2
$\begingroup$

It has been proven that the CDF of the eigenvalue distribution of random Gaussian matrix converges to a uniform disk circular law. Is it true for the PDF of the limiting eigenvalue distribution? In general, convergence in distribution doesn't say anything about the PDF, and I wonder whether it's also the case here. Thanks.

$\endgroup$
4
  • $\begingroup$ If I'm understanding the question right you're asking about a distribution function for processes, which do not allow such a distribution function. The spectral empirical law has point-masses at the eingenvaules of the random matrix and thus does not have a PDF. The eigenvalues $\lambda_1,...,\lambda_n$ have a PDF in $\mathbb{C}^n$, which I believe is known since Wigner, but I'm not sure what meaningful limiting properties can be found when the dimension $n$ goes to infinity. $\endgroup$
    – Tardis
    Nov 1, 2021 at 22:14
  • $\begingroup$ you may want to clarify what it is that you wish to compute; the average of $\sum_{i=1}^n \delta(\lambda-\lambda_i)$ tends to a constant for $|\lambda|<1$ in the limit $n\rightarrow\infty$, when the $\lambda_i$'s are eigenvalues of a complex matrix with i.i.d. Gaussian matrix elements; if the matrix elements are real, then this limit only holds when ${\rm Im}\,\lambda\neq 0$. Some $\sqrt n$ eigenvalues accumulate on the real axis. $\endgroup$ Nov 2, 2021 at 12:57
  • $\begingroup$ Hi, guys thanks for the comments. @Tardis I'm not familiar with random matrix theory enough to understand the first point you raised, but I believe I want the (limiting) joint pdf of $\lambda_1,\ldots, \lambda_n$. What I want to compute ultimately is $\lim_{n\to\infty}\frac{1}{n}E[tr((I-W)^{-T}(I-W)^{-1})]$, where $W$ is a 0 mean random Gaussian matrix with variance $a/n$ (where a is small enough s.t the inverse exists with high probability). I want to mimic what is done in mathoverflow.net/questions/359693/…. $\endgroup$ Nov 2, 2021 at 18:51
  • $\begingroup$ I want to know can I directly calculate the expectation using integration with the uniform measure over the unit ball (like the circular law). Now I do realize that (I-W)^{-T}(I-W)^{-1} doesn't have eigenvalue of $1/(1-\lambda)^{2}$ where $\lambda$ is an eigenvalue of $W$ (and I don't know what that matrix's eigenvalue is). But suppose I have a normal matrix $A$ whose eigenvalue is $f(\lambda)$, and $\lambda$ is an eigenvalue of $W$, will $\int f(\lambda)^2 p(\lambda)d\lambda$ give me $E[tr(A^TA)]$, where $p(\lambda)=1/(\pi a^2)$ is a uniform distribution over the disk? $\endgroup$ Nov 2, 2021 at 18:58

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy