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.