Permanental rank versus determinantal rank of random matrices over finite fields
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arXiv
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2025
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| author | Ghasemi, Fatemeh Gross, Gal Kopparty, Swastik |
| author_facet | Ghasemi, Fatemeh Gross, Gal Kopparty, Swastik |
| contents | 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. |
| format | Preprint |
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arxiv_https___arxiv_org_abs_2512_03221 |
| institution | arXiv |
| publishDate | 2025 |
| record_format | arxiv |
| spellingShingle | Permanental rank versus determinantal rank of random matrices over finite fields Ghasemi, Fatemeh Gross, Gal Kopparty, Swastik Computational Complexity Combinatorics Probability 68Q17 (Primary) 15A15, 15B33, 15B52 (Secondary) F.1.m 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. |
| title | Permanental rank versus determinantal rank of random matrices over finite fields |
| topic | Computational Complexity Combinatorics Probability 68Q17 (Primary) 15A15, 15B33, 15B52 (Secondary) F.1.m |
| url | https://arxiv.org/abs/2512.03221 |