Cayley-Hamilton theorem and Nakayama’s lemma

Originally published at 狗和留美者不得入内. You can comment here or there.

The Cayley-Hamilton theorem states that every square matrix over a commutative ring $A$ satisfies its own characteristic equation. That is, with $I_n$ the $n \times n$ identity matrix, the characteristic polynomial of $A$

$p(\lambda) = \det (\lambda I_n - A)$

is such that $p(A) = 0$. I recalled that in a post a while ago, I mentioned that for any matrix $A$, $A(\mathrm{adj}(A)) = (\det A) I_n$, a fact that is not hard to visualize based on calculation of determinants via minors, which is in fact much of what brings the existence of this adjugate to reason in some sense. This can be used to prove the Cayley-Hamilton theorem.

So we have

$(\lambda I_n - A)\mathrm{adj}(\lambda I_n - A) = p(\lambda)I_n$,

where $p$ is the characteristic polynomial of $A$. The adjugate in the above is a matrix of polynomials in $t$ with coefficients that are matrices which are polynomials in $A$, which we can represent in the form $\displaystyle\sum_{i=0}^{n-1}t^i B_i$.

We have

\displaystyle {\begin{aligned}p(\lambda)I_{n} &= (\lambda I_n - A)\displaystyle\sum_{i=0}^{n-1}\lambda^i B_i \\ &= \displaystyle\sum_{i=0}^{n-1}\lambda^{i+1}B_{i}-\sum _{i=0}^{n-1}\lambda^{i}AB_{i} \\ &= \lambda^{n}B_{n-1}+\sum _{i=1}^{n-1}\lambda^{i}(B_{i-1}-AB_{i})-AB_{0}.\end{aligned}}

Equating coefficients gives us

$B_{n-1} = I_n, \qquad B_{i-1} - AB_i = c_i I_n, 1 \leq i \leq n-1, \qquad -AB_0 = c_0I_0$.

With this, we have

$A^n + c_{n-1}A^{n-1} + \cdots + c_1A + c_0I_n = A^nB_{n-1} + \displaystyle\sum_{i=1}^{n-1} (A^iB_{i-1} - A^{i+1}B_i) - AB_0 = 0$,

with the RHS telescoping and annihilating itself to $0$.

There is generalized version of this for a module over a ring, which goes as follows.

Cayley-Hamilton theorem (for modules) Let $A$ be a commutative ring with unity, $M$ a finitely generated $A$-module, $I$ an ideal of $A$, $\phi$ an endomorphism of $M$ with $\phi M \subset IM$.

Proof: It’s mostly the same. Let $\{m_i\} \subset M$ be a generating set. Then for every $i$, $\phi(m_i) \in IM$, with $\phi(m_i) = \displaystyle\sum_{j=1}^n a_{ij}m_j$, with the $a_{ij}$s in $I$. This means by closure properties of ideals the polynomial coefficients in the above will stay in $I$.     ▢

From this follows easily a statement of Nakayama’s lemma, ubiquitous in commutative algebra.

Nakayama’s lemma  Let $I$ be an ideal in $R$, and $M$ a finitely-generated module over $R$. If $IM = M$, then there exists an $r \in R$ with $r \equiv 1 \pmod{I}$, such that $rM = 0$.

Proof: With reference to the Cayley-Hamilton theorem, take $\phi = I_M$, the identity map on $M$, and define the polynomial $p$ as above. Then

$rI_M = p(I_M) = (1 + c_{n-1} + c_{n-2} + \cdots + c_0)I_M = 0$

both annihilates the $c_i$s, coefficients residing in $I$, so that $r \equiv 1 \pmod{I}$ and gives the zero map on $M$ in order for $rM = 0$.     ▢

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