# Spectral norm of random matrix

Suppose $$A$$ is a $$n \times n$$ random matrix with centered Gaussian (real) i.i.d. entries with variance $$\frac{\sigma^2}{n}$$.

What to we know about the spectral norm $$s(A)$$ of $$A$$, that is $$\sqrt{\rho(A^t A)}$$? Here, $$\rho(\cdot)$$ denotes the largest eigenvalue of a matrix.

In particular, if $$A$$ is symmetric, we know that $$s(A)$$ is precisely equal to $$\rho(A)$$ and from the circular law it implies that $$s(A)$$ converges to $$\sigma$$ as $$n \to \infty$$.

Is this last statement true for non-symmetric $$A$$?

Actually, neither statement is true. The circular law does not control the spectral radius: it only predicts that the majority of eigenvalues lie in the disc, while the spectral radius is concerned with the most extreme eigenvalues. There could still be as many as $o(n)$ eigenvalues lying outside of the disc, and so it is not necessarily true that $\rho(A)$ is a.s. equal to $\sigma$ in the limit.
I believe that almost surely $s(A) \to 2\sigma$ in both the symmetric and non-symmetric cases. For the non-symmetric case in particular, see Theorem 2.1 in this survey article by Rudelson and Vershynin (from their ICM invited talk).