It is known that a random matrix $A \in \mathbb{R}^{n \times n}$ whose entries are i.i.d. and follow a standard Gaussian distribution, the following upper-bound holds

$$ \mathbb{E} [\| A \|] \leq \sqrt{2n \log (2n) } $$

by the Bernstein inequality. Does anyone know how the result would change if the Gaussian entries had a variance $\sigma$ and did not follow the standard Gaussian, i.e., variance = 1?

  • $\begingroup$ My guess: The new random matrix become $\|\sigma A \|$, and $\mathbb{E}[\|\sigma A\|] = \sigma \mathbb{E}[\|A\|]$. Then apply the result. $\endgroup$ – BGM Aug 7 '17 at 17:40

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