Consider a random matrix $A \in \mathbb{R}^{n\times n}$ with i.i.d. entries, with symmetric law and finite variance. I am curious about the behavior of $$\mathrm{Tr}( (A^T A + \lambda \mathrm{Id})^{-1})$$ when the size of the matrix $n$ goes to infinity. Here, 
$\lambda > 0$ is fixed and ensures that $A^T A + \lambda \mathrm{Id}$ is invertible as a positive definite matrix.


Typically, I am wondering if this quantity behaves asymptotically like $n^{\gamma}$ for some $\gamma$. 



Trace of inverse of random positive-definite matrix in high dimension?
If the elements of the $n\times n$ matrix $A$ are independent identically distributed random variables with mean 0 and variance $\sigma^2$, the $n$ eigenvalues $x\_i$ of $A^{\rm T}A$ have for large $n$ the [Marchenko-Pastur distribution](https://en.wikipedia.org/wiki/Marchenko%E2%80%93Pastur_distribution)
$$P(x)=\frac{ \sqrt{4 n\sigma^2-x}}{2 \pi n \sigma^2 \sqrt{x}},\;\;0<x<4n\sigma^2.$$
The expectation value of $Y={\rm Tr}\,(A^{\rm T}A+\lambda)^{-1}$ then becomes
$$E[Y]=n\int\_{0}^{4n\sigma^2}(x+\lambda)^{-1}P(x)dx=\frac{\sqrt{4 n \sigma^2/\lambda+1}-1}{2 \sigma^2}.$$
So I would conclude a growth as $n^{\gamma}$ with $\gamma=1/2$.


