# Trace inverse of random PSD matrix?

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

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 $$m).
For $$m,n\gg 1$$, and $$m/n\equiv r\in (0,1)$$ fixed, an integration over the Marchenko–Pastur distribution gives the answer (with $$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}$$, see https://mathoverflow.net/a/332889/11260
• Could you show the integration bounds? I.e more explicitly how the integration is calculated. In addition, how would the calculation change if we had a diagonal matrix $D \in \mathbb{R}^{n\times n}$ $$E_{A}(\mathrm{Tr}( (DA^T AD + \lambda \mathrm{Id})^{-1})),$$ so you are unable to exchange the transposes on A.
• I added the integral formula; if you insert a matrix $D$, even if it is diagonal, the answer is no longer only a function of the eigenvalues of $AA^\top$, so the MP distribution does not help; this has nothing to do with the order $AA^\top$ or $A^\top A$, its whether the eigenvectors enter in addition to the eigenvalues. Jun 15, 2022 at 18:43