑐 𝐼_{𝑛})$ and $𝑝 \mapsto 𝑝 (𝐴)$ respect sums and scalar multiples, and for $𝑝 = 𝑥^{𝑘}$ , using Remark 7.1.11,

𝑞 (𝐴 + 𝑐 𝐼_{𝑛}) = ({(𝑥 - 𝑐)}^{𝑘}) (𝐴 + 𝑐 𝐼_{𝑛}) = {((𝐴 + 𝑐 𝐼_{𝑛}) - 𝑐 𝐼_{𝑛})}^{𝑘} = 𝐴^{𝑘} .

So $𝑞 (𝐴 + 𝑐 𝐼_{𝑛}) =$ $𝑝 (𝐴)$ for every $𝑝$ . Substituting $𝑥 - 𝑐$ is a bijection of $𝐹 [𝑥]$ (its inverse substitutes $𝑥 + 𝑐$ ) that preserves degrees and leading coefficients. Hence it maps the monic annihilating polynomials of $𝐴$ bijectively onto those of $𝐴 + 𝑐 𝐼_{𝑛}$ , preserving degrees, and the one of least degree goes to the one of least degree: $𝑚_{𝐴 + 𝑐 𝐼_{𝑛}} (𝑥) =$ $𝑚_{𝐴} (𝑥 - 𝑐)$ .

For $𝑁$ , Example 7.6.5(b) gives $𝑚_{𝑁} = 𝑥^{𝑛}$ , so $𝑚_{𝜆 𝐼_{𝑛} + 𝑁} =$ ${(𝑥 - 𝜆)}^{𝑛}$ .

7.6.2Roots of the minimal polynomial

In each part of Example 7.6.5, the roots of $𝑚_{𝑇}$ in $𝐹$ were exactly the eigenvalues of $𝑇$ : $0$ and $1$ for a projection, $0$ for the shift, none for the real rotation, $\pm 𝑖$ for the complex one, and the cube roots of $1$ in $𝐹$ for $𝑍$ . This always happens.

Theorem 7.6.7 (The roots of the minimal polynomial are the eigenvalues). Let

𝜆 \in 𝐹

. Then

𝜆

is an eigenvalue of

𝑇

if and only if

𝑚_{𝑇} (𝜆) = 0

One direction is Proposition 7.2.28. For the other, peel the factor $𝑥 - 𝜆$ off $𝑚_{𝑇}$ . What is left has smaller degree, so it does not annihilate $𝑇$ ; it produces some nonzero vector, and $𝑇 - 𝜆 id$ kills that vector.

Proof.routine

Uses. Proposition 7.2.28 (Polynomials act on eigenvectors by their values), Theorem 7.1.20 (Factor theorem), Theorem 7.6.2 (Existence and uniqueness of the minimal polynomial), Proposition 7.2.27 (Evaluating polynomials at an operator), Definition 7.2.6 (Eigenvalue, eigenvector)

Techniques. Extremal choice and minimal counterexamples (Technique 1.5.9)

Hint 1. The direction “eigenvalue

\Rightarrow

root” was proved in Section 7.2. Where?

Hint 2. If

𝑚_{𝑇} (𝜆) = 0

, write

𝑚_{𝑇} = (𝑥 - 𝜆) 𝑞

. Why is

𝑞 (𝑇) \neq 0

Hint 3. Choose

𝐯

with

𝑞 (𝑇) 𝐯 \neq 𝟎

, and apply

𝑇 - 𝜆 {id}_{𝑉}

𝑞 (𝑇) 𝐯

Sketch. If

𝜆

is an eigenvalue, then

𝑚_{𝑇} (𝜆) = 0

by Proposition 7.2.28, as

𝑚_{𝑇} (𝑇) = 0

. Conversely, if

𝑚_{𝑇} (𝜆) = 0

, the factor theorem gives

𝑚_{𝑇} = (𝑥 - 𝜆) 𝑞

with

𝑞

monic of degree

deg 𝑚_{𝑇} - 1

, so

𝑞 (𝑇) \neq 0

by minimality. For

𝐯

with

𝐰 = 𝑞 (𝑇) 𝐯 \neq 𝟎

, we get

(𝑇 - 𝜆 {id}_{𝑉}) 𝐰 =

𝑚_{𝑇} (𝑇) 𝐯 =

𝟎

, so

𝐰

is an eigenvector for

𝜆

If $𝜆$ is an eigenvalue of $𝑇$ , then $𝑚_{𝑇} (𝜆) = 0$ by Proposition 7.2.28, because $𝑚_{𝑇} (𝑇) = 0$ .

Conversely, suppose $𝑚_{𝑇} (𝜆) = 0$ . By the factor theorem (Theorem 7.1.20), $𝑚_{𝑇} = (𝑥 - 𝜆) 𝑞$ for some $𝑞 \in 𝐹 [𝑥]$ of degree $deg 𝑚_{𝑇} - 1$ , and $𝑞$ is monic because $𝑚_{𝑇}$ is. Since $𝑞 \neq 𝟎$ has smaller degree than $𝑚_{𝑇}$ , Theorem 7.6.2 gives $𝑞 (𝑇) \neq 0$ . Choose $𝐯 \in 𝑉$ with $𝐰 = 𝑞 (𝑇) 𝐯 \neq 𝟎$ . By Proposition 7.2.27(a),

(𝑇 - 𝜆 {id}_{𝑉}) 𝐰 = (𝑇 - 𝜆 {id}_{𝑉}) 𝑞 (𝑇) 𝐯 = 𝑚_{𝑇} (𝑇) 𝐯 = 𝟎 .

So $𝑇 𝐰 = 𝜆 𝐰$ with $𝐰 \neq 𝟎$ , and $𝜆$ is an eigenvalue of $𝑇$ (Definition 7.2.6). □

So $𝑚_{𝑇}$ and the characteristic polynomial $𝑝_{𝑇}$ have the same roots in $𝐹$ , namely the eigenvalues (Corollary 7.3.13). They can differ in the multiplicities of those roots: $𝑚_{{id}_{𝑉}} = 𝑥 - 1$ , while $𝑝_{{id}_{𝑉}} = {(𝑥 - 1)}^{𝑛}$ .

Corollary 7.6.8 (The shape of the minimal polynomial over $ℂ$ ). Suppose $𝐹 = ℂ$ , and let $𝜆_{1}, \dots, 𝜆_{𝑚}$ be the distinct eigenvalues of $𝑇$ . Then

𝑚_{𝑇} = {(𝑥 - 𝜆_{1})}^{𝑘_{1}} {(𝑥 - 𝜆_{2})}^{𝑘_{2}} \dots {(𝑥 - 𝜆_{𝑚})}^{𝑘_{𝑚}}

for some integers $𝑘_{1}, \dots, 𝑘_{𝑚} \geq 1$ .

Proof.routine

Uses. Corollary 7.1.35 (Factorization over $ℂ$ ), Theorem 7.6.7 (The roots of the minimal polynomial are the eigenvalues)

Hint. Factor the monic polynomial

𝑚_{𝑇}

completely over

ℂ

. Which numbers can appear in the factors, and which must?

Sketch. By Corollary 7.1.35,

𝑚_{𝑇}

is a product of factors

𝑥 - 𝜇

over its roots

𝜇

. By Theorem 7.6.7, the roots are exactly

𝜆_{1}, \dots, 𝜆_{𝑚}

, so each occurs at least once; group equal factors.

Since

𝑚_{𝑇}

is monic of degree at least

1

, Corollary 7.1.35 writes it as

(𝑥 - 𝜇_{1}) \dots (𝑥 - 𝜇_{𝑑})

, where

𝜇_{1}, \dots, 𝜇_{𝑑}

are its roots listed with multiplicity. By Theorem 7.6.7, the set of roots is

{𝜆_{1}, \dots, 𝜆_{𝑚}}

. Grouping equal factors gives the stated form, with

𝑘_{𝑖} \geq 1

the number of

𝑗

with

𝜇_{𝑗} = 𝜆_{𝑖}

. □

Two quick consequences show what the minimal polynomial is good for.

Corollary 7.6.9 (Inverses are polynomials). $𝑇$ is invertible if and only if $𝑚_{𝑇} (0) \neq 0$ . In that case, write $𝑚_{𝑇} = 𝑥 𝑠 + 𝑐_{0}$ with $𝑐_{0} = 𝑚_{𝑇} (0)$ and $𝑠 \in 𝐹 [𝑥]$ . Then

𝑇^{- 1} = - 𝑐_{0}^{- 1} 𝑠 (𝑇),

a polynomial in $𝑇$ of degree $deg 𝑚_{𝑇} - 1$ .

Proof.routine

Uses. Theorem 7.2.9 (Eigenvalues and invertibility), Theorem 7.6.7 (The roots of the minimal polynomial are the eigenvalues), Theorem 7.1.20 (Factor theorem), Proposition 7.2.27 (Evaluating polynomials at an operator)

Hint 1.

𝑇

is invertible if and only if

0

is not an eigenvalue. Use Theorem 7.6.7.

Hint 2. Evaluate

𝑚_{𝑇} = 𝑥 𝑠 + 𝑐_{0}

𝑇

and solve for

{id}_{𝑉}

Sketch. Since

𝑉

is finite-dimensional,

𝑇

is invertible exactly when

0

is not an eigenvalue, that is, when

𝑚_{𝑇} (0) \neq 0

. Then

0 =

𝑚_{𝑇} (𝑇) =

𝑇 𝑠 (𝑇) +

𝑐_{0} {id}_{𝑉}

gives

𝑇 (- 𝑐_{0}^{- 1} 𝑠 (𝑇)) =

{id}_{𝑉}

, and

𝑠 (𝑇)

commutes with

𝑇

By Theorem 7.2.9 (with $𝜆 = 0$ , using $dim 𝑉 < \infty$ ), $𝑇$ is invertible if and only if $0$ is not an eigenvalue of $𝑇$ , which by Theorem 7.6.7 happens if and only if $𝑚_{𝑇} (0) \neq 0$ .

Suppose $𝑐_{0} = 𝑚_{𝑇} (0) \neq 0$ . By Theorem 7.1.20 with $𝑐 = 0$ , $𝑚_{𝑇} = 𝑥 𝑠 + 𝑐_{0}$ with $deg 𝑠 = deg 𝑚_{𝑇} - 1$ . Evaluating at $𝑇$ (Proposition 7.2.27(a)),

0 = 𝑚_{𝑇} (𝑇) = 𝑇 𝑠 (𝑇) + 𝑐_{0} {id}_{𝑉}, so 𝑇 (- 𝑐_{0}^{- 1} 𝑠 (𝑇)) = {id}_{𝑉} .

By Proposition 7.2.27(b), $𝑠 (𝑇)$ commutes with $𝑇$ , so also $(- 𝑐_{0}^{- 1} 𝑠 (𝑇)) 𝑇 =$ ${id}_{𝑉}$ , and $- 𝑐_{0}^{- 1} 𝑠 (𝑇)$ is the inverse of $𝑇$ . □

For example, $𝐴 = (\begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix})$ has $𝐴^{2} =$ $(\begin{matrix} 5 & 3 \\ 3 & 2 \end{matrix}) =$ $3 𝐴 -$ $𝐼_{2}$ , and $𝐴$ is not a scalar matrix, so $𝑚_{𝐴} = 𝑥^{2} - 3 𝑥 + 1$ . Here $𝑚_{𝐴} (0) = 1$ and $𝑚_{𝐴} = 𝑥 (𝑥 - 3) + 1$ , so $𝐴^{- 1} =$ $- (𝐴 - 3 𝐼_{2}) =$ $(\begin{matrix} 1 & - 1 \\ - 1 & 2 \end{matrix})$ , as one checks by multiplying.

Proposition 7.6.10 (Invariant subspaces and similarity).

(a)If $𝑈 \neq {𝟎}$ is a $𝑇$ -invariant subspace of $𝑉$ , then $𝑚_{𝑇 ↾_{𝑈}}$ divides $𝑚_{𝑇}$ , where $𝑇 ↾_{𝑈}$ is regarded as an operator on $𝑈$ .
(b)If $𝑉 = 𝑈 \oplus 𝑊$ with $𝑈$ and $𝑊$ both $𝑇$ -invariant and nonzero, then for every $𝑝 \in 𝐹 [𝑥]$ ,
$𝑝 (𝑇) = 0 ⟺ 𝑝 (𝑇 ↾_{𝑈}) = 0 and 𝑝 (𝑇 ↾_{𝑊}) = 0 .$
(c)For every basis $ℬ︀$ of $𝑉$ , $𝑚_{{[𝑇]}_{ℬ︀ ℬ︀}} = 𝑚_{𝑇}$ . In particular, similar matrices have the same minimal polynomial.

Proof.routine

Uses. Theorem 7.6.2 (Existence and uniqueness of the minimal polynomial), Proposition 7.2.27 (Evaluating polynomials at an operator), Definition 4.2.1 (Restriction), Proposition 2.4.12 (Matrices of sums and scalar multiples), Theorem 4.5.25 (Similar matrices are one operator in two bases)

Techniques. Restrict to a subspace (Technique 4.2.5)

Hint 1. First show that

𝑝 (𝑇)

restricted to an invariant

𝑈

𝑝 (𝑇 ↾_{𝑈})

. Start with powers.

Hint 2. For (c), use

{[𝑝 (𝑇)]}_{ℬ︀ ℬ︀} =

𝑝 ({[𝑇]}_{ℬ︀ ℬ︀})

and the injectivity of

𝑆 \mapsto {[𝑆]}_{ℬ︀ ℬ︀}

Sketch. For invariant

𝑈

𝑇^{𝑘} ↾_{𝑈} = {(𝑇 ↾_{𝑈})}^{𝑘}

by induction, so

𝑝 (𝑇) ↾_{𝑈} =

𝑝 (𝑇 ↾_{𝑈})

. Then (a):

𝑚_{𝑇} (𝑇 ↾_{𝑈}) =

𝑚_{𝑇} (𝑇) ↾_{𝑈} =

0

, so

𝑚_{𝑇 ↾_{𝑈}} ∣ 𝑚_{𝑇}

. (b): write

𝐯 = 𝐮 + 𝐰

. (c):

𝑝 (𝑇) = 0

if and only if

𝑝 ({[𝑇]}_{ℬ︀ ℬ︀}) =

{[𝑝 (𝑇)]}_{ℬ︀ ℬ︀} =

0

, so

𝑇

and

{[𝑇]}_{ℬ︀ ℬ︀}

have the same annihilating polynomials; similar matrices are matrices of one operator.

Let $𝑈$ be $𝑇$ -invariant. For $𝐮 \in 𝑈$ , induction on $𝑘$ gives $𝑇^{𝑘} 𝐮 =$ ${(𝑇 ↾_{𝑈})}^{𝑘} 𝐮$ : both sides are $𝐮$ for $𝑘 = 0$ , and $𝑇^{𝑘 + 1} 𝐮 =$ $𝑇 {(𝑇 ↾_{𝑈})}^{𝑘} 𝐮 =$ ${(𝑇 ↾_{𝑈})}^{𝑘 + 1} 𝐮$ because ${(𝑇 ↾_{𝑈})}^{𝑘} 𝐮 \in 𝑈$ . Taking linear combinations, $𝑝 (𝑇) 𝐮 =$ $𝑝 (𝑇 ↾_{𝑈}) 𝐮$ for every $𝑝 \in 𝐹 [𝑥]$ and $𝐮 \in 𝑈$ .

(a) By the above, $𝑚_{𝑇} (𝑇 ↾_{𝑈}) 𝐮 =$ $𝑚_{𝑇} (𝑇) 𝐮 =$ $𝟎$ for every $𝐮 \in 𝑈$ . So $𝑚_{𝑇}$ annihilates $𝑇 ↾_{𝑈}$ , and $𝑚_{𝑇 ↾_{𝑈}} ∣ 𝑚_{𝑇}$ by Theorem 7.6.2.

(b) If $𝑝 (𝑇) = 0$ , then $𝑝 (𝑇 ↾_{𝑈}) 𝐮 =$ $𝑝 (𝑇) 𝐮 =$ $𝟎$ for all $𝐮 \in 𝑈$ , and likewise on $𝑊$ . Conversely, if both restrictions are annihilated, write $𝐯 \in 𝑉$ as $𝐮 + 𝐰$ with $𝐮 \in 𝑈$ , $𝐰 \in 𝑊$ . Then $𝑝 (𝑇) 𝐯 =$ $𝑝 (𝑇 ↾_{𝑈}) 𝐮 +$ $𝑝 (𝑇 ↾_{𝑊}) 𝐰 =$ $𝟎$ .

(c) By Proposition 7.2.27(c), ${[𝑝 (𝑇)]}_{ℬ︀ ℬ︀} =$ $𝑝 ({[𝑇]}_{ℬ︀ ℬ︀})$ , and $𝑆 \mapsto {[𝑆]}_{ℬ︀ ℬ︀}$ is injective (Proposition 2.4.12). So $𝑝 (𝑇) = 0$ if and only if $𝑝 ({[𝑇]}_{ℬ︀ ℬ︀}) = 0$ . Thus $𝑇$ and ${[𝑇]}_{ℬ︀ ℬ︀}$ have the same annihilating polynomials, hence the same minimal polynomial by the uniqueness in Theorem 7.6.2. If $𝐴 \sim 𝐵$ , then $𝐴$ and $𝐵$ are matrices of a single operator $𝑇$ (Theorem 4.5.25), so $𝑚_{𝐴} = 𝑚_{𝑇} = 𝑚_{𝐵}$ . □

Remark 7.6.11 (Neither polynomial determines the other). The matrices

𝐼_{2}

and

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

have the same characteristic polynomial

{(𝑥 - 1)}^{2}

. But

𝑚_{𝐼_{2}} = 𝑥 - 1

, while

𝐽 - 𝐼_{2} \neq 0

and

{(𝐽 - 𝐼_{2})}^{2} = 0

, so

𝑚_{𝐽} = {(𝑥 - 1)}^{2}

. In the other direction,

diag (2, 2, 3)

and

diag (2, 3, 3)

have the same minimal polynomial

(𝑥 - 2) (𝑥 - 3)

but different characteristic polynomials. Even the two together do not determine a matrix up to similarity (Exercise 7.6.43); the Jordan form of Section 10.3 does.

7.6.3Diagonalizability

Proposition 7.4.26 shows that a diagonalizable operator is killed by a product of distinct linear factors. The converse is the most useful property of the minimal polynomial: to show that $𝑇$ is diagonalizable, it is enough to find one polynomial identity of the right shape, without looking for a single eigenvector.

Intuition (Why distinct factors should split the space). If

𝑉 = 𝐸_{𝜆_{1}} \oplus \dots \oplus 𝐸_{𝜆_{𝑚}}

, each summand is killed by its own factor

𝑇 - 𝜆_{𝑖} id

, so the product of all the factors kills

𝑉

. Running this backwards: if

(𝑇 - 𝑐_{1} id) \dots (𝑇 - 𝑐_{𝑘} id) =

0

with distinct

𝑐_{𝑖}

, we want to write each

𝐯

as a sum of vectors

𝐯_{𝑖}

with

(𝑇 - 𝑐_{𝑖} id) 𝐯_{𝑖} =

𝟎

. Polynomials in

𝑇

are the only operators we have, so we look for polynomials

𝐿_{𝑖}

with

𝐿_{1} + \dots + 𝐿_{𝑘} = 1

and

(𝑥 - 𝑐_{𝑖}) 𝐿_{𝑖}

divisible by the product

𝑞 =

(𝑥 - 𝑐_{1}) \dots (𝑥 - 𝑐_{𝑘})

. Then

𝐯 = \sum 𝐿_{𝑖} (𝑇) 𝐯

, and

𝐿_{𝑖} (𝑇) 𝐯

is killed by

𝑇 - 𝑐_{𝑖} id

. The Lagrange polynomials of the nodes

𝑐_{1}, \dots, 𝑐_{𝑘}

(Section 7.1.4) are exactly such polynomials. If a factor is repeated, as in

{(𝑥 - 𝑐)}^{2}

, no such decomposition can work, and the shift shows it does not.

Proposition 7.6.12 (Distinct linear factors force a diagonalization). Let $𝑐_{1}, \dots, 𝑐_{𝑘} \in 𝐹$ be distinct, and suppose $𝑞 =$ $(𝑥 - 𝑐_{1}) (𝑥 - 𝑐_{2}) \dots (𝑥 - 𝑐_{𝑘})$ annihilates $𝑇$ . Then

𝑉 = 𝐸_{𝑐_{1}} \oplus 𝐸_{𝑐_{2}} \oplus \dots \oplus 𝐸_{𝑐_{𝑘}},

where $𝐸_{𝑐} =$ $ker (𝑇 - 𝑐 {id}_{𝑉})$ (some of these may be ${𝟎}$ ). In particular, $𝑇$ is diagonalizable, and its eigenvalues are among $𝑐_{1}, \dots, 𝑐_{𝑘}$ .

Proof (Lagrange interpolation in $𝑇$ ).standard

Uses. Proposition 7.1.29 (Lagrange polynomials in closed form), Corollary 7.1.23 (Polynomials and their values), Proposition 7.2.27 (Evaluating polynomials at an operator), Corollary 7.2.23 (Eigenspaces are independent), Theorem 4.4.13 (Bases and dimension of a sum), Theorem 7.4.2 (Diagonalizable means an eigenbasis)

Techniques. Lagrange interpolation in $𝑇$ (Technique 7.6.13), Use an annihilating polynomial (Technique 7.6.17)

Hint 1. Let

𝐿_{1}, \dots, 𝐿_{𝑘}

be the Lagrange polynomials for the nodes

𝑐_{1}, \dots, 𝑐_{𝑘}

. What is

𝐿_{1} + \dots + 𝐿_{𝑘}

Hint 2.

(𝑥 - 𝑐_{𝑖}) 𝐿_{𝑖}

is a constant multiple of

𝑞

. What does that say about

im 𝐿_{𝑖} (𝑇)

Hint 3. Write

𝐯 = \sum_{𝑖} 𝐿_{𝑖} (𝑇) 𝐯

. Why is the resulting sum of eigenspaces direct?

Sketch. Let

𝐿_{𝑖} =

\prod_{𝑗 \neq 𝑖} (𝑥 - 𝑐_{𝑗}) / (𝑐_{𝑖} - 𝑐_{𝑗})

. Then

\sum_{𝑖} 𝐿_{𝑖} = 1

, since both sides have degree less than

𝑘

and agree at the

𝑘

nodes. Also

(𝑥 - 𝑐_{𝑖}) 𝐿_{𝑖} =

𝑎_{𝑖} 𝑞

for a constant

𝑎_{𝑖}

, so

(𝑇 - 𝑐_{𝑖} id) 𝐿_{𝑖} (𝑇) =

0

and

im 𝐿_{𝑖} (𝑇) \subseteq 𝐸_{𝑐_{𝑖}}

. So

𝐯 =

\sum 𝐿_{𝑖} (𝑇) 𝐯 \in

𝐸_{𝑐_{1}} +

\dots +

𝐸_{𝑐_{𝑘}}

, and the sum is direct by Corollary 7.2.23. Concatenating bases of the summands gives an eigenbasis.

If $𝑘 = 1$ , then $𝑇 - 𝑐_{1} {id}_{𝑉} = 0$ , so $𝑉 = 𝐸_{𝑐_{1}}$ and every basis is an eigenbasis. Let $𝑘 \geq 2$ , and let

𝐿_{𝑖} = \prod_{𝑗 \neq 𝑖} \frac{𝑥 - 𝑐_{𝑗}}{𝑐_{𝑖} - 𝑐_{𝑗}} (𝑖 = 1, \dots, 𝑘)

be the Lagrange polynomials of the nodes $𝑐_{1}, \dots, 𝑐_{𝑘}$ (Proposition 7.1.29), so $𝐿_{𝑖} (𝑐_{𝑗}) = 1$ if $𝑖 = 𝑗$ and $0$ otherwise.

First, $𝐿_{1} + \dots + 𝐿_{𝑘} = 1$ . Both sides have degree at most $𝑘 - 1$ , and at each node $𝑐_{𝑗}$ the left side takes the value $\sum_{𝑖} 𝐿_{𝑖} (𝑐_{𝑗}) = 1$ . By Corollary 7.1.23(a), they are equal.

Second, $(𝑥 - 𝑐_{𝑖}) 𝐿_{𝑖} =$ $𝑎_{𝑖} 𝑞$ with $𝑎_{𝑖} =$ $\prod_{𝑗 \neq 𝑖} {(𝑐_{𝑖} - 𝑐_{𝑗})}^{- 1}$ , since multiplying $𝐿_{𝑖}$ by $𝑥 - 𝑐_{𝑖}$ supplies the one missing factor of $𝑞$ .

Let $𝑃_{𝑖} = 𝐿_{𝑖} (𝑇)$ . Evaluating the two identities at $𝑇$ (Proposition 7.2.27(a)) gives

𝑃_{1} + \dots + 𝑃_{𝑘} = {id}_{𝑉}, (𝑇 - 𝑐_{𝑖} {id}_{𝑉}) 𝑃_{𝑖} = 𝑎_{𝑖} 𝑞 (𝑇) = 0 .

The second says $im 𝑃_{𝑖} \subseteq 𝐸_{𝑐_{𝑖}}$ . So every $𝐯 \in 𝑉$ satisfies

𝐯 = 𝑃_{1} 𝐯 + \dots + 𝑃_{𝑘} 𝐯 \in 𝐸_{𝑐_{1}} + \dots + 𝐸_{𝑐_{𝑘}},

and $𝑉 = 𝐸_{𝑐_{1}} + \dots + 𝐸_{𝑐_{𝑘}}$ . The $𝑐_{𝑖}$ are distinct, so this sum is direct (Corollary 7.2.23). Concatenating bases of the nonzero $𝐸_{𝑐_{𝑖}}$ gives a basis of $𝑉$ (Theorem 4.4.13) consisting of eigenvectors of $𝑇$ , so $𝑇$ is diagonalizable (Theorem 7.4.2). Every eigenvalue is a root of $𝑞$ by Proposition 7.2.28, so it is one of the $𝑐_{𝑖}$ . □

The operators $𝑃_{𝑖} = 𝐿_{𝑖} (𝑇)$ in this proof are worth keeping. $𝑃_{𝑖}$ maps $𝑉$ into $𝐸_{𝑐_{𝑖}}$ , and on each $𝐸_{𝑐_{𝑗}}$ it acts as the scalar $𝐿_{𝑖} (𝑐_{𝑗})$ (Proposition 7.2.28), which is $1$ for $𝑗 = 𝑖$ and $0$ otherwise. So $𝑃_{𝑖}$ is the projection of $𝑉$ onto $𝐸_{𝑐_{𝑖}}$ along the sum of the other eigenspaces, and it is a polynomial in $𝑇$ . The same argument, without the Lagrange formula, is Exercise 7.4.49.

