Limit law of eigenvalue of random matrix with mean different to 0

Question

If $X$ denotes a $m \times n$ random matrix whose entries are independent identically distributed random variables with mean $\mu$ and $\sigma^2 < \infty$, let

$$Y = X X^T$$

with $X^T$ the transpose of $X$. Let $\lambda_1 , \lambda_2 ,\ldots, \lambda_m$ be the eigenvalues of $Y$ (viewed as random variables).

If $\mu = 0$, it is known that the law of $\lambda$ converges to Marchenko–Pastur distribution: https://en.wikipedia.org/wiki/Marchenko%E2%80%93Pastur_distribution

My question is that in the case $\mu \neq 0$ what is the limit distribution of $\lambda$?

Answer 1

The addition of the same constant $\mu$ to all elements of $X$ (so that their mean becomes $\mu$) is a rank-one perturbation of the matrix, which has no effect on the distribution of the eigenvalues of $XX^T$ in the limit $n,m\rightarrow\infty$ at fixed $n/m$ --- this limiting distribution remains the Marchenko-Pastur distribution. The rank-one perturbation introduces a single outlier, but the bulk of the spectrum is unchanged. See for example The singular values and vectors of low rank perturbations of large rectangular random matrices (2011).