Properties of Matrix Polynomials

Question

I just read the following property of matrix polynomials,

If $f,g$ are polynomials, $A$ is an $n \times n$ matrix, $f(A)$ and $g(A)$ commute.

The "proof" of this statement says "follows directly from the definition."

Why is this true? Do the properties of polynomials in $\mathbb{F}[z]$ automatically apply for polynomials in $\mathbb{F}[A]$? Why?

Answer 1

Let $f,\,g\in \mathbb{F}[z]$, where

$$f(t)=a_nz^n+a_{n-1}z^{n-1}+\dotsb+a_0,$$

$$g(t)=b_mz^m+b_{m-1}z^{m-1}+\dotsb+b_0,$$ $a_i,\,b_i\in\mathbb{F}$. Now define for square-matrix $A$

$$f(A)=a_nA^n+a_{n-1}A^{n-1}+\dotsb+a_0I,$$

$$g(A)=b_mA^m+b_{m-1}A^{m-1}+\dotsb+b_0I.$$

Now form the product: \begin{align*} f(A)g(A)&=\sum_{i=0}^{n} a_iA^{i}\sum_{j=0}^{m}b_jA^{j}\\ &=\sum_{i=0}^{n} \sum_{j=0}^{m}a_iA^{i}b_jA^{j}\\ &=\sum_{i=0}^{n} \sum_{j=0}^{m}a_ib_jA^{i+j}\\ &=\sum_{j=0}^{m}\sum_{i=0}^{n} b_ja_iA^{j+i}\\ &=\sum_{j=0}^{m}\sum_{i=0}^{n} b_jA^{j}a_iA^{i}\\ &=\sum_{j=0}^{m} b_jA^{j}\sum_{i=0}^{n}a_iA^{i}\\ &=g(A)f(A) \end{align*}