# The largest eigenvalue of a singular random matrix

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 $$m$$?

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

• 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?
– GCru
Nov 4, 2022 at 8:41
• @GCru This is a straightforward fact. Suppose $Y\succeq Z$. Let $v$ be a unit eigenvector corresponding to the largest eigenvalue of $Z$. Then $\lambda_\max(Y)\ge v^TYv\ge v^TZv=\lambda_\max(Z)$. Nov 4, 2022 at 8:44
• Aha! Thanks again @user1551. I guess this cannot be applied here, because the matrices are not Hermitian.
– GCru
Nov 4, 2022 at 9:49