= Invertibility forced by a polynomial product identity
{title2=$AB=p(A),\ p(0)\ne0\Rightarrow A^{-1}=(B-q(A))/p(0)$}
Let $A,B$ be finite square <matrices> over a <field> and write $p(t)=c+tq(t)$ with $c\ne0$. If $AB=p(A)$, then $A(B-q(A))=cI$. Taking <determinants> proves $A$ is an <invertible matrix>, and rearrangement gives $B=q(A)+cA^{-1}$. Thus $B$ commutes with $A$. If $Av=\lambda v$, then $\lambda\ne0$ 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.
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