Suppose that $G$ is an $n\times n$ Gaussian random matrix of i.i.d. entries $N(0,1/n)$ and $D$ is an $n\times n$ deterministic diagonal elements. I'd like to know if there have been results on the singular values of the matrix product $GD$.

There is classical result on the operator norm of $GD$, which says that $$ \|GD\|_{op} \approx \|D\|_{op} \pm \frac{\|D\|_F}{\sqrt{n}}, $$ where $\|D\|_F$ is the Frobenius norm of $D$.

I'd like to know if there are similar results on other singular values, or on the trace norm $\|GD\|_\ast$, preferably a lower bound.


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 $G$, independent of of $U,V$, and follow the Wishart distribution). Now you ask about the eigenvalues of $(D_1VD^2V^*D_1)^{1/2}$. The limit of the empirical measure can be computed by free probability methods. It is the free multiplicative convolution of the law of $D^2$ and that of $D_1^2$, pushed forward by square-root.

  • $\begingroup$ sorry I am new to free probability. What does 'free multiplicative product' mean? Can you point me to some reading? Do you mean the S-transform of the product is the product of the S-transform of the distributions $D^2$ and $D_1^2$? $\endgroup$ – user58955 Jan 7 '17 at 13:59
  • $\begingroup$ Yes. There are many sources to red about it, for example en.wikipedia.org/wiki/… You can also read more details in wisdom.weizmann.ac.il/~zeitouni/cupbook.pdf $\endgroup$ – ofer zeitouni Jan 7 '17 at 19:12
  • $\begingroup$ Thanks! btw, do you know any non-asymptotic analysis along this? By 'non-asymptotic' I mean not considering the limiting distribution but giving a quantitative bound for large $n$. The bounds on the operator norm I had in the problem statement, which holds with high probability, is a non-asymptotic bound. Do we expect something like that for singular values or trace norms of $GD$? $\endgroup$ – user58955 Jan 8 '17 at 2:17
  • $\begingroup$ There are concentration inequalities that quantify the convergence. See again our book for an introduction and references. $\endgroup$ – ofer zeitouni Jan 8 '17 at 5:14
  • $\begingroup$ Er, I mean $D$ is a deterministic $n\times n$ matrix, where $n$ is a 'fixed' large number. There is no limiting distribution for $D$ as $n\to\infty$. $\endgroup$ – user58955 Jan 13 '17 at 1:40

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