characteristic polynomial

Characteristic polynomial and Cayley-Hamilton theorem. See also determinant.

Let $R$ be a commutative ring, and let $A$ be an $n\times n$ matrix with entries in $R$. Then there exists an $n\times n$ matrix $\tilde{A}$ with entries in $R$ such that $A\tilde{A}=\tilde{A}A=\mathrm{det}(A)\cdot {I}_{n}$.

We may as well take $R$ to be the polynomial ring $\mathbb{Z}[{a}_{ij}{]}_{1\le i,j\le n}$, since we are then free to interpret the indeterminates ${a}_{ij}$ however we like along a ring map $\mathbb{Z}[{a}_{ij}]\to R$. Let $A$ denote the corresponding generic matrix.

Guided by Cramer’s rule, put

$${\tilde{A}}_{ji}=\mathrm{det}({a}_{1},\dots ,{e}_{i},\dots {a}_{n}),$$

the ${a}_{i}$ being columns of $A$ and ${e}_{i}$, the column vector with $1$ in the ${i}^{\mathrm{th}}$ row and $0$’s elsewhere, appearing as the ${j}^{\mathrm{th}}$ column. If we pretend $A$ is invertible, then we know $A\tilde{A}=\mathrm{det}(A)\cdot {I}_{n}=\tilde{A}A$ by Cramer’s rule. We claim this holds for general $A$.

Indeed, we can interpret this as a polynomial equation in $\u2102[{a}_{ij}]$ and check it there. As an equation between polynomial functions on the space of matrices $A\in {\mathrm{Mat}}_{n}(\u2102)=\mathrm{Spec}(\u2102[{a}_{ij}])$, it holds on the dense subset ${\mathrm{GL}}_{n}(\u2102)\hookrightarrow {\mathrm{Mat}}_{n}(\u2102)$. Therefore, by continuity, it holds on all of ${\mathrm{Mat}}_{n}(\u2102)$. But a polynomial function equation with coefficients in $\u2102$ implies the corresponding polynomial identity, and the proof is complete.

(**Cayley-Hamilton**) Let $V$ be a finitely generated free module over a commutative ring $R$, and let $f:V\to V$ be an $R$-module map. Let $p(t)\in R[t]$ be the **characteristic polynomial** $\mathrm{det}(t\cdot {1}_{V}-f)$ of $f$, and let ${\varphi}_{f}:R[t]\to {\mathrm{Mod}}_{R}(V,V)$ be the unique $R$-algebra map sending $t$ to $f$. Then $p(f)\u2254{\varphi}_{f}(p)$ is the zero map $0:V\to V$.

Via ${\varphi}_{f}$, regard $V$ as an $R[t]$-module, and with regard to some $R$-basis $\{{v}_{i}{\}}_{1\le i\le n}$ of $V$, represent $f$ by a matrix $A$. Now consider $t\cdot {I}_{n}-A$ as an $n\times n$ matrix $B(t)$ with entries in $R[t]$. By definition of the module structure, this matrix $B(t)$, seen as acting on ${V}^{n}$, annihilates the length $n$ column vector $c$ whose ${i}^{\mathrm{th}}$ row entry is ${v}_{i}$.

By the previous lemma, there is $\tilde{B}(t)$ such that $\tilde{B}(t)B(t)$ is $\mathrm{det}(t\cdot {I}_{n}-A)$ times the identity matrix. It follows that

$$\mathrm{det}(t\cdot {I}_{n}-A)c=\tilde{B}(t)B(t)c=\tilde{B}(t)0=0$$

i.e., $\mathrm{det}(t\cdot {I}_{n}-A)\cdot {v}_{i}=0$ for each $i$. Since the ${v}_{i}$ form an $R$-basis, the $R[t]$-scalar $\mathrm{det}(t\cdot {I}_{n}-A)$ annihilates the $R[t]$-module $V$, as was to be shown.

