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?

  • $\begingroup$ In the symmetric case, an estimate of the type you ask about has recently been derived by Nguyen, Tao and Vu. arxiv.org/pdf/1504.00396v1.pdf I am not aware of a similar result in the non-Hermitian case $\endgroup$ Nov 7, 2015 at 21:41

1 Answer 1


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 $[0,1]$).

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    $\begingroup$ This is still in the determinantal setup..... $\endgroup$ Nov 8, 2015 at 9:38
  • $\begingroup$ Right... As for universality, it looks like the Tao–Vu four moment theorem only gives that at most $O(n^{1-c})$ eigenvalues are repeated (see Corollary 18 here: arxiv.org/pdf/1206.1893v6.pdf). $\endgroup$
    – Nick Cook
    Nov 8, 2015 at 21:41

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