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 $\lambda\_i(A)<m^2$, 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.



Sum of Square of the Eigenvalues of Wishart Matrix
I will assume $m=\alpha\_d d$ 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>


