$*$-convergence of Schur-Hadamard products of independent non-symmetric random matrices.

Let $\{x_{\alpha}\}_{\alpha\in\mathbb{Z}}$ and $\{y_{\alpha}\}_{\alpha\in\mathbb{Z}}$ be two independent sets of zero mean, unit variance random variables with uniformly bounded moments of all orders. Consider a non-symmetric Toeplitz matrix $X_n=((x_{i-j}))_{1\le i,j\le n}$ and a Hankel matrix $Y_n=((y_{i+j}))_{1\le i,j\le n}$ and let $M_n=X_n\odot Y_n$ be their elementwise/Schur-Hadamard product. We show that $n^{-1/2}M_n$, as an element of the $*$-probability space $(\mathcal{M}_n(L^{\infty,-}(\Omega,\mathbb{P})), \frac{1}{n}\mathbb{E}\mathrm{tr})$, converges in $*$-distribution to a circular variable. This gives a matrix model for circular variables with only $O(n)$ bits of randomness. As a direct corollary, we recover a result of Bose and Mukherjee (2014) that the empirical spectral measure of the $n^{-1/2}$-scaled Schur-Hadamard product of symmetric Toeplitz and Hankel matrices converges weakly almost surely to the semi-circular law. Based on numerical evidence, we conjecture that the circular law $\mu_{\mathrm{circ}}$, i.e. the uniform measure on the unit disk of $\mathbb{C}$, also the Brown measure of $n^{-1/2}M_n$, is in fact the limiting spectral measure of $n^{-1/2}M_n$. If true, this would be an interesting example where a random matrix with only $O(n)$ bits of randomness has the circular law as its limiting spectral measure (all the standard examples have $\Omega(n^2)$ bits of randomness). More generally, we prove similar results for structured random matrices of the form $A=((a_{L(i,j)}))$ with link function $L:\mathbb{Z}_+^2\rightarrow\mathbb{Z}^d$. Given two such ensembles of matrices with link functions $L_X$ and $L_Y$, we show that $*$-convergence to a circular variable holds for their Schur-Hadamard product if the map $$(i,j) \mapsto (L_X(i,j), L_Y(i,j))$$ is injective and some mild regularity assumptions on $L_X$ and $L_Y$ are satisfied.