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Given two freely independent random hermitian matrices $A$ and $B$ following laws $\mu, \nu$, one can compute the empirical spectral distribution of $AB$ by their free multiplicative convolution $\mu\boxtimes\nu$ using the $S$-transform. Is there a way to compute the empirical spectral distribution of other products of $A$ and $B$, such as $AB^{-1}$?

(I am new to random matrix theory so the question might sound naive.)

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2 Answers 2

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There is the possibility of dealing with any non-commutative rational function in A and B, by using more general (operator-valued) versions of free probability. Usually the corresponding equations cannot be solved analytically, but they are nice fixed-point equations which can be addressed numerically. See for example, chapter 10 in the book Mingo, Speicher: Free Probability and Random Matrices or the article Helton, Mai, Speicher: Applications of Realizations (aka Linearizations) to Free Probability

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  • $\begingroup$ Thank you Prof Speicher for the references! For the relatively simple case of $AB^{-1}$, does the standard method work (see discussions above)? $\endgroup$
    – Shadumu
    Aug 24, 2021 at 15:28
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    $\begingroup$ @Shadumu In the case $AB^{-1}$ the standard method with the S-transform works; actually the S-transform S' of $B^{-1}$ can be expressed in terms of the S-transform S of B via $S'(z)=1/S(-1-z)$; this is shown in Prop. 3.13. of the paper: U. Haagerup, H. Schultz, Brown measures of unbounded operators affiliated with a finite von Neumann algebra. Math. Scand. 100(2) $\endgroup$ Aug 24, 2021 at 21:16
  • $\begingroup$ @RolandSpeicher Thanks a lot for the reference! $\endgroup$
    – Shadumu
    Aug 25, 2021 at 10:25
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If we denote the spectral density of $B$ by $\rho_{B} (\lambda )$, with $\lambda $ labeling the spectral values, then $\rho_{B^{-1} } (\lambda ) = (1/\lambda^{2} ) \rho_{B} (1/\lambda )$; also $B^{-1} $ and $A$ are freely independent, and you can convolve them multiplicatively using the $S$-transform.

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  • $\begingroup$ Thanks! I can think of doing the S-tranform by first finding the resolvent/green function $G_B(z)=(1+w)/z$ from the density $\rho_B$ and then use $zwS(w)=1+w$ to solve for $S(w)$. Is the resolvent $G_{B^{-1}}(z)$ still convergent for $\rho_{B^{-1}}$? How is $G_{B^{-1}}(z)$ related to $G_B(z)$? $\endgroup$
    – Shadumu
    Aug 23, 2021 at 18:22
  • $\begingroup$ In general, there are no guarantees for the convergence of $G_{B^{-1} } (z)$ - if $B$ has eigenvalues near zero, then $B^{-1} $ will be correspondingly singular. And vice versa. I don't think there's a simple algebraic relation between $G_{B^{-1} } (z)$ and $G_{B} (z)$, but you can think of substituting $\lambda \rightarrow 1/\lambda $ in $G_{B^{-1} } (z)$ - then you're integrating over the same density as in $G_{B} (z)$, but with a changed kernel ... $\endgroup$ Aug 23, 2021 at 19:55
  • $\begingroup$ Thanks. Are you aware of any references that properly treat such a case? $\endgroup$
    – Shadumu
    Aug 23, 2021 at 20:01
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    $\begingroup$ Not sure whether I properly treated these things, but I played with them in hep-th/9604012 and hep-th/9609216. Note that I edited my answer - I was originally missing the $(1/\lambda^{2} )$ measure factor in $\rho_{B^{-1} } $. $\endgroup$ Aug 23, 2021 at 20:07

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