Cayley-Hamilton Theorem

Theorem
Let $$A$$ be a commutative ring.

Let $$N=(a_{ij})$$ be an $$n\times n$$ matrix with entries in $$A$$.

Let $$I_n$$ denote the $$n\times n$$ identity matrix.

Let $$p_N \left({x}\right)$$ be the determinant $$\det \left({x \cdot I_n - N}\right)$$.

Then $$p_N \left({N}\right) = \mathbf 0$$ as an $$n \times n$$ zero matrix.

We have $$p_N \left({x}\right)$$ defined. But what set does $x$ belong to? And as $x$ behaves here as a scalar (which is apparent from the notation), what does it means when $x$ is a matrix, as it is in $p_N \left({N}\right)$?

Proof
$$N$$ defines a homomorphism of the free $A$-module $$A^n$$, as follows: let $$e_k=(0,\ldots,0,1,0,\ldots,0)\in A^n$$ have $$1$$ in the $$k^{\text{th}}$$ position, and $$0$$ elsewhere. Then define a homomorphism $$\phi$$ such that


 * $$e_k\mapsto e_kN=(a_{k1},\ldots,a_{kn})$$

Let $$\text{End}(A^n)$$ be the ring of endomorphisms of $$A^n$$, and let $$R\subseteq A^n$$ be the subring generated by scalar multiples of $$I_n$$ and $$\phi$$. Viewing $$A^n$$ as an $$R$$-module in the obvious way, we obtain the equations


 * $$\phi e_k=\sum_{i=1}^na_{ki}e_i,\qquad k=1,\ldots,n.$$

If $$\delta_{ij}$$ is the Kronecker delta, we have


 * $$\sum_{i=1}^n(\phi\delta_{ki}-a_{ki})e_i=0\qquad k=1,\ldots,n$$

Now let $$\Delta=(\phi\delta_{ki}-a_{ki})$$, an $$n\times n$$ matrix with coefficients in $$R$$.

We claim that $$\det(\Delta)=0$$ in $$R$$. Let $$\text{adj}(\Delta)=(b_{kj})$$ be the adjugate matrix of $$\Delta$$, so the usual rule for expanding a determinant along a row or a column is written


 * $$\text{adj}(\Delta)\cdot\Delta=\Delta\cdot\text{adj}(\Delta)=\det(\Delta)\cdot I_n.$$

Clearly $$\det(\Delta)\in R$$, so it is sufficient to prove that $$\det(\Delta)\cdot e_j=0$$ for $$j=1,\ldots,n$$. We have


 * $$\sum_{k=1}^nb_{kj}\sum_{i=1}^n(\phi\delta_{ki}-a_{ki})e_i=0.$$

Here we see that the coefficient before $$e_i$$ is $$\det(\Delta)\delta_{ji}$$, so $$\det(\Delta)e_j=0$$ for $$j=1,\ldots,n$$. This establishes the claim.

Now define a map from $$A[x]$$, the ring of polynomial forms in $x$ over $A$, to $$R$$ by $$f(x)\mapsto f(N)$$.

This is a homomorphism, so the map commutes with the operations used to calculate determinants.

Moreover, $$p_N(x)$$ maps to $$p_N(N)=\det(\Delta)$$, since $$\det(\Delta)=0$$, this proves the theorem.