 Invertibility forced by a polynomial product identity

Let $A,B$ be finite square matrices over a field and write $p(t)=c+tq(t)$ with $c\ne 0$. If $AB=p(A)$, then $A(B-q(A))=cI$. Taking determinants proves $A$ is an invertible matrix, and rearrangement gives $B=q(A)+c{A}^{-1}$. Thus $B$ commutes with $A$. If $Av=\lambda v$, then $\lambda \ne 0$ and $Bv=(p(\lambda )/\lambda )v$, proving an explicit eigenvalue transfer. The argument avoids assuming commutation before proving it; it depends on square finite-dimensional matrices, where a right inverse is two-sided.