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Suppose $D$ is a diagonal matrix of size $n \times n$ with diagonal elements $D_{ii}$ which are independent standard centered Gaussian random variables. Then consider a matrix $J$ such that its elements $J_{ij}$ are independent centered Gaussian variables with variance $\sigma^2/n$.

The question is: what is the limiting eigenvalue distribution of $A=D+J$?

In particular, has this distribution a bounded support ?

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  • $\begingroup$ I presume by "limiting" you mean the limit $n\rightarrow\infty$. The eigenvalues of $D$ are of order unity, independent of $n$, the perturbation by $M$ is of order $\sigma/\sqrt n\rightarrow 0$, so the limiting eigenvalue distribution of $A=D+M$ is just the original eigenvalue distribution of $D$. $\endgroup$
    – Carlo Beenakker
    Commented Nov 7, 2013 at 10:18
  • $\begingroup$ @Carlo, I think you are missing the scaling here, see my answer below. The entries of D are standard Gaussian, so same order of magnitude as eigenvalues of J. $\endgroup$
    – ofer zeitouni
    Commented Nov 7, 2013 at 11:55
  • $\begingroup$ @oferzeitouni --- indeed, I stand corrected, thanks. $\endgroup$
    – Carlo Beenakker
    Commented Nov 8, 2013 at 1:38

2 Answers 2

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You did not specify whether J is assumed symmetric or not. If it is, the limit of empirical values of eigenvalues of J alone is the semicircle, while in the non-symmetric it is the circular law.

If J is symmetric, the answer is the free convolution of Gaussian with the semicircle law. Google free convolution...

If J is not symmetric, the limit should be computable using Brown measures, although I am not sure it is done explicitly anywhere, and there are technical issues to overcome. See http://arxiv.org/abs/math/9912242 for some examples of computations in related problems (at the level of the limit) and http://arxiv.org/pdf/0909.2214.pdf for an example where a proof of convergence is given.

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  • $\begingroup$ I was indeed thinking about the not symmetric case. Thanks for these references. Do you think the support of the limit distribution has any chance to be bounded ? $\endgroup$
    – user16215
    Commented Nov 8, 2013 at 16:18
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    $\begingroup$ I suspect that it will not be bounded, because of the unboundedness of the Gaussian entries. $\endgroup$
    – ofer zeitouni
    Commented Nov 8, 2013 at 16:40
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A very similar problem was studied by Charles Bordenave, Pietro Caputo, and Djalil Chafai in:

http://arxiv.org/abs/1202.0644

The support of the distribution is not bounded.

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