# Sum of Square of the Eigenvalues of Wishart Matrix

Let $$A\in\mathbb{R}^{m\times d}$$ matrix with iid standard normal entries, and $$m\geqslant d$$, and define $$S=A^T A$$.

I want to have a tight upper bound for $$\sum_{k=1}^d \lambda_k^2$$, where $$\lambda_1,\dots,\lambda_d$$ are the eigenvalues of $$S$$.

What I tried:

• We know that (see e.g. Corollary 5.35 in Vershynin's notes), for $$A\in\mathbb{R}^{m\times d}$$, for any $$t\geqslant 0$$, with probability at least $$1-2\exp(-\Omega(t^2))$$, it holds: $$\sqrt{m}-\sqrt{d}-t \leqslant \sigma_{min}(A)\leqslant \sigma_{max}(A)\leqslant \sqrt{m}+\sqrt{d}+t.$$ Simply ignoring $$\sqrt{d},t$$ terms (say I am in the regime $$m\gg d,t$$), this yields $$\lambda_i(A), and thus, the sum above is upper bounded by $$m^2d$$.
• We also have the following: $$\sum_{k=1}^d (\lambda_k - m) = \sum_{i =1}^m \sum_{j=1}^d (A_{ij}^2-1),$$ which is sum of sub-exponential random variables, and thus, by a Bernstein-type bound, $$\sum_{k=1}^d \lambda_k \leqslant md+\omega(\sqrt{md})$$, for some function $$\omega(\sqrt{md})$$ growing faster than $$\sqrt{md}$$.
• The sum above is simply the trace of $$S^2=A^TAA^TA$$.

I'm new to random matrix business, so any help is greatly appreciated.

## 1 Answer

I will assume $$m=\alpha_d ds$$ with $$\alpha_d\to \alpha \in [1,\infty)$$ independent of $$d$$. The case $$\alpha\to\infty$$ is actually easier.

Define $$Z=d^{-1} m^{-2} \sum_{i=1}^d \lambda_i^2$$. Then $$Z$$ converges a.s. to $$\int x^2 d\mu_\alpha(x)$$ where $$\mu_\alpha$$ is the Pastur-Marchenko distribution of parameter $$\lambda=1/\alpha$$, see https://en.wikipedia.org/wiki/Marchenko%E2%80%93Pastur_distribution

• Ofer, thanks for the answer. Sorry, if it is a trivial question: How to handle the case $\alpha\to+\infty$ (more precisely, when $d\to+\infty$ and $d=o(m)$)? – kawa Jun 17 '19 at 20:13
• And also why the normalization is by $d^{-1}m^{-2}$? I understand that MP law requires $\frac1m$ normalization, and you get an extra for $d$, but why $d^2$? – kawa Jun 17 '19 at 20:19
• That is the limiting case, in which the limit empirical measure is a dirac at 1.... – ofer zeitouni Jun 17 '19 at 20:19
• and the normalization is as in the wikipedia page I quoted. You get $1/d$ from the empirical measure and $1/m^2$ from the normalization of the entries (you are taking the square of the MP matrix...) – ofer zeitouni Jun 17 '19 at 20:22