Technique 7.6.13 (Lagrange interpolation in $𝑇$ ). If $𝑞 (𝑇) = 0$ for $𝑞 =$ $(𝑥 - 𝑐_{1}) \dots (𝑥 - 𝑐_{𝑘})$ with distinct $𝑐_{𝑖}$ , evaluate the Lagrange polynomials $𝐿_{𝑖}$ of the nodes $𝑐_{𝑖}$ at $𝑇$ . The operators $𝑃_{𝑖} = 𝐿_{𝑖} (𝑇)$ satisfy $\sum 𝑃_{𝑖} = id$ , $𝑃_{𝑖} 𝑃_{𝑗} = 0$ for $𝑖 \neq 𝑗$ (as $𝑞 ∣ 𝐿_{𝑖} 𝐿_{𝑗}$ ), and $𝑃_{𝑖}^{2} = 𝑃_{𝑖}$ , and $𝑃_{𝑖}$ is the projection onto $𝐸_{𝑐_{𝑖}}$ along the other eigenspaces. More generally, for any values $𝑑_{1}, \dots, 𝑑_{𝑘}$ , the interpolating polynomial $𝑝 = \sum 𝑑_{𝑖} 𝐿_{𝑖}$ gives an operator $𝑝 (𝑇)$ that acts as $𝑑_{𝑖}$ on $𝐸_{𝑐_{𝑖}}$ .

Reach for it to write spectral projections as polynomials in $𝑇$ , to build operators with a prescribed action on each eigenspace (square roots, inverses, operators commuting with $𝑇$ ), or to decompose a vector along the eigenspaces without finding an eigenbasis. The canonical example is the first proof of Proposition 7.6.12. Later: the spectral decomposition (Section 8.2).

Used in 2 places

Proposition 7.6.12, Exercise 7.6.45.

The second proof avoids interpolation. It peels off one eigenspace at a time, like the second proof of Theorem 7.5.13, and the only polynomial fact it uses is that $𝑞 (𝑐_{𝑘}) \neq 0$ after the factor $𝑥 - 𝑐_{𝑘}$ is removed.

Proof (induction on the number of factors).standard

Uses. Proposition 7.2.30 (Kernels and images of commuting operators are invariant), Proposition 7.2.28 (Polynomials act on eigenvectors by their values), Theorem 4.2.11 (Rank–nullity), Theorem 4.4.7 (Criteria for direct sums), Theorem 4.4.13 (Bases and dimension of a sum), Theorem 7.4.2 (Diagonalizable means an eigenbasis)

Techniques. Dimension counting (Technique 4.2.24), Look for invariant subspaces among kernels and images of commuting operators (Technique 7.2.31), Restrict to a subspace (Technique 4.2.5)

Hint 1. Induct on

𝑘

. Let

𝑁 = 𝑇 - 𝑐_{𝑘} {id}_{𝑉}

and

𝑈 = im 𝑁

. Which polynomial annihilates

𝑇 ↾_{𝑈}

Hint 2. Show

𝐸_{𝑐_{𝑘}} \cap 𝑈 = {𝟎}

: for

𝐯 = 𝑁 𝐰

𝐸_{𝑐_{𝑘}}

, compute

𝑞^{'} (𝑇) 𝐯

in two ways, where

𝑞^{'} =

(𝑥 - 𝑐_{1}) \dots (𝑥 - 𝑐_{𝑘 - 1})

Hint 3. Rank–nullity gives

dim 𝐸_{𝑐_{𝑘}} +

dim 𝑈 =

𝑛

, so

𝑉 = 𝐸_{𝑐_{𝑘}} \oplus 𝑈

Sketch. By induction on

𝑘

;

𝑘 = 1

means

𝑇 = 𝑐_{1} {id}_{𝑉}

. Let

𝑁 = 𝑇 - 𝑐_{𝑘} {id}_{𝑉}

𝑈 = im 𝑁

(which is

𝑇

-invariant), and

𝑞^{'} = 𝑞 / (𝑥 - 𝑐_{𝑘})

. Then

𝑞^{'} (𝑇 ↾_{𝑈}) = 0

, since

𝑞^{'} (𝑇) 𝑁 𝐰 =

𝑞 (𝑇) 𝐰

, so by induction

𝑈

is the direct sum of the eigenspaces of

𝑇 ↾_{𝑈}

. If

𝐯 = 𝑁 𝐰 \in 𝐸_{𝑐_{𝑘}}

, then

𝑞^{'} (𝑐_{𝑘}) 𝐯 =

𝑞^{'} (𝑇) 𝐯 =

𝑞 (𝑇) 𝐰 =

𝟎

with

𝑞^{'} (𝑐_{𝑘}) \neq 0

, so

𝐸_{𝑐_{𝑘}} \cap 𝑈 = {𝟎}

; with rank–nullity,

𝑉 = 𝐸_{𝑐_{𝑘}} \oplus 𝑈

We prove the statement for all finite-dimensional spaces, by induction on $𝑘$ . If $𝑘 = 1$ , then $𝑇 = 𝑐_{1} {id}_{𝑉}$ and $𝑉 = 𝐸_{𝑐_{1}}$ .

Let $𝑘 \geq 2$ , and assume the statement for $𝑘 - 1$ distinct factors. Let $𝑁 = 𝑇 - 𝑐_{𝑘} {id}_{𝑉}$ , $𝑈 = im 𝑁$ , and $𝑞^{'} =$ $(𝑥 - 𝑐_{1}) \dots (𝑥 - 𝑐_{𝑘 - 1})$ , so $𝑞 = 𝑞^{'} (𝑥 - 𝑐_{𝑘})$ . The subspace $𝑈$ is $𝑇$ -invariant by Proposition 7.2.30(a). For $𝐮 = 𝑁 𝐰 \in 𝑈$ ,

𝑞^{'} (𝑇) 𝐮 = 𝑞^{'} (𝑇) (𝑇 - 𝑐_{𝑘} {id}_{𝑉}) 𝐰 = 𝑞 (𝑇) 𝐰 = 𝟎,

so $𝑞^{'}$ annihilates $𝑇 ↾_{𝑈}$ . If $𝑈 \neq {𝟎}$ , the induction hypothesis applied to $𝑇 ↾_{𝑈} \in L (𝑈)$ shows that $𝑈$ is the direct sum of the eigenspaces of $𝑇 ↾_{𝑈}$ , so $𝑈$ has a basis of eigenvectors of $𝑇 ↾_{𝑈}$ , which are eigenvectors of $𝑇$ . If $𝑈 = {𝟎}$ , then $𝑇 = 𝑐_{𝑘} {id}_{𝑉}$ and there is nothing to prove.

Next, $𝐸_{𝑐_{𝑘}} \cap 𝑈 = {𝟎}$ . Let $𝐯 \in 𝐸_{𝑐_{𝑘}} \cap 𝑈$ , say $𝐯 = 𝑁 𝐰$ . As above, $𝑞^{'} (𝑇) 𝐯 =$ $𝑞 (𝑇) 𝐰 =$ $𝟎$ . On the other hand, if $𝐯 \neq 𝟎$ , it is an eigenvector for $𝑐_{𝑘}$ , and Proposition 7.2.28 gives $𝑞^{'} (𝑇) 𝐯 =$ $𝑞^{'} (𝑐_{𝑘}) 𝐯$ , where $𝑞^{'} (𝑐_{𝑘}) =$ $\prod_{𝑗 < 𝑘} (𝑐_{𝑘} - 𝑐_{𝑗}) \neq$ $0$ . Then $𝑞^{'} (𝑐_{𝑘}) 𝐯 = 𝟎$ forces $𝐯 = 𝟎$ , a contradiction.

By rank–nullity (Theorem 4.2.11) applied to $𝑁$ , $dim 𝐸_{𝑐_{𝑘}} +$ $dim 𝑈 =$ $dim ker 𝑁 +$ $dim im 𝑁 =$ $𝑛$ . So the sum $𝐸_{𝑐_{𝑘}} + 𝑈$ is direct (Theorem 4.4.7) of dimension $𝑛$ (Theorem 4.4.13(c)), hence equal to $𝑉$ (Proposition 4.1.21), and $𝑉 = 𝐸_{𝑐_{𝑘}} \oplus 𝑈$ . Concatenating a basis of $𝐸_{𝑐_{𝑘}}$ with an eigenbasis of $𝑈$ gives an eigenbasis of $𝑉$ , so $𝑇$ is diagonalizable (Theorem 7.4.2). The eigenspace decomposition follows: the eigenvalues of $𝑇$ are roots of $𝑞$ (Proposition 7.2.28), so the eigenbasis lies in $𝐸_{𝑐_{1}} \cup \dots \cup 𝐸_{𝑐_{𝑘}}$ , which gives $𝑉 = 𝐸_{𝑐_{1}} + \dots + 𝐸_{𝑐_{𝑘}}$ , and this sum is direct by Corollary 7.2.23. □

Remark 7.6.14 (Coprime factors). The identity

\sum 𝐿_{𝑖} = 1

in the first proof is a special case of Bézout’s identity: if

𝑝

and

𝑠

are coprime, then

𝑢 𝑝 + 𝑣 𝑠 = 1

for some

𝑢, 𝑣 \in 𝐹 [𝑥]

(Corollary 7.1.48). The same argument then shows that if

(𝑝 𝑠) (𝑇) = 0

, then

𝑉 =

ker 𝑝 (𝑇) \oplus

ker 𝑠 (𝑇)

, with both summands

𝑇

-invariant (Exercise 7.2.55). Applied to the factorization

𝑚_{𝑇} =

{(𝑥 - 𝜆_{1})}^{𝑘_{1}} \dots {(𝑥 - 𝜆_{𝑚})}^{𝑘_{𝑚}}

over

ℂ

, whose factors are pairwise coprime, it splits

𝑉

into the subspaces

{ker (𝑇 - 𝜆_{𝑖} id)}^{𝑘_{𝑖}}

even when

𝑇

is not diagonalizable. That is the generalized eigenspace decomposition of Section 10.2.

Theorem 7.6.15 (Diagonalizability and the minimal polynomial). The following are equivalent.

(I) $𝑇$ is diagonalizable.
(II) $𝑚_{𝑇} =$ $(𝑥 - 𝜆_{1}) (𝑥 - 𝜆_{2}) \dots (𝑥 - 𝜆_{𝑚})$ for some distinct $𝜆_{1}, \dots, 𝜆_{𝑚} \in 𝐹$ .
(III)Some product $(𝑥 - 𝑐_{1}) \dots (𝑥 - 𝑐_{𝑘})$ of distinct linear factors, with $𝑐_{1}, \dots, 𝑐_{𝑘} \in 𝐹$ , annihilates $𝑇$ .

In that case, the $𝜆_{𝑖}$ in (II) are exactly the distinct eigenvalues of $𝑇$ .

Proof.standard

Uses. Proposition 7.4.26 (Polynomials that kill a diagonalizable operator), Theorem 7.6.2 (Existence and uniqueness of the minimal polynomial), Theorem 7.6.7 (The roots of the minimal polynomial are the eigenvalues), Proposition 7.1.22 (Counting roots), Proposition 7.6.12 (Distinct linear factors force a diagonalization)

Techniques. Proving TFAE (Technique 2.7.1)

Hint 1. For (I)

\Rightarrow

(II), Proposition 7.4.26 gives an annihilating polynomial. How does it compare with

𝑚_{𝑇}

Hint 2. Both

𝑚_{𝑇}

and that polynomial have every eigenvalue as a root. Show that each divides the other.

Sketch. (I)

\Rightarrow

(II):

𝑞 = \prod (𝑥 - 𝜆_{𝑖})

over the distinct eigenvalues annihilates

𝑇

, so

𝑚_{𝑇} ∣ 𝑞

. Each

𝜆_{𝑖}

is a root of

𝑚_{𝑇}

, so

𝑞 ∣ 𝑚_{𝑇}

by Proposition 7.1.22. Both are monic, so

𝑚_{𝑇} = 𝑞

. (II)

\Rightarrow

(III) is immediate, and (III)

\Rightarrow

(I) is Proposition 7.6.12.

We prove (I) $\Rightarrow$ (II) $\Rightarrow$ (III) $\Rightarrow$ (I).

(I) $\Rightarrow$ (II). Let $𝜆_{1}, \dots, 𝜆_{𝑚}$ be the distinct eigenvalues of $𝑇$ and $𝑞 =$ $(𝑥 - 𝜆_{1}) \dots (𝑥 - 𝜆_{𝑚})$ . By Proposition 7.4.26, $𝑞 (𝑇) = 0$ , so $𝑚_{𝑇} ∣ 𝑞$ (Theorem 7.6.2). By Theorem 7.6.7, each $𝜆_{𝑖}$ is a root of $𝑚_{𝑇}$ , so $𝑞 ∣ 𝑚_{𝑇}$ by Proposition 7.1.22. Two monic polynomials that divide each other are equal (their quotient is a constant, which is $1$ by comparing leading coefficients). So $𝑚_{𝑇} = 𝑞$ .

(II) $\Rightarrow$ (III). Take $𝑐_{𝑖} = 𝜆_{𝑖}$ ; $𝑚_{𝑇}$ annihilates $𝑇$ .

(III) $\Rightarrow$ (I). This is Proposition 7.6.12.

Finally, by Theorem 7.6.7 the roots of $𝑚_{𝑇}$ are the eigenvalues, so in (II) the $𝜆_{𝑖}$ are exactly the distinct eigenvalues. □

Corollary 7.6.16 (Restrictions of diagonalizable operators). If

𝑇

is diagonalizable and

𝑈 \neq {𝟎}

𝑇

-invariant, then

𝑇 ↾_{𝑈} \in L (𝑈)

is diagonalizable.

Proof.routine

Uses. Theorem 7.6.15 (Diagonalizability and the minimal polynomial), Proposition 7.6.10 (Invariant subspaces and similarity)

Hint. Which polynomial annihilates

𝑇 ↾_{𝑈}

Sketch.

𝑚_{𝑇}

is a product of distinct linear factors and annihilates

𝑇 ↾_{𝑈}

(Proposition 7.6.10(a)), so (III) of Theorem 7.6.15 holds for

𝑇 ↾_{𝑈}

By Theorem 7.6.15,

𝑚_{𝑇}

is a product of distinct linear factors. By the proof of Proposition 7.6.10(a),

𝑚_{𝑇} (𝑇 ↾_{𝑈}) = 0

. So

𝑇 ↾_{𝑈}

satisfies condition (III) of Theorem 7.6.15, and it is diagonalizable. (Compare Exercise 7.4.50, whose first solution argues directly with Lagrange polynomials.) □

Technique 7.6.17 (Use an annihilating polynomial). When an operator satisfies a polynomial identity $𝑞 (𝑇) = 0$ , factor $𝑞$ and read off:

every eigenvalue of $𝑇$ is a root of $𝑞$ (Proposition 7.2.28), and the minimal polynomial divides $𝑞$ ;
if $𝑞$ is a product of distinct linear factors, $𝑇$ is diagonalizable (Theorem 7.6.15);
if $𝑞 (0) \neq 0$ , then $𝑇$ is invertible and $𝑇^{- 1}$ is a polynomial in $𝑇$ (Corollary 7.6.9);
powers $𝑇^{𝑘}$ reduce to polynomials of degree less than $deg 𝑞$ , by dividing $𝑥^{𝑘}$ by $𝑞$ .

Reach for it whenever the hypothesis is an identity such as $𝐴^{2} = 𝐴$ , $𝐴^{3} = 𝐴$ , $𝐴^{𝑘} = 𝐼$ , or $𝑇^{2} = 3 𝑇 - 2 id$ , and whenever the characteristic polynomial is available, since it always annihilates $𝑇$ (Theorem 7.6.20). Watch the field: $𝑥^{2} -$ $1 =$ $(𝑥 - 1) (𝑥 + 1)$ has distinct roots only if $1 + 1 \neq 0$ , and $𝑥^{2} + 1$ splits over $ℂ$ and over $𝔽_{5}$ but not over $ℝ$ or $𝔽_{3}$ . The canonical example is Theorem 7.6.15.

Used in 16 places

Proposition 7.6.12, Exercise 7.6.19, Exercise 7.6.27, Exercise 7.6.41, Exercise 7.6.42, Exercise 7.6.43, Exercise 7.6.45, Exercise 7.6.46, Exercise 7.6.50, Exercise 7.7.4, Exercise 7.7.7, Exercise 7.7.8, Exercise 7.7.10, Problem 7.14, Problem 7.16, Problem 7.17.

Example 7.6.18 (Diagonalizability from an identity).

(a)Projections. If $𝑃^{2} = 𝑃$ , then $𝑥 (𝑥 - 1)$ annihilates $𝑃$ , and $0 \neq 1$ in every field. So every projection is diagonalizable, with eigenvalues in ${0, 1}$ , over every field.
(b)Involutions. If $𝑇^{2} = {id}_{𝑉}$ , then $(𝑥 - 1) (𝑥 + 1)$ annihilates $𝑇$ . When $1 + 1 \neq 0$ , the roots $1$ and $- 1$ are distinct, and $𝑇$ is diagonalizable with $𝑉 = 𝐸_{1} \oplus 𝐸_{- 1}$ . Over $𝔽_{2}$ this fails: $𝐽 = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}) \in 𝔽_{2}^{2 \times 2}$ satisfies $𝐽^{2} = (\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}) = 𝐼_{2}$ , but $𝐽 \neq 𝐼_{2}$ , so $𝑚_{𝐽} =$ $𝑥^{2} -$ $1 =$ ${(𝑥 - 1)}^{2}$ in $𝔽_{2} [𝑥]$ , and $𝐽$ is not diagonalizable.
(c)Nilpotent and diagonalizable. If $𝑇^{𝑘} = 0$ and $𝑇$ is diagonalizable, then every eigenvalue is a root of $𝑥^{𝑘}$ , so $0$ is the only eigenvalue, and (II) of Theorem 7.6.15 gives $𝑚_{𝑇} = 𝑥$ . So $𝑇 = 0$ .
(d)Roots of unity. The cyclic permutation $𝑍$ of Example 7.6.5(d) has $𝑚_{𝑍} = 𝑥^{3} - 1$ . Over $ℂ$ the three cube roots of $1$ are distinct (Theorem B.1.26), so $𝑍$ is diagonalizable over $ℂ$ . Over $𝔽_{7}$ , $𝑥^{3} -$ $1 =$ $(𝑥 - 1) (𝑥 - 2) (𝑥 - 4)$ (as $2^{3} = 8$ and $4^{3} = 64$ are $1$ in $𝔽_{7}$ ), so $𝑍$ is diagonalizable over $𝔽_{7}$ . Over $𝔽_{3}$ , ${(𝑥 - 1)}^{3} =$ $𝑥^{3} -$ $3 𝑥^{2} +$ $3 𝑥 -$ $1 =$ $𝑥^{3} -$ $1$ , so $𝑚_{𝑍} = {(𝑥 - 1)}^{3}$ and $𝑍$ is not diagonalizable. Over $ℝ$ , $𝑥^{3} -$ $1 =$ $(𝑥 - 1) (𝑥^{2} + 𝑥 + 1)$ does not split, and $𝑍$ is not diagonalizable.

Exercise 7.6.19.routine For each matrix in $ℚ^{𝑛 \times 𝑛}$ , find a polynomial of degree at most $2$ that annihilates it, find its minimal polynomial, and decide whether it is diagonalizable.

𝐴 = (\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}), 𝐵 = (\begin{matrix} 1 & - 1 \\ 1 & - 1 \end{matrix}), 𝐶 = (\begin{matrix} 3 & - 2 \\ 4 & - 3 \end{matrix}) .

Solution.

Uses. Theorem 7.6.15 (Diagonalizability and the minimal polynomial), Theorem 7.6.2 (Existence and uniqueness of the minimal polynomial)

Techniques. Use an annihilating polynomial (Technique 7.6.17)

Hint 1. Compute the squares.

Hint 2. None of the three is a scalar matrix, so each minimal polynomial has degree at least

2

Sketch.

𝐴^{2} = 3 𝐴

𝐵^{2} = 0

𝐶^{2} = 𝐼_{2}

. None is scalar, so

𝑚_{𝐴} = 𝑥 (𝑥 - 3)

𝑚_{𝐵} = 𝑥^{2}

, and

𝑚_{𝐶} = (𝑥 - 1) (𝑥 + 1)

. Hence

𝐴

and

𝐶

are diagonalizable and

𝐵

is not.

Every entry of $𝐴^{2}$ is $1 + 1 + 1 = 3$ , so $𝐴^{2} = 3 𝐴$ , and $𝑥^{2} - 3 𝑥 = 𝑥 (𝑥 - 3)$ annihilates $𝐴$ . Next, $𝐵^{2} =$ $(\begin{matrix} 1 - 1 & - 1 + 1 \\ 1 - 1 & - 1 + 1 \end{matrix}) =$ $0$ , and $𝐶^{2} =$ $(\begin{matrix} 9 - 8 & - 6 + 6 \\ 12 - 12 & - 8 + 9 \end{matrix}) =$ $𝐼_{2}$ .

None of $𝐴$ , $𝐵$ , $𝐶$ is a scalar multiple of the identity, so no monic polynomial of degree $1$ annihilates them, and each minimal polynomial has degree $2$ . Since the minimal polynomial divides the quadratic found above (Theorem 7.6.2) and both are monic of degree $2$ , they are equal: $𝑚_{𝐴} = 𝑥 (𝑥 - 3)$ , $𝑚_{𝐵} = 𝑥^{2}$ , $𝑚_{𝐶} = (𝑥 - 1) (𝑥 + 1)$ . By Theorem 7.6.15, $𝐴$ (eigenvalues $0$ and $3$ ) and $𝐶$ (eigenvalues $1$ and $- 1$ ) are diagonalizable over $ℚ$ , and $𝐵$ is not, since $𝑥^{2}$ has a repeated root.

7.6.4The Cayley–Hamilton theorem

In every example so far, $𝑚_{𝑇}$ divided the characteristic polynomial $𝑝_{𝑇}$ : for the shift both are $𝑥^{𝑛}$ , for a projection $𝑥 (𝑥 - 1)$ divides $𝑥^{𝑟} {(𝑥 - 1)}^{𝑛 - 𝑟}$ , and for $diag (2, 2, 3)$ , $(𝑥 - 2) (𝑥 - 3)$ divides ${(𝑥 - 2)}^{2} (𝑥 - 3)$ . By Theorem 7.6.2, this is the statement that $𝑝_{𝑇}$ annihilates $𝑇$ .

Theorem 7.6.20 (Cayley–Hamilton).

𝑝_{𝑇} (𝑇) = 0

. Equivalently,

𝑚_{𝑇}

divides

𝑝_{𝑇}

. In particular,

𝑝_{𝐴} (𝐴) = 0

for every

𝐴 \in 𝐹^{𝑛 \times 𝑛}

, and

deg 𝑚_{𝑇} \leq 𝑛

Intuition (Four ways in). For a diagonal matrix $𝐷 =$ $diag (𝑑_{1}, \dots, 𝑑_{𝑛})$ the theorem is clear: $𝑝_{𝐷} =$ $(𝑥 - 𝑑_{1}) \dots (𝑥 - 𝑑_{𝑛})$ , and $𝑝_{𝐷} (𝐷) =$ $diag (𝑝_{𝐷} (𝑑_{1}), \dots, 𝑝_{𝐷} (𝑑_{𝑛})) =$ $0$ because each $𝑑_{𝑖}$ is a root. The four proofs below get from this case to the general one in different ways.

Triangular form. Over $ℂ$ , $𝑇$ is upper triangular in a basis whose flag is $𝑉_{0} \subseteq 𝑉_{1} \subseteq \dots \subseteq 𝑉_{𝑛}$ , and $𝑝_{𝑇} =$ $(𝑥 - 𝜆_{1}) \dots (𝑥 - 𝜆_{𝑛})$ with the $𝜆_{𝑗}$ on the diagonal. Each factor $𝑇 - 𝜆_{𝑗} id$ pushes $𝑉_{𝑗}$ down into $𝑉_{𝑗 - 1}$ , so applying all $𝑛$ factors pushes $𝑉$ down to ${𝟎}$ .
One Krylov subspace at a time. To show that $𝑝_{𝑇} (𝑇)$ kills $𝐯$ , only the Krylov subspace of $𝐯$ matters. There $𝑇$ acts by a companion matrix, which visibly satisfies its own characteristic polynomial, and that polynomial divides $𝑝_{𝑇}$ .
The adjugate. The tempting one-liner “ $𝑝_{𝐴} (𝐴) =$ $det (𝐴 𝐼 - 𝐴) =$ $det 0 =$ $0$ ” is wrong (Exercise 7.3.47). Its idea can be rescued: substitute $𝐴$ not into a determinant, but into the matrix identity $(𝑥 𝐼 - 𝐴) adj (𝑥 𝐼 - 𝐴) =$ $𝑝_{𝐴} 𝐼$ .
Perturbation. Diagonalizable matrices are “most” matrices. Nudge $𝐴$ to a diagonalizable matrix, where the theorem is easy, and let the nudge go to zero. Because everything is polynomial, “go to zero” can be done with algebra instead of limits.

The first proof rests on a lemma that is worth having on its own.

Lemma 7.6.21 (A triangular operator is killed by its diagonal). Let $ℬ︀ = 𝐛_{1}, \dots, 𝐛_{𝑛}$ be a basis with ${[𝑇]}_{ℬ︀}$ upper triangular, with diagonal entries $𝜆_{1}, \dots, 𝜆_{𝑛}$ , and let $𝑉_{𝑗} =$ $span (𝐛_{1}, \dots, 𝐛_{𝑗})$ . Then $(𝑇 - 𝜆_{𝑗} {id}_{𝑉}) (𝑉_{𝑗}) \subseteq$ $𝑉_{𝑗 - 1}$ for each $𝑗$ , and

(𝑇 - 𝜆_{1} {id}_{𝑉}) (𝑇 - 𝜆_{2} {id}_{𝑉}) \dots (𝑇 - 𝜆_{𝑛} {id}_{𝑉}) = 0 .

Proof.routine

Uses. Proposition 7.5.2 (Upper-triangular matrices and invariant subspaces), Definition 2.3.1 (Matrix of a linear map), Proposition 7.2.4 (Invariance is checked on a spanning list)

Techniques. Build a flag of invariant subspaces (Technique 7.5.4)

Hint 1. Check

(𝑇 - 𝜆_{𝑗} {id}_{𝑉}) 𝐛_{𝑖} \in

𝑉_{𝑗 - 1}

for

𝑖 < 𝑗

and for

𝑖 = 𝑗

separately.

