Permanental rank versus determinantal rank of random matrices over finite fields
Fatemeh Ghasemi, Gal Gross, Swastik Kopparty
TL;DR
This work studies the permanental rank of random finite-field matrices, proving that for k up to O(√n) the probability that all k×k submatrices have zero permanent is essentially governed by the presence of a zero column, yielding Pr[Z] ≈ k/q^n and revealing a sharp separation from the determinant’s behavior. The authors develop a structural framework around permanull subspaces and jointly-permanull lists, culminating in a codimension-1 classification via the cofactor-expansion of the permanent and a graph-theoretic lemma, which underpins the probabilistic bound. They further explore permanull subspaces in large and arbitrary characteristic, introducing the permanental polynomial as a tool to characterize permanullness and enable algorithmic testing, thereby broadening the understanding of when permanents vanish in random and structured settings. Overall, the paper establishes a clear structural distinction between permanents and determinants and advances the program of separating algebraic complexity classes by examining permanents 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.