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Intuitively, the fact that the density vanishes at the edge already hints that the spacing at the edge changes (which indeed it does), and that the asymptotics are different; this is confirmed in the …

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 …

It depends what do you mean by "look very different". I assume that you mean that the empirical measure is close to that of the MP law. The answer below assumes this is what you meant. Short answer …

In case $d$ increases linearly in $k$, say $k=\alpha d$, you can compute the limiting spectral measure of your matrix (you are dealing with the product of two wishart matrices, so you can compute the …

First, I understand the question as one asking about $$ det(I+a_i v_iv_i^*(I+\sum_{j\neq i} a_jv_jv_j^*)^{-1}) =det(I+\sum_ia_iv_iv_i^*)/det(I+\sum_{j\neq i} a_j v_j v_j^*)=:A/B$$ Since $A$ does not …

Actually, just to get the semicircle is not hard. Take an $n$-by-$n$ Jacobi matrix whose on diagonal entries are $0$ and $i$th entry on the off diagonal is $\sqrt{i/n}$. The limit ESD will be the semi …

I believe this follows from the standard estimates that show the convergence. For example, look at the proof in Anderson-Guionnet-Zeitouni's book, page 24. Indeed, from the last display there, you get …

I do not understand the role of $k$ in your question, so I will just ignore it. If you take your matrix to consist of independent Gaussian entries (normalized by $\sqrt{n}$), this will give a precise …

In general you need more than second moment - you need fourth moment finite, otherwise the top eigenvalue can run off to infinity in your scaling. For this and more see the book of Bai and Silverstein …

For $a_n=b_n$ with $\|a_n\|=\|b_n\|$, the result is standard (and can be found for example in papers of Bai and Silverstein, usually as a technical lemma in the appendix...): let $\rho$ be the limit …

Expanding a bit on Yemon Choi's comment: concentration is indeed the key. First, $\|B\|_F^2$ is simply the sum of the squares of singular values of $A$, and $E\|B\|_F^2=m(m-1)n+mn^2$. On the other ha …

To amplify on Brendan's answer: You are dealing with a matrix of zero mean iid's plus a rank one perturbation: $M=W+ A$ where $A=c11^T$ and $c=EY$. The spectral norm of $W$ is asymptotically $2\sqrt …

For standard Gaussians, and with the matrix $W/n$, the proof of the LDP given by Ben Arous-Guionnet adapts to the Wishart setup. However, you will have different scalings and so the non-commutative en …

This falls within the domain of free probability, at least when the dimension is large. That is, if the matrices are say unitarily invariant and independent, and if we denote by $tr_N$ the trace divid …

You can rephrase your question as follows: Let $U,V$ be random (=Haar distributed) independent unitaries. You can write $G=UD_1 V$ where $D_1$ is diagonal (entries are the singular values of $G$, in …