Hint 2. For the product, show by downward induction on

𝑗

that

(𝑇 - 𝜆_{𝑗} {id}_{𝑉}) \dots (𝑇 - 𝜆_{𝑛} {id}_{𝑉})

maps

𝑉

into

𝑉_{𝑗 - 1}

Sketch. For

𝑖 < 𝑗

(𝑇 - 𝜆_{𝑗} id) 𝐛_{𝑖} \in

𝑉_{𝑖} \subseteq

𝑉_{𝑗 - 1}

since

𝑉_{𝑖}

is invariant; and

(𝑇 - 𝜆_{𝑗} id) 𝐛_{𝑗} =

\sum_{𝑖 < 𝑗} 𝑎_{𝑖 𝑗} 𝐛_{𝑖} \in

𝑉_{𝑗 - 1}

. So

𝑇 - 𝜆_{𝑗} id

maps

𝑉_{𝑗}

into

𝑉_{𝑗 - 1}

. Applying

𝑇 -

𝜆_{𝑛} id,

𝑇 -

𝜆_{𝑛 - 1} id,

\dots,

𝑇 -

𝜆_{1} id

in turn moves

𝑉 = 𝑉_{𝑛}

into

𝑉_{𝑛 - 1}

, then

𝑉_{𝑛 - 2}

, and finally into

𝑉_{0} = {𝟎}

Write ${[𝑇]}_{ℬ︀} = (𝑎_{𝑖 𝑗})$ , so $𝑎_{𝑗 𝑗} = 𝜆_{𝑗}$ and, by Proposition 7.5.2, $𝑇 𝐛_{𝑗} =$ $𝜆_{𝑗} 𝐛_{𝑗} +$ $\sum_{𝑖 < 𝑗} 𝑎_{𝑖 𝑗} 𝐛_{𝑖}$ . Fix $𝑗$ . For $𝑖 < 𝑗$ , $𝐛_{𝑖} \in 𝑉_{𝑖}$ and $𝑉_{𝑖}$ is $𝑇$ -invariant (Proposition 7.5.2), so $(𝑇 - 𝜆_{𝑗} {id}_{𝑉}) 𝐛_{𝑖} \in$ $𝑉_{𝑖} \subseteq$ $𝑉_{𝑗 - 1}$ . For $𝑖 = 𝑗$ ,

(𝑇 - 𝜆_{𝑗} {id}_{𝑉}) 𝐛_{𝑗} = \sum_{𝑖 < 𝑗} 𝑎_{𝑖 𝑗} 𝐛_{𝑖} \in 𝑉_{𝑗 - 1} .

So $𝑇 - 𝜆_{𝑗} {id}_{𝑉}$ maps the spanning list $𝐛_{1}, \dots, 𝐛_{𝑗}$ of $𝑉_{𝑗}$ into the subspace $𝑉_{𝑗 - 1}$ , hence maps $𝑉_{𝑗}$ into $𝑉_{𝑗 - 1}$ .

Now let $𝑄_{𝑗} =$ $(𝑇 - 𝜆_{𝑗} {id}_{𝑉}) \dots (𝑇 - 𝜆_{𝑛} {id}_{𝑉})$ . We show $𝑄_{𝑗} (𝑉) \subseteq 𝑉_{𝑗 - 1}$ by downward induction on $𝑗$ . For $𝑗 = 𝑛$ , $𝑄_{𝑛} (𝑉) =$ $(𝑇 - 𝜆_{𝑛} {id}_{𝑉}) (𝑉_{𝑛}) \subseteq$ $𝑉_{𝑛 - 1}$ . If $𝑄_{𝑗 + 1} (𝑉) \subseteq 𝑉_{𝑗}$ , then $𝑄_{𝑗} (𝑉) =$ $(𝑇 - 𝜆_{𝑗} {id}_{𝑉}) (𝑄_{𝑗 + 1} (𝑉)) \subseteq$ $(𝑇 - 𝜆_{𝑗} {id}_{𝑉}) (𝑉_{𝑗}) \subseteq$ $𝑉_{𝑗 - 1}$ . For $𝑗 = 1$ this says $𝑄_{1} (𝑉) \subseteq 𝑉_{0} = {𝟎}$ , so $𝑄_{1} = 0$ . □

Proof (via triangular form).standard

Uses. Theorem 7.5.13 (Upper-triangular form), Theorem 7.5.17 (Triangularizable means the characteristic polynomial splits), Proposition 7.5.9 (Multiplicities on the diagonal), Lemma 7.6.21 (A triangular operator is killed by its diagonal), Proposition 7.2.27 (Evaluating polynomials at an operator), Definition 7.3.1 (Determinant over a commutative ring)

Techniques. Prove it for triangular matrices first (Technique 7.5.23), Build a flag of invariant subspaces (Technique 7.5.4)

Hint 1. Over

ℂ

, choose a basis in which

𝑇

is upper triangular. What is

𝑝_{𝑇}

in terms of the diagonal?

Hint 2. Write

𝑝_{𝑇} (𝑇)

as a product of the factors

𝑇 - 𝜆_{𝑗} {id}_{𝑉}

in the order of Lemma 7.6.21.

Hint 3. For a real matrix

𝐴

, regard it as a complex matrix. Do

𝑝_{𝐴}

and

𝑝_{𝐴} (𝐴)

change?

Sketch. If

𝑝_{𝑇}

splits (always over

ℂ

𝑇

is upper triangular in some basis with diagonal

𝜆_{1}, \dots, 𝜆_{𝑛}

, and

𝑝_{𝑇} =

(𝑥 - 𝜆_{1}) \dots (𝑥 - 𝜆_{𝑛})

. Then

𝑝_{𝑇} (𝑇) =

(𝑇 - 𝜆_{1} id) \dots (𝑇 - 𝜆_{𝑛} id) =

0

by Lemma 7.6.21. A real matrix has the same characteristic polynomial and the same

𝑝_{𝐴} (𝐴)

when regarded as a complex matrix, so the complex case covers it.

Suppose first that $𝑝_{𝑇}$ splits over $𝐹$ , as it always does when $𝐹 = ℂ$ . By Theorem 7.5.17 (or Theorem 7.5.13 when $𝐹 = ℂ$ ), $𝑉$ has a basis $ℬ︀$ in which ${[𝑇]}_{ℬ︀}$ is upper triangular, with diagonal entries $𝜆_{1}, \dots, 𝜆_{𝑛}$ , and then $𝑝_{𝑇} =$ $(𝑥 - 𝜆_{1}) \dots (𝑥 - 𝜆_{𝑛})$ by Proposition 7.5.9. By Proposition 7.2.27(a) and Lemma 7.6.21,

𝑝_{𝑇} (𝑇) = (𝑇 - 𝜆_{1} {id}_{𝑉}) (𝑇 - 𝜆_{2} {id}_{𝑉}) \dots (𝑇 - 𝜆_{𝑛} {id}_{𝑉}) = 0 .

Now let $𝐴 \in ℝ^{𝑛 \times 𝑛}$ , and regard it as an element of $ℂ^{𝑛 \times 𝑛}$ . Its characteristic polynomial is the same polynomial in both views, because $det (𝑥 𝐼_{𝑛} - 𝐴)$ is computed by the Leibniz formula from the entries of $𝐴$ alone (Definition 7.3.1). The matrix $𝑝_{𝐴} (𝐴)$ is also the same, since it is computed from the entries of $𝐴$ and the coefficients of $𝑝_{𝐴}$ by sums and products. By the complex case, $𝑝_{𝐴} (𝐴) = 0$ . For a real operator $𝑇$ , apply this to $𝐴 = {[𝑇]}_{ℬ︀ ℬ︀}$ and use ${[𝑝_{𝑇} (𝑇)]}_{ℬ︀ ℬ︀} =$ $𝑝_{𝑇} (𝐴)$ (Proposition 7.2.27(c), Definition 7.3.12).

The same argument works over any subfield of $ℂ$ , such as $ℚ$ . Over a general field it needs a larger field in which $𝑝_{𝑇}$ splits; such a field always exists (Section 14.2), but the next two proofs avoid the question. □

The second proof works over every field. All of its work was done in Section 7.3, where Proposition 7.3.26 identified the action of $𝑇$ on a Krylov subspace with a companion matrix; Exercise 7.3.51 already carried it out.

Proof (one Krylov subspace at a time).routine

Uses. Proposition 7.2.35 (A basis of a Krylov subspace), Proposition 7.3.26 (Krylov subspaces and the characteristic polynomial), Proposition 7.2.27 (Evaluating polynomials at an operator)

Techniques. Krylov sequences must become dependent (Technique 7.3.28)

Hint 1. It suffices to show

𝑝_{𝑇} (𝑇) 𝐯 = 𝟎

for every

𝐯 \neq 𝟎

. Which polynomial kills

𝐯

Hint 2. The Krylov polynomial

𝑝

𝐯

kills

𝐯

, and by Proposition 7.3.26 it divides

𝑝_{𝑇}

Sketch. For

𝐯 \neq 𝟎

, let

𝑝

be the Krylov polynomial of

𝐯

(Proposition 7.2.35), so

𝑝 (𝑇) 𝐯 = 𝟎

. By Proposition 7.3.26,

𝑝_{𝑇} = 𝑔 𝑝

for some

𝑔

, so

𝑝_{𝑇} (𝑇) 𝐯 =

𝑔 (𝑇) 𝑝 (𝑇) 𝐯 =

𝟎

Let $𝐯 \in 𝑉$ . If $𝐯 = 𝟎$ , then $𝑝_{𝑇} (𝑇) 𝐯 = 𝟎$ . Otherwise let $𝑝$ be the monic polynomial of Proposition 7.2.35 for $𝐯$ , so $𝑝 (𝑇) 𝐯 = 𝟎$ . By Proposition 7.3.26(c), $𝑝_{𝑇} = 𝑔 𝑝$ for some $𝑔 \in 𝐹 [𝑥]$ . By Proposition 7.2.27(a),

𝑝_{𝑇} (𝑇) 𝐯 = 𝑔 (𝑇) (𝑝 (𝑇) 𝐯) = 𝑔 (𝑇) 𝟎 = 𝟎 .

So $𝑝_{𝑇} (𝑇)$ sends every vector to $𝟎$ , and $𝑝_{𝑇} (𝑇) = 0$ . □

The third proof is pure algebra: it uses only that $𝐹 [𝑥]$ is a commutative ring, so it works verbatim for matrices with entries in any commutative ring, such as $ℤ$ . It turns the naive substitution into a correct argument.

Proof (via the adjugate).hard

Uses. Definition 7.3.4 (Characteristic polynomial), Proposition 7.3.2 (Determinants over a commutative ring), Definition 5.5.6 (Cofactors, cofactor matrix, adjugate), Proposition 7.1.4 (Degree of a product), Remark 7.1.11 (Polynomials in a matrix)

Techniques. Polynomial identity arguments (Technique 5.5.24)

Hint 1. In

{𝐹 [𝑥]}^{𝑛 \times 𝑛}

(𝑥 𝐼_{𝑛} - 𝐴) adj (𝑥 𝐼_{𝑛} - 𝐴) =

𝑝_{𝐴} 𝐼_{𝑛}

by Proposition 7.3.2(f). The entries of

adj (𝑥 𝐼_{𝑛} - 𝐴)

have degree at most

𝑛 - 1

; collect powers of

𝑥

to write

adj (𝑥 𝐼_{𝑛} - 𝐴) =

\sum_{𝑘 = 0}^{𝑛 - 1} 𝑥^{𝑘} 𝐵_{𝑘}

with

𝐵_{𝑘} \in 𝐹^{𝑛 \times 𝑛}

Hint 2. Expand the left side and compare the coefficients of each

𝑥^{𝑘}

with those of

𝑝_{𝐴} 𝐼_{𝑛} =

\sum_{𝑘} 𝑐_{𝑘} 𝑥^{𝑘} 𝐼_{𝑛}

. You get

𝑛 + 1

matrix equations.

Hint 3. Multiply the equation for

𝑥^{𝑘}

on the left by

𝐴^{𝑘}

and add them all up. The left sides telescope.

Sketch. Write

adj (𝑥 𝐼_{𝑛} - 𝐴) =

\sum_{𝑘 = 0}^{𝑛 - 1} 𝑥^{𝑘} 𝐵_{𝑘}

and

𝑝_{𝐴} =

\sum_{𝑘 = 0}^{𝑛} 𝑐_{𝑘} 𝑥^{𝑘}

. Comparing coefficients in

(𝑥 𝐼_{𝑛} - 𝐴) \sum_{𝑘} 𝑥^{𝑘} 𝐵_{𝑘} =

𝑝_{𝐴} 𝐼_{𝑛}

gives

𝐵_{𝑛 - 1} = 𝐼_{𝑛}

𝐵_{𝑘 - 1} -

𝐴 𝐵_{𝑘} =

𝑐_{𝑘} 𝐼_{𝑛}

for

1 \leq 𝑘 \leq 𝑛 - 1

, and

- 𝐴 𝐵_{0} = 𝑐_{0} 𝐼_{𝑛}

. Multiplying the

𝑥^{𝑘}

equation by

𝐴^{𝑘}

and summing, the left sides telescope to

0

and the right sides add up to

𝑝_{𝐴} (𝐴)

It suffices to treat a matrix $𝐴 \in 𝐹^{𝑛 \times 𝑛}$ ; for an operator, take $𝐴 = {[𝑇]}_{ℬ︀ ℬ︀}$ as in the first proof. Write $𝑝_{𝐴} =$ $𝑐_{𝑛} 𝑥^{𝑛} +$ $𝑐_{𝑛 - 1} 𝑥^{𝑛 - 1} +$ $\dots +$ $𝑐_{0}$ , with $𝑐_{𝑛} = 1$ , and let $𝑀 =$ $𝑥 𝐼_{𝑛} -$ $𝐴 \in$ ${𝐹 [𝑥]}^{𝑛 \times 𝑛}$ . By Proposition 7.3.2(f) (for $𝑛 \geq 2$ ; for $𝑛 = 1$ , $adj 𝑀 = (1)$ and the identity is clear),

𝑀 adj (𝑀) = (det 𝑀) 𝐼_{𝑛} = 𝑝_{𝐴} 𝐼_{𝑛} .

(7.6.1)

Each entry of $adj (𝑀)$ is, up to sign, the determinant of an $(𝑛 - 1) \times (𝑛 - 1)$ submatrix of $𝑀$ (Definition 5.5.6). Each Leibniz term of such a determinant is a product of $𝑛 - 1$ entries of degree at most $1$ , so every entry of $adj (𝑀)$ is $𝟎$ or has degree at most $𝑛 - 1$ (Proposition 7.1.4). Collecting, in each entry, the coefficient of $𝑥^{𝑘}$ , we get matrices $𝐵_{0},$ $\dots,$ $𝐵_{𝑛 - 1} \in$ $𝐹^{𝑛 \times 𝑛}$ with

adj (𝑀) = 𝐵_{0} + 𝑥 𝐵_{1} + \dots + 𝑥^{𝑛 - 1} 𝐵_{𝑛 - 1},

an identity in ${𝐹 [𝑥]}^{𝑛 \times 𝑛}$ , in which $𝑥^{𝑘} 𝐵_{𝑘}$ means the matrix with entries $𝑥^{𝑘} {(𝐵_{𝑘})}_{𝑖 𝑗}$ . Since $𝑥$ commutes with everything in $𝐹 [𝑥]$ , the left side of (7.6.1) is

(𝑥 𝐼_{𝑛} - 𝐴) \sum_{𝑘 = 0}^{𝑛 - 1} 𝑥^{𝑘} 𝐵_{𝑘} = \sum_{𝑘 = 0}^{𝑛 - 1} 𝑥^{𝑘 + 1} 𝐵_{𝑘} - \sum_{𝑘 = 0}^{𝑛 - 1} 𝑥^{𝑘} 𝐴 𝐵_{𝑘} .

Two matrices with entries in $𝐹 [𝑥]$ are equal exactly when, for every $𝑘$ , the matrices of coefficients of $𝑥^{𝑘}$ are equal. Comparing with the right side $\sum_{𝑘 = 0}^{𝑛} 𝑐_{𝑘} 𝑥^{𝑘} 𝐼_{𝑛}$ :

\begin{aligned} 𝑥^{𝑛} : & 𝐵_{𝑛 - 1} = 𝐼_{𝑛}, \\ 𝑥^{𝑘} for 1 \leq 𝑘 \leq 𝑛 - 1 : & 𝐵_{𝑘 - 1} - 𝐴 𝐵_{𝑘} = 𝑐_{𝑘} 𝐼_{𝑛}, \\ 𝑥^{0} : & - 𝐴 𝐵_{0} = 𝑐_{0} 𝐼_{𝑛} . \end{aligned}

Multiply the equation for $𝑥^{𝑘}$ on the left by $𝐴^{𝑘}$ , for $𝑘 = 0, 1, \dots, 𝑛$ , and add. The right sides add up to $\sum_{𝑘} 𝑐_{𝑘} 𝐴^{𝑘} =$ $𝑝_{𝐴} (𝐴)$ . The left sides add up to

\sum_{𝑘 = 1}^{𝑛} 𝐴^{𝑘} 𝐵_{𝑘 - 1} - \sum_{𝑘 = 0}^{𝑛 - 1} 𝐴^{𝑘 + 1} 𝐵_{𝑘} = \sum_{𝑘 = 1}^{𝑛} 𝐴^{𝑘} 𝐵_{𝑘 - 1} - \sum_{𝑗 = 1}^{𝑛} 𝐴^{𝑗} 𝐵_{𝑗 - 1} = 0,

where the first sum collects the terms $𝐴^{𝑘} 𝐵_{𝑘 - 1}$ (from $𝑥^{𝑘}$ with $1 \leq 𝑘 \leq 𝑛$ ) and the second the terms $- 𝐴^{𝑘 + 1} 𝐵_{𝑘}$ (from $𝑥^{𝑘}$ with $0 \leq 𝑘 \leq 𝑛 - 1$ ), reindexed by $𝑗 = 𝑘 + 1$ . So $𝑝_{𝐴} (𝐴) = 0$ . □

Remark 7.6.22 (Where the naive substitution goes wrong). The bogus argument of Exercise 7.3.47 substitutes

𝐴

for

𝑥

inside a determinant, which turns scalars

𝑎_{𝑖 𝑗}

into nothing in particular. The adjugate proof substitutes only into identities between polynomials with matrix coefficients, and only after multiplying each coefficient by

𝐴^{𝑘}

on the left. That step is safe here because the coefficients

𝐼_{𝑛}

and

- 𝐴

of the left factor

𝑥 𝐼_{𝑛} - 𝐴

commute with

𝐴

. For a factor whose coefficients did not commute with

𝐴

, “evaluate the product at

𝐴

” and “multiply the evaluations” could give different answers.

The fourth proof introduces a technique that recurs whenever a statement is easy for “generic” matrices.

Technique 7.6.23 (Perturbation and density). To prove an identity for every matrix, first prove it for matrices in general position (diagonalizable, invertible, with distinct eigenvalues), where it is easy. Then move an arbitrary matrix $𝐴$ into general position along a path $𝐴_{𝑡}$ that depends polynomially on a parameter $𝑡$ , with $𝐴_{0} = 𝐴$ and $𝐴_{𝑡}$ in general position for all but finitely many $𝑡$ . If both sides of the identity are polynomials in the entries, the difference is a polynomial in $𝑡$ that vanishes at infinitely many points, hence identically, and in particular at $𝑡 = 0$ . With analysis available, the same idea reads: the general-position matrices are dense, and both sides are continuous.

Reach for it when a statement is clear for diagonalizable or invertible matrices and both sides are polynomial in the entries, over $ℂ$ or another infinite field. It fails over finite fields, where there is no room to perturb. The canonical example is the fourth proof of Theorem 7.6.20. Compare Technique 5.3.18, which is the same move with $𝐴 + 𝑡 𝐼_{𝑛}$ to reach invertible matrices.

Used in 2 places

Lemma 7.6.21, Exercise 7.7.14.

Proof (by perturbation, over $ℂ$ ).hard

Uses. Proposition 7.4.23 (Powers and polynomials of a diagonalizable matrix), Proposition 7.3.11 (Similar matrices have the same characteristic polynomial), Corollary 7.5.14 (Triangularizing a matrix), Proposition 7.5.8 (The diagonal of a triangular matrix), Corollary 7.4.5 (Distinct eigenvalues give a diagonalization), Theorem 5.3.3 (The Leibniz formula), Corollary 7.1.23 (Polynomials and their values)

Techniques. Perturbation and density (Technique 7.6.23), $det$ is a polynomial in the entries (Technique 5.3.18)

Hint 1. Prove

𝑝_{𝐴} (𝐴) = 0

first for diagonalizable

𝐴 = 𝑆 𝐷 𝑆^{- 1}

Hint 2. Triangularize

𝐴 = 𝑆 𝑅 𝑆^{- 1}

and add

𝑡 diag (1, 2, \dots, 𝑛)

𝑅

. For which

𝑡

are the new diagonal entries distinct?

Hint 3. Each entry of

𝑝_{𝐴_{𝑡}} (𝐴_{𝑡})

is a polynomial in

𝑡

. How many roots does it have?

Sketch. For

𝐴 = 𝑆 𝐷 𝑆^{- 1}

with

𝐷

diagonal,

𝑝_{𝐴} = 𝑝_{𝐷}

and

𝑝_{𝐴} (𝐴) =

𝑆 diag (𝑝_{𝐷} (𝑑_{𝑖})) 𝑆^{- 1} =

0

. In general write

𝐴 = 𝑆 𝑅 𝑆^{- 1}

with

𝑅

triangular and let

𝐴_{𝑡} =

𝑆 (𝑅 + 𝑡 diag (1, \dots, 𝑛)) 𝑆^{- 1}

. Except for finitely many

𝑡

𝐴_{𝑡}

has

𝑛

distinct eigenvalues, so

𝑝_{𝐴_{𝑡}} (𝐴_{𝑡}) = 0

. Each entry of

𝑝_{𝐴_{𝑡}} (𝐴_{𝑡})

is a polynomial in

𝑡

with infinitely many roots, hence

𝟎

; evaluate at

𝑡 = 0

Diagonalizable matrices. Let $𝐴 = 𝑆 𝐷 𝑆^{- 1}$ with $𝐷 =$ $diag (𝑑_{1}, \dots, 𝑑_{𝑛})$ . By Proposition 7.3.11, $𝑝_{𝐴} =$ $𝑝_{𝐷} =$ $(𝑥 - 𝑑_{1}) \dots (𝑥 - 𝑑_{𝑛})$ , and by Proposition 7.4.23,

𝑝_{𝐴} (𝐴) = 𝑆 diag (𝑝_{𝐴} (𝑑_{1}), \dots, 𝑝_{𝐴} (𝑑_{𝑛})) 𝑆^{- 1} = 𝑆 0 𝑆^{- 1} = 0 .

Perturbing. Let $𝐴 \in ℂ^{𝑛 \times 𝑛}$ be arbitrary. By Corollary 7.5.14, $𝐴 = 𝑆 𝑅 𝑆^{- 1}$ with $𝑆$ invertible and $𝑅$ upper triangular with diagonal entries $𝑟_{11}, \dots, 𝑟_{𝑛 𝑛}$ . For $𝑡 \in ℂ$ let $𝐸 =$ $𝑆 diag (1, 2, \dots, 𝑛) 𝑆^{- 1}$ and

𝐴_{𝑡} = 𝑆 (𝑅 + 𝑡 diag (1, 2, \dots, 𝑛)) 𝑆^{- 1} = 𝐴 + 𝑡 𝐸 .

The matrix $𝑅 + 𝑡 diag (1, \dots, 𝑛)$ is upper triangular with diagonal entries $𝑟_{𝑗 𝑗} + 𝑗 𝑡$ . For $𝑖 < 𝑗$ , $𝑟_{𝑖 𝑖} +$ $𝑖 𝑡 =$ $𝑟_{𝑗 𝑗} +$ $𝑗 𝑡$ holds for exactly one $𝑡$ , namely $𝑡 =$ $(𝑟_{𝑖 𝑖} - 𝑟_{𝑗 𝑗}) / (𝑗 - 𝑖)$ . So outside a finite set of values of $𝑡$ , these diagonal entries are distinct, $𝐴_{𝑡}$ has $𝑛$ distinct eigenvalues (Proposition 7.5.8(b), as $𝐴_{𝑡}$ is similar to a triangular matrix), $𝐴_{𝑡}$ is diagonalizable (Corollary 7.4.5), and $𝑝_{𝐴_{𝑡}} (𝐴_{𝑡}) = 0$ by the first step.

Passing to $𝑡 = 0$ . Each entry of $𝐴_{𝑡} = 𝐴 + 𝑡 𝐸$ is a polynomial in $𝑡$ of degree at most $1$ . By the Leibniz formula (Theorem 5.3.3), each coefficient of $𝑝_{𝐴_{𝑡}} =$ $det (𝑥 𝐼_{𝑛} - 𝐴_{𝑡})$ is a sum of products of entries of $𝐴_{𝑡}$ and constants, hence a polynomial in $𝑡$ ; so is each entry of every power $𝐴_{𝑡}^{𝑘}$ . Therefore each entry of the matrix $𝑝_{𝐴_{𝑡}} (𝐴_{𝑡})$ is a polynomial $𝑓_{𝑖 𝑗} (𝑡) \in ℂ [𝑡]$ . We showed that $𝑓_{𝑖 𝑗}$ vanishes at all but finitely many $𝑡 \in ℂ$ , that is, at infinitely many points, so $𝑓_{𝑖 𝑗} = 𝟎$ (Corollary 7.1.23(b)). In particular $𝑓_{𝑖 𝑗} (0) = 0$ for all $𝑖, 𝑗$ , and $𝐴_{0} = 𝐴$ , so $𝑝_{𝐴} (𝐴) = 0$ . □

Each proof shows something the others do not. The triangular proof is the geometric one: it shows each factor $𝑇 - 𝜆_{𝑗} id$ doing its share of the work, but it needs the eigenvalues to lie in $𝐹$ . The Krylov proof reduces everything to companion matrices and works over every field. The adjugate proof never mentions eigenvalues or vectors, and holds over every commutative ring. The perturbation proof shows that the diagonalizable case, which is nearly trivial, already implies the general one.

Corollary 7.6.24 (The minimal polynomial divides the characteristic polynomial).

(a) $𝑚_{𝑇} ∣ 𝑝_{𝑇}$ . Hence $deg 𝑚_{𝑇} \leq 𝑛$ , the operators ${id}_{𝑉}, 𝑇, \dots, 𝑇^{𝑛}$ are linearly dependent, and $dim 𝐹 [𝑇] \leq 𝑛$ .
(b) $𝑚_{𝑇}$ and $𝑝_{𝑇}$ have the same roots in $𝐹$ , namely the eigenvalues of $𝑇$ .
(c)If $𝑝_{𝑇} =$ ${(𝑥 - 𝜆_{1})}^{𝑎_{1}} \dots {(𝑥 - 𝜆_{𝑚})}^{𝑎_{𝑚}}$ with distinct $𝜆_{𝑖} \in 𝐹$ (as always happens over $ℂ$ ), then
$𝑚_{𝑇} = {(𝑥 - 𝜆_{1})}^{𝑘_{1}} \dots {(𝑥 - 𝜆_{𝑚})}^{𝑘_{𝑚}} with 1 \leq 𝑘_{𝑖} \leq 𝑎_{𝑖} .$

Proof.standard

Uses. Theorem 7.6.20 (Cayley–Hamilton), Theorem 7.6.2 (Existence and uniqueness of the minimal polynomial), Proposition 7.6.4 (The minimal polynomial is the first dependence among the powers), Theorem 7.6.7 (The roots of the minimal polynomial are the eigenvalues), Corollary 7.3.13 (Eigenvalues of an operator), Theorem 7.1.20 (Factor theorem), Corollary 7.1.7 (No zero divisors, cancellation, and units)

Hint 1. (a) and (b) are direct. For (c), show that a monic divisor of a product of linear factors is a product of some of those factors.

