Permanental rank versus determinantal rank of random matrices over finite fields

Abstract: This paper is motivated by basic complexity and probability questions about permanents of random matrices over finite fields, and in particular, about properties separating the permanent and the determinant. Fix $q = p^m$ some power of an odd prime, and let $k \leq n$ both be growing. For a uniformly random $n \times k$ matrix $A$ over $\mathbb{F}_q$, we study the probability that all $k \times k$ submatrices of $A$ have zero permanent; namely that $A$ does not have full "permanental rank". When $k = n$, this is simply the probability that a random square matrix over $\mathbb{F}_q$ has zero permanent, which we do not understand. We believe that the probability in this case is $\frac{1}{q} + o(1)$, which would be in contrast to the case of the determinant, where the answer is $\frac{1}{q} + Ω_q(1)$. Our main result is that when $k$ is $O(\sqrt{n})$, the probability that a random $n \times k$ matrix does not have full permanental rank is essentially the same as the probability that the matrix has a $0$ column, namely $(1 +o(1)) \frac{k}{q^n}$. In contrast, for determinantal (standard) rank the analogous probability is $Θ(\frac{q^k}{q^n})$. At the core of our result are some basic linear algebraic properties of the permanent that distinguish it from the determinant.