# What does a product of many Gaussian matrices converge to?

Let $$A$$ be a product of $$n$$ $$d\times d$$ matrices with IID standard Gaussian entries and consider the value of $$g(x)=x f(x)$$ where $$f(x)$$ is the density of squared singular values of $$A/\|A\|$$.

Is anything known about $$g(x)$$ in the $$n\to \infty$$, $$d\to \infty$$ limit, where $$d$$ limit taken first?

The effect of taking $$n\to \infty$$ limit first is better understood as such product is almost surely rank-1. Plotting $$g(x)$$ heuristically by relying on formula 3.61 from Ipsen's thesis yields the chart below: Notebook

Empirically, I see the same density when considering matrices with Gaussian, symmetric Bernoulli and symmetric uniform entries so I'm wondering if the large $$n$$ limit has a name. • This is a slightly different problem (matrices are rectangular and main results is when outer dimension is small) but may be helpful cs.cmu.edu/afs/cs/user/dwoodruf/www/lw21.pdf Aug 8 at 8:49
• Motivated by your added "free-probability" tag: To use multiplicative free convolution via the S-transform, the mean of the eigenvalue distribution would have to be positive - one could imagine shifting the semicircular distribution by a small amount $\epsilon$. Then, the S-transform can be given and taken to the $n$-th power. Re-extracting the corresponding eigenvalue distribution doesn't seem straightforward - I haven't made further progress in that direction. If you're interested in such a partial/failed attempt, I'll be happy to post it ... maybe someone sees how to take it further. Aug 9 at 2:22
• @MichaelEngelhardt S-transform of $f(x)$ is $\frac{1}{(1+z)^n}$, right? (Eq 32 of Burda). What's not clear to me is 1) why does $\frac{1}{(1+z)^\infty}$, give $1/x$ behavior at 0 and 2) how to get S-transform of $g(x)$ Aug 9 at 2:55
• This formula of Burda is for the product of Wishart matrices, i.e., squares of Gaussian matrices. That seems to me to be different from what you're considering. For a Gaussian matrix, the S-transform is $1/\sqrt{z}$, and you'll have trouble inverting something like $1/z^{n/2}$. Hence the need for a shift by $\epsilon$ ... Aug 9 at 3:23
• @MichaelEngelhardt I've checked formula 35 which is density that Burda gets by inverting $\frac{1}{(1+z)^2}$ S-transform, and it matches Ipsen's formula 3.61 for eigenvalues of $XYY'X'$, which is squared singular values of $XY$, Gaussian $X,Y$ Aug 9 at 6:05