Let $k$ be a field and let $A$ be an $n \times n$ matrix with entries in $k$ (so that the powers $A_{i}$ are defined).
If $f (x)$ = $c_0$ + $c_1x$ +···+ $c_m x^m$ ∈ $k[x]$, define $f (A)$ = $c_0 I$ + $c_1A$ +···+ $c_m A^m$.
Q: If $f (x) = p(x)q(x)$ ∈ $k[x]$ and if $A$ is an $n \times n$ matrix over $k$, prove that $f (A) = p(A)q(A).$
A(?): Given any commutative ring $R$ and $r∈R$, the evaluation map $f↦f(r)$ from $R[x]$ to $R$ is a ring homomorphism.
In particular it is true for $R=k[A]$, which is a commutative subalgebra of $M_n(k)$. After you have the evaluation homomorphism from $R[x]→R[A]$, you can restrict it to the subring $k[x]$ of $R[x]$, then it becomes a ring homomorphism from $k[x]$ to $R[A]$. Is this right? Thank for your help.
Multiplication and addition of matrices that are sums of powers of a given $A$ satisfy the same formal rules (associativity and distributivity) as multiplication and addition in $k[x]$ (sums of powers of a formal variable $x$).
Therefore whatever is true for polynomials that depends only on the language of $+$ and $\times$ also holds for sums of powers of $A$. The formal way to state it is to say that
$$\begin{align}k[x]& \longrightarrow \mathcal{M}_{n\times n}(k)\\P(x)&\longmapsto P(A)\end{align}$$
is a ring homomorphism.
To address your question in the way it is stated, first say that by distributivity it is enough to consider the case where $p$ and $q$ are monomials, and in that case it follows from the fact that $A^iA^j=A^{i+j}$ for all integers $i$, $j$.