Consider a random matrix $A \in \mathbb{R}^{m\times n}$ with i.i.d. entries, with mean zero and variance 1 and $m <n $. I am interested in the expectation of $$E\_{A}(\mathrm{Tr}( (A^T A + \lambda \mathrm{Id})^{-1})).$$ This question is similar to [Trace of inverse of random positive-definite matrix in high dimension?](https://mathoverflow.net/a/332889/483386), except $A$ is non-symmetric. I am unfamiliar with Marchenko-Pastur distribution, so any references would be great!



Trace inverse of random PSD matrix?
I think the product $A^\top A$ in the OP should read $AA^\top$, to avoid a trivial contribution from zero eigenvalues (assuming $A\in\mathbb{R}^{m\times n}$ and $m<n$).


For $m,n\gg 1$, and $m/n\equiv r\in (0,1)$ fixed, an integration over the [Marchenko–Pastur distribution](https://en.wikipedia.org/wiki/Marchenko%E2%80%93Pastur_distribution) gives the answer (with $x\_\pm=(1\pm\sqrt{r})^2$)
$$\lim\_{m,n\rightarrow\infty}\mathbb{E}[m^{-1}\mathrm{Tr}\,(n^{-1}AA^\top + \lambda I)^{-1})]=\int\_{x\_-}^{x\_+} \frac{1}{x+\lambda}\frac{\sqrt{\left(x\_+-x\right) \left(x-x\_-\right)}}{2 \pi r x}\,dx$$
$$\qquad=\frac{1}{2\lambda r}\left(\sqrt{\lambda^2+2 \lambda (1+r)+(1-r)^2}-\lambda+r-1\right).$$
The rescaling of $AA^\top$ by a factor $1/n$ is needed for a $\lambda$-dependent answer in the large $n$ limit. If you do not rescale, then
$$\lim\_{m,n\rightarrow\infty}\mathbb{E}[\mathrm{Tr}\,(AA^\top + \lambda I)^{-1})]=\frac{r}{1-r}.$$
This diverges for $r=1$, in that case the trace grows as $\sqrt{n}$, see <https://mathoverflow.net/a/332889/11260>


