# nLab characteristic polynomial

## Characteristic polynomial and Cayley-Hamilton theorem.

###### Lemma

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 $A \tilde{A} = \tilde{A} A = \det(A) \cdot I_n$.

###### Proof

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

Guided by Cramer’s rule, put

$\tilde{A}_{j i} = \det(a_1, \ldots, e_i, \ldots a_n),$

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

Indeed, we can interpret this as a polynomial equation in $\mathbb{C}[a_{i j}]$ and check it there. As an equation between polynomial functions on the space of matrices $A \in Mat_n(\mathbb{C}) = Spec(\mathbb{C}[a_{i j}])$, it holds on the dense subset $GL_n(\mathbb{C}) \hookrightarrow Mat_n(\mathbb{C})$. Therefore, by continuity, it holds on all of $Mat_n(\mathbb{C})$. But a polynomial function equation with coefficients in $\mathbb{C}$ implies the corresponding polynomial identity, and the proof is complete.

###### Theorem

(Cayley-Hamilton) Let $V$ be a finitely generated free module over a commutative ring $R$, and let $f \colon V \to V$ be an $R$-module map. Let $p(t) \in R[t]$ be the characteristic polynomial $\det(t \cdot 1_V - f)$ of $f$, and let $\phi_f \colon R[t] \to Mod_R(V, V)$ be the unique $R$-algebra map sending $t$ to $f$. Then $p(f) \coloneqq \phi_f(p)$ is the zero map $0 \colon V \to V$.

###### Proof

Via $\phi_f$, regard $V$ as an $R[t]$-module, and with regard to some $R$-basis $\{v_i\}_{1 \leq i \leq n}$ 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 $i^{th}$ row entry is $v_i$.

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

$\det(t \cdot I_n - A) c = \tilde{B}(t) B(t) c = \tilde{B}(t) 0 = 0$

i.e., $\det(t \cdot I_n - A) \cdot v_i = 0$ for each $i$. Since the $v_i$ form an $R$-basis, the $R[t]$-scalar $\det(t \cdot I_n - A)$ 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 $f \colon V \to V$ be a module endomorphism, and suppose $\pi \colon R^n \to V$ is an epimorphism. Since $R^n$ is projective, the map $f \circ \pi$ can be lifted through $\pi$ to a map $A \colon R^n \to R^n$. Let $P(t)$ be the characteristic polynomial of $A$.

###### Proposition

$P(f) = 0$.

###### Proof

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 $i \geq 0$. 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.

###### Lemma

Suppose $V$ is a finitely generated $R$-module, and $g \colon V \to V$ is a module map such that $g(V) \subseteq I V$ for some ideal $I$ of $R$. Then there is a polynomial $p(t) = t^n + a_1 t^{n-1} + \ldots + a_n$, with all $a_i \in I$, such that $p(g) = 0$.

###### Proof

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

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

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.

###### Proposition

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

###### Proof

Regard $V$ as a finitely generated $R[t]$-module via $\phi_f \colon R[t] \to Mod_R(V, V)$. Since $f$ is assumed surjective, we have $I V = V$ for the ideal $I = (t)$ of $R[t]$. Now take $g = 1_V$ as in the preceding lemma, a module map over the ring $R' = R[t]$. By the lemma, we see that $g^n + a_1 g^{n-1} + \ldots + a_n = 0$ where $a_i \in (t)$, in other words the $R[t]$-scalar

$(1 + a_1 + \ldots + a_n)1_V = 0$

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.

## Referemces

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

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

Revised on August 27, 2014 14:37:15 by Urs Schreiber (82.136.246.44)