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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 $m$?

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

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  • $\begingroup$ Thank you @user1551. All makes sense, except I do not follow the step $Y \succeq x_1 x_1^T/n$ implies $\lambda_{\max}(Y) \ge \lambda_{\max}( x_1 x_1^T/n)$. Could you please clarify or give a reference? $\endgroup$
    – GCru
    Nov 4, 2022 at 8:41
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    $\begingroup$ @GCru This is a straightforward fact. Suppose $Y\succeq Z$. Let $v$ be a unit eigenvector corresponding to the largest eigenvalue of $Z$. Then $\lambda_\max(Y)\ge v^TYv\ge v^TZv=\lambda_\max(Z)$. $\endgroup$
    – user1551
    Nov 4, 2022 at 8:44
  • $\begingroup$ Aha! Thanks again @user1551. I guess this cannot be applied here, because the matrices are not Hermitian. $\endgroup$
    – GCru
    Nov 4, 2022 at 9:49

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