# Free multiplicative convolution of two random matrices

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.)

• Thank you Prof Speicher for the references! For the relatively simple case of $AB^{-1}$, does the standard method work (see discussions above)? Aug 24, 2021 at 15:28
• @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) Aug 24, 2021 at 21:16
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.
• 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 $G_B(z)$? Aug 23, 2021 at 18:22
• 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 ... Aug 23, 2021 at 19:55
• 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 $\rho_{B^{-1} }$. Aug 23, 2021 at 20:07