Hint 2. Induct on the number of linear factors. If

𝑞 = 𝑠 𝑡

and

𝜇

is a root of

𝑞

, then

𝜇

is a root of

𝑠

or of

𝑡

; cancel

𝑥 - 𝜇

Sketch. (a) By Cayley–Hamilton and Theorem 7.6.2,

𝑚_{𝑇} ∣ 𝑝_{𝑇}

; then Proposition 7.6.4 gives the rest. (b) Both root sets are the set of eigenvalues. (c) A monic divisor

𝑠

of a product

𝑞

𝑁

linear factors is a subproduct: some root

𝜇

𝑞

is a root of

𝑠

or of the cofactor; cancel

𝑥 - 𝜇

and induct on

𝑁

. So

𝑚_{𝑇} =

\prod {(𝑥 - 𝜆_{𝑖})}^{𝑘_{𝑖}}

with

𝑘_{𝑖} \leq 𝑎_{𝑖}

, and

𝑘_{𝑖} \geq 1

by (b).

(a) By Theorem 7.6.20, $𝑝_{𝑇}$ annihilates $𝑇$ , so $𝑚_{𝑇} ∣ 𝑝_{𝑇}$ by Theorem 7.6.2, and $deg 𝑚_{𝑇} \leq$ $deg 𝑝_{𝑇} =$ $𝑛$ . By Proposition 7.6.4, $𝑇^{deg 𝑚_{𝑇}}$ is a combination of lower powers, so ${id}_{𝑉}, \dots, 𝑇^{𝑛}$ are dependent, and $dim 𝐹 [𝑇] =$ $deg 𝑚_{𝑇} \leq$ $𝑛$ .

(b) By Theorem 7.6.7 and Corollary 7.3.13, the roots of $𝑚_{𝑇}$ and of $𝑝_{𝑇}$ in $𝐹$ are both exactly the eigenvalues of $𝑇$ .

(c) We first show: if $𝑞 =$ $(𝑥 - 𝜇_{1}) \dots (𝑥 - 𝜇_{𝑁})$ with $𝜇_{𝑗} \in 𝐹$ (repetitions allowed) and $𝑠$ is a monic divisor of $𝑞$ , then $𝑠$ is the product of some of the factors $𝑥 - 𝜇_{𝑗}$ . We induct on $𝑁$ . If $𝑁 = 0$ , then $𝑞 = 1 = 𝑠 𝑡$ , so $𝑠$ is a nonzero constant, and $𝑠 = 1$ since it is monic. Let $𝑁 \geq 1$ and write $𝑞 = 𝑠 𝑡$ . Since $𝑞 (𝜇_{𝑁}) = 0$ , we get $𝑠 (𝜇_{𝑁}) = 0$ or $𝑡 (𝜇_{𝑁}) = 0$ , because $𝐹$ has no zero divisors. In the first case, $𝑠 = (𝑥 - 𝜇_{𝑁}) 𝑠^{'}$ by Theorem 7.1.20, and cancelling $𝑥 - 𝜇_{𝑁}$ (Corollary 7.1.7) in $(𝑥 - 𝜇_{𝑁}) (𝑥 - 𝜇_{1}) \dots (𝑥 - 𝜇_{𝑁 - 1}) =$ $(𝑥 - 𝜇_{𝑁}) 𝑠^{'} 𝑡$ shows that the monic polynomial $𝑠^{'}$ divides $(𝑥 - 𝜇_{1}) \dots (𝑥 - 𝜇_{𝑁 - 1})$ ; by induction $𝑠^{'}$ is a product of some of these factors, and so is $𝑠$ . In the second case, $𝑡 = (𝑥 - 𝜇_{𝑁}) 𝑡^{'}$ , and cancelling shows that $𝑠$ divides $(𝑥 - 𝜇_{1}) \dots (𝑥 - 𝜇_{𝑁 - 1})$ ; again the induction hypothesis applies.

Apply this with $𝑞 = 𝑝_{𝑇}$ and $𝑠 = 𝑚_{𝑇}$ , which divides $𝑝_{𝑇}$ by (a). So $𝑚_{𝑇} =$ $\prod_{𝑖} {(𝑥 - 𝜆_{𝑖})}^{𝑘_{𝑖}}$ with $0 \leq 𝑘_{𝑖} \leq 𝑎_{𝑖}$ , and $𝑘_{𝑖} \geq 1$ because each $𝜆_{𝑖}$ is a root of $𝑚_{𝑇}$ by (b). □

Example 7.6.25 (Finding a minimal polynomial from the characteristic polynomial). Let

𝐷 = (\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{matrix}), 𝐽 = (\begin{matrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{matrix}) .

Both have $𝑝 = {(𝑥 - 2)}^{2} (𝑥 - 3)$ . By Corollary 7.6.24(c), each minimal polynomial is $(𝑥 - 2) (𝑥 - 3)$ or ${(𝑥 - 2)}^{2} (𝑥 - 3)$ , so it suffices to test the smaller candidate. For $𝐷$ , $(𝐷 - 2 𝐼) (𝐷 - 3 𝐼) =$ $diag (0, 0, 1) diag (- 1, - 1, 0) =$ $0$ , so $𝑚_{𝐷} = (𝑥 - 2) (𝑥 - 3)$ . For $𝐽$ ,

(𝐽 - 2 𝐼) (𝐽 - 3 𝐼) = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} - 1 & 1 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 0 \end{matrix}) = (\begin{matrix} 0 & - 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) \neq 0,

so $𝑚_{𝐽} =$ ${(𝑥 - 2)}^{2} (𝑥 - 3)$ , and $𝐽$ is not diagonalizable (Theorem 7.6.15).

Example 7.6.26 (Powers without a diagonalization). Let $𝐴 = (\begin{matrix} 3 & 1 \\ - 1 & 1 \end{matrix})$ . Then $tr 𝐴 = 4$ and $det 𝐴 = 4$ , so $𝑝_{𝐴} =$ $𝑥^{2} -$ $4 𝑥 +$ $4 =$ ${(𝑥 - 2)}^{2}$ , and $𝐴 \neq 2 𝐼_{2}$ , so $𝑚_{𝐴} = {(𝑥 - 2)}^{2}$ and $𝐴$ is not diagonalizable. The method of Section 7.4.3 is not available, but Cayley–Hamilton is. By the binomial theorem in the commutative ring $𝐹 [𝑥]$ ,

𝑥^{𝑘} = {(2 + (𝑥 - 2))}^{𝑘} = 2^{𝑘} + 𝑘 2^{𝑘 - 1} (𝑥 - 2) + {(𝑥 - 2)}^{2} 𝑔_{𝑘}

for some $𝑔_{𝑘} \in 𝐹 [𝑥]$ . Evaluating at $𝐴$ and using ${(𝐴 - 2 𝐼_{2})}^{2} = 0$ ,

𝐴^{𝑘} = 2^{𝑘} 𝐼_{2} + 𝑘 2^{𝑘 - 1} (𝐴 - 2 𝐼_{2}) = 2^{𝑘 - 1} (\begin{matrix} 2 + 𝑘 & 𝑘 \\ - 𝑘 & 2 - 𝑘 \end{matrix}) .

In general, $𝐴^{𝑘} = 𝑟 (𝐴)$ , where $𝑟$ is the remainder of $𝑥^{𝑘}$ on division by $𝑝_{𝐴}$ (or by $𝑚_{𝐴}$ ), so $𝐴^{𝑘}$ is always a combination of $𝐼, 𝐴, \dots, 𝐴^{𝑛 - 1}$ .

Exercise 7.6.27.routine Let

𝐴 = (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix})

. Use the Cayley–Hamilton theorem to write

𝐴^{- 1}

and

𝐴^{3}

in the form

𝑎 𝐼_{2} + 𝑏 𝐴

, and compute both.

Solution.

Uses. Theorem 7.6.20 (Cayley–Hamilton), Proposition 7.3.18 (The first and last coefficients), Corollary 7.6.9 (Inverses are polynomials)

Techniques. Use an annihilating polynomial (Technique 7.6.17)

Hint 1.

𝑝_{𝐴} =

𝑥^{2} -

(tr 𝐴) 𝑥 +

det 𝐴

Hint 2. From

𝐴^{2} = 5 𝐴 + 2 𝐼_{2}

, factor out

𝐴

to find

𝐴^{- 1}

, and multiply by

𝐴

to find

𝐴^{3}

Sketch.

𝑝_{𝐴} = 𝑥^{2} - 5 𝑥 - 2

, so

𝐴^{2} = 5 𝐴 + 2 𝐼_{2}

. Then

𝐴 (𝐴 - 5 𝐼_{2}) = 2 𝐼_{2}

gives

𝐴^{- 1} = \frac{1}{2} (𝐴 - 5 𝐼_{2})

, and

𝐴^{3} =

5 𝐴^{2} +

2 𝐴 =

27 𝐴 +

10 𝐼_{2}

Here $tr 𝐴 = 5$ and $det 𝐴 = 4 - 6 = - 2$ , so $𝑝_{𝐴} = 𝑥^{2} - 5 𝑥 - 2$ (Proposition 7.3.18), and Cayley–Hamilton gives $𝐴^{2} = 5 𝐴 + 2 𝐼_{2}$ . Then $𝐴 (𝐴 - 5 𝐼_{2}) = 2 𝐼_{2}$ , and since $𝐴 - 5 𝐼_{2}$ commutes with $𝐴$ ,

𝐴^{- 1} = \frac{1}{2} (𝐴 - 5 𝐼_{2}) = (\begin{matrix} - 2 & 1 \\ 3 / 2 & - 1 / 2 \end{matrix}) .

Next, $𝐴^{3} =$ $𝐴 \cdot$ $𝐴^{2} =$ $5 𝐴^{2} +$ $2 𝐴 =$ $5 (5 𝐴 + 2 𝐼_{2}) +$ $2 𝐴 =$ $27 𝐴 +$ $10 𝐼_{2} =$ $(\begin{matrix} 37 & 54 \\ 81 & 118 \end{matrix})$ .

7.6.5Spectral mapping

On an eigenvector, $𝑝 (𝑇)$ acts as the scalar $𝑝 (𝜆)$ (Proposition 7.2.28). So every $𝑝 (𝜆)$ , for $𝜆$ an eigenvalue of $𝑇$ , is an eigenvalue of $𝑝 (𝑇)$ . Over $ℂ$ there are no others, and even the multiplicities match.

Theorem 7.6.28 (Spectral mapping theorem). Let $𝑉$ be a finite-dimensional complex vector space with $dim 𝑉 = 𝑛 \geq 1$ , let $𝑇 \in L (𝑉)$ and $𝑝 \in ℂ [𝑥]$ , and let $𝜆_{1}, \dots, 𝜆_{𝑛}$ be the eigenvalues of $𝑇$ counted with multiplicity.

(a)The eigenvalues of $𝑝 (𝑇)$ are exactly the numbers $𝑝 (𝜆)$ with $𝜆$ an eigenvalue of $𝑇$ .
(b) $𝑝_{𝑝 (𝑇)} =$ $(𝑥 - 𝑝 (𝜆_{1})) \dots (𝑥 - 𝑝 (𝜆_{𝑛}))$ . That is, the eigenvalues of $𝑝 (𝑇)$ counted with multiplicity are $𝑝 (𝜆_{1}),$ $\dots,$ $𝑝 (𝜆_{𝑛})$ .

Proof (via triangular form).standard

Uses. Theorem 7.5.13 (Upper-triangular form), Proposition 7.5.9 (Multiplicities on the diagonal), Proposition 7.2.27 (Evaluating polynomials at an operator), Definition 2.4.1 (Matrix product), Corollary 7.3.13 (Eigenvalues of an operator)

Techniques. Prove it for triangular matrices first (Technique 7.5.23), Compute entry by entry (Technique 2.4.15)

Hint 1. Triangularize

𝑇

. What does

𝑝

do to an upper-triangular matrix?

Hint 2. Show that a product of upper-triangular matrices is upper triangular, with diagonal entries the products of the diagonal entries.

Sketch. In a basis with

𝑅 = {[𝑇]}_{ℬ︀}

upper triangular with diagonal

𝜆_{1}, \dots, 𝜆_{𝑛}

, products and sums of upper-triangular matrices are upper triangular with the diagonal computed entrywise, so

{[𝑝 (𝑇)]}_{ℬ︀} = 𝑝 (𝑅)

is upper triangular with diagonal

𝑝 (𝜆_{1}),

\dots,

𝑝 (𝜆_{𝑛})

. Then Proposition 7.5.9 gives (b), and (a) follows by taking roots.

By Theorem 7.5.13, there is a basis $ℬ︀$ with $𝑅 = {[𝑇]}_{ℬ︀}$ upper triangular, and its diagonal entries are $𝜆_{1}, \dots, 𝜆_{𝑛}$ (Proposition 7.5.9).

If $𝑋, 𝑌 \in ℂ^{𝑛 \times 𝑛}$ are upper triangular, then ${(𝑋 𝑌)}_{𝑖 𝑗} =$ $\sum_{𝑙} 𝑥_{𝑖 𝑙} 𝑦_{𝑙 𝑗}$ , and $𝑥_{𝑖 𝑙} 𝑦_{𝑙 𝑗} \neq 0$ requires $𝑖 \leq 𝑙 \leq 𝑗$ (Definition 2.4.1). For $𝑖 > 𝑗$ there is no such $𝑙$ , and for $𝑖 = 𝑗$ only $𝑙 = 𝑖$ , so $𝑋 𝑌$ is upper triangular with ${(𝑋 𝑌)}_{𝑖 𝑖} =$ $𝑥_{𝑖 𝑖} 𝑦_{𝑖 𝑖}$ . Sums and scalar multiples of upper-triangular matrices are upper triangular, with diagonals added and scaled. By induction on $𝑘$ , $𝑅^{𝑘}$ is upper triangular with diagonal entries $𝜆_{𝑗}^{𝑘}$ , and so $𝑝 (𝑅)$ is upper triangular with diagonal entries $𝑝 (𝜆_{1}),$ $\dots,$ $𝑝 (𝜆_{𝑛})$ .

Since ${[𝑝 (𝑇)]}_{ℬ︀} = 𝑝 (𝑅)$ (Proposition 7.2.27(c)), Proposition 7.5.9 applied to $𝑝 (𝑇)$ gives (b). By Corollary 7.3.13, the eigenvalues of $𝑝 (𝑇)$ are the roots of $𝑝_{𝑝 (𝑇)}$ , which by (b) are $𝑝 (𝜆_{1}),$ $\dots,$ $𝑝 (𝜆_{𝑛})$ ; this is (a). □

The second proof gives only (a), but it uses no triangularization, only the factorization of one polynomial and the fact that polynomials in $𝑇$ commute.

Proof (factor $𝑝 - 𝜇$ ).standard

Uses. Corollary 7.1.35 (Factorization over $ℂ$ ), Proposition 7.2.27 (Evaluating polynomials at an operator), Theorem 7.2.9 (Eigenvalues and invertibility), Theorem 7.3.27 (Existence of eigenvalues over $ℂ$ )

Techniques. Look for invariant subspaces among kernels and images of commuting operators (Technique 7.2.31)

Hint 1. Fix

𝜇 \in ℂ

. When is

𝑝 (𝑇) - 𝜇 {id}_{𝑉}

not injective? Factor

𝑝 - 𝜇

over

ℂ

Hint 2. A product of commuting operators is injective if and only if each factor is. Why does commuting matter?

Sketch. If

𝑝

is constant the claim is clear, since

𝑇

has an eigenvalue. Otherwise

𝑝 -

𝜇 =

𝑎 (𝑥 - 𝑟_{1}) \dots (𝑥 - 𝑟_{𝑑})

, so

𝑝 (𝑇) -

𝜇 id =

𝑎 (𝑇 - 𝑟_{1} id) \dots (𝑇 - 𝑟_{𝑑} id)

. These factors commute, so the product fails to be injective exactly when some factor does, that is, when some root

𝑟_{𝑖}

𝑝 - 𝜇

is an eigenvalue of

𝑇

, that is, when

𝜇 = 𝑝 (𝜆)

for some eigenvalue

𝜆

We prove (a). Let $𝜇 \in ℂ$ . If $𝑝 = 𝑐$ is constant, then $𝑝 (𝑇) = 𝑐 {id}_{𝑉}$ , whose only eigenvalue is $𝑐$ , and $𝑝 (𝜆) = 𝑐$ for every eigenvalue $𝜆$ of $𝑇$ , of which there is at least one (Theorem 7.3.27). So assume $deg 𝑝 = 𝑑 \geq 1$ . Then $𝑝 - 𝜇$ has degree $𝑑$ and some leading coefficient $𝑎 \neq 0$ , and by Corollary 7.1.35,

𝑝 - 𝜇 = 𝑎 (𝑥 - 𝑟_{1}) (𝑥 - 𝑟_{2}) \dots (𝑥 - 𝑟_{𝑑}),

where $𝑟_{1}, \dots, 𝑟_{𝑑}$ are the roots of $𝑝 - 𝜇$ , that is, the solutions of $𝑝 (𝑟) = 𝜇$ . By Proposition 7.2.27(a), $𝑝 (𝑇) -$ $𝜇 {id}_{𝑉} =$ $𝑎 𝑆_{1} 𝑆_{2} \dots 𝑆_{𝑑}$ with $𝑆_{𝑖} = 𝑇 - 𝑟_{𝑖} {id}_{𝑉}$ , and these factors commute with each other (Proposition 7.2.27(b)).

We claim that $𝑆_{1} \dots 𝑆_{𝑑}$ is injective if and only if every $𝑆_{𝑖}$ is. If every $𝑆_{𝑖}$ is injective, so is their composite. Conversely, suppose some $𝑆_{𝑖}$ is not injective, say $𝑆_{𝑖} 𝐯 = 𝟎$ with $𝐯 \neq 𝟎$ . Since the factors commute, $𝑆_{1} \dots 𝑆_{𝑑} =$ $(\prod_{𝑗 \neq 𝑖} 𝑆_{𝑗}) 𝑆_{𝑖}$ , which sends $𝐯$ to $𝟎$ , so the composite is not injective.

Hence, by Theorem 7.2.9, $𝜇$ is an eigenvalue of $𝑝 (𝑇)$ if and only if $𝑝 (𝑇) - 𝜇 {id}_{𝑉}$ is not injective, if and only if some $𝑇 - 𝑟_{𝑖} {id}_{𝑉}$ is not injective, if and only if some $𝑟_{𝑖}$ is an eigenvalue of $𝑇$ . Since the $𝑟_{𝑖}$ are exactly the solutions of $𝑝 (𝑟) = 𝜇$ , this says that $𝜇 = 𝑝 (𝜆)$ for some eigenvalue $𝜆$ of $𝑇$ . □

The second proof never used $dim 𝑉 < \infty$ except to know that $𝑇$ has an eigenvalue when $𝑝$ is constant, so part (a) holds for operators on any complex vector space, with nonconstant $𝑝$ .

Non-example 7.6.29 (Spectral mapping over

ℝ

). Over any field, every

𝑝 (𝜆)

is an eigenvalue of

𝑝 (𝑇)

, by Proposition 7.2.28. The converse needs

ℂ

. The rotation

𝑅 = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}) \in ℝ^{2 \times 2}

has no real eigenvalues, but

𝑅^{2} = - 𝐼_{2}

has the eigenvalue

- 1

. In the second proof, the polynomial

𝑥^{2} - (- 1) = 𝑥^{2} + 1

cannot be factored over

ℝ

Corollary 7.6.30 (Trace, determinant, and invertibility of $𝑝 (𝑇)$ ). In the setting of Theorem 7.6.28,

tr 𝑝 (𝑇) = 𝑝 (𝜆_{1}) + \dots + 𝑝 (𝜆_{𝑛}), det 𝑝 (𝑇) = 𝑝 (𝜆_{1}) \dots 𝑝 (𝜆_{𝑛}),

and $𝑝 (𝑇)$ is invertible if and only if $𝑝 (𝜆) \neq 0$ for every eigenvalue $𝜆$ of $𝑇$ .

Proof.routine

Uses. Theorem 7.6.28 (Spectral mapping theorem), Theorem 7.5.22 (Trace and determinant from eigenvalues), Theorem 7.2.9 (Eigenvalues and invertibility)

Hint. Apply Theorem 7.5.22 to the operator

𝑝 (𝑇)

Sketch. By Theorem 7.6.28(b), the eigenvalues of

𝑝 (𝑇)

with multiplicity are the

𝑝 (𝜆_{𝑗})

, so Theorem 7.5.22 gives the trace and determinant.

𝑝 (𝑇)

is invertible exactly when

0

is not among its eigenvalues.

By Theorem 7.6.28(b),

𝑝 (𝜆_{1}),

\dots,

𝑝 (𝜆_{𝑛})

are the eigenvalues of

𝑝 (𝑇)

counted with multiplicity, and Theorem 7.5.22 applied to

𝑝 (𝑇)

gives the two formulas. By Theorem 7.2.9,

𝑝 (𝑇)

is invertible if and only if

0

is not an eigenvalue of

𝑝 (𝑇)

, which by Theorem 7.6.28(a) means

𝑝 (𝜆) \neq 0

for every eigenvalue

𝜆

𝑇

. □

For example, $𝐴 = (\begin{matrix} 0 & 1 \\ - 2 & 3 \end{matrix})$ has $𝑝_{𝐴} =$ $𝑥^{2} -$ $3 𝑥 +$ $2 =$ $(𝑥 - 1) (𝑥 - 2)$ , so its eigenvalues are $1$ and $2$ . For $𝑝 = 𝑥^{2} + 𝑥 + 1$ , the matrix $𝑝 (𝐴)$ has eigenvalues $𝑝 (1) = 3$ and $𝑝 (2) = 7$ , so $tr 𝑝 (𝐴) = 10$ and $det 𝑝 (𝐴) = 21$ . Directly, $𝑝 (𝐴) =$ $𝐴^{2} +$ $𝐴 +$ $𝐼_{2} =$ $(\begin{matrix} - 1 & 4 \\ - 8 & 11 \end{matrix})$ , with trace $10$ and determinant $- 11 + 32 = 21$ .

7.6.6Commuting operators

If $𝑆 𝑇 = 𝑇 𝑆$ , then each eigenspace of $𝑇$ is mapped into itself by $𝑆$ (Proposition 7.2.30(b)). Inside an eigenspace of $𝑇$ , every nonzero vector is an eigenvector of $𝑇$ , so an eigenvector of $𝑆$ found there serves both operators. Over $ℂ$ , this is enough to find common eigenvectors, and then common triangular and diagonal forms.

Theorem 7.6.31 (Commuting operators have a common eigenvector). Let

𝑉

be a finite-dimensional complex vector space with

𝑉 \neq {𝟎}

, and let

𝑇_{1}, \dots, 𝑇_{𝑘} \in L (𝑉)

commute pairwise, that is,

𝑇_{𝑖} 𝑇_{𝑗} = 𝑇_{𝑗} 𝑇_{𝑖}

for all

𝑖, 𝑗

. Then some nonzero

𝐯 \in 𝑉

is an eigenvector of every

𝑇_{𝑖}

. More generally, every nonzero subspace

𝑊

that is invariant under all of the

𝑇_{𝑖}

contains such a

𝐯

Proof (one operator at a time).standard

Uses. Theorem 7.3.27 (Existence of eigenvalues over $ℂ$ ), Proposition 7.2.30 (Kernels and images of commuting operators are invariant), Definition 7.2.1 (Invariant subspace), Definition 7.2.13 (Eigenspace)

Techniques. Look for invariant subspaces among kernels and images of commuting operators (Technique 7.2.31), Restrict to a subspace (Technique 4.2.5)

Hint 1. Prove the “more generally” statement by induction on

𝑘

. For

𝑘 = 1

, restrict

𝑇_{1}

𝑊

Hint 2. Let

𝜆

be an eigenvalue of

𝑇_{𝑘} ↾_{𝑊}

. Which subspace of

𝑊

consists of eigenvectors of

𝑇_{𝑘}

(and

𝟎

), and why is it invariant under

𝑇_{1}, \dots, 𝑇_{𝑘 - 1}

Sketch. Induct on

𝑘

. For

𝑘 = 1

𝑇_{1} ↾_{𝑊}

is an operator on a nonzero complex space and has an eigenvector. For

𝑘 \geq 2

, let

𝜆

be an eigenvalue of

𝑇_{𝑘} ↾_{𝑊}

and

𝑊^{'} =

𝑊 \cap

𝐸_{𝜆} (𝑇_{𝑘}) \neq

{𝟎}

. Each

𝑇_{𝑖}

preserves

𝑊

, and preserves

𝐸_{𝜆} (𝑇_{𝑘})

because it commutes with

𝑇_{𝑘}

, so it preserves

𝑊^{'}

. By induction,

𝑊^{'}

contains a common eigenvector of

𝑇_{1}, \dots, 𝑇_{𝑘 - 1}

, and every nonzero vector of

𝑊^{'}

is an eigenvector of

𝑇_{𝑘}

We prove the second statement by induction on $𝑘$ ; the first is the case $𝑊 = 𝑉$ . Let $𝑊 \neq {𝟎}$ be invariant under every $𝑇_{𝑖}$ .

If $𝑘 = 1$ , then $𝑇_{1} ↾_{𝑊}$ is an operator on the nonzero finite-dimensional complex space $𝑊$ , so it has an eigenvector $𝐯 \in 𝑊$ (Theorem 7.3.27), and $𝑇_{1} 𝐯 = 𝑇_{1} ↾_{𝑊} 𝐯$ is a multiple of $𝐯$ .

Let $𝑘 \geq 2$ , and assume the statement for $𝑘 - 1$ operators. Let $𝜆$ be an eigenvalue of $𝑇_{𝑘} ↾_{𝑊}$ and

𝑊^{'} = 𝑊 \cap 𝐸_{𝜆} (𝑇_{𝑘}) = {𝐰 \in 𝑊 | 𝑇_{𝑘} 𝐰 = 𝜆 𝐰},

which contains an eigenvector of $𝑇_{𝑘} ↾_{𝑊}$ , so $𝑊^{'} \neq {𝟎}$ . For $𝑖 < 𝑘$ , $𝑇_{𝑖}$ commutes with $𝑇_{𝑘}$ , so $𝐸_{𝜆} (𝑇_{𝑘})$ is $𝑇_{𝑖}$ -invariant (Proposition 7.2.30(b)); $𝑊$ is $𝑇_{𝑖}$ -invariant by hypothesis; so $𝑊^{'}$ is $𝑇_{𝑖}$ -invariant. By the induction hypothesis applied to $𝑇_{1}, \dots, 𝑇_{𝑘 - 1}$ and $𝑊^{'}$ , some nonzero $𝐯 \in 𝑊^{'}$ is an eigenvector of $𝑇_{1}, \dots, 𝑇_{𝑘 - 1}$ . Since $𝐯 \in 𝐸_{𝜆} (𝑇_{𝑘})$ , it is also an eigenvector of $𝑇_{𝑘}$ . □

Intuition (A smallest invariant subspace). The first proof shrinks

𝑊

once for each operator. The second goes straight to the end: take a nonzero subspace that is invariant under all the operators and as small as possible. Each operator must act on it as a scalar, since otherwise one of its eigenspaces inside would be a smaller invariant subspace. This extremal argument never counts the operators, so it works for any set of commuting operators, finite or not.

