All,
Suppose $A \in Mat(n, \mathbb{F}_{q})$ for $q$ prime, $q \geq 5$, and $n \geq 2^{q-2}$ . Let $\pi_A(x)$ be the permanental polynomial for $A$. That is,
\begin{equation*}
\pi_{A}(x)=per(xI-A).
\end{equation*}
My question is this: if $x=0$ is a root of $\pi_A$ of multiplicity at least $q-2$, does this imply that $A$ is not invertible?
I am working on a problem recreationally and I am unable to proceed without the answer to the above question being in the affirmative. Intuitively, I believe it to be the case but have nothing stronger than intuition at the moment. Unfortunately, I have found no useful literature on the subject of permanental roots of matrices over finite fields. Any insight is greatly appreciated. For example, is there any implication following the fact that $x=0$ is a root of $\pi_A$ of "large" multiplicity?
