We will assume that $k$ is a field.

Definition. Let $A$ be a $k$-algebra. A representation of $A$ is a $k$-algebra homomorphism $A \longrightarrow M_m(k),$ for some integer $m \geq 1.$ If $\ker \phi =(0),$ then $\phi$ is called faithful.

If $\phi$ is a representation of a $k$-algebra $A$ and $a \in A,$ then $\phi(a)$ is a matrix and so it has a characteristic polynomial. Things become interesting when $A$ is a finite dimensional central simple $k$-algebra. In this case, the characteristic polynomial of $\phi(a)$ is always a power of the reduced characteristic polynomial of $a.$ This fact justifies the name “reduced characteristic polynomial”!

Theorem. Let $A$ be a finite dimensional central simple $k$-algebra of degree $n$ and let $a \in A.$ Let $\phi : A \longrightarrow M_m(k)$ be a representation of $A$ and suppose that $p(x)$ is the characteristic polynomial of $\phi(a).$ Then $m =rn$ for some integer $r \geq 1$ and $p(x) = (\text{Prd}_A(a,x))^r.$

Proof. Let $K$ be a splitting field of $A.$ We now define $\psi : A \otimes_k K \longrightarrow M_m(K)$ by $\psi(b \otimes_k \alpha) = \alpha \phi(b)$ for all $b \in A$ and $\alpha \in K$ and then extend it linearly to all $A \otimes_k K.$ We now show that $\psi$ is a $K$-algebra homomorphism. Clearly $\psi$ is additive because $\phi$ is so. Now let $\alpha, \beta \in K$ and $b,c \in A.$ Then

$\psi(\beta (b \otimes_k \alpha)) = \psi(b \otimes_k \alpha \beta)= \beta \alpha \phi(b)=\beta \psi(b \otimes_k \alpha)$

and

$\psi((b \otimes_k \alpha)(c \otimes_k \beta))=\psi(bc \otimes_k \alpha \beta)=\alpha \beta \phi(bc) = \alpha \phi(b) \beta \phi(c)$

$=\psi(b \otimes_k \alpha) \psi(c \otimes_k \beta).$

So, by Lemma 2, $m=rn$ for some integer $r \geq 1$ and

$p(x)=\det(xI - \phi(a))=\det(xI - \psi(a \otimes_k 1))=(\text{Prd}_A(a,x))^r. \ \Box$

Example. Let $A$ be a finite dimensional central simple $k$-algebra of degree $n.$ So $\dim_k A = n^2$ and  we have a faithful representation $\phi : A \longrightarrow \text{End}_k(A) \cong M_{n^2}(k)$ defined by $\phi(a)(b)=ab,$ for all $a,b \in A.$ Let $a \in A$ and suppose that $p(x)$ is the characteristic polynomial of $\phi(a).$ Then $p(x) = (\text{Prd}_A(a,x))^n,$ by the above theorem.