The Cayley-Hamilton theorem easily generalizes to finitely generated $R$-modules (not necessarily free) as follows. Let $f:V\to V$ be a module endomorphism, and suppose $\pi :{R}^{n}\to V$ is an epimorphism. Since ${R}^{n}$ is projective, the map $f\circ \pi $ can be lifted through $\pi $ to a map $A:{R}^{n}\to {R}^{n}$. Let $P(t)$ be the characteristic polynomial of $A$.

$P(f)=0$.

Write $P(t)={\sum}_{i}{a}_{i}{t}^{i}$. We already know $P(A)=0$. From $f\circ \pi =\pi \circ A$, it follows that ${f}^{i}\circ \pi =\pi \circ {A}^{i}$ for any $i\ge 0$. Hence $P(f)\circ \pi =\pi \circ P(A)=0$. Since $\pi $ is epic, $P(f)=0$ follows.

We give an interesting and perhaps surprising consequence of the Cayley-Hamilton theorem below, after establishing a lemma close in spirit to Nakayama's lemma.

Suppose $V$ is a finitely generated $R$-module, and $g:V\to V$ is a module map such that $g(V)\subseteq IV$ for some ideal $I$ of $R$. Then there is a polynomial $p(t)={t}^{n}+{a}_{1}{t}^{n-1}+\dots +{a}_{n}$, with all ${a}_{i}\in I$, such that $p(g)=0$.

For some finite $n\ge 0$, we have a surjective map ${R}^{n}\to M$, and by hypothesis we have a surjective map ${I}^{n}\to \mathrm{im}(g)$ in

$$\begin{array}{ccccc}& & & & {I}^{n}\\ & & & & \downarrow \\ {R}^{n}& \to & M& \stackrel{g}{\to}& \mathrm{im}(g)\end{array}$$

By projectivity of ${R}^{n}$, we can lift the bottom composite to a map ${R}^{n}\to {I}^{n}$ making the diagram commute. Let $A$ be the $R$-module map ${R}^{n}\to {I}^{n}\hookrightarrow {R}^{n}$, regarded as a matrix. Then the characteristic polynomial of $A$ satisfies the conclusion, by the Cayley-Hamilton theorem.

Let $V$ be a finitely generated module over a commutative ring $R$, and let $f:V\to V$ be a surjective module map. Then $f$ is an isomorphism.

Regard $V$ as a finitely generated $R[t]$-module via ${\varphi}_{f}:R[t]\to {\mathrm{Mod}}_{R}(V,V)$. Since $f$ is assumed surjective, we have $IV=V$ for the ideal $I=(t)$ of $R[t]$. Now take $g={1}_{V}$ as in the preceding lemma, a module map over the ring $R\prime =R[t]$. By the lemma, we see that ${g}^{n}+{a}_{1}{g}^{n-1}+\dots +{a}_{n}=0$ where ${a}_{i}\in (t)$, in other words the $R[t]$-scalar

$$(1+{a}_{1}+\dots +{a}_{n}){1}_{V}=0$$

as an operator on $V$. Write ${a}_{i}={b}_{i}(t)t$ for polynomials ${b}_{i}(t)\in R[t]$. Now we may rewrite the previous displayed equation as

$${1}_{V}(v)=-(\sum _{i}{b}_{i}(t))t\cdot v$$

for all $v\in V$, which translates into saying that ${1}_{V}=-{\sum}_{i}{b}_{i}(f)f$, i.e., that $-{\sum}_{i}{b}_{i}(f)$ is a retraction of $f$. Since $f$ is epic, we now see $f$ is an isomorphism.

The proof of the Cayley-Hamilton theorem follows the treatment in

- Serge Lange,
*Algebra*(${3}^{\mathrm{rd}}$ edition), Addison-Wesley, 1993.

Revised on April 19, 2013 21:16:14
by Todd Trimble
(67.81.93.26)