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Mar 26, 2021 at 11:02 comment added dohmatob Makes sense. Thanks.
Mar 25, 2021 at 0:49 comment added ofer zeitouni An individual eigenvalue does not affect the limit of empirical measure. As an example, take a sequence where $n-1$ eigenvalues are 0 and one is 1. Then, the empirical measure is $(1-1/n)\delta_0+\delta_1/n$. Then, the limit (under the weak topology) when $n\to\infty$ is $\delta_0$ and does not see the eigenvalue at $1$.
Mar 24, 2021 at 13:03 comment added dohmatob @oferzeitouni I'm even more confused. It's probably a problem with terminology on my part. By "only the limits are the same" do you mean "only the extreme eigenvalues" are the same ? Otherwise I don't understand the difference between limit of empirical (eigenvalue) measure and limiting spectral distribution (LSD). Thanks in advance.
Mar 10, 2021 at 22:03 comment added ofer zeitouni Your question is not clear, so my comment referred to the fact that if you are interested in the limit of the empirical measure, as the dimension grows, it is the same for $YY^T$ and $XX^T$, for the reason I mentioned (which is actually a proof). It is not the same LSD, only the limits are the same. This is a much weaker statement than your conjecture.
Mar 9, 2021 at 21:05 answer added dohmatob timeline score: 1
Mar 9, 2021 at 16:28 comment added dohmatob @oferzeitouni Thanks for the comment. Do you mean $YY^\top$ and $XX^\top + aa^\top$ have the same LSD and that the latter can be directly related to the LSD of $XX^\top$ ?
Mar 9, 2021 at 11:50 comment added ofer zeitouni It is a finite rank perturbation, since $YY^T =XX^T+ AX^T+X^TA+AA^T$ where $A_{ij}=a$; so you won't feel any change in the limit of the emprirical measure (due to interleaving)
Mar 8, 2021 at 13:16 history edited dohmatob CC BY-SA 4.0
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Mar 8, 2021 at 13:09 history asked dohmatob CC BY-SA 4.0