Proof (a minimal common invariant subspace).standard

Uses. Theorem 7.3.27 (Existence of eigenvalues over $ℂ$ ), Proposition 7.2.30 (Kernels and images of commuting operators are invariant), Proposition 4.1.21 (Dimensions of subspaces)

Techniques. Extremal choice and minimal counterexamples (Technique 1.5.9), Look for invariant subspaces among kernels and images of commuting operators (Technique 7.2.31)

Hint 1. Among the nonzero subspaces of

𝑊

invariant under every

𝑇_{𝑖}

, choose one,

𝑊_{0}

, of smallest dimension.

Hint 2. Let

𝜆

be an eigenvalue of

𝑇_{𝑖} ↾_{𝑊_{0}}

. Show that

𝑊_{0} \cap 𝐸_{𝜆} (𝑇_{𝑖})

is again invariant under every

𝑇_{𝑗}

. What does minimality force?

Sketch. Pick a nonzero subspace

𝑊_{0} \subseteq 𝑊

, invariant under all

𝑇_{𝑖}

, of least dimension. For each

𝑖

, take an eigenvalue

𝜆

𝑇_{𝑖} ↾_{𝑊_{0}}

; then

𝑊_{0} \cap 𝐸_{𝜆} (𝑇_{𝑖})

is nonzero, invariant under all

𝑇_{𝑗}

, and contained in

𝑊_{0}

, so it equals

𝑊_{0}

. Thus every

𝑇_{𝑖}

acts on

𝑊_{0}

as a scalar, and every nonzero vector of

𝑊_{0}

is a common eigenvector.

Consider the nonzero subspaces of $𝑊$ that are invariant under every $𝑇_{𝑖}$ . There is at least one, $𝑊$ itself, so we may choose one, $𝑊_{0}$ , of smallest dimension. Fix $𝑖$ . By Theorem 7.3.27, $𝑇_{𝑖} ↾_{𝑊_{0}}$ has an eigenvalue $𝜆$ , so $𝑊_{1} =$ $𝑊_{0} \cap$ $𝐸_{𝜆} (𝑇_{𝑖})$ is nonzero. For each $𝑗$ , $𝐸_{𝜆} (𝑇_{𝑖})$ is $𝑇_{𝑗}$ -invariant (Proposition 7.2.30(b)), so $𝑊_{1}$ is $𝑇_{𝑗}$ -invariant. By the choice of $𝑊_{0}$ , $dim 𝑊_{1} \geq dim 𝑊_{0}$ , and as $𝑊_{1} \subseteq 𝑊_{0}$ , we get $𝑊_{1} = 𝑊_{0}$ (Proposition 4.1.21). That is, $𝑇_{𝑖} 𝐰 = 𝜆 𝐰$ for every $𝐰 \in 𝑊_{0}$ .

So each $𝑇_{𝑖}$ acts on $𝑊_{0}$ as a scalar, and every nonzero $𝐯 \in 𝑊_{0}$ is an eigenvector of every $𝑇_{𝑖}$ . The argument used neither the number of operators nor their order, so it applies to any set of pairwise commuting operators on $𝑉$ . □

Both hypotheses are needed. Over $ℝ$ even a single operator, a rotation, may have no eigenvector. And for non-commuting operators the conclusion fails: the eigenvectors of $𝑁 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ are the nonzero multiples of $𝐞_{1}$ , and those of $𝑁^{𝖳}$ the nonzero multiples of $𝐞_{2}$ .

Theorem 7.6.32 (Simultaneous triangularization). Let

𝑉

be a finite-dimensional complex vector space with

dim 𝑉 \geq 1

, and let

𝑇_{1}, \dots, 𝑇_{𝑘} \in L (𝑉)

commute pairwise. Then there is a basis

ℬ︀

𝑉

such that

{[𝑇_{𝑖}]}_{ℬ︀}

is upper triangular for every

𝑖

Proof.standard

Uses. Theorem 7.6.31 (Commuting operators have a common eigenvector), Proposition 7.2.7 (Invariant lines are spanned by eigenvectors), Proposition 4.6.32 (Operators on a quotient), Theorem 4.6.27 (Dimension of a quotient), Proposition 4.6.28 (Lifting a basis of a quotient), Proposition 7.5.2 (Upper-triangular matrices and invariant subspaces)

Techniques. Induction on dimension via an invariant subspace (Technique 7.5.12), Build a flag of invariant subspaces (Technique 7.5.4), Pass to a quotient (Technique 4.6.31)

Hint 1. Induct on

𝑛 = dim 𝑉

, as in the proof of Theorem 7.5.13. The bottom of the flag should be a common eigenvector.

Hint 2. Pass to

𝑉 / 𝑈

with

𝑈 = span (𝐛_{1})

. Do the quotient operators

𝑇_{𝑖} / 𝑈

still commute?

Hint 3. Lift a basis that triangularizes all

𝑇_{𝑖} / 𝑈

, and put

𝐛_{1}

in front.

Sketch. Induct on

𝑛

. Take a common eigenvector

𝐛_{1}

(Theorem 7.6.31);

𝑈 = span (𝐛_{1})

is invariant under every

𝑇_{𝑖}

. The quotient operators commute, since

(𝑇_{𝑖} / 𝑈) (𝑇_{𝑗} / 𝑈) (𝐯 + 𝑈) =

𝑇_{𝑖} 𝑇_{𝑗} 𝐯 +

𝑈

. By induction,

𝑉 / 𝑈

has a basis

𝐛_{2} + 𝑈, \dots, 𝐛_{𝑛} + 𝑈

triangularizing all of them. Then

𝐛_{1}, \dots, 𝐛_{𝑛}

is a basis of

𝑉

, and

𝑇_{𝑖} 𝐛_{𝑗} \in

span (𝐛_{1}, \dots, 𝐛_{𝑗})

because this holds modulo

𝑈 = span (𝐛_{1})

We induct on $𝑛 = dim 𝑉$ . For $𝑛 = 1$ every basis works.

Let $𝑛 \geq 2$ and assume the theorem in dimension $𝑛 - 1$ . By Theorem 7.6.31 there is a common eigenvector $𝐛_{1}$ of $𝑇_{1}, \dots, 𝑇_{𝑘}$ , and $𝑈 = span (𝐛_{1})$ is invariant under every $𝑇_{𝑖}$ (Proposition 7.2.7). Let $𝑇_{𝑖} =$ $𝑇_{𝑖} / 𝑈 \in$ $L (𝑉 / 𝑈)$ be the quotient operators (Proposition 4.6.32(a)). They commute: for $𝐯 \in 𝑉$ ,

𝑇_{𝑖} 𝑇_{𝑗} (𝐯 + 𝑈) = 𝑇_{𝑖} (𝑇_{𝑗} 𝐯 + 𝑈) = 𝑇_{𝑖} 𝑇_{𝑗} 𝐯 + 𝑈 = 𝑇_{𝑗} 𝑇_{𝑖} 𝐯 + 𝑈 = 𝑇_{𝑗} 𝑇_{𝑖} (𝐯 + 𝑈) .

Since $dim 𝑉 / 𝑈 = 𝑛 - 1$ (Theorem 4.6.27), the induction hypothesis and Proposition 7.5.2 give $𝐛_{2}, \dots, 𝐛_{𝑛} \in 𝑉$ such that $𝐛_{2} + 𝑈, \dots, 𝐛_{𝑛} + 𝑈$ is a basis of $𝑉 / 𝑈$ and, for all $𝑖$ and all $𝑗 \geq 2$ ,

𝑇_{𝑖} (𝐛_{𝑗} + 𝑈) \in span (𝐛_{2} + 𝑈, \dots, 𝐛_{𝑗} + 𝑈) .

By Proposition 4.6.28, $ℬ︀ = 𝐛_{1}, 𝐛_{2}, \dots, 𝐛_{𝑛}$ is a basis of $𝑉$ . Fix $𝑖$ and $𝑗 \geq 2$ , and write $𝑇_{𝑖} (𝐛_{𝑗} + 𝑈) =$ $\sum_{𝑙 = 2}^{𝑗} 𝑐_{𝑙} (𝐛_{𝑙} + 𝑈) =$ $(\sum_{𝑙 = 2}^{𝑗} 𝑐_{𝑙} 𝐛_{𝑙}) +$ $𝑈$ . Since also $𝑇_{𝑖} (𝐛_{𝑗} + 𝑈) =$ $𝑇_{𝑖} 𝐛_{𝑗} +$ $𝑈$ , the difference $𝑇_{𝑖} 𝐛_{𝑗} -$ $\sum_{𝑙 = 2}^{𝑗} 𝑐_{𝑙} 𝐛_{𝑙}$ lies in $𝑈 = span (𝐛_{1})$ , so $𝑇_{𝑖} 𝐛_{𝑗} \in$ $span (𝐛_{1}, \dots, 𝐛_{𝑗})$ . Also $𝑇_{𝑖} 𝐛_{1} \in span (𝐛_{1})$ . By Proposition 7.5.2, ${[𝑇_{𝑖}]}_{ℬ︀}$ is upper triangular for every $𝑖$ . □

With an inner product, the basis can be taken orthonormal: all the $𝑇_{𝑖}$ preserve the flag of $ℬ︀$ , and Gram–Schmidt produces an orthonormal basis with the same flag, exactly as in the proof of Theorem 7.5.18 (Technique 7.5.4). So commuting complex matrices $𝐴_{1}, \dots, 𝐴_{𝑘}$ can be written $𝐴_{𝑖} = 𝑈 𝑅_{𝑖} 𝑈^{*}$ with one unitary $𝑈$ and upper-triangular $𝑅_{𝑖}$ . The converse fails: upper-triangular matrices need not commute (Exercise 7.6.51).

Corollary 7.6.33 (Eigenvalues of sums and products of commuting operators). Let

𝑉

be a finite-dimensional complex vector space with

dim 𝑉 = 𝑛 \geq 1

, and let

𝑆, 𝑇 \in L (𝑉)

commute. Then the eigenvalues of

𝑆

and of

𝑇

, counted with multiplicity, can be listed as

𝜆_{1}, \dots, 𝜆_{𝑛}

and

𝜇_{1}, \dots, 𝜇_{𝑛}

so that the eigenvalues of

𝑆 + 𝑇

and of

𝑆 𝑇

, counted with multiplicity, are

𝜆_{1} +

𝜇_{1},

\dots,

𝜆_{𝑛} +

𝜇_{𝑛}

and

𝜆_{1} 𝜇_{1}, \dots, 𝜆_{𝑛} 𝜇_{𝑛}

. In particular, every eigenvalue of

𝑆 + 𝑇

𝜆 + 𝜇

for some eigenvalue

𝜆

𝑆

and some eigenvalue

𝜇

𝑇

Proof.routine

Uses. Theorem 7.6.32 (Simultaneous triangularization), Proposition 7.5.9 (Multiplicities on the diagonal), Theorem 2.4.2 (The matrix of a composite), Proposition 2.4.12 (Matrices of sums and scalar multiples)

Hint. Triangularize

𝑆

and

𝑇

in one basis. What are the diagonals of

[𝑆 + 𝑇]

and

[𝑆 𝑇]

Sketch. In a common triangularizing basis,

[𝑆 + 𝑇] = [𝑆] + [𝑇]

and

[𝑆 𝑇] = [𝑆] [𝑇]

are upper triangular, with diagonal entries

𝜆_{𝑗} + 𝜇_{𝑗}

and

𝜆_{𝑗} 𝜇_{𝑗}

(as in the proof of Theorem 7.6.28). Read the eigenvalues off the diagonals with Proposition 7.5.9.

By Theorem 7.6.32, there is a basis

ℬ︀

in which

{[𝑆]}_{ℬ︀}

and

{[𝑇]}_{ℬ︀}

are both upper triangular; let their diagonal entries be

𝜆_{1}, \dots, 𝜆_{𝑛}

and

𝜇_{1}, \dots, 𝜇_{𝑛}

. By Proposition 7.5.9, these are the eigenvalues of

𝑆

and of

𝑇

counted with multiplicity. Now

{[𝑆 + 𝑇]}_{ℬ︀} =

{[𝑆]}_{ℬ︀} +

{[𝑇]}_{ℬ︀}

(Proposition 2.4.12) and

{[𝑆 𝑇]}_{ℬ︀} =

{[𝑆]}_{ℬ︀} {[𝑇]}_{ℬ︀}

(Theorem 2.4.2) are upper triangular, with diagonal entries

𝜆_{𝑗} + 𝜇_{𝑗}

and

𝜆_{𝑗} 𝜇_{𝑗}

respectively (see the proof of Theorem 7.6.28). Applying Proposition 7.5.9 to

𝑆 + 𝑇

and to

𝑆 𝑇

gives the claim. □

Without commuting, nothing of the kind holds: $𝑁 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ and $𝑁^{𝖳}$ both have $0$ as their only eigenvalue, while $𝑁 + 𝑁^{𝖳} = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ has eigenvalues $\pm 1$ and $𝑁 𝑁^{𝖳} = diag (1, 0)$ has the eigenvalue $1$ .

For diagonalization, commuting is also necessary, since diagonal matrices commute.

Theorem 7.6.34 (Simultaneous diagonalization). Let

𝑉

be a finite-dimensional vector space over any field

𝐹

, with

dim 𝑉 \geq 1

, and let

𝑇_{1}, \dots, 𝑇_{𝑘} \in L (𝑉)

be diagonalizable. Then

𝑉

has a basis consisting of vectors that are eigenvectors of every

𝑇_{𝑖}

(equivalently, a basis

ℬ︀

in which every

{[𝑇_{𝑖}]}_{ℬ︀}

is diagonal) if and only if

𝑇_{𝑖} 𝑇_{𝑗} = 𝑇_{𝑗} 𝑇_{𝑖}

for all

𝑖, 𝑗

Proof.standard

Uses. Theorem 7.4.2 (Diagonalizable means an eigenbasis), Theorem 2.4.2 (The matrix of a composite), Proposition 2.4.12 (Matrices of sums and scalar multiples), Corollary 7.6.16 (Restrictions of diagonalizable operators), Proposition 7.2.30 (Kernels and images of commuting operators are invariant), Corollary 7.2.23 (Eigenspaces are independent), Theorem 4.4.13 (Bases and dimension of a sum)

Techniques. Look for invariant subspaces among kernels and images of commuting operators (Technique 7.2.31), Glue bases of the summands (Technique 4.4.14)

Hint 1. For “only if”, compute

{[𝑇_{𝑖} 𝑇_{𝑗}]}_{ℬ︀}

and

{[𝑇_{𝑗} 𝑇_{𝑖}]}_{ℬ︀}

Hint 2. For “if”, induct on

𝑘

, proving: every nonzero subspace

𝑊

invariant under all

𝑇_{𝑖}

has a basis of common eigenvectors.

Hint 3. Split

𝑊

into the eigenspaces of

𝑇_{𝑘} ↾_{𝑊}

. Why is each piece invariant under

𝑇_{1}, \dots, 𝑇_{𝑘 - 1}

, and why is

𝑇_{𝑘} ↾_{𝑊}

diagonalizable?

Sketch. If every

{[𝑇_{𝑖}]}_{ℬ︀}

is diagonal, the matrices commute, hence so do the operators. Conversely, induct on

𝑘

with the stronger statement for invariant subspaces

𝑊

𝑇_{𝑘} ↾_{𝑊}

is diagonalizable (Corollary 7.6.16), so

𝑊

is the direct sum of the pieces

𝑊 \cap 𝐸_{𝜇} (𝑇_{𝑘})

. Each piece is invariant under

𝑇_{1}, \dots, 𝑇_{𝑘 - 1}

because they commute with

𝑇_{𝑘}

, so by induction it has a basis of common eigenvectors of

𝑇_{1}, \dots, 𝑇_{𝑘 - 1}

, which are also eigenvectors of

𝑇_{𝑘}

. Concatenate.

Only if. Let $ℬ︀$ be a basis with every ${[𝑇_{𝑖}]}_{ℬ︀}$ diagonal (Theorem 7.4.2). Diagonal matrices commute, so by Theorem 2.4.2, ${[𝑇_{𝑖} 𝑇_{𝑗}]}_{ℬ︀} =$ ${[𝑇_{𝑖}]}_{ℬ︀} {[𝑇_{𝑗}]}_{ℬ︀} =$ ${[𝑇_{𝑗}]}_{ℬ︀} {[𝑇_{𝑖}]}_{ℬ︀} =$ ${[𝑇_{𝑗} 𝑇_{𝑖}]}_{ℬ︀}$ . Since $𝑆 \mapsto {[𝑆]}_{ℬ︀}$ is injective (Proposition 2.4.12), $𝑇_{𝑖} 𝑇_{𝑗} = 𝑇_{𝑗} 𝑇_{𝑖}$ .

If. We show by induction on $𝑘$ : every nonzero subspace $𝑊$ that is invariant under all of $𝑇_{1}, \dots, 𝑇_{𝑘}$ has a basis of common eigenvectors of $𝑇_{1}, \dots, 𝑇_{𝑘}$ . The theorem is the case $𝑊 = 𝑉$ .

For $𝑘 = 1$ , $𝑇_{1} ↾_{𝑊}$ is diagonalizable by Corollary 7.6.16, and its eigenvectors are eigenvectors of $𝑇_{1}$ .

Let $𝑘 \geq 2$ . By Corollary 7.6.16, $𝑇_{𝑘} ↾_{𝑊}$ is diagonalizable. Let $𝜇_{1}, \dots, 𝜇_{𝑟}$ be its distinct eigenvalues and $𝑊_{𝑙} =$ $𝑊 \cap$ $𝐸_{𝜇_{𝑙}} (𝑇_{𝑘})$ , the eigenspaces of $𝑇_{𝑘} ↾_{𝑊}$ . Grouping an eigenbasis of $𝑇_{𝑘} ↾_{𝑊}$ by eigenvalue shows $𝑊 = 𝑊_{1} + \dots + 𝑊_{𝑟}$ , and the sum is direct by Corollary 7.2.23. For $𝑖 < 𝑘$ , $𝑇_{𝑖}$ commutes with $𝑇_{𝑘}$ , so $𝐸_{𝜇_{𝑙}} (𝑇_{𝑘})$ is $𝑇_{𝑖}$ -invariant (Proposition 7.2.30(b)), and so is $𝑊$ ; hence each $𝑊_{𝑙}$ is invariant under $𝑇_{1}, \dots, 𝑇_{𝑘 - 1}$ . Each $𝑊_{𝑙}$ is nonzero, so by the induction hypothesis it has a basis of common eigenvectors of $𝑇_{1}, \dots, 𝑇_{𝑘 - 1}$ , and these lie in $𝐸_{𝜇_{𝑙}} (𝑇_{𝑘})$ , so they are eigenvectors of $𝑇_{𝑘}$ as well. Concatenating these bases gives a basis of $𝑊$ (Theorem 4.4.13) of common eigenvectors of $𝑇_{1}, \dots, 𝑇_{𝑘}$ . □

This is the statement of Exercise 7.4.51; the minimal polynomial supplies the one step that needed work there, that a diagonalizable operator stays diagonalizable on an invariant subspace. The theorem extends to an arbitrary set $Ω$ of commuting diagonalizable operators: $span (Ω)$ is a subspace of the finite-dimensional space $L (𝑉)$ , so it is spanned by finitely many elements $𝑇_{1}, \dots, 𝑇_{𝑘}$ of $Ω$ (keep adding elements of $Ω$ outside the span of those chosen; the list stays independent, so this stops). A basis that diagonalizes $𝑇_{1}, \dots, 𝑇_{𝑘}$ diagonalizes every linear combination of them, hence all of $Ω$ .

Example 7.6.35 (A common eigenbasis). Let

𝐴 = (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{matrix}), 𝐵 = (\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 5 \end{matrix}) in ℚ^{3 \times 3} .

Both are block diagonal with blocks of sizes $2$ and $1$ , and $𝐴$ is a scalar on each block, so $𝐴 𝐵 = 𝐵 𝐴$ . Both are diagonalizable: $𝐴$ is diagonal, and $𝐵$ has the distinct eigenvalues $1, - 1, 5$ . The eigenspaces of $𝐴$ are $𝐸_{1} (𝐴) =$ $span (𝐞_{1}, 𝐞_{2})$ and $𝐸_{2} (𝐴) = span (𝐞_{3})$ , and $𝐵$ preserves both. On $𝐸_{1} (𝐴)$ , $𝐵$ swaps $𝐞_{1}$ and $𝐞_{2}$ , with eigenvectors $𝐞_{1} + 𝐞_{2}$ (eigenvalue $1$ ) and $𝐞_{1} - 𝐞_{2}$ (eigenvalue $- 1$ ). So in the basis $(1, 1, 0),$ $(1, - 1, 0),$ $(0, 0, 1)$ ,

𝐴 \mapsto diag (1, 1, 2), 𝐵 \mapsto diag (1, - 1, 5) .

Not every eigenbasis of $𝐴$ works: $𝐞_{1}, 𝐞_{2}, 𝐞_{3}$ is one, but $𝐞_{1}$ is not an eigenvector of $𝐵$ . The common basis has to be chosen inside the eigenspaces of $𝐴$ , using $𝐵$ , which is how the proof proceeds.

7.6.7Gershgorin’s disk theorem

The eigenvalues of $𝐴 \in ℂ^{𝑛 \times 𝑛}$ are the roots of a polynomial of degree $𝑛$ , and for $𝑛 \geq 5$ there is in general no formula for them. Often a location is enough: is $0$ an eigenvalue, that is, is $𝐴$ singular? Do all eigenvalues have negative real part, so that the solutions of $𝐱^{'} = 𝐴 𝐱$ decay (Proposition 7.4.34)? If the off-diagonal entries of $𝐴$ were $0$ , the eigenvalues would be the diagonal entries. Gershgorin’s theorem says that off-diagonal entries can move the eigenvalues away from the diagonal entries, but only as far as the size of the off-diagonal entries in a row allows.

Theorem 7.6.36 (Gershgorin’s disk theorem). Let $𝐴 =$ $(𝑎_{𝑖 𝑗}) \in$ $ℂ^{𝑛 \times 𝑛}$ . For $𝑖 = 1, \dots, 𝑛$ , let $𝑅_{𝑖} =$ $\sum_{𝑗 \neq 𝑖} | 𝑎_{𝑖 𝑗} |$ and

𝐷_{𝑖} = {𝑧 \in ℂ | | 𝑧 - 𝑎_{𝑖 𝑖} | \leq 𝑅_{𝑖}},

the closed disk with center $𝑎_{𝑖 𝑖}$ and radius $𝑅_{𝑖}$ . Then every eigenvalue of $𝐴$ lies in $𝐷_{1} \cup 𝐷_{2} \cup \dots \cup 𝐷_{𝑛}$ . More precisely, if $𝐴 𝐯 = 𝜆 𝐯$ with $𝐯 \neq 𝟎$ , and $𝑖$ is an index with $| 𝑣_{𝑖} | =$ ${max}_{𝑗} | 𝑣_{𝑗} |$ , then $𝜆 \in 𝐷_{𝑖}$ .

Technique 7.6.37 (Look at the largest coordinate). To bound an eigenvalue $𝜆$ with $𝐴 𝐯 = 𝜆 𝐯$ , look at the row of this equation belonging to a coordinate $𝑣_{𝑖}$ of largest absolute value. After dividing by $𝑣_{𝑖}$ , every other coordinate enters only through the ratio $| 𝑣_{𝑗} / 𝑣_{𝑖} | \leq 1$ , so the row becomes an inequality involving the entries of $𝐴$ alone.

Reach for it when you need a bound on eigenvalues in terms of entries, or to show that a matrix with a dominant diagonal, or with nonnegative rows summing to $1$ , is invertible or has its eigenvalues in a given region. The canonical example is Theorem 7.6.36. Later: $| 𝜆 | \leq 1$ for stochastic matrices (Section 11.1) and the Perron–Frobenius theorem (Section 11.2).

Used in 3 places

Theorem 7.6.36, Exercise 7.6.54, Exercise 7.6.55.

Proof.standard

Uses. Definition 2.3.8 (Matrix–vector product), Theorem B.1.11 (Triangle inequality in $ℂ$ ), Proposition B.1.8 (Properties of the modulus)

Techniques. Look at the largest coordinate (Technique 7.6.37)

Hint 1. Write out row

𝑖

𝐴 𝐯 = 𝜆 𝐯

and move the diagonal term to the right.

Hint 2. Take absolute values, use the triangle inequality, and bound each

| 𝑣_{𝑗} |

| 𝑣_{𝑖} |

Sketch. Row

𝑖

𝐴 𝐯 = 𝜆 𝐯

gives

(𝜆 - 𝑎_{𝑖 𝑖}) 𝑣_{𝑖} =

\sum_{𝑗 \neq 𝑖} 𝑎_{𝑖 𝑗} 𝑣_{𝑗}

. Taking absolute values and using

| 𝑣_{𝑗} | \leq | 𝑣_{𝑖} |

gives

| 𝜆 -

𝑎_{𝑖 𝑖} | | 𝑣_{𝑖} | \leq

𝑅_{𝑖} | 𝑣_{𝑖} |

, and

| 𝑣_{𝑖} | > 0

since

𝐯 \neq 𝟎

Let $𝐴 𝐯 = 𝜆 𝐯$ with $𝐯 \neq 𝟎$ , and choose $𝑖$ with $| 𝑣_{𝑖} | =$ ${max}_{𝑗} | 𝑣_{𝑗} |$ ; then $| 𝑣_{𝑖} | > 0$ . Row $𝑖$ of $𝐴 𝐯 = 𝜆 𝐯$ reads $\sum_{𝑗} 𝑎_{𝑖 𝑗} 𝑣_{𝑗} =$ $𝜆 𝑣_{𝑖}$ (Definition 2.3.8), that is,

(𝜆 - 𝑎_{𝑖 𝑖}) 𝑣_{𝑖} = \sum_{𝑗 \neq 𝑖} 𝑎_{𝑖 𝑗} 𝑣_{𝑗} .

By Proposition B.1.8(d) and the triangle inequality (Theorem B.1.11),

| 𝜆 - 𝑎_{𝑖 𝑖} | | 𝑣_{𝑖} | \leq \sum_{𝑗 \neq 𝑖} | 𝑎_{𝑖 𝑗} | | 𝑣_{𝑗} | \leq \sum_{𝑗 \neq 𝑖} | 𝑎_{𝑖 𝑗} | | 𝑣_{𝑖} | = 𝑅_{𝑖} | 𝑣_{𝑖} | .

Dividing by $| 𝑣_{𝑖} | > 0$ gives $| 𝜆 - 𝑎_{𝑖 𝑖} | \leq 𝑅_{𝑖}$ , that is, $𝜆 \in 𝐷_{𝑖}$ . □

Example 7.6.38 (Three disks). Let

𝐴 = (\begin{matrix} 3 & - 2 & 0 \\ 2 & 0 & 0 \\ 0 & 1 & - 4 \end{matrix}) .

