Limiting eigenvalue distribution of $YY^\top$ where $Y_{ij} = X_{ij} + a$ and $X$ has iid rows from an isotropic log-concave distribution Let $a \in \mathbb R$ be a determinstic scalar and let $X$ be and $n \times d$  such that the $n \times n$ psd random matrix $S=XX^T$ has limiting eigenvalue distribution $\mu$, when $n,d \to \infty$ with $n/d = \gamma \in (0,\infty)$. Consider the psd random matrix $T=YY^\top$, where $Y$ is the $n \times d$ random matrix with entries $Y_{ij} = X_{ij} + a$.

Question. What is the limiting eigenvalue distribution $\lambda$ of $T$ ?

The example I have in mind is when $X$ has iid rows drawn from an isotropic log-concave distribution in $\mathbb R^d$. For gaussian of uniform on unit-sphere in $\mathbb R^d$.
Notes

*

*My guess is that the Stieltjes transform of $\lambda$ would be some how given implicitly via the Stieltjes transform of $\mu$.

*I'm only interesting in being able to integrate w.r.t $\lambda$. For example, I'm interested computing sums like $\sum_i \dfrac{\ell_i}{(\ell_i + \lambda)^2}$, where $\ell_1 \ge \ell_2 \ge \ldots $ are the eigenvalues of $T$.

 A: A heuristic argument based on Ofer's comment
Note that $T := YY^\top = XX^\top + AX^\top +X^\top A + AA^\top$, where $A:=a 1_n 1_d^\top$ is the all-ones matrix of shape $n \times d$, times the scalar $a \in \mathbb R$. Note that $AA^\top=da^21_n1_n^\top$.

Conjecture. $\lambda_i(YY^\top) \to \lambda_i(XX^\top) + \delta_{i=1}nda^2$.

I've empirically observed that the above conjecture is likely to be true. Below, I sketch some ideas which point in the direction of a possible proof.


Fact 1: $\mbox{LSD}(XX^\top + AX^\top +X^\top A + AA^\top) \to \mbox{LSD}(XX^\top + AA^\top)$, in some sense (probably weakly).

The above fact (it seems) was implied by Ofer's comment. I've observed this empirically, but I'm not quite sure why it should be true. Maybe this is a consequence of some classical / well-known lemma from free probability theory ?
Also, using a result due to G. Golub and V. Loan (see Lemma 1 of this monograph), we deduce that

Fact 2: There exist $q_1,\ldots,q_n \ge 0$ with $\sum_{i=1}^n q_i = 1$, such that $\lambda_i(XX^\top+AA^\top) = \lambda_i(XX^\top + \alpha^2d 1_n1_n^\top) = \lambda_i(XX^\top) + q_i nda^2$.

Moreover, we have the formula $q_i = \dfrac{(1_n^\top u_i)(1_n^\top v_i)}{nu_i^\top v_i}$ (provided $u_i^\top v_i \ne 0$), where $u_1,\ldots,u_n$ are the eigenvectors of $XX^\top$ and $v_1,\ldots,v_n$ are the eigenvector of $v_i$. Intuitively, one whould expect $u_i$ to be pretty much orthogonal to $v_i$ for $i \ne 1$. As a consequnce, one would expect $q \to \delta_{i=1} \in \ell^2$, in some sense.
Putting all the bits together (and of course making some of the above arguments rigorous) should prove the above claim / conjecture.
