The largest eigenvalue of a singular random matrix

Question

Let $X$ be a $m \times n$ random matrix whose entries are i.i.d. random variables with mean $0$ and finite variance $\sigma^2$. Let $$ Y_n =\frac{1}{n} X X^T.$$ Then according to the Marchenko-Pastur theorem, as $m, n \rightarrow \infty$ and $m/n \rightarrow c \in (0,\infty)$ the largest eigenvalue is bounded by $$ \lambda_{\max} =\sigma^2 \left (1+ \sqrt{\frac{m}{n}} \right )^2. $$

Now let's assume $n$ remains fixed while $m \rightarrow \infty $. Can one formally show that the largest eigenvalue is not bounded and that its rate of increase is related to $m$?

Answer 1

Denote the $j$-th column of $X$ by $x_j$. Since $Y\succeq \frac{1}{n}x_1x_1^T$, we have $\lambda_\max(Y)\ge\lambda_\max(\frac{1}{n}x_1x_1^T)=\frac1n\|x_1\|_2^2$. Therefore $E[\lambda_\max(Y)]\ge\frac1nE\|x_1\|_2^2=\frac{m\sigma^2}{n}$.

Also, $\lambda_\max(Y)=\|Y\|_2\le\frac1n\sum_j\|x_jx_j^T\|_2=\frac1n\sum_j\|x_j\|_2^2$. Hence $E[\lambda_\max(Y)]\le m\sigma^2$.

In short, $E[\lambda_\max(Y)]=\Theta(m)$.