A simple proof of almost sure convergence for the largest singular value of a product of Gaussian matrices

TL;DR

This paper provides a straightforward proof that the largest singular value squared of a product of Gaussian matrices converges almost surely to a specific deterministic limit, extending known results for a single matrix case.

Contribution

It offers a simple, accessible proof of almost sure convergence for the largest singular value of Gaussian matrix products, avoiding advanced free probability methods.

Findings

Largest singular value squared converges almost surely to a deterministic limit

Generalizes known results from single to multiple Gaussian matrices

Proof is simple and does not rely on free probability techniques

Abstract

Let $m \geq 1$ and consider the product of $m$ independent $n \times n$ matrices $\mathbf{W} = \mathbf{W}_1 \dots \mathbf{W}_m$, each $\mathbf{W}_{i}$ with i.i.d. normalised $\mathcal{N}(0, n^{-1/2})$ entries. It is shown in Penson et al. (2011) that the empirical distribution of the squared singular values of $\mathbf{W}$ converges to a deterministic distribution compactly supported on $[0, u_m]$, where $u_m = \frac{{(m+1)}^{m+1}}{m^m}$. This generalises the well-known case of $m=1$, corresponding to the Marchenko-Pastur distribution for square matrices. Moreover, for $m=1$, it was first shown by Geman (1980) that the largest squared singular value almost surely converges to the right endpoint (the so-called ``soft edge'') of the support, i.e. $s_1^2(\mathbf{W}) \xrightarrow{a.s.} u_1$. Herein, we present a proof for the general case $s_1^2(\mathbf{W}) \xrightarrow{a.s.} u_m$ for $m\geq 1$. Although we do not claim novelty for our result, the proof is simple and does not require familiarity with modern techniques of free probability.