Number of invertible {0,1} real matrices?

Question

This question is inspired from here, where it was asked what possible determinants an $n \times n$ matrix with entries in {0,1} can have over $\mathbb{R}$.

My question is: how many such matrices have non-zero determinant?

If we instead view the matrix as over $\mathbb{F}_2$ instead of $\mathbb{R}$, then the answer is

$(2^n-1)(2^n-2)(2^n-2^2) \dots (2^n-2^{n-1}).$

This formula generalizes to all finite fields $\mathbb{F}_q$, which leads us to the more general question of how many $n \times n$ matrices with entries in { $0, \dots, q-1$ } have non-zero determinant over $\mathbb{R}$?

Answer 1

See Sloane, A046747 for the number of singular (0,1)-matrices. It doesn't seem like there's an exact formula, but it's conjectured that the probability that a random (0,1)-matrix is singular is asymptotic to $n^2/2^n$.

Over $F_2$ the probability that a random matrix is nonsingular, as $n \to \infty$, approaches the product $(1/2)(3/4)(7/8)\cdots = 0.2887880951$, and so the probability that a random large matrix is singular is only around 71 percent. I should note that a matrix is singular over $F_2$ if its real determinant is even, so this tells us that determinants of 0-1 matrices are more likely to be even than odd.

Answer 2

As Michael noted, the conjectured bound for the probability a random $(0,1)$ matrix is singular is $(1+o(1)) n^2 2^{-n} $. This corresponds to the natural lower bound coming from the observation that if a matrix has two equal rows or columns it is automatically singular.

The best bound for a long time for this problem was $(\frac{1}{\sqrt{2}} + o(1) )^n$, due to Bourgain, Vu, and Wood. Corollary 3.3 in their paper also gives a bound of $(\frac{1}{\sqrt{q}}+o(1))^n$ in the case where entries are uniformly chosen from $\{0, 1, \dots, q-1\}$ (here the conjectured bound would be around $n^2 q^{-n})$

Even showing that the determinant is almost surely non-zero is not easy (this was first proven by Komlos in 1967, and the reference is given in Michael's Sloane link).

Update: Konstantin Tikhimorov has uploaded a preprint giving a bound of $(\frac{1}{2}+o(1))^n$ on the singularity probability in the $(0,1)$ case, matching the lower bound up to the $o(1)$ in the exponent) (the result stated in his paper is for $\pm 1$ matrices, but there's a bijection showing that the singularity probability of an $n \times n$ $\pm 1$ matrix is the same as that of an $(n-1) \times (n-1)$ matrix with entries uniformly from $\{0,1\}$).

Answer 3

Lurking around MO, I found a question which is related to the second part of my question. Namely, Greg Martin and Erick B. Wong prove that assuming that the entries of an $n \times n$ matrix are chosen randomly with respect to a uniform distribution from the set $\{-k, \dots, k\}$, then the probability that the resulting matrix will be singular is $\ll k^{-2 + \epsilon}$.

See this MO question (where the above paragraph is plagarized from) and also here for the link to the Martin & Wong paper.