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Consider the following sequence of random matrices.

Let $D_N = N\ diag(d_1, d_2, ..., d_N) / \sum_{n=1}^N d_n$ where $d_n=1/n$. So $Tr(D_N)=N$, but $D_N$ has eigenvalues that follow a $n^{-1}$ power law.

Let $R_N$ be a random matrix with independent standard Gaussian entries, of size $N \times M$, where $M/N$ converges to a constant. Let $X_N = R_N^T D_N R_N$.

This sequence could be conceived as the covariance matrix of a random projection of a Gaussian vector which does not have a limiting spectral density.

What can we say about the eigenvalues of $X_N$, for $N$ large?

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  • $\begingroup$ You could use a quantative version of the Marchenko-Pastur law, called local law ([Anisotropic local laws for random matrices by Knowles and Yin] or [Local law for random Gram matrices by Alt, Erdös and Krüger]). For each $N$ the eigenvalue distribution of $X_N$ should still be close to the distribution, that one gets from the Marchenko-Pastur equation with the $n^{-1}$ power law as population distribution. $\endgroup$ Commented May 30 at 15:58
  • $\begingroup$ Thanks Ben! I will read those papers. In the meantime a question: if I understand correctly, using Marchenko-Pastur with non-white covariance works in a large-N limit where the eigenvalues can be considered randomly sampled from a distribution. If that measure is the $n^-1$ power law for fixed $N$, then it means there will be multiple eigenvalues for each value of $n$. But the actual matrix in question only has one eigenvalue for each $n$. Is that a problem? $\endgroup$ Commented Jun 2 at 11:32
  • $\begingroup$ Sorry for not going into more detail before. You will specifically need equation (3.11) from Theorem 3.6 in the mentioned Paper by Knowles and Yin. The $m_N$ written there is the Stieltjes transform of the spectral distribution of your $X_N$. The $m$ is the Stieltjes transform of a "non-asymptotic" reference distribution $\varrho$ that depends on $N$ (see Lemma 2.2). The $\pi$ in Lemma 2.2 is in your case $\frac{1}{N} \sum\limits_{n=1}^N \delta_{\frac{N}{nS}}$, where $S = \sum\limits_{n=1}^N \frac{1}{n}$. $\endgroup$ Commented Jun 2 at 14:54
  • $\begingroup$ It may be that you need to change the factor $\frac{N}{S}$ that you multiply to each population eigenvalue in order to get usable results. $\endgroup$ Commented Jun 2 at 14:58
  • $\begingroup$ fantastic, thanks! will read that ref in more detail $\endgroup$ Commented Jun 2 at 17:11

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