The disks are $𝐷_{1}$ with center $3$ and radius $2$ , $𝐷_{2}$ with center $0$ and radius $2$ , and $𝐷_{3}$ with center $- 4$ and radius $1$ . Expanding $det (𝑥 𝐼_{3} - 𝐴)$ along the third column gives $𝑝_{𝐴} =$ $(𝑥 + 4) (𝑥^{2} - 3 𝑥 + 4)$ , so the eigenvalues are $- 4$ and $(3 \pm 𝑖 \sqrt{7}) / 2$ . The complex pair satisfies ${| 𝜆 |}^{2} = (9 + 7) / 4 = 4$ and ${| 𝜆 - 3 |}^{2} = 9 / 4 + 7 / 4 = 4$ : each lies on the boundary of both $𝐷_{1}$ and $𝐷_{2}$ , at the two points where the circles cross. So the disks cannot be shrunk in general. Here Gershgorin cannot decide whether $𝐴$ is invertible, because $0 \in 𝐷_{2}$ ; in fact $det 𝐴 = - 16$ .

Figure 7.6.1. The Gershgorin disks of the matrix in Example 7.6.38 and its three eigenvalues (red). The eigenvalue

- 4

is the center of

𝐷_{3}

; the other two sit where the circles bounding

𝐷_{1}

and

𝐷_{2}

meet.

Corollary 7.6.39 (Diagonally dominant matrices are invertible). Let

𝐴 \in ℂ^{𝑛 \times 𝑛}

be strictly diagonally dominant:

| 𝑎_{𝑖 𝑖} | >

\sum_{𝑗 \neq 𝑖} | 𝑎_{𝑖 𝑗} |

for every

𝑖

. Then

𝐴

is invertible. The same holds for real or rational matrices.

Proof.routine

Uses. Theorem 7.6.36 (Gershgorin’s disk theorem), Theorem 7.2.9 (Eigenvalues and invertibility), Theorem 5.4.7 (Invertibility and the determinant)

Hint. Is

0

in any Gershgorin disk?

Sketch.

| 0 - 𝑎_{𝑖 𝑖} | > 𝑅_{𝑖}

, so

0 \notin 𝐷_{𝑖}

for every

𝑖

, and

0

is not an eigenvalue. A real matrix that is invertible over

ℂ

has nonzero determinant, so it is invertible over

ℝ

For every

𝑖

| 0 -

𝑎_{𝑖 𝑖} | =

| 𝑎_{𝑖 𝑖} | >

𝑅_{𝑖}

, so

0 \notin 𝐷_{𝑖}

. By Theorem 7.6.36,

0

is not an eigenvalue of

𝐴

, so

𝐴

is invertible (Theorem 7.2.9). If

𝐴

has real (or rational) entries, then

det 𝐴 \neq 0

by Theorem 5.4.7, and the determinant does not depend on whether

𝐴

is regarded as real or complex, so

𝐴

is invertible over

ℝ

(or

ℚ

) as well. □

Since $𝐴^{𝖳}$ has the same characteristic polynomial as $𝐴$ (Exercise 7.3.17), the theorem also holds with columns in place of rows, and every eigenvalue lies in the intersection of the union of the row disks with the union of the column disks.

Aside (Counting eigenvalues in the disks). A sharper form of the theorem says: if the union of some

𝑘

of the disks is disjoint from the union of the other

𝑛 - 𝑘

, then it contains exactly

𝑘

eigenvalues, counted with multiplicity. The proof follows the eigenvalues of

𝐴_{𝑡} =

diag (𝑎_{11}, \dots, 𝑎_{𝑛 𝑛}) +

𝑡 (𝐴 - diag (𝑎_{11}, \dots, 𝑎_{𝑛 𝑛}))

𝑡

runs from

0

1

. The disks of

𝐴_{𝑡}

have radii

𝑡 𝑅_{𝑖}

, so they only grow, and the eigenvalues, being roots of a polynomial whose coefficients depend continuously on

𝑡

, move continuously and cannot jump from one group of disks to the other. Making “the roots move continuously” precise takes some analysis, which we do not do here. For Example 7.6.38 it says that

𝐷_{3}

contains exactly one eigenvalue. For a real matrix, an isolated disk containing exactly one eigenvalue

𝜆

also contains

𝜆

(the disk is symmetric about

ℝ

, and

𝜆

is an eigenvalue by Proposition 7.1.38), so

𝜆

is real.

Exercises

Exercise 7.6.40.routine Find the characteristic and minimal polynomials of each matrix in $ℚ^{3 \times 3}$ , and decide whether it is diagonalizable.

𝐴 = (\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}), 𝐵 = (\begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}), 𝐶 = (\begin{matrix} 2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{matrix}) .

Solution.

Uses. Corollary 7.6.24 (The minimal polynomial divides the characteristic polynomial), Theorem 7.6.15 (Diagonalizability and the minimal polynomial), Proposition 7.3.15 (Block-triangular matrices)

Hint 1. The minimal polynomial divides

𝑝

and has the same roots. List the candidates and test them from the smallest degree up.

Hint 2. For

𝐶

, the characteristic polynomial factors along the blocks.

Sketch.

𝑝_{𝐴} = {(𝑥 - 1)}^{3}

𝐴 - 𝐼 \neq 0 = {(𝐴 - 𝐼)}^{2}

, so

𝑚_{𝐴} = {(𝑥 - 1)}^{2}

𝑝_{𝐵} = 𝑥^{3}

𝐵 \neq 0 = 𝐵^{2}

, so

𝑚_{𝐵} = 𝑥^{2}

𝑝_{𝐶} = 𝑥 {(𝑥 - 2)}^{2}

and

𝐶 (𝐶 - 2 𝐼) = 0

, so

𝑚_{𝐶} = 𝑥 (𝑥 - 2)

. Only

𝐶

is diagonalizable.

The matrix $𝐴$ . It is upper triangular with diagonal $1, 1, 1$ , so $𝑝_{𝐴} = {(𝑥 - 1)}^{3}$ . By Corollary 7.6.24(c), $𝑚_{𝐴}$ is ${(𝑥 - 1)}^{𝑘}$ for some $1 \leq 𝑘 \leq 3$ . Now $𝐴 -$ $𝐼 =$ $(\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) \neq$ $0$ and ${(𝐴 - 𝐼)}^{2} = 0$ , so $𝑚_{𝐴} = {(𝑥 - 1)}^{2}$ . It has a repeated root, so $𝐴$ is not diagonalizable (Theorem 7.6.15).

The matrix $𝐵$ . It is strictly upper triangular, so $𝑝_{𝐵} = 𝑥^{3}$ . Since $𝐵 \neq 0$ and $𝐵^{2} = 0$ (the only nonzero row of $𝐵$ is the first, and it picks out the third row of $𝐵$ , which is zero), $𝑚_{𝐵} = 𝑥^{2}$ , and $𝐵$ is not diagonalizable.

The matrix $𝐶$ . It is block diagonal with blocks $(2)$ and $𝐾 = (\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix})$ , so $𝑝_{𝐶} =$ $(𝑥 - 2) 𝑝_{𝐾} =$ $(𝑥 - 2) (𝑥^{2} - 2 𝑥) =$ $𝑥 {(𝑥 - 2)}^{2}$ (Proposition 7.3.15). The candidates are $𝑥 (𝑥 - 2)$ and $𝑥 {(𝑥 - 2)}^{2}$ . Since $𝐾^{2} = 2 𝐾$ ,

𝐶 (𝐶 - 2 𝐼) = (\begin{matrix} 2 \cdot 0 & 0 \\ 0 & 𝐾 (𝐾 - 2 𝐼_{2}) \end{matrix}) = 0,

so $𝑚_{𝐶} = 𝑥 (𝑥 - 2)$ , and $𝐶$ is diagonalizable, with eigenvalues $0, 2, 2$ .

Exercise 7.6.41.standard

(a)Let $𝑛 \geq 2$ and let $𝐽 \in 𝐹^{𝑛 \times 𝑛}$ be the matrix with every entry $1$ . Show that $𝐽^{2} = 𝑛 𝐽$ , where $𝑛$ means $1 + \dots + 1$ ( $𝑛$ terms) in $𝐹$ . Find $𝑚_{𝐽}$ , and decide when $𝐽$ is diagonalizable. (Distinguish whether $𝑛 = 0$ in $𝐹$ .)
(b)Let $𝑅 = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ . Find $𝑚_{𝑅}$ over $𝔽_{2}$ , $𝔽_{3}$ , and $𝔽_{5}$ , and decide in each case whether $𝑅$ is diagonalizable over that field. When it is, find an eigenbasis.

Solution.

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

Techniques. Use an annihilating polynomial (Technique 7.6.17), Try small cases and small fields (Technique 1.2.38)

Hint 1. Every entry of

𝐽^{2}

is a sum of

𝑛

ones. Is

𝐽

a scalar matrix?

Hint 2.

𝑅^{2} = - 𝐼_{2}

over every field. Factor

𝑥^{2} + 1

over each

𝔽_{𝑝}

by trying all

𝑝

values.

Sketch. (a)

𝐽^{2} = 𝑛 𝐽

, and

𝐽

is not scalar, so

𝑚_{𝐽} = 𝑥 (𝑥 - 𝑛)

. If

𝑛 \neq 0

𝐹

, the roots

0, 𝑛

are distinct and

𝐽

is diagonalizable; if

𝑛 = 0

𝐹

𝑚_{𝐽} = 𝑥^{2}

and it is not. (b)

𝑚_{𝑅} = 𝑥^{2} + 1

in each case. Over

𝔽_{2}

it is

{(𝑥 + 1)}^{2}

: not diagonalizable. Over

𝔽_{3}

it has no roots: no eigenvalues. Over

𝔽_{5}

it is

(𝑥 - 2) (𝑥 - 3)

: diagonalizable, with eigenvectors

(1, 3)

for

2

and

(1, 2)

for

3

(a) The $(𝑖, 𝑗)$ entry of $𝐽^{2}$ is $\sum_{𝑙 = 1}^{𝑛} 1 \cdot 1 = 𝑛$ , so $𝐽^{2} = 𝑛 𝐽$ , and $𝑥^{2} -$ $𝑛 𝑥 =$ $𝑥 (𝑥 - 𝑛)$ annihilates $𝐽$ . Since $𝑛 \geq 2$ , $𝐽$ has nonzero off-diagonal entries, so it is not a scalar matrix and $deg 𝑚_{𝐽} \geq 2$ . Hence $𝑚_{𝐽} = 𝑥 (𝑥 - 𝑛)$ (Theorem 7.6.2). If $𝑛 \neq 0$ in $𝐹$ , the roots $0$ and $𝑛$ are distinct and $𝐽$ is diagonalizable, with eigenvalues $0$ and $𝑛$ (Theorem 7.6.15). If $𝑛 = 0$ in $𝐹$ (for instance $𝐹 = 𝔽_{𝑝}$ with $𝑝$ dividing $𝑛$ ), then $𝑚_{𝐽} = 𝑥^{2}$ and $𝐽$ is not diagonalizable; it is a nonzero matrix with $𝐽^{2} = 0$ . For example, over $𝔽_{2}$ , ${(\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix})}^{2} = 0$ .

(b) Over every field $𝑅^{2} = (\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix}) = - 𝐼_{2}$ , and $𝑅$ is not scalar, so $𝑚_{𝑅} = 𝑥^{2} + 1$ .

Over $𝔽_{2}$ , $𝑥^{2} +$ $1 =$ $𝑥^{2} -$ $1 =$ ${(𝑥 - 1)}^{2}$ , with a repeated root, so $𝑅$ is not diagonalizable over $𝔽_{2}$ .

Over $𝔽_{3}$ , $𝑥^{2} + 1$ takes the values $1, 2, 2$ at $0, 1, 2$ , so it has no roots, $𝑅$ has no eigenvalues (Theorem 7.6.7), and $𝑅$ is not diagonalizable.

Over $𝔽_{5}$ , $2^{2} + 1 = 5 = 0$ and $3^{2} + 1 = 10 = 0$ , so $𝑥^{2} +$ $1 =$ $(𝑥 - 2) (𝑥 - 3)$ with distinct roots, and $𝑅$ is diagonalizable. If $𝑅 (1, 𝑎) =$ $(- 𝑎, 1) =$ $𝜆 (1, 𝑎)$ , then $𝜆 = - 𝑎$ and $1 = 𝜆 𝑎 = - 𝑎^{2}$ . For $𝑎 = 3$ : $𝜆 = - 3 = 2$ , and $𝑅 (1, 3) =$ $(- 3, 1) =$ $(2, 1) =$ $2 (1, 3)$ in $𝔽_{5}$ . For $𝑎 = 2$ : $𝜆 = - 2 = 3$ , and $𝑅 (1, 2) =$ $(- 2, 1) =$ $(3, 1) =$ $3 (1, 2)$ . So $(1, 3), (1, 2)$ is an eigenbasis, with eigenvalues $2, 3$ .

Exercise 7.6.42.standard Let $𝐴 =$ $(\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix}) \in$ $ℚ^{3 \times 3}$ .

(a)Find $𝑝_{𝐴}$ and $𝑚_{𝐴}$ .
(b)Show that $𝐴^{𝑘} =$ $(𝑘 - 1) 𝐴^{2} +$ $(2 - 𝑘) 𝐴$ for every $𝑘 \geq 1$ .
(c)Let $𝐵 = 𝐴 + 𝐼_{3}$ . Use (a) and Exercise 7.6.6 to find $𝑚_{𝐵}$ , and write $𝐵^{- 1}$ as a polynomial in $𝐵$ . Compute $𝐵^{- 1}$ .

Solution.

Uses. Corollary 7.6.24 (The minimal polynomial divides the characteristic polynomial), Theorem 7.6.20 (Cayley–Hamilton), Corollary 7.6.9 (Inverses are polynomials), Exercise 7.6.6

Techniques. Use an annihilating polynomial (Technique 7.6.17)

Hint 1.

𝐴

is triangular. For

𝑚_{𝐴}

, test the candidate

𝑥 (𝑥 - 1)

Hint 2. For (b), Cayley–Hamilton gives

𝐴^{3} = 2 𝐴^{2} - 𝐴

. Induct on

𝑘

Hint 3.

𝑚_{𝐵} (𝑥) =

𝑚_{𝐴} (𝑥 - 1)

. Expand it, and use Corollary 7.6.9.

Sketch. (a)

𝑝_{𝐴} = 𝑥 {(𝑥 - 1)}^{2}

and

𝐴 (𝐴 - 𝐼) \neq 0

, so

𝑚_{𝐴} = 𝑝_{𝐴}

. (b)

𝐴^{3} = 2 𝐴^{2} - 𝐴

; if

𝐴^{𝑘} =

(𝑘 - 1) 𝐴^{2} +

(2 - 𝑘) 𝐴

, then

𝐴^{𝑘 + 1} =

(𝑘 - 1) 𝐴^{3} +

(2 - 𝑘) 𝐴^{2} =

𝑘 𝐴^{2} +

(1 - 𝑘) 𝐴

. (c)

𝑚_{𝐵} =

(𝑥 - 1) {(𝑥 - 2)}^{2} =

𝑥^{3} -

5 𝑥^{2} +

8 𝑥 -

4

, so

𝐵^{- 1} =

\frac{1}{4} (𝐵^{2} - 5 𝐵 + 8 𝐼_{3})

(a) $𝐴$ is upper triangular with diagonal $1, 1, 0$ , so $𝑝_{𝐴} =$ $𝑥 {(𝑥 - 1)}^{2} =$ $𝑥^{3} -$ $2 𝑥^{2} +$ $𝑥$ . By Corollary 7.6.24(c), $𝑚_{𝐴}$ is $𝑥 (𝑥 - 1)$ or $𝑥 {(𝑥 - 1)}^{2}$ . Now

𝐴 (𝐴 - 𝐼_{3}) = (\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix}) (\begin{matrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & - 1 \end{matrix}) = (\begin{matrix} 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) \neq 0,

so $𝑚_{𝐴} =$ $𝑥 {(𝑥 - 1)}^{2} =$ $𝑝_{𝐴}$ .

(b) By Cayley–Hamilton, $𝐴^{3} - 2 𝐴^{2} + 𝐴 = 0$ . We induct on $𝑘$ . For $𝑘 = 1$ the formula reads $𝐴 = 0 \cdot 𝐴^{2} + 𝐴$ . If $𝐴^{𝑘} =$ $(𝑘 - 1) 𝐴^{2} +$ $(2 - 𝑘) 𝐴$ , then

𝐴^{𝑘 + 1} = (𝑘 - 1) 𝐴^{3} + (2 - 𝑘) 𝐴^{2} = (𝑘 - 1) (2 𝐴^{2} - 𝐴) + (2 - 𝑘) 𝐴^{2} = 𝑘 𝐴^{2} + (1 - 𝑘) 𝐴,

which is the formula for $𝑘 + 1$ . (Equivalently, $(𝑘 - 1) 𝑥^{2} +$ $(2 - 𝑘) 𝑥$ is the remainder of $𝑥^{𝑘}$ on division by $𝑥 {(𝑥 - 1)}^{2}$ .)

(c) By Exercise 7.6.6(b), $𝑚_{𝐵} (𝑥) =$ $𝑚_{𝐴} (𝑥 - 1) =$ $(𝑥 - 1) {(𝑥 - 2)}^{2} =$ $𝑥^{3} -$ $5 𝑥^{2} +$ $8 𝑥 -$ $4$ . Since $𝑚_{𝐵} (0) = - 4 \neq 0$ , $𝐵$ is invertible, and writing $𝑚_{𝐵} =$ $𝑥 (𝑥^{2} - 5 𝑥 + 8) -$ $4$ , Corollary 7.6.9 gives

𝐵^{- 1} = \frac{1}{4} (𝐵^{2} - 5 𝐵 + 8 𝐼_{3}) .

With $𝐵 = (\begin{matrix} 2 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{matrix})$ and $𝐵^{2} = (\begin{matrix} 4 & 4 & 4 \\ 0 & 4 & 3 \\ 0 & 0 & 1 \end{matrix})$ ,

𝐵^{- 1} = \frac{1}{4} (\begin{matrix} 2 & - 1 & - 1 \\ 0 & 2 & - 2 \\ 0 & 0 & 4 \end{matrix}) = (\begin{matrix} 1 / 2 & - 1 / 4 & - 1 / 4 \\ 0 & 1 / 2 & - 1 / 2 \\ 0 & 0 & 1 \end{matrix}) .

Multiplying by $𝐵$ confirms this.

Exercise 7.6.43.standard Let $𝐴, 𝐵 \in 𝐹^{𝑛 \times 𝑛}$ . Prove or disprove each statement.

(a)If $𝑝_{𝐴} = 𝑝_{𝐵}$ , then $𝑚_{𝐴} = 𝑚_{𝐵}$ .
(b)If $𝑝_{𝐴} = 𝑝_{𝐵}$ and $𝑚_{𝐴} = 𝑚_{𝐵}$ , then $𝐴$ and $𝐵$ are similar.
(c)If $𝐴^{2}$ is diagonalizable, then $𝐴$ is diagonalizable.
(d)If $𝐹 = ℂ$ , $𝐴$ is invertible, and $𝐴^{2}$ is diagonalizable, then $𝐴$ is diagonalizable.

Solution.

Uses. Remark 7.6.11 (Neither polynomial determines the other), Theorem 7.6.15 (Diagonalizability and the minimal polynomial), Proposition 4.5.27 (Similarity invariants)

Techniques. Finding counterexamples and minimal examples (Technique 1.4.18), Rule out similarity with an invariant (Technique 4.5.29), Use an annihilating polynomial (Technique 7.6.17)

Hint 1. For (b), look at

4 \times 4

matrices with

𝐴^{2} = 0

. Which similarity invariant can still tell them apart?

Hint 2. For (c), think of a nilpotent matrix.

Hint 3. For (d), if

𝑚_{𝐴^{2}} =

(𝑥 - 𝜇_{1}) \dots (𝑥 - 𝜇_{𝑚})

, which polynomial annihilates

𝐴

? Are its roots distinct?

Sketch. (a) False:

𝐼_{2}

and

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

. (b) False: with

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

, the block matrices

diag (𝑁, 0)

and

diag (𝑁, 𝑁)

both have

𝑝 = 𝑥^{4}

and

𝑚 = 𝑥^{2}

but ranks

1

and

2

. (c) False:

𝑁^{2} = 0

. (d) True:

𝐴

is killed by

\prod (𝑥^{2} - 𝜇_{𝑖})

, whose

2 𝑚

roots

\pm \sqrt{𝜇_{𝑖}}

are distinct because the

𝜇_{𝑖}

are distinct and nonzero.

(a) False. $𝐼_{2}$ and $𝐽 = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ both have characteristic polynomial ${(𝑥 - 1)}^{2}$ , but $𝑚_{𝐼_{2}} = 𝑥 - 1$ and $𝑚_{𝐽} = {(𝑥 - 1)}^{2}$ (Remark 7.6.11).

(b) False. Let $𝑁 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ , and in $𝐹^{4 \times 4}$ let $𝐴 = (\begin{matrix} 𝑁 & 0 \\ 0 & 0 \end{matrix})$ and $𝐵 = (\begin{matrix} 𝑁 & 0 \\ 0 & 𝑁 \end{matrix})$ in block form. Both are strictly upper triangular, so $𝑝_{𝐴} = 𝑝_{𝐵} = 𝑥^{4}$ . Both are nonzero with $𝐴^{2} = 𝐵^{2} = 0$ , so $𝑚_{𝐴} = 𝑚_{𝐵} = 𝑥^{2}$ . But $rank 𝐴 = 1$ and $rank 𝐵 = 2$ , and similar matrices have the same rank (Proposition 4.5.27), so $𝐴$ and $𝐵$ are not similar.

(c) False. $𝑁^{2} = 0$ is diagonal, but $𝑁$ is not diagonalizable, since $𝑚_{𝑁} = 𝑥^{2}$ .

(d) True. Since $𝐴^{2}$ is diagonalizable, $𝑚_{𝐴^{2}} =$ $(𝑥 - 𝜇_{1}) \dots (𝑥 - 𝜇_{𝑚})$ with distinct $𝜇_{𝑖} \in ℂ$ (Theorem 7.6.15), and every $𝜇_{𝑖}$ is an eigenvalue of $𝐴^{2}$ . As $𝐴$ is invertible, so is $𝐴^{2}$ , so every $𝜇_{𝑖} \neq 0$ . Choose $𝜈_{𝑖} \in ℂ$ with $𝜈_{𝑖}^{2} = 𝜇_{𝑖}$ . Then

𝑞 = (𝑥^{2} - 𝜇_{1}) \dots (𝑥^{2} - 𝜇_{𝑚}) = \prod_{𝑖 = 1}^{𝑚} (𝑥 - 𝜈_{𝑖}) (𝑥 + 𝜈_{𝑖})

satisfies $𝑞 (𝐴) =$ $𝑚_{𝐴^{2}} (𝐴^{2}) =$ $0$ . Its $2 𝑚$ roots $\pm 𝜈_{𝑖}$ are distinct: $𝜈_{𝑖} \neq - 𝜈_{𝑖}$ because $𝜈_{𝑖} \neq 0$ , and for $𝑖 \neq 𝑗$ the roots $\pm 𝜈_{𝑖}$ and $\pm 𝜈_{𝑗}$ have different squares $𝜇_{𝑖} \neq 𝜇_{𝑗}$ . By Theorem 7.6.15(III), $𝐴$ is diagonalizable. (Invertibility was needed: in (c), $𝜇_{1} = 0$ and $𝑥^{2}$ has a double root.)

Exercise 7.6.44.standard Find the error in the following argument.

Claim. For every $𝐴 \in ℂ^{𝑛 \times 𝑛}$ , $𝑝_{𝐴} (𝐴) = 0$ .

“Proof.” Let $𝜆$ be an eigenvalue of $𝐴$ and $𝐯$ an eigenvector for $𝜆$ . By Proposition 7.2.28, $𝑝_{𝐴} (𝐴) 𝐯 =$ $𝑝_{𝐴} (𝜆) 𝐯 =$ $𝟎$ , since $𝜆$ is a root of $𝑝_{𝐴}$ . So $𝑝_{𝐴} (𝐴)$ sends every eigenvector of $𝐴$ to $𝟎$ . A linear map that sends every vector of a spanning set to $𝟎$ is zero, so $𝑝_{𝐴} (𝐴) = 0$ .

Which matrices does the argument handle correctly? Explain why the fourth proof of Theorem 7.6.20 can be read as a repair of it.

Solution.

Uses. Proposition 7.2.28 (Polynomials act on eigenvectors by their values), Theorem 7.4.2 (Diagonalizable means an eigenbasis), Theorem 7.6.20 (Cayley–Hamilton)

Hint 1. Every step is correct except one. Do the eigenvectors of

𝐴

always span

ℂ^{𝑛}

Hint 2. Try

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

Sketch. The eigenvectors of

𝐴

span

ℂ^{𝑛}

only if

𝐴

is diagonalizable; for

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

they span only

span (𝐞_{1})

, and the argument says nothing about

𝑝_{𝑁} (𝑁) 𝐞_{2}

. So the argument proves Cayley–Hamilton exactly for diagonalizable matrices. The perturbation proof extends the diagonalizable case to all matrices.

The computation $𝑝_{𝐴} (𝐴) 𝐯 =$ $𝑝_{𝐴} (𝜆) 𝐯 =$ $𝟎$ is correct, and so is the principle that a linear map vanishing on a spanning set is zero. The error is the unstated claim that the eigenvectors of $𝐴$ span $ℂ^{𝑛}$ . They do if and only if $ℂ^{𝑛}$ has a basis of eigenvectors, that is, if and only if $𝐴$ is diagonalizable (Theorem 7.4.2). For $𝑁 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ the eigenvectors are the nonzero multiples of $𝐞_{1}$ , and the argument gives no information about $𝑝_{𝑁} (𝑁) 𝐞_{2}$ . (Here $𝑝_{𝑁} (𝑁) = 𝑁^{2} = 0$ anyway, but the argument does not show it.)

So the argument is a valid proof for diagonalizable $𝐴$ . The fourth proof of Theorem 7.6.20 starts from exactly this case and then reaches every matrix: it moves $𝐴$ to diagonalizable matrices $𝐴_{𝑡}$ , and uses that the entries of $𝑝_{𝐴_{𝑡}} (𝐴_{𝑡})$ are polynomials in $𝑡$ that vanish for all but finitely many $𝑡$ , hence also at $𝑡 = 0$ .

Exercise 7.6.45.standard Let $𝑇 \in L (𝑉)$ satisfy $𝑇^{2} = 𝑇 + 2 {id}_{𝑉}$ .

(a)Suppose $1 + 1 + 1 \neq 0$ in $𝐹$ . Show that $𝑇$ is diagonalizable with eigenvalues in ${2, - 1}$ , so that $𝑉 = 𝐸_{2} \oplus 𝐸_{- 1}$ , and write the projection of $𝑉$ onto $𝐸_{2}$ along $𝐸_{- 1}$ as a polynomial in $𝑇$ .
(b)Find $𝐴 \in 𝔽_{3}^{2 \times 2}$ with $𝐴^{2} = 𝐴 + 2 𝐼_{2}$ that is not diagonalizable.

Solution.

Uses. Proposition 7.6.12 (Distinct linear factors force a diagonalization), Theorem 7.6.15 (Diagonalizability and the minimal polynomial)

Techniques. Use an annihilating polynomial (Technique 7.6.17), Lagrange interpolation in $𝑇$ (Technique 7.6.13)

Hint 1. Factor

𝑥^{2} - 𝑥 - 2

. When are its roots distinct?

Hint 2. The projection onto

𝐸_{2}

𝐿 (𝑇)

, where

𝐿

is the Lagrange polynomial with

𝐿 (2) = 1

and

𝐿 (- 1) = 0

Hint 3. Over

𝔽_{3}

𝑥^{2} - 𝑥 - 2 = {(𝑥 + 1)}^{2}

. Look for

𝐴 = - 𝐼_{2} + 𝑁

with

𝑁^{2} = 0

Sketch. (a)

𝑥^{2} -

𝑥 -

2 =

(𝑥 - 2) (𝑥 + 1)

, and

2 - (- 1) = 3 \neq 0

, so the roots are distinct; Proposition 7.6.12 gives

𝑉 = 𝐸_{2} \oplus 𝐸_{- 1}

, and the projection is

𝐿 (𝑇) =

\frac{1}{3} (𝑇 + {id}_{𝑉})

. (b) Over

𝔽_{3}

the polynomial is

{(𝑥 + 1)}^{2}

, and

𝐴 = (\begin{matrix} 2 & 1 \\ 0 & 2 \end{matrix})

works.

(a) The polynomial $𝑞 =$ $𝑥^{2} -$ $𝑥 -$ $2 =$ $(𝑥 - 2) (𝑥 + 1)$ annihilates $𝑇$ . Its roots $2$ and $- 1$ differ by $3 \neq 0$ , so they are distinct, and Proposition 7.6.12 gives $𝑉 = 𝐸_{2} \oplus 𝐸_{- 1}$ ; $𝑇$ is diagonalizable with eigenvalues in ${2, - 1}$ . The Lagrange polynomial of the node $2$ (with the other node $- 1$ ) is $𝐿 = (𝑥 + 1) / 3$ , so $𝑃 =$ $𝐿 (𝑇) =$ $\frac{1}{3} (𝑇 + {id}_{𝑉})$ . Indeed, $𝑃$ acts on $𝐸_{2}$ as $(2 + 1) / 3 = 1$ and on $𝐸_{- 1}$ as $(- 1 + 1) / 3 = 0$ , which is the projection onto $𝐸_{2}$ along $𝐸_{- 1}$ .

(b) In $𝔽_{3}$ , $- 1 = 2$ and $- 2 = 1$ , so $𝑥^{2} -$ $𝑥 -$ $2 =$ $𝑥^{2} +$ $2 𝑥 +$ $1 =$ ${(𝑥 + 1)}^{2}$ . Let $𝐴 =$ $(\begin{matrix} 2 & 1 \\ 0 & 2 \end{matrix}) =$ $- 𝐼_{2} +$ $(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ . Then

𝐴^{2} = (\begin{matrix} 4 & 4 \\ 0 & 4 \end{matrix}) = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}), 𝐴 + 2 𝐼_{2} = (\begin{matrix} 4 & 1 \\ 0 & 4 \end{matrix}) = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}),

so $𝐴^{2} = 𝐴 + 2 𝐼_{2}$ . Since $𝐴 \neq - 𝐼_{2}$ , $𝑚_{𝐴} = {(𝑥 + 1)}^{2}$ , and $𝐴$ is not diagonalizable (Theorem 7.6.15).

Exercise 7.6.46.standard

(a)Find all $𝐴 \in 𝔽_{2}^{2 \times 2}$ with $𝐴^{2} + 𝐴 + 𝐼_{2} = 0$ .
(b)For one such $𝐴$ , show that the four matrices $0, 𝐼_{2}, 𝐴, 𝐴 + 𝐼_{2}$ are closed under addition and multiplication, and that each nonzero one has an inverse among them.

Solution.

Uses. Theorem 7.6.7 (The roots of the minimal polynomial are the eigenvalues), Theorem 7.6.2 (Existence and uniqueness of the minimal polynomial), Theorem 7.6.20 (Cayley–Hamilton), Proposition 7.3.18 (The first and last coefficients)

Techniques. Use an annihilating polynomial (Technique 7.6.17)

Hint 1.

𝑞 = 𝑥^{2} + 𝑥 + 1

has no root in

𝔽_{2}

. What does that say about

𝑚_{𝐴}

, given that

𝑚_{𝐴} ∣ 𝑞

Hint 2. So

𝑚_{𝐴} = 𝑞

, and Cayley–Hamilton forces

𝑝_{𝐴} = 𝑞

. Translate

𝑝_{𝐴} = 𝑥^{2} + 𝑥 + 1

into conditions on

tr 𝐴

and

det 𝐴

Sketch. (a)

𝑞

has no roots in

𝔽_{2}

, so

𝑚_{𝐴}

, a divisor of

𝑞

of degree at least

1

, cannot be linear:

𝑚_{𝐴} = 𝑞

. By Cayley–Hamilton

𝑞 ∣ 𝑝_{𝐴}

, so

𝑝_{𝐴} = 𝑞

, that is,

tr 𝐴 = 1

and

det 𝐴 = 1

. Conversely these give

𝐴^{2} + 𝐴 + 𝐼_{2} = 0

. The solutions are

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

and

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

. (b)

𝐴^{2} = 𝐴 + 𝐼_{2}

𝐴 (𝐴 + 𝐼_{2}) = 𝐼_{2}

, and

{(𝐴 + 𝐼_{2})}^{2} = 𝐴

(a) Let $𝑞 =$ $𝑥^{2} +$ $𝑥 +$ $1 \in$ $𝔽_{2} [𝑥]$ ; $𝑞 (0) = 𝑞 (1) = 1$ , so $𝑞$ has no roots in $𝔽_{2}$ . Suppose $𝑞 (𝐴) = 0$ . Then $𝑚_{𝐴} ∣ 𝑞$ . If $𝑚_{𝐴}$ had degree $1$ , say $𝑚_{𝐴} = 𝑥 - 𝑐$ , then $𝑐$ would be a root of $𝑞$ . So $𝑚_{𝐴} = 𝑞$ . By Theorem 7.6.20, $𝑚_{𝐴} ∣ 𝑝_{𝐴}$ , and both are monic of degree $2$ , so $𝑝_{𝐴} = 𝑞$ . By Proposition 7.3.18, $𝑝_{𝐴} =$ $𝑥^{2} -$ $(tr 𝐴) 𝑥 +$ $det 𝐴$ , so $tr 𝐴 = - 1 = 1$ and $det 𝐴 = 1$ . Conversely, if $tr 𝐴 = det 𝐴 = 1$ , then $𝑝_{𝐴} = 𝑞$ , and Cayley–Hamilton gives $𝑞 (𝐴) = 0$ .

So we need $𝐴 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ with $𝑎 + 𝑑 = 1$ and $𝑎 𝑑 - 𝑏 𝑐 = 1$ . The first condition gives $(𝑎, 𝑑) = (1, 0)$ or $(0, 1)$ ; in both cases $𝑎 𝑑 = 0$ , so $𝑏 𝑐 = - 1 = 1$ , forcing $𝑏 = 𝑐 = 1$ . The solutions are

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

(b) From $𝐴^{2} + 𝐴 + 𝐼_{2} = 0$ and $- 1 = 1$ in $𝔽_{2}$ we get $𝐴^{2} = 𝐴 + 𝐼_{2}$ . Sums: $𝐴 + 𝐴 = 0$ , $𝐴 + 𝐼_{2}$ is in the list, and $𝐴 + (𝐴 + 𝐼_{2}) = 𝐼_{2}$ , $𝐼_{2} + (𝐴 + 𝐼_{2}) = 𝐴$ ; the rest are clear. Products: $𝐴^{2} = 𝐴 + 𝐼_{2}$ , $𝐴 (𝐴 + 𝐼_{2}) =$ $𝐴^{2} +$ $𝐴 =$ $𝐼_{2}$ , and ${(𝐴 + 𝐼_{2})}^{2} =$ $𝐴^{2} +$ $𝐼_{2} =$ $𝐴$ ; products with $0$ and $𝐼_{2}$ are clear. So $𝐴^{- 1} = 𝐴 + 𝐼_{2}$ , ${(𝐴 + 𝐼_{2})}^{- 1} = 𝐴$ , and $𝐼_{2}^{- 1} = 𝐼_{2}$ . These four matrices form a field with four elements, $𝔽_{4}$ (Section 14.3).

Exercise 7.6.47.hard A vector $𝐯$ is cyclic for $𝑇$ if $𝐾_{𝑇} (𝐯) = 𝑉$ (Definition 7.2.33).

(a)Show that if $𝑇$ has a cyclic vector, then $𝑚_{𝑇} = 𝑝_{𝑇}$ .
(b)Suppose $𝑇$ is diagonalizable. Show that $𝑇$ has a cyclic vector if and only if $𝑇$ has $𝑛$ distinct eigenvalues.

Solution.

Uses. Proposition 7.2.35 (A basis of a Krylov subspace), Theorem 7.6.20 (Cayley–Hamilton), Theorem 7.6.15 (Diagonalizability and the minimal polynomial), Proposition 7.1.22 (Counting roots), Proposition 7.2.28 (Polynomials act on eigenvectors by their values)

Techniques. Krylov sequences must become dependent (Technique 7.3.28)

Hint 1. For (a): if

𝐯

is cyclic,

𝐯,

𝑇 𝐯,

\dots,

𝑇^{𝑛 - 1} 𝐯

is a basis, so no nonzero polynomial of degree less than

𝑛

kills

𝐯

Hint 2. For “

\Rightarrow

” in (b), combine (a) with Theorem 7.6.15.

Hint 3. For “

⟸

”, try the sum

𝐯

of an eigenbasis. What is

𝑞 (𝑇) 𝐯

in that basis?

Sketch. (a) By Proposition 7.2.35, no nonzero polynomial of degree less than

𝑛

kills a cyclic

𝐯

, so

deg 𝑚_{𝑇} \geq 𝑛

; with

𝑚_{𝑇} ∣ 𝑝_{𝑇}

𝑚_{𝑇} = 𝑝_{𝑇}

. (b) If

𝑇

is diagonalizable and cyclic,

𝑚_{𝑇} = \prod (𝑥 - 𝜆_{𝑖})

over the distinct eigenvalues has degree

𝑛

. Conversely, with an eigenbasis

𝐯_{𝑖}

for distinct

𝜆_{𝑖}

and

𝐯 = \sum 𝐯_{𝑖}

𝑞 (𝑇) 𝐯 =

\sum 𝑞 (𝜆_{𝑖}) 𝐯_{𝑖}

vanishes only if

𝑞

has the

𝑛

roots

𝜆_{𝑖}

, so no nonzero

𝑞

of degree less than

𝑛

kills

𝐯

, and

𝐯, \dots, 𝑇^{𝑛 - 1} 𝐯

is independent.

(a) Let $𝐯$ be cyclic. Then $𝐯 \neq 𝟎$ and $dim 𝐾_{𝑇} (𝐯) = 𝑛$ , so by Proposition 7.2.35 the Krylov sequence first becomes dependent at $𝑑 = 𝑛$ , and no nonzero polynomial of degree less than $𝑛$ sends $𝐯$ to $𝟎$ . Since $𝑚_{𝑇} (𝑇) 𝐯 = 𝟎$ , $deg 𝑚_{𝑇} \geq 𝑛$ . By Theorem 7.6.20, $𝑚_{𝑇} ∣ 𝑝_{𝑇}$ , which is monic of degree $𝑛$ , so $𝑚_{𝑇} = 𝑝_{𝑇}$ .

(b) Suppose $𝑇$ is diagonalizable with a cyclic vector. By (a), $deg 𝑚_{𝑇} = 𝑛$ , and by Theorem 7.6.15, $𝑚_{𝑇}$ is the product of $𝑥 - 𝜆$ over the distinct eigenvalues $𝜆$ . So $𝑇$ has $𝑛$ distinct eigenvalues.

Conversely, let $𝐯_{1}, \dots, 𝐯_{𝑛}$ be eigenvectors for distinct eigenvalues $𝜆_{1}, \dots, 𝜆_{𝑛}$ ; they form a basis (Corollary 7.4.5). Let $𝐯 = 𝐯_{1} + \dots + 𝐯_{𝑛}$ . For $𝑞 \in 𝐹 [𝑥]$ , Proposition 7.2.28 gives

𝑞 (𝑇) 𝐯 = 𝑞 (𝜆_{1}) 𝐯_{1} + \dots + 𝑞 (𝜆_{𝑛}) 𝐯_{𝑛},

which is $𝟎$ only if $𝑞 (𝜆_{𝑖}) = 0$ for every $𝑖$ . A nonzero polynomial of degree less than $𝑛$ has fewer than $𝑛$ roots (Proposition 7.1.22), so no such polynomial kills $𝐯$ . A linear relation among $𝐯,$ $𝑇 𝐯,$ $\dots,$ $𝑇^{𝑛 - 1} 𝐯$ is a polynomial of degree less than $𝑛$ killing $𝐯$ , so these $𝑛$ vectors are independent, hence a basis of $𝑉$ , and $𝐾_{𝑇} (𝐯) = 𝑉$ .

Exercise 7.6.48.standard Let

𝐯

be a cyclic vector for

𝑇

(as in Exercise 7.6.47). Show that

𝑆 \in L (𝑉)

commutes with

𝑇

if and only if

𝑆 = 𝑞 (𝑇)

for some

𝑞 \in 𝐹 [𝑥]

. Deduce that the operators commuting with

𝑇

form a subspace of

L (𝑉)

of dimension

𝑛

Solution.

Uses. Definition 7.2.33 (Krylov subspace), Proposition 7.2.35 (A basis of a Krylov subspace), Proposition 7.2.27 (Evaluating polynomials at an operator), Theorem 2.2.1 (Linear maps are determined by a basis), Proposition 7.6.4 (The minimal polynomial is the first dependence among the powers), Exercise 7.6.47

Techniques. Check it on a basis (Technique 2.2.7)

Hint 1.

𝑆 𝐯

lies in

𝐾_{𝑇} (𝐯) = 𝑉

, so

𝑆 𝐯 = 𝑞 (𝑇) 𝐯

for some

𝑞

Hint 2. Compare

𝑆

and

𝑞 (𝑇)

on the basis

𝐯,

𝑇 𝐯,

\dots,

𝑇^{𝑛 - 1} 𝐯

, using

𝑆 𝑇^{𝑘} = 𝑇^{𝑘} 𝑆

Sketch. Polynomials in

𝑇

commute with

𝑇

. Conversely, write

𝑆 𝐯 = 𝑞 (𝑇) 𝐯

; then

𝑆 𝑇^{𝑘} 𝐯 =

𝑇^{𝑘} 𝑆 𝐯 =

𝑞 (𝑇) 𝑇^{𝑘} 𝐯

, so

𝑆

and

𝑞 (𝑇)

agree on a basis. The commutant is therefore

𝐹 [𝑇]

, of dimension

deg 𝑚_{𝑇} = 𝑛

If $𝑆 = 𝑞 (𝑇)$ , then $𝑆 𝑇 = 𝑇 𝑆$ by Proposition 7.2.27(b). Conversely, suppose $𝑆 𝑇 = 𝑇 𝑆$ . Since $𝑆 𝐯 \in 𝑉 = 𝐾_{𝑇} (𝐯)$ , there is $𝑞 \in 𝐹 [𝑥]$ with $𝑆 𝐯 = 𝑞 (𝑇) 𝐯$ (Definition 7.2.33). For every $𝑘 \geq 0$ , $𝑆 𝑇^{𝑘} = 𝑇^{𝑘} 𝑆$ (by induction on $𝑘$ ), and $𝑞 (𝑇)$ commutes with $𝑇^{𝑘}$ , so

𝑆 (𝑇^{𝑘} 𝐯) = 𝑇^{𝑘} (𝑆 𝐯) = 𝑇^{𝑘} 𝑞 (𝑇) 𝐯 = 𝑞 (𝑇) (𝑇^{𝑘} 𝐯) .

So $𝑆$ and $𝑞 (𝑇)$ agree on the basis $𝐯,$ $𝑇 𝐯,$ $\dots,$ $𝑇^{𝑛 - 1} 𝐯$ of $𝑉$ (Proposition 7.2.35), and hence $𝑆 = 𝑞 (𝑇)$ (Theorem 2.2.1).

So the operators commuting with $𝑇$ are exactly the elements of $𝐹 [𝑇]$ . By Proposition 7.6.4, $dim 𝐹 [𝑇] = deg 𝑚_{𝑇}$ , and $𝑚_{𝑇} = 𝑝_{𝑇}$ has degree $𝑛$ by Exercise 7.6.47(a).

Exercise 7.6.49.hard Let

𝑉

be a finite-dimensional complex vector space and

𝑇 \in L (𝑉)

. Show that

𝑇

is diagonalizable if and only if every

𝑇

-invariant subspace

𝑈

𝑉

has a

𝑇

-invariant complement, that is, a

𝑇

-invariant

𝑊

with

𝑉 = 𝑈 \oplus 𝑊

. Does “

⟸

” hold over

ℝ

Solution.

Uses. Corollary 7.6.16 (Restrictions of diagonalizable operators), Corollary 7.2.23 (Eigenspaces are independent), Proposition 4.4.16 (Complements exist), Theorem 4.4.13 (Bases and dimension of a sum), Theorem 7.3.27 (Existence of eigenvalues over $ℂ$ )

Techniques. Glue bases of the summands (Technique 4.4.14), Extremal choice and minimal counterexamples (Technique 1.5.9)

Hint 1. For “

\Rightarrow

”:

𝑇 ↾_{𝑈}

is diagonalizable, so

𝑈

is the direct sum of the pieces

𝑈 \cap 𝐸_{𝜆}

. Complete each piece inside

𝐸_{𝜆}

Hint 2. For “

⟸

”: apply the hypothesis to

𝑈 = 𝐸_{𝜆_{1}} \oplus \dots \oplus 𝐸_{𝜆_{𝑚}}

, the sum of all eigenspaces. What happens if its invariant complement is nonzero?

Sketch. (

\Rightarrow

)

𝑈 = \oplus_{𝜆} (𝑈 \cap 𝐸_{𝜆})

by Corollary 7.6.16. Choose a complement

𝑊_{𝜆}

𝑈 \cap 𝐸_{𝜆}

inside

𝐸_{𝜆}

and let

𝑊 = \sum 𝑊_{𝜆}

;

𝑇

acts on

𝑊_{𝜆}

𝜆

, so

𝑊

is invariant, and concatenated bases show

𝑉 = 𝑈 \oplus 𝑊

. (

⟸

) Let

𝑈

be the sum of all eigenspaces and

𝑊

an invariant complement. If

𝑊 \neq {𝟎}

𝑇 ↾_{𝑊}

has an eigenvector, which lies in

𝑈 \cap 𝑊 = {𝟎}

. So

𝑈 = 𝑉

. Over

ℝ

, a rotation of

ℝ^{2}

is a counterexample.

$(\Rightarrow)$ Let $𝜆_{1}, \dots, 𝜆_{𝑚}$ be the distinct eigenvalues of $𝑇$ , so $𝑉 = 𝐸_{𝜆_{1}} \oplus \dots \oplus 𝐸_{𝜆_{𝑚}}$ , and let $𝑈$ be $𝑇$ -invariant. If $𝑈 = {𝟎}$ , take $𝑊 = 𝑉$ . Otherwise $𝑇 ↾_{𝑈}$ is diagonalizable (Corollary 7.6.16), and each of its eigenvectors lies in some $𝑈_{𝑖} = 𝑈 \cap 𝐸_{𝜆_{𝑖}}$ , so $𝑈 = 𝑈_{1} + \dots + 𝑈_{𝑚}$ . For each $𝑖$ , choose a complement $𝑊_{𝑖}$ of $𝑈_{𝑖}$ inside $𝐸_{𝜆_{𝑖}}$ (Proposition 4.4.16), and let $𝑊 = 𝑊_{1} + \dots + 𝑊_{𝑚}$ . Since $𝑇$ acts on $𝑊_{𝑖} \subseteq 𝐸_{𝜆_{𝑖}}$ as multiplication by $𝜆_{𝑖}$ , $𝑊$ is $𝑇$ -invariant.

Choose bases ${𝒞︀}_{𝑖}$ of $𝑈_{𝑖}$ and ${𝒟︀}_{𝑖}$ of $𝑊_{𝑖}$ . Since $𝐸_{𝜆_{𝑖}} = 𝑈_{𝑖} \oplus 𝑊_{𝑖}$ , the list ${𝒞︀}_{𝑖}, {𝒟︀}_{𝑖}$ is a basis of $𝐸_{𝜆_{𝑖}}$ , and since $𝑉$ is the direct sum of the $𝐸_{𝜆_{𝑖}}$ , all these lists together form a basis of $𝑉$ (Theorem 4.4.13(b)). The vectors from the ${𝒞︀}_{𝑖}$ span $𝑈$ , and those from the ${𝒟︀}_{𝑖}$ span $𝑊$ . A basis of $𝑉$ split into two parts, one spanning $𝑈$ and the other spanning $𝑊$ , shows that $𝑈 + 𝑊 = 𝑉$ and that each vector has only one decomposition, so $𝑉 = 𝑈 \oplus 𝑊$ (Theorem 4.4.13(b)).

$(⟸)$ Let $𝑈 = 𝐸_{𝜆_{1}} + \dots + 𝐸_{𝜆_{𝑚}}$ , the sum of all eigenspaces, which is direct (Corollary 7.2.23) and $𝑇$ -invariant. By hypothesis $𝑉 = 𝑈 \oplus 𝑊$ with $𝑊$ invariant. If $𝑊 \neq {𝟎}$ , then $𝑇 ↾_{𝑊}$ has an eigenvector $𝐰$ (Theorem 7.3.27), and $𝐰$ is an eigenvector of $𝑇$ , so $𝐰 \in 𝑈 \cap 𝑊 = {𝟎}$ , a contradiction. So $𝑊 = {𝟎}$ , $𝑉 = 𝑈$ , and concatenating bases of the eigenspaces gives an eigenbasis.

Over $ℝ$ , “ $⟸$ ” fails: the rotation $𝑅 = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ of $ℝ^{2}$ has no invariant lines, so its only invariant subspaces are ${𝟎}$ and $ℝ^{2}$ , which are complements of each other. Yet $𝑅$ is not diagonalizable over $ℝ$ .

Exercise 7.6.50.hard Let

𝐴 \in ℝ^{𝑛 \times 𝑛}

satisfy

𝐴^{3} = 𝐴 + 𝐼_{𝑛}

. Show that

det 𝐴 > 0

Solution.

Uses. Proposition 7.2.28 (Polynomials act on eigenvectors by their values), Theorem 7.5.22 (Trace and determinant from eigenvalues), Proposition 7.1.38 (Conjugate roots), Proposition B.1.8 (Properties of the modulus)

Techniques. Use an annihilating polynomial (Technique 7.6.17)

Hint 1. Every complex eigenvalue of

𝐴

is a root of

𝑓 = 𝑥^{3} - 𝑥 - 1

. The determinant is the product of the complex eigenvalues with multiplicity.

Hint 2. Show that every real root of

𝑓

is positive. What do the nonreal eigenvalues contribute to the product?

Sketch. The complex eigenvalues are roots of

𝑓 = 𝑥^{3} - 𝑥 - 1

. A real root

𝑟

satisfies

𝑟 (𝑟^{2} - 1) = 1

; if

𝑟 < 0

this needs

- 1 < 𝑟 < 0

, but then

𝑟 (𝑟^{2} - 1) =

| 𝑟 | (1 - 𝑟^{2}) <

1

. So real eigenvalues are positive. Nonreal eigenvalues pair up with their conjugates, with equal multiplicity, contributing

{| 𝜆 |}^{2} > 0

. So

det 𝐴

, the product, is positive.

Regard $𝐴$ as a complex matrix, and let $𝜆_{1}, \dots, 𝜆_{𝑛}$ be its eigenvalues counted with multiplicity, so that $det 𝐴 = 𝜆_{1} \dots 𝜆_{𝑛}$ (Theorem 7.5.22). Since $𝑓 = 𝑥^{3} - 𝑥 - 1$ annihilates $𝐴$ , every $𝜆_{𝑗}$ is a root of $𝑓$ (Proposition 7.2.28).

Real eigenvalues are positive. Let $𝑟 \in ℝ$ with $𝑓 (𝑟) = 0$ , that is, $𝑟 (𝑟^{2} - 1) = 1$ . Clearly $𝑟 \neq 0$ . If $𝑟 < 0$ , then $𝑟^{2} - 1 < 0$ is needed for the product to be positive, so $- 1 < 𝑟 < 0$ ; but then $𝑟 (𝑟^{2} - 1) =$ $| 𝑟 | (1 - 𝑟^{2}) \leq$ $| 𝑟 | <$ $1$ , a contradiction. So $𝑟 > 0$ .

Nonreal eigenvalues pair up. The polynomial $𝑝_{𝐴}$ has real coefficients, so by Proposition 7.1.38 each nonreal $𝜆$ and its conjugate $𝜆 \neq 𝜆$ occur in the list $𝜆_{1}, \dots, 𝜆_{𝑛}$ equally often. Grouping each such $𝜆$ with a copy of $𝜆$ , the pair contributes $𝜆 𝜆 = {| 𝜆 |}^{2} > 0$ to the product (Proposition B.1.8).

So $det 𝐴$ is a product of positive real eigenvalues and positive numbers ${| 𝜆 |}^{2}$ , and $det 𝐴 > 0$ .

Exercise 7.6.51.standard Let $𝐴, 𝐵 \in ℂ^{𝑛 \times 𝑛}$ .

(a)Suppose $𝐴 𝐵 = 𝐵 𝐴$ and $𝐴^{𝑘} = 𝐵^{𝑘} = 0$ for some $𝑘$ . Show that $𝐴 + 𝐵$ and $𝐴 𝐵$ are nilpotent.
(b)Show by example that (a) fails without $𝐴 𝐵 = 𝐵 𝐴$ .
(c)Suppose $𝐴$ and $𝐵$ are both upper triangular. Show that $𝐴 𝐵 - 𝐵 𝐴$ is nilpotent. Deduce that $𝐴 𝐵 - 𝐵 𝐴$ is nilpotent whenever $𝐴$ and $𝐵$ can be triangularized by the same basis, and give two upper-triangular matrices that do not commute.

Solution.

Uses. Corollary 7.6.33 (Eigenvalues of sums and products of commuting operators), Proposition 7.2.28 (Polynomials act on eigenvectors by their values), Theorem 7.6.20 (Cayley–Hamilton), Proposition 7.5.9 (Multiplicities on the diagonal), Corollary 7.1.35 (Factorization over $ℂ$ )

Techniques. Prove it for triangular matrices first (Technique 7.5.23)

Hint 1. A nilpotent matrix has

0

as its only eigenvalue. Use Corollary 7.6.33.

Hint 2. A complex matrix whose only eigenvalue is

0

has

𝑝 = 𝑥^{𝑛}

. Then apply Cayley–Hamilton.

