Cayley–Hamilton theorem

The aim of this essay is to prove the Cayley–Hamilton theorem naturally.

Definition. Let $M_{n\times n}(R)$$M_{n×n}(R)$ be the matrix ring over R and $M_{n\times n}(R)[T]$$M_{n×n}(R)[T]$ be the polynomial ring over $M_{n\times n}(R).$$M_{n×n}(R).$

For $p_A(T)I_n=\det (T{I}_n-A)I_n\in M_{n\times n}(R)[T]$$p_A(T)I_n = \det(TI_n-A)I_n ∈ M_{n×n}(R)[T]$, we have the following decomposition:

$$\begin{array}{}\text{(1)}& (T{I}_n-A)\mathrm{adj}(T{I}_n-A)=p_A(T)I_n=\mathrm{adj}(T{I}_n-A)(T{I}_n-A)\end{array}$$

Let ${\displaystyle \mathrm{adj}(T{I}_n-A)=\sum _{i=0}^n{B}_i{T}^i.}$$\displaystyle\mathrm{adj}(TI_n-A)=\sum_{i=0}^nB_i T^i.$

But the evaluation map at A is not a ring homomorphism in $M_{n\times n}(R)[T],$$M_{n×n}(R)[T],$ since this is a non-commutative ring.

For example, the identity

$$\begin{array}{}\text{(2)}& (IT+A{)}^2=I·T^2+2A·T+A^2\end{array}$$

is true in $M_{n\times n}(R)[T]$$M_{n×n}(R)[T]$, but what if you evaluate this identity at B such that $AB-BA\ne O$$AB-BA ≠ O$​?

Another example is, Let $R[T]$$R[T]$ be a polynomial ring and $R$$R$ is not commutative. Consider $f(T)=AT,g(T)=BT$$f(T)=AT,g(T)=BT$, by definition, $fg(T)=AB{T}^2$$fg(T)=ABT^2$. But $e{v}_C(f)e{v}_C(g)=ACBC\ne AB{C}^2$$ev_C(f)ev_C(g)=ACBC\ne ABC^2$ in general. However, if $C\in Z(R)$$C\in Z(R)$, the evaluation map will be a ring homomorphism. Let us prove it.

Proposition. Let $R[T]$$R[T]$ be a polynomial ring and $R$$R$ is not commutative. Let $\alpha \in Z(R)$$\alpha\in Z(R)$, denote the evaluation map at $\alpha$$\alpha$ by $e{v}_{\alpha }$$ev_{\alpha}$. Then $e{v}_{\alpha }$$ev_\alpha$ is a ring homomorphism.

Proof. Let ${\displaystyle f(T)=\sum _{i=0}^n{a}_i{T}^i,g(T)=\sum _{j=0}^m{b}_j{T}^j}$$\displaystyle f(T)=\sum_{i=0}^n a_i T^i,g(T)=\sum_{j=0}^m b_j T^j$. Then ${\displaystyle fg(T)=\sum _{k=0}^{n+m}\left(\sum _{i+j=k}{a}_i{b}_j\right)T^k}$$\displaystyle fg(T)=\sum_{k=0}^{n+m}\left(\sum_{i+j=k}a_ib_j \right)T^k$.

Now consider $e{v}_{\alpha }(fg)$$ev_{\alpha}(fg)$ and $e{v}_{\alpha }(f)e{v}_{\alpha }(g)$$ev_\alpha(f) ev_{\alpha}(g)$​.

We have $e{v}_{\alpha }({\displaystyle fg)=\sum _{k=0}^{n+m}\left(\sum _{i+j=k}{a}_i{b}_j\right){\alpha }^k}$$ev_{\alpha}(\displaystyle fg)=\sum_{k=0}^{n+m}\left(\sum_{i+j=k}a_ib_j \right)\alpha^k$, and ${\displaystyle e{v}_{\alpha }(f)e{v}_{\alpha }(g)=\left(\sum _{i=0}^n{a}_i{\alpha }^i\right)\left(\sum _{j=0}^m{b}_j{\alpha }^j\right)=\sum _{k=0}^{n+m}\left(\sum _{i+j=k}{a}_i{b}_j\right){\alpha }^k}$$\displaystyle ev_\alpha(f)ev_\alpha(g)=\left(\sum_{i=0}^{n}a_i\alpha^i\right)\left(\sum_{j=0}^m b_j \alpha^j\right)=\sum_{k=0}^{n+m}\left(\sum_{i+j=k}a_ib_j \right)\alpha^k$ since $\alpha \in Z(R)$$\alpha\in Z(R)$. Hence $e{v}_{\alpha }(fg)=e{v}_{\alpha }(f)e{v}_{\alpha }(g)$$ev_{\alpha}(fg)=ev_{\alpha}(f)ev_\alpha(g)$ $◻$$\square$​

Now, we want to find a proper ring such that $e{v}_A$$ev_A$ is a ring homomorphism and prove Cayley–Hamilton theorem directly.

The proper choice is $Z(A)[T]$$Z(A)[T]$. Since $A$$A$ in the centre of $Z[A]$$Z[A]$ by definition.

Easy to see that $A,I_n\in Z(A)$$A,I_n\in Z(A)$. Now we need to prove each $B_i\in Z(A)$$B_i\in{Z}(A)$​.

This follows from ${\displaystyle (I_n-A)\sum _{i=0}^n{B}_i{T}^i={\displaystyle \sum _{i=0}^n{B}_i{T}^i(I_n-A)⟹A{\displaystyle \sum _{i=0}^n{B}_i{T}^i={\displaystyle \sum _{i=0}^n{B}_i{T}^{i}A⟹A{B}_i=B_{i}A}}}}$$\displaystyle (I_n-A)\sum_{i=0}^nB_i T^i=\displaystyle \sum_{i=0}^nB_i T^i(I_n-A)\implies A\displaystyle\sum_{i=0}^nB_i T^i=\displaystyle\sum_{i=0}^nB_i T^iA\implies AB_i=B_i A$.

Hence we have:

$$\begin{array}{}\text{(3)}& p_A(A)=e{v}_A(p_A(T))=e{v}_A((T{I}_n-A)\mathrm{adj}(T{I}_n-A))={ev}_A(T{I}_n-A){ev}_A(\mathrm{adj}(T{I}_n-A))=O\end{array}$$