The factorization of the permanent of a matrix with minimal rank in prime characteristic

It is known that any square matrix A of size n over a field of prime characteristic p that has rank less than n/(p− 1) has a permanent that is zero. We give a new proof involving the invariant Xp. There are always matrices of any larger rank with non-zero permanents. It is shown that when the rank of A is exactly n/(p − 1), its permanent may be factorized into two functions involving Xp. Let n be any positive integer. Let A = (aij) be an n×n matrix over a field F . The permanent of A is a polynomial of degree n: per(A) := Σγ∈SnΠ n i=1 ai,γ(i), where Sn is the symmetric group of all permutations of N := {1, . . . , n}. If F has characteristic two (p = 2), the permanent equals the determinant. See [2] and [3] for some of the theory of permanents. Valiant [5] showed that the complexity of computing per(A) is #P-hard, and this is true also for computing integer matrices modulo any integer except powers of two. Barvinok [1] showed that if the rank(A) is fixed then the permanent function is of polynomial complexity in n. It was known [6] that matrices of small rank have zero permanent modulo a prime p. In fact, Yu observed a stronger result, that the perrank of a matrix in characteristic p is at most (p − 1)rank(A). The perrank is the maximum size of a square submatrix of A with non-zero permanent. We give an alternate proof of the permanent bound that involves the invariant Xp defined below. We remark that for integer matrices A, rankp(A) ≤ rank(A), since the rank of a matrix is determined by the size of the largest square submatrices having non-zero determinant. A corollary is that a matrix over the integers with a non-zero permanent has a rank that is at least n/(p− 1), where p is the smallest prime that does not divide the permanent. Let F be a field of characteristic p. Let q = p, h ≥ 1. We use some of the theory of [4], and in particular the following results. Definition 1. (See [4, Def. 3.1].) The invariant Xq is defined for any m × m(q − 1) matrix A = (aij) over F by Xq(A) = Σai1,1.ai2,2 . . . aim(q−1),m(q−1), where the sum is over all vectors (i1, . . . , im(q−1)), with ij ∈M , each element of M being repeated precisely q − 1 times. Date: 22 March 2011. 2000 Mathematics Subject Classification. 05B20, 15A15, 11A07, 11C20, 03D15, 65Y20.