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This is a repost from mathstackexchange, as I think asking this question is more appropriate here.

Coming from statistical physics, I am interested in the (real) spectrum of the following sum, and ultimately $l_p$ norms of the spectrum, \begin{equation} \sum_{n=1}^m c_n X_n, \end{equation} where $c_n$ are non-random real numbers and $X_n$ are independent Wishart matrices of size $m\times m$. Independent in the sense of elementwise independence. The weights $c_n$ are nice in the sense that the above sum is uniformly bounded with respect to $n$. Let the degree of freedom of the Wishart matrices be $p$. I am interested in the limit large $m$ and the cases where $p/m\to 0$ and $p/m\to$ constant > 0. I know that in the latter case, the spectrum of $X_n$ will be approximated by the Marchenko-Pastur distribution.

I am new to operator valued probability theory. I did some searching and could find results on spectra of A+B, where A and B are random (free) $m\times m$ matrices. Most of them where in the limit of large dimension $m$. I could not find something related to summing over orders of the dimension of the matrices $m$. Are there any results in that direction, maybe some kind of central limit theorem? If yes, where can I find them?

I am grateful for any advice and literature hints!

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    $\begingroup$ I am not completely sure I understand your parameters (there are many $p$'s), but I guess that, as in Rudelson's "Random vectors in the isotropic position." (JFA 1999)", you can obtain powerful estimates for the Schatten $p$-norms (=$\ell_p$ norm of the spectrum) by symetrization and using the non-commutative Khintchine inequalities. $\endgroup$
    – Mikael de la Salle
    Commented Nov 27, 2020 at 15:42
  • $\begingroup$ Thanks @MikaeldelaSalle! Sorry for the confusion. The $p$ in $l_p$ is of course independent of the degree of freedom $p$. In Rudelson's paper, p64/65 where he states the Khintchine inequality, to what sequence are the Rademacher functions applied and what is $\|\cdot\|_{L^p(Q,\mu,C^n_p)}$ space? $\endgroup$
    – jamblejoe
    Commented Nov 27, 2020 at 17:17
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    $\begingroup$ I do not have the paper under my eyes, but I guess that $L^p(Q,\mu,C_p^n)$ is the space of size-n matrix-valued random variables on a probability space $(\Omega,\mu)$, for the norm $\|X\|^p = \mathbf{E} Tr( |X|^p)$. $\endgroup$
    – Mikael de la Salle
    Commented Nov 27, 2020 at 17:26
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    $\begingroup$ The idea of symmetrization is that, for independant mean $0$ matrices $Y_i$, up to a factor $2$, the $L^p(\Omega,\mu,C_p^n)$-norm of $\sum_i Y_i$ is of the order the norm of $\sum_i (Y_i - Y_i')$, where $Y'_i$ are independant copies of $Y_i$. By independance, this last quantity as the same norm as $\sum_i \varepsilon_i (Y_i-Y'_i)$ for independant Rademacher $\varepsilon_i$. This is how rademacher functions come into the picture and how NC Khintchine can be used. $\endgroup$
    – Mikael de la Salle
    Commented Nov 27, 2020 at 17:30

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This problem is addressed in On the Exact and Approximate Eigenvalue Distribution for Sum of Wishart Matrices. The approximation replaces the weighted sum of Wishart matrices by one equivalent Wishart matrix.

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  • $\begingroup$ Thank you for the quick answer! I had a quick look into the reference you gave. There, the coefficients are assumed to be positive. I will see if I can generalize to real coefficients. $\endgroup$
    – jamblejoe
    Commented Nov 27, 2020 at 16:49
  • $\begingroup$ Another question: In the reference complex Wishart matrices are considered. I have not understood the proofs yet, so I can not say, in what sense they hold for real Wishart matrices as well. Or is this easily seen? $\endgroup$
    – jamblejoe
    Commented Nov 27, 2020 at 16:51
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    $\begingroup$ real or complex Wishart matrices is not an essential difference, the formulas will carry over; negative weights, however, is an essential difference; if you allow for negative weights the resulting matrix may no longer be positive definite, so this approximation of the weighted sum by an equivalent Wishart matrix will fail. $\endgroup$
    – Carlo Beenakker
    Commented Nov 27, 2020 at 20:24
  • $\begingroup$ Thanks! I guess one can still make some use of that result by splitting the $c_j$'s into positive and negative and reduce the problem to the difference of just two wishart matrices. $\endgroup$
    – jamblejoe
    Commented Nov 27, 2020 at 21:42

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