TITLE: Smallest eigenvalue gap of a non-symmetric random matrix
QUESTION [6 upvotes]: The question: Let $A$ be the matrix whose each element is an independently generated random variable which is uniform on $[0,1]$. One can see that the eigenvalues of $A$ will be distinct almost surely. Let $\delta = \min_{i,j} |\lambda_i(A) - \lambda_j(A)|$ be the smallest distance between any two eigenvalues of $A$ in the complex plane. What sort of lower bounds does $\delta$ satisfy with high probability? 
Background: looking at the literature, it seems that for some symmetric random matrices $A$ the question has been settled, in the sense that the exact distribution of $\delta$ is known. This does not, however, seem to apply to the non-symmetric case above, and it seems reasonable to guess that the exact distribution of $\delta$ for the random matrix I am asking about is an open question. 
However, here I am asking for something much less than an exact distribution of $\delta$ -- for example, can one make a statement along the lines of $P(\delta \geq 1/n^{10}) \geq 1-1/n^{0.1}$, or something like that?

REPLY [4 votes]: This paper treats the case of iid complex Gaussian entries:
http://arxiv.org/abs/1207.4240
In particular, the smallest eigenvalue gap is shown to be of order $n^{-3/4}$ when the entries are normalized to have variance $1/n$. One might expect this to extend to more general entry distributions, but as far as I am aware this is unknown. 
Edit: As to whether one can show $P(\delta\ge n^{-C})\rightarrow 1$ for some $C>0$, for general (possibly discrete) entry distributions it seems difficult to even show $P(\delta> 0)\rightarrow 1$. However, as is the case for controlling the smallest singular value, it may be easier to lower bound $\delta$ when the entries have a bounded density (perhaps this is why you posed the question for the uniform distribution on $[0,1]$).