Singular matrices with integer entries

Question

I am motivated by the following paper by Greg Martin and Erick B. Wong:

http://www.math.ubc.ca/~gerg/papers/downloads/AAIMHNIE.pdf

Here the authors prove that assuming that the entries of an $n \times n$ matrix are chosen randomly with respect to a uniform distribution from the set {$-k, -k + 1 \cdots, -1, 0, 1, \cdots, k-1, k$}, then the probability that the resulting matrix will be singular is $\ll k^{-2 + \epsilon}$ (lemma 1 in the above paper).

What I am interested in is a more refined case. Suppose that $H, D$ are positive integers, and suppose that $(a_h, b_h, c_h, d_h)$, $1 \leq h \leq H$, are tuples of integers. Consider the monomials $m_r(a, b, c, d) = a^u b^v c^w d^x$, where $u + v + w + x = D$, and $u,v,w,x$ are non-negative integers. Here we allow $(u,v,w,x)$ to range over all possible choices. Let $R$ be the number of such monomials, which by construction is $\displaystyle \binom{D+3}{3} = \frac{(D+3)(D+2)(D+1)}{6}$, and consider the $H \times R$ matrix where the $hr$-th entry is the monomial $m_r(a_h, b_h, c_h, d_h)$. Finally, we can assume that $R < H$ to make the problem interesting.

Now suppose that the integers $a_h, b_h, c_h, d_h$, $1 \leq h \leq H$ are chosen uniformly from the set {$-k, \cdots, -1, 0, 1, \cdots, k$} say, then what is the probability that the resulting matrix will have rank at most $R-1$?

This amounts to showing that every sub $R \times R$ matrix is singular, and so is related to the original paper cited.

Answer 1

One can actually do better and get an asymptotic formula for the number of matrices of fixed rank. See the following paper:

Katznelson, Yonatan R. Integral matrices of fixed rank. Proc. Amer. Math. Soc. 120 (1994), no. 3, 667–675.

The main result is as follows:

Let $N(T; n, m, k)$ denote the number of integral $n \times m$ matrices of rank $k$, and norm at most $T$.

Theorem 1. For $n \geq m > k \geq 1$ and as $T \to \infty$:

(1) for $n>m$, $N(T; n, m,k) = a(n, m,k)T^{nk} + O(T^{nk-1}).$

(2) for $n=m$, $N(T;n,n,k) = p(n,k)T^{nk}\log T + O(T^{nk}).$

The methods of proof are pretty elementary (at least according to the author).

Answer 2

I'd guess that the probability of having even one singular $R \times R$ minor is about equal to the probability of having a tuple which is either 0 or a multiple of another one. What is the probability that a given $R \times R$ minor is non-singular? I'd suspect that the probability is similar to that from the original paper for $n=4.$ (so at least $1-\frac{C}{k^2}.$)

The article you cite mentions that the actual probability is much lower and not much higher than the probability of an all $0$ row. This does not surprise me. Suppose that you have chosen $0 \le r \lt n$ rows and that the rank so far is $r.$ What could go wrong with the next row? It might be all $0$ or plus or minus a previous row. For it to be some other non-zero rational multiple of a previous row seems pretty unlikely and to be a linear combination even less so.