Hint 3. For (c), compare the diagonals of

𝐴 𝐵

and

𝐵 𝐴

Sketch. (a) All eigenvalues of

𝐴

and

𝐵

are

0

, so by Corollary 7.6.33 all eigenvalues of

𝐴 + 𝐵

and

𝐴 𝐵

are

0

; then

𝑝 = 𝑥^{𝑛}

and Cayley–Hamilton gives

{(𝐴 + 𝐵)}^{𝑛} =

0 =

{(𝐴 𝐵)}^{𝑛}

. (b)

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

and

𝑁^{𝖳}

. (c)

𝐴 𝐵

and

𝐵 𝐴

have the same diagonal

𝑎_{𝑖 𝑖} 𝑏_{𝑖 𝑖}

, so

𝐴 𝐵 - 𝐵 𝐴

is strictly upper triangular, with

𝑝 = 𝑥^{𝑛}

. Example:

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

and

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

(a) If $𝐴^{𝑘} = 0$ , every eigenvalue of $𝐴$ is a root of $𝑥^{𝑘}$ (Proposition 7.2.28), so $0$ is the only eigenvalue of $𝐴$ ; likewise for $𝐵$ . By Corollary 7.6.33, every eigenvalue of $𝐴 + 𝐵$ is a sum $0 + 0$ and every eigenvalue of $𝐴 𝐵$ is a product $0 \cdot 0$ . So $𝑝_{𝐴 + 𝐵}$ is a monic polynomial of degree $𝑛$ whose only complex root is $0$ , that is, $𝑝_{𝐴 + 𝐵} = 𝑥^{𝑛}$ (Corollary 7.1.35), and Cayley–Hamilton gives ${(𝐴 + 𝐵)}^{𝑛} = 0$ . In the same way ${(𝐴 𝐵)}^{𝑛} = 0$ .

(A direct argument also works: since $𝐴$ and $𝐵$ commute, the binomial theorem gives ${(𝐴 + 𝐵)}^{2 𝑘} =$ $\sum_{𝑗} (\binom{2 𝑘}{𝑗}) 𝐴^{𝑗} 𝐵^{2 𝑘 - 𝑗}$ , and in each term $𝑗 \geq 𝑘$ or $2 𝑘 - 𝑗 \geq 𝑘$ . And ${(𝐴 𝐵)}^{𝑘} =$ $𝐴^{𝑘} 𝐵^{𝑘} =$ $0$ .)

(b) $𝑁 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ and $𝑁^{𝖳}$ satisfy $𝑁^{2} = {(𝑁^{𝖳})}^{2} = 0$ , but $𝑁 + 𝑁^{𝖳} = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ squares to $𝐼_{2}$ , and $𝑁 𝑁^{𝖳} = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ is a nonzero projection; neither is nilpotent.

(c) As in the proof of Theorem 7.6.28, $𝐴 𝐵$ and $𝐵 𝐴$ are upper triangular with the same diagonal entries $𝑎_{𝑖 𝑖} 𝑏_{𝑖 𝑖}$ . So $𝐶 = 𝐴 𝐵 - 𝐵 𝐴$ is upper triangular with zero diagonal, $𝑝_{𝐶} = 𝑥^{𝑛}$ by Proposition 7.5.9, and $𝐶^{𝑛} = 0$ by Cayley–Hamilton. If $𝐴 = 𝑆 𝑅 𝑆^{- 1}$ and $𝐵 = 𝑆 𝑅^{'} 𝑆^{- 1}$ with $𝑅, 𝑅^{'}$ upper triangular, then $𝐴 𝐵 -$ $𝐵 𝐴 =$ $𝑆 (𝑅 𝑅^{'} - 𝑅^{'} 𝑅) 𝑆^{- 1}$ , and ${(𝐴 𝐵 - 𝐵 𝐴)}^{𝑛} =$ $𝑆 {(𝑅 𝑅^{'} - 𝑅^{'} 𝑅)}^{𝑛} 𝑆^{- 1} =$ $0$ .

For $𝑋 = (\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix})$ and $𝑌 = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})$ , $𝑋 𝑌 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ and $𝑌 𝑋 = 0$ . So matrices can be triangular in the same basis without commuting; the converse of Theorem 7.6.32 fails.

Exercise 7.6.52.standard Let

𝐴, 𝐵 \in ℝ^{𝑛 \times 𝑛}

with

𝐴 𝐵 = 𝐵 𝐴

. Show that

det (𝐴^{2} + 𝐵^{2}) \geq 0

. Find

𝐴, 𝐵 \in ℝ^{2 \times 2}

with

det (𝐴^{2} + 𝐵^{2}) < 0

Solution.

Uses. Theorem 5.4.10 (Multiplicativity), Proposition 7.3.2 (Determinants over a commutative ring), Proposition B.1.8 (Properties of the modulus)

Hint 1. Factor

𝐴^{2} + 𝐵^{2}

over

ℂ

, the way

𝑎^{2} +

𝑏^{2} =

(𝑎 + 𝑖 𝑏) (𝑎 - 𝑖 𝑏)

. Where is

𝐴 𝐵 = 𝐵 𝐴

used?

Hint 2. How are

det (𝐴 + 𝑖 𝐵)

and

det (𝐴 - 𝑖 𝐵)

related when

𝐴, 𝐵

are real?

Sketch. Since

𝐴 𝐵 = 𝐵 𝐴

(𝐴 + 𝑖 𝐵) (𝐴 - 𝑖 𝐵) =

𝐴^{2} +

𝐵^{2}

. As

𝐴 - 𝑖 𝐵

is the entrywise conjugate of

𝐴 + 𝑖 𝐵

det (𝐴 - 𝑖 𝐵) =

det (𝐴 + 𝑖 𝐵)

, so

det (𝐴^{2} + 𝐵^{2}) =

{| det (𝐴 + 𝑖 𝐵) |}^{2} \geq

0

. For

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

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

det (𝐴^{2} + 𝐵^{2}) = - 1

Regard $𝐴$ and $𝐵$ as complex matrices. Then

(𝐴 + 𝑖 𝐵) (𝐴 - 𝑖 𝐵) = 𝐴^{2} - 𝑖 𝐴 𝐵 + 𝑖 𝐵 𝐴 + 𝐵^{2} = 𝐴^{2} + 𝐵^{2},

because $𝐴 𝐵 = 𝐵 𝐴$ . Since $𝐴$ and $𝐵$ are real, $𝐴 - 𝑖 𝐵$ is the entrywise complex conjugate of $𝐴 + 𝑖 𝐵$ . Complex conjugation $ℂ \to ℂ$ respects sums and products and fixes $1$ , so Proposition 7.3.2(g) gives $det (𝐴 - 𝑖 𝐵) =$ $det (𝐴 + 𝑖 𝐵)$ . By Theorem 5.4.10 and Proposition B.1.8,

det (𝐴^{2} + 𝐵^{2}) = det (𝐴 + 𝑖 𝐵) det (𝐴 + 𝑖 𝐵) = {| det (𝐴 + 𝑖 𝐵) |}^{2} \geq 0 .

Without commuting this fails. Let $𝐴 = (\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix})$ and $𝐵 = (\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix})$ . Then $𝐴^{2} = 𝐴$ and $𝐵^{2} = 𝐵$ , so $𝐴^{2} + 𝐵^{2} = (\begin{matrix} 2 & 1 \\ 1 & 0 \end{matrix})$ , with determinant $- 1$ . (Here $𝐴 𝐵 =$ $(\begin{matrix} 2 & 0 \\ 0 & 0 \end{matrix}) \neq$ $(\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}) =$ $𝐵 𝐴$ .)

Exercise 7.6.53.standard Let

𝐴 = (\begin{matrix} 2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{matrix}), 𝐵 = (\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{matrix}) .

(a)Check that $𝐴 𝐵 = 𝐵 𝐴$ and that $𝐴$ and $𝐵$ are diagonalizable over $ℝ$ .
(b)Find a basis of $ℝ^{3}$ consisting of common eigenvectors of $𝐴$ and $𝐵$ .
(c)Is there such a basis of $ℚ^{3}$ ?

Solution.

Uses. Theorem 7.6.34 (Simultaneous diagonalization), Proposition 7.2.30 (Kernels and images of commuting operators are invariant), Corollary 7.4.5 (Distinct eigenvalues give a diagonalization)

Hint 1. Find the eigenspaces of

𝐴

first.

𝐵

must preserve them.

Hint 2. On the plane

𝐸_{2} (𝐴)

, use the basis

𝐞_{1}

𝐟 = 𝐞_{2} + 𝐞_{3}

and write down the

2 \times 2

matrix of

𝐵

Hint 3. For (c), what are the eigenvalues of

𝐵

Sketch. (a) Both products equal

(\begin{matrix} 0 & 2 & 2 \\ 2 & 0 & 0 \\ 2 & 0 & 0 \end{matrix})

𝐸_{2} (𝐴) =

span (𝐞_{1}, 𝐞_{2} + 𝐞_{3})

and

𝐸_{0} (𝐴) =

span (𝐞_{2} - 𝐞_{3})

;

𝐵

has the three distinct eigenvalues

0, \pm \sqrt{2}

. (b) On

𝐸_{2} (𝐴)

𝐵

has matrix

(\begin{matrix} 0 & 2 \\ 1 & 0 \end{matrix})

, with eigenvectors

\pm \sqrt{2} 𝐞_{1} + 𝐟

. The basis is

(\sqrt{2}, 1, 1),

(- \sqrt{2}, 1, 1),

(0, 1, - 1)

. (c) No:

𝐵

has irrational eigenvalues, so it is not diagonalizable over

ℚ

(a) Computing,

𝐴 𝐵 = (\begin{matrix} 0 & 2 & 2 \\ 2 & 0 & 0 \\ 2 & 0 & 0 \end{matrix}) = 𝐵 𝐴 .

For $𝐴$ : $𝐴 𝐞_{1} = 2 𝐞_{1}$ , $𝐴 (0, 1, 1) = (0, 2, 2)$ , and $𝐴 (0, 1, - 1) = 𝟎$ . These three vectors form a basis, so $𝐴$ is diagonalizable with $𝐸_{2} (𝐴) =$ $span (𝐞_{1}, 𝐞_{2} + 𝐞_{3})$ and $𝐸_{0} (𝐴) =$ $span (𝐞_{2} - 𝐞_{3})$ . For $𝐵$ , part (b) produces three eigenvectors for the distinct eigenvalues $\sqrt{2}, - \sqrt{2}, 0$ , so $𝐵$ is diagonalizable over $ℝ$ (Corollary 7.4.5).

(b) By Proposition 7.2.30(b), $𝐵$ preserves the eigenspaces of $𝐴$ . Let $𝐟 = 𝐞_{2} + 𝐞_{3}$ . Then $𝐵 𝐞_{1} = 𝐟$ and $𝐵 𝐟 = 2 𝐞_{1}$ , so on $𝐸_{2} (𝐴)$ , in the basis $𝐞_{1}, 𝐟$ , $𝐵$ has matrix $(\begin{matrix} 0 & 2 \\ 1 & 0 \end{matrix})$ , with characteristic polynomial $𝑥^{2} - 2$ . One checks

𝐵 (\sqrt{2} 𝐞_{1} + 𝐟) = \sqrt{2} 𝐟 + 2 𝐞_{1} = \sqrt{2} (\sqrt{2} 𝐞_{1} + 𝐟), 𝐵 (- \sqrt{2} 𝐞_{1} + 𝐟) = - \sqrt{2} (- \sqrt{2} 𝐞_{1} + 𝐟) .

On $𝐸_{0} (𝐴)$ , $𝐵 (𝐞_{2} - 𝐞_{3}) = 𝟎$ . So in the basis $(\sqrt{2}, 1, 1),$ $(- \sqrt{2}, 1, 1),$ $(0, 1, - 1)$ of $ℝ^{3}$ ,

𝐴 \mapsto diag (2, 2, 0), 𝐵 \mapsto diag (\sqrt{2}, - \sqrt{2}, 0) .

(c) No. If $𝐯 \in ℚ^{3}$ is an eigenvector of $𝐵$ with eigenvalue $𝜆$ , then $𝜆 = {(𝐵 𝐯)}_{𝑖} / 𝑣_{𝑖}$ for any $𝑖$ with $𝑣_{𝑖} \neq 0$ , so $𝜆 \in ℚ$ . The eigenvalues of $𝐵$ are $\pm \sqrt{2}$ and $0$ , so the only rational eigenvectors of $𝐵$ lie in the line $𝐸_{0} (𝐵)$ , and $ℚ^{3}$ has no basis of eigenvectors of $𝐵$ . Simultaneous diagonalization needs each operator to be diagonalizable over the field at hand.

Exercise 7.6.54.standard Let $𝐺 =$ $(\begin{matrix} 6 & 1 & - 1 \\ 1 & 5 & 2 \\ 0 & - 1 & 4 \end{matrix}) \in$ $ℝ^{3 \times 3}$ .

(a)Show that every complex eigenvalue $𝜆$ of $𝐺$ satisfies $Re 𝜆 \geq 2$ . Deduce that $𝐺$ is invertible.
(b)Show that $det 𝐺 > 0$ without computing it.
(c)Show that $| 𝜆 | \leq 7$ for every eigenvalue, and explain why the row disks alone give only $| 𝜆 | \leq 8$ .

Solution.

Uses. Theorem 7.6.36 (Gershgorin’s disk theorem), Theorem 7.5.22 (Trace and determinant from eigenvalues), Proposition 7.1.38 (Conjugate roots), Exercise 7.3.17

Techniques. Look at the largest coordinate (Technique 7.6.37)

Hint 1. Find the three disks. How far left does each reach?

Hint 2. The determinant is the product of the complex eigenvalues. What can a real eigenvalue be, and what does a conjugate pair contribute?

Hint 3. Apply Gershgorin to

𝐺^{𝖳}

, which has the same eigenvalues.

Sketch. (a) The disks have centers

6, 5, 4

and radii

2, 3, 1

, so they lie in

Re 𝑧 \geq 4, 2, 3

. (b) Real eigenvalues are

\geq 2

, nonreal ones pair with their conjugates, so

det 𝐺

is a product of positive numbers. (c) The column disks have radii

1, 2, 3

, so

| 𝜆 | \leq 6 + 1, 5 + 2, 4 + 3

; the row disks give

6 + 2, 5 + 3, 4 + 1

(a) The row disks are $𝐷_{1}$ : center $6$ , radius $1 + 1 = 2$ ; $𝐷_{2}$ : center $5$ , radius $1 + 2 = 3$ ; $𝐷_{3}$ : center $4$ , radius $0 + 1 = 1$ . A point $𝑧$ of the disk with center $𝑐 \in ℝ$ and radius $𝑟$ has $Re 𝑧 \geq$ $𝑐 -$ $| 𝑧 -$ $𝑐 | \geq$ $𝑐 -$ $𝑟$ , which is $4$ , $2$ , $3$ for the three disks. By Theorem 7.6.36 every eigenvalue lies in one of them, so $Re 𝜆 \geq 2$ . In particular $0$ is not an eigenvalue, and $𝐺$ is invertible.

(b) By Theorem 7.5.22, $det 𝐺 = 𝜆_{1} 𝜆_{2} 𝜆_{3}$ , the product of the complex eigenvalues with multiplicity. A real eigenvalue is at least $2$ by (a). A nonreal eigenvalue $𝜆$ occurs as often as $𝜆$ (Proposition 7.1.38, as $𝑝_{𝐺}$ has real coefficients), and each such pair contributes ${| 𝜆 |}^{2} > 0$ . So $det 𝐺 > 0$ . (In fact $det 𝐺 = 129$ .)

(c) $𝐺^{𝖳}$ has the same characteristic polynomial (Exercise 7.3.17), hence the same eigenvalues, and its row disks are the column disks of $𝐺$ : centers $6, 5, 4$ and radii $1 + 0 = 1$ , $1 + 1 = 2$ , $1 + 2 = 3$ . A point of the disk with center $𝑐$ and radius $𝑟$ has $| 𝑧 | \leq | 𝑐 | + 𝑟$ , which here is $7$ in all three cases. So $| 𝜆 | \leq 7$ . The row disks give $6 + 2 = 8$ , $5 + 3 = 8$ , and $4 + 1 = 5$ , so only $| 𝜆 | \leq 8$ .

Exercise 7.6.55.hard Let $𝐴 \in ℝ^{𝑛 \times 𝑛}$ have nonnegative entries, with every row summing to $1$ .

(a)Show that every complex eigenvalue $𝜆$ of $𝐴$ satisfies $| 𝜆 | \leq 1$ , and that $1$ is an eigenvalue.
(b)Suppose moreover that $𝑎_{𝑖 𝑖} > 0$ for every $𝑖$ . Show that $1$ is the only eigenvalue with $| 𝜆 | = 1$ .
(c)Show that (b) fails without the assumption $𝑎_{𝑖 𝑖} > 0$ .

Solution.

Uses. Theorem 7.6.36 (Gershgorin’s disk theorem), Theorem B.1.11 (Triangle inequality in $ℂ$ )

Techniques. Look at the largest coordinate (Technique 7.6.37)

Hint 1. The disk

𝐷_{𝑖}

has center

𝑎_{𝑖 𝑖}

and radius

1 - 𝑎_{𝑖 𝑖}

. Which vector does

𝐴

fix?

Hint 2. If

| 𝜆 | = 1

and

𝜆 \in 𝐷_{𝑖}

, then

1 =

| 𝜆 | \leq

| 𝜆 -

𝑎_{𝑖 𝑖} | +

𝑎_{𝑖 𝑖} \leq

1

. When is the triangle inequality an equality?

Sketch. (a)

𝐷_{𝑖}

has center

𝑎_{𝑖 𝑖} \geq 0

and radius

1 - 𝑎_{𝑖 𝑖}

, so it lies in the closed unit disk;

𝐴 (1, \dots, 1) = (1, \dots, 1)

. (b) If

| 𝜆 | = 1

, equality holds in

| 𝜆 | \leq

| 𝜆 -

𝑎_{𝑖 𝑖} | +

𝑎_{𝑖 𝑖}

, so

𝜆 - 𝑎_{𝑖 𝑖}

is a nonnegative multiple of the positive number

𝑎_{𝑖 𝑖}

, and

𝜆

is a positive real with

| 𝜆 | = 1

. (c)

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

has eigenvalue

- 1

(a) Since the entries are nonnegative and row $𝑖$ sums to $1$ , $𝑅_{𝑖} =$ $\sum_{𝑗 \neq 𝑖} 𝑎_{𝑖 𝑗} =$ $1 -$ $𝑎_{𝑖 𝑖}$ . By Theorem 7.6.36, $𝜆 \in 𝐷_{𝑖}$ for some $𝑖$ , and by the triangle inequality

| 𝜆 | \leq | 𝜆 - 𝑎_{𝑖 𝑖} | + | 𝑎_{𝑖 𝑖} | \leq (1 - 𝑎_{𝑖 𝑖}) + 𝑎_{𝑖 𝑖} = 1 .

The row sums say $𝐴 𝟏 = 𝟏$ for $𝟏 = (1, \dots, 1)$ , so $1$ is an eigenvalue.

(b) Suppose $| 𝜆 | = 1$ , and let $𝑖$ be as in (a). Then both inequalities in (a) are equalities. Equality in the triangle inequality for $𝑧_{1} = 𝜆 - 𝑎_{𝑖 𝑖}$ and $𝑧_{2} = 𝑎_{𝑖 𝑖}$ means that $𝑧_{1} = | 𝑧_{1} | 𝑢$ and $𝑧_{2} = | 𝑧_{2} | 𝑢$ for some $𝑢 \in ℂ$ with $| 𝑢 | = 1$ (Theorem B.1.11(b)). Since $𝑧_{2} = 𝑎_{𝑖 𝑖} > 0$ , $𝑢 = 1$ , so $𝜆 - 𝑎_{𝑖 𝑖}$ is a nonnegative real number. Then $𝜆$ is a real number with $𝜆 \geq 𝑎_{𝑖 𝑖} > 0$ and $| 𝜆 | = 1$ , so $𝜆 = 1$ .

(c) $𝐴 = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ has nonnegative entries and row sums $1$ , and $𝐴 (1, - 1) = - (1, - 1)$ , so $- 1$ is an eigenvalue of absolute value $1$ .

Exercise 7.6.56.challenge Let

𝐴, 𝐵 \in ℂ^{𝑛 \times 𝑛}

, and suppose

𝐶 = 𝐴 𝐵 - 𝐵 𝐴

commutes with

𝐴

. Show that

𝐶

is nilpotent. Show that the statement fails over

𝔽_{𝑝}

Solution.

Uses. Proposition 2.4.28 (Cyclicity of the trace), Corollary 7.6.30 (Trace, determinant, and invertibility of $𝑝 (𝑇)$ ), Theorem 5.5.21 (The Vandermonde determinant), Theorem 7.6.20 (Cayley–Hamilton), Corollary 7.1.35 (Factorization over $ℂ$ )

Techniques. Cyclicity of trace (Technique 2.4.29), Extremal choice and minimal counterexamples (Technique 1.5.9)

Hint 1. Show that

tr (𝐶^{𝑘}) = 0

for every

𝑘 \geq 1

. Write

𝐶^{𝑘} =

𝐶^{𝑘 - 1} 𝐴 𝐵 -

𝐶^{𝑘 - 1} 𝐵 𝐴

and move

𝐴

past

𝐶^{𝑘 - 1}

Hint 2. If

𝜆_{1}, \dots, 𝜆_{𝑛}

are the eigenvalues of

𝐶

, then

tr (𝐶^{𝑘}) =

\sum_{𝑗} 𝜆_{𝑗}^{𝑘}

. Suppose some are nonzero, group the distinct nonzero ones, and get a Vandermonde system.

Hint 3. Over

𝔽_{𝑝}

, think of the

𝑝 \times 𝑝

matrices of

𝑥

and

𝑑 / 𝑑 𝑥

on polynomials of degree less than

𝑝

Sketch. For

𝑘 \geq 1

tr (𝐶^{𝑘 - 1} 𝐴 𝐵) =

tr (𝐴 𝐶^{𝑘 - 1} 𝐵) =

tr (𝐶^{𝑘 - 1} 𝐵 𝐴)

, using

𝐶 𝐴 = 𝐴 𝐶

and cyclicity, so

tr (𝐶^{𝑘}) = 0

. With distinct nonzero eigenvalues

𝜇_{1}, \dots, 𝜇_{𝑟}

of multiplicities

𝑚_{𝑙}

, this gives

\sum_{𝑙} 𝑚_{𝑙} 𝜇_{𝑙}^{𝑘} = 0

for

𝑘 = 1, \dots, 𝑟

, an invertible Vandermonde-type system forcing

𝑚_{𝑙} = 0

. So

0

is the only eigenvalue,

𝑝_{𝐶} = 𝑥^{𝑛}

, and

𝐶^{𝑛} = 0

. Over

𝔽_{𝑝}

𝐴 𝐵 - 𝐵 𝐴 = 𝐼_{𝑝}

is possible.

Traces of powers vanish. Let $𝑘 \geq 1$ . Since $𝐴$ commutes with $𝐶$ , it commutes with $𝐶^{𝑘 - 1}$ , so

tr (𝐶^{𝑘 - 1} 𝐴 𝐵) = tr (𝐴 𝐶^{𝑘 - 1} 𝐵) = tr ((𝐶^{𝑘 - 1} 𝐵) 𝐴)

by cyclicity of the trace (Proposition 2.4.28). Hence $tr (𝐶^{𝑘}) =$ $tr (𝐶^{𝑘 - 1} 𝐴 𝐵) -$ $tr (𝐶^{𝑘 - 1} 𝐵 𝐴) =$ $0$ .

All eigenvalues are $0$ . Let $𝜆_{1}, \dots, 𝜆_{𝑛}$ be the eigenvalues of $𝐶$ counted with multiplicity. By Corollary 7.6.30 with $𝑝 = 𝑥^{𝑘}$ , $tr (𝐶^{𝑘}) =$ $𝜆_{1}^{𝑘} +$ $\dots +$ $𝜆_{𝑛}^{𝑘}$ . Suppose some $𝜆_{𝑗} \neq 0$ , let $𝜇_{1}, \dots, 𝜇_{𝑟}$ be the distinct nonzero values among them ( $𝑟 \geq 1$ ), and let $𝑚_{𝑙} \geq 1$ be the number of $𝑗$ with $𝜆_{𝑗} = 𝜇_{𝑙}$ . Then for $𝑘 = 1, \dots, 𝑟$ ,

0 = tr (𝐶^{𝑘}) = 𝑚_{1} 𝜇_{1}^{𝑘} + \dots + 𝑚_{𝑟} 𝜇_{𝑟}^{𝑘} .

This is $𝑀 𝐦 = 𝟎$ with $𝐦 = (𝑚_{1}, \dots, 𝑚_{𝑟})$ (as elements of $ℂ$ ) and $𝑀 = {(𝜇_{𝑙}^{𝑘})}_{𝑘, 𝑙}$ . Factoring $𝜇_{𝑙}$ out of column $𝑙$ leaves the Vandermonde matrix $(𝜇_{𝑙}^{𝑘 - 1})$ , so $det 𝑀 =$ $𝜇_{1} \dots 𝜇_{𝑟} \prod_{𝑙 < 𝑙^{'}} (𝜇_{𝑙^{'}} - 𝜇_{𝑙}) \neq$ $0$ (Theorem 5.5.21), as the $𝜇_{𝑙}$ are distinct and nonzero. So $𝐦 = 𝟎$ , contradicting $𝑚_{1} \geq 1$ . Hence every $𝜆_{𝑗} = 0$ , $𝑝_{𝐶} = 𝑥^{𝑛}$ (Corollary 7.1.35), and $𝐶^{𝑛} = 0$ by Cayley–Hamilton.

Over $𝔽_{𝑝}$ . On $𝑉 = 𝔽_{𝑝} {[𝑥]}_{< 𝑝}$ , with basis $1, 𝑥, \dots, 𝑥^{𝑝 - 1}$ , let $𝐷$ be differentiation and $𝑋$ the operator $𝑓 \mapsto 𝑥 𝑓$ followed by deleting the $𝑥^{𝑝}$ term. For $𝑗 < 𝑝 - 1$ , $(𝐷 𝑋 - 𝑋 𝐷) 𝑥^{𝑗} =$ $(𝑗 + 1) 𝑥^{𝑗} -$ $𝑗 𝑥^{𝑗} =$ $𝑥^{𝑗}$ . For $𝑗 = 𝑝 - 1$ , $𝑋 𝑥^{𝑝 - 1} = 𝟎$ and $𝐷 𝑋 𝑥^{𝑝 - 1} = 𝟎$ , while $𝑋 𝐷 𝑥^{𝑝 - 1} =$ $(𝑝 - 1) 𝑥^{𝑝 - 1} =$ $- 𝑥^{𝑝 - 1}$ ; so $(𝐷 𝑋 - 𝑋 𝐷) 𝑥^{𝑝 - 1} =$ $𝑥^{𝑝 - 1}$ as well. With $𝐴 = 𝐷$ , $𝐵 = 𝑋$ , we get $𝐶 = 𝐴 𝐵 - 𝐵 𝐴 = {id}_{𝑉}$ , which commutes with $𝐴$ and is not nilpotent. The proof breaks at the last step: over $𝔽_{𝑝}$ the multiplicities $𝑚_{𝑙}$ are elements of $𝔽_{𝑝}$ , and $𝑚_{𝑙} = 𝑝$ is $0$ there. (Here $𝐶 = {id}_{𝑉}$ has $tr (𝐶^{𝑘}) = 𝑝 = 0$ for all $𝑘$ .)

More practice

The practice bank has 17 exercises on this section’s topics.