Let $A\in \mathbb{R}^{n\times n}$ be a random matrix whose entries are i.i.d. $\mathcal{N}(0,1)$. I know that $\mathbb{E}\|A\|\lesssim \sqrt{n}$, where $\|\cdot\|$ is the operator norm. But I have no idea how to show that $\mathbb{E}\|A\|^4\lesssim n^2$. Jensen's inequality does not work here. May I have some hint/reference on this problem? Thank you!
Determine the order of operator norm of Gaussian random matrix raised to the 4th power
probability theoryrandom matrices
Best Answer
I would say you need some "big guns" to get a result this tight. The main ingredients are:
Concentration inequality of Theorem 4.4.5 from Vershynin: $P( \|A\| > K\sqrt{n} +t ) \le 2exp(-t^2)$. This takes some effort to prove (although Vershynin's textbook is excellent and makes it very transparent!).
Standard representation of moments of positive RV: $E|X|^p = \int_0^\infty P(|X|>t) pt^{p-1}dt$.
With these two ingredients,
$$ E\|A\|^4 = \int_0^\infty P(\|A\|>t) 4t^{3}dt = \int_0^{K\sqrt{n}} P(\|A\|>t) 4t^{3}dt + \int_{K\sqrt{n}}^\infty P(\|A\|>t) 4t^{3}dt \le K_2 n^2 + Constant, $$ where we used in the last line that $P(\|A\|>t) \le 1$, and that $\int_{K\sqrt{n}}^\infty P(\|A\|>t) 4t^{3}dt$ is bounded by applying the concentration inequality.