TITLE: Singularity of sparse random matrices
QUESTION [8 upvotes]: The following topic came up in conversation with my office-mate Lionel: Let $p$ be a fixed prime, $c$ a fixed positive real parameter and $n$ a large number. Consider a random $(0,1)$ matrix with entries in $Z/p$, where the probability of a $0$ is $1-\frac{c}{n}$ and that of a $1$ is $\frac{c}{n}$. As $n\rightarrow \infty$, what is the probability that this matrix is singular? UPDATE: As moonface points out, this probability is $1$. Furthermore, his argument points out that we should expect the corank to be something like $a*n$. So I'll ask more generally what we can say about the behavior of the rank as $n\rightarrow\infty$.
Actually, in our motivating example, the matrix is symmetric and the distribution on the diagonal is different than in the rest of the matrix. I left these details out, but please mention it if this point would strongly effect your answer.

REPLY [5 votes]: A few words of caution about random matrices over finite fields: it takes very little deviation from the zero distribution to force the spectral properties of a random matrix to look like one chosen with a uniform distribution.
For example, take the field to be Z/pZ and consider the distribution where each entry is 0 with probability 1 - (log n)^2/n and 1 with probability (log n)^2/n. The probability that the random matrix is non-singular then converges to (1 - 1/p)(1 - 1/p^2)(1 - 1/p^3)...(1-1/p^n), which is the same probability as a uniformly chosen matrix from M(n,Z/pZ) being non-singular.  In other words, the singularity probability will converge to a constant about 1/p, not to zero or to one, and the probability that the matrix has corank k is about 1/p^k.  This phenomena will hold as long as the distribution you choose is not concentrated more than about 1 - log n/n; otherwise, we do indeed expect the matrix to be almost surely singular as n -> infty, as is the case for the random matrix in the original post.
The key point seems to be bounding the probability that a row is all zeros or that the matrix has multiple equal rows.  If you can show that the probability of such an event goes to zero as n -> infty, then I would expect the asymptotics I mentioned above to take over.