Let $A$ be an $n\times n$ Hermitian matrix, with well-separated eigenvalues $\lambda_1 > \lambda_2 ... > \lambda_n$, with $|\lambda_i-\lambda_j|>\epsilon$, for all $i \neq j$. Let $G$ be a Gaussian matrix, i.e. each $G_{i,j}$ is distributed ${\cal N}(0,1)$. What can be said about the distribution of the eigenvalues of $A+ f(\epsilon) \cdot G$, where $f$ is some function $f(\epsilon)<<\epsilon$. Note that the strength of the perturbation is significantly smaller than the inter-eigenvalue distance, so eigenvalue repulsion can be made arbitrarily weak.

I would like to know whether there are properties that hold for ANY such matrix $A$:

(1) Are the eigenvalues of $A+G$ distributed approximately independently?

(2) Can the variance of the distribution of each eigenvalue of $A+G$ be lower-bounded by some function of $\epsilon$?

If the answer is negative, is there ANY perturbation technique that can yield properties (1) and (2) for any such matrix $A$?

  • $\begingroup$ independent eigenvalues --- no: there is no way to avoid eigenvalue repulsion; in particular, take $A=0$, then the eigenvalues of $A+G$ are those of $G$, which show a linear repulsion (Gaussian Orthogonal Ensemble) $\endgroup$ – Carlo Beenakker Oct 22 '13 at 9:41
  • $\begingroup$ Thanks for the comment. Please check out modified question. $\endgroup$ – Lior Eldar Oct 22 '13 at 12:22

Since $P(G)\propto{\rm exp}\bigl(-\frac{1}{2}{\rm Tr}\,GG^{\rm T}\bigr)$ is invariant under orthogonal transformations, if $A$ is real Hermitian you can work in a basis where $A$ is diagonal, with the $\lambda_n$'s on the diagonal. The first order correction to the $n$-th eigenvalue is then just $\delta\lambda_{n}= f(\epsilon)G_{nn}$, and the answer to both of your questions is Yes. I would think the same is true if $A$ is complex Hermitian.

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  • $\begingroup$ Although the OP may have meant for $G$ to be a GOE matrix, that's not what the question actually said. That shouldn't actually change the answer, but it makes the analysis less trivial. $\endgroup$ – Mark Meckes Oct 22 '13 at 20:42
  • $\begingroup$ @MarkMeckes --- you mean that the analysis depends on whether $G$ is symmetric or not? I'm probably confused, but where did I make that assumption? $\endgroup$ – Carlo Beenakker Oct 22 '13 at 21:05
  • $\begingroup$ No, I was confused. I thought the OP meant for $G$ to be symmetric matrix with N(0,1) entries on and above the diagonal, which is not quite the GOE and is not orthogonally invariant. But I see now that the OP didn't say that $G$ should be symmetric, so there's no problem with what you said. $\endgroup$ – Mark Meckes Oct 23 '13 at 8:02

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