A heuristic argument based on Ofer's comment
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Note that $T := YY^\top = XX^\top + AX^\top +X^\top A + AA^\top$, where $A:=\alpha 1_n 1_d^\top$ is the all-ones matrix of shape $n \times d$. Note that $AA^\top=\alpha^2 d1_n1_n^\top$.

>**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 some due to some free probability lemma ?

Also, from [Lemma 1 of this monograph][1], we deduce that

>**Fact 2** $q_1,\ldots,q_n \ge 0$ that depend measurably on $X$, with $\sum_{i=1}^n q_i = 1$, such that $\lambda_i(XX^\top+AA^\top) = \lambda_i(XX^\top + \alpha^2d 1_n1_n^\top) = \lambda_i(XX^\top) + q_i\alpha^2nd$.

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 $q \to (1,0,\ldots,0)$, in some sense.

>Putting things together, one would expect $\lambda_i(YY^\top) \to \lambda_i(XX^\top) + \delta_{i=1}\alpha^2 nd$.


  [1]: http://moz-extension://b8ab8d95-45e6-4909-998e-8b13a85d8ca1/enhanced-reader.html?openApp&pdf=https%3A%2F%2Fhal.archives-ouvertes.fr%2Fhal-01902562%2Fdocument