Assume $A\in \mathbb{R}^{n\times n}$ with each entry being i.i.d. bounded r.v. in $[a,b]$, is $\Vert A\Vert\_2$ is sub-Gaussian?


Intuitively, since $\{A\_{ij}\}\_{i,j=1,...,n}$ is bounded, then
$$\Vert A \Vert\_2 = \sup\_{\Vert v \Vert = 1} \vert v^TA^TAv\vert = \sup\_{\Vert v \Vert = 1}\vert\sum\_{i,j}v\_iv\_j(\sum\_k A\_{ki}A\_{kj})\vert\leq \max(a^2,b^2)$$


Then $\Vert A\Vert\_2$ is bounded so that it is sub-Gaussian. I doubt whether there exists any problem in my guess above because sub-Gaussian is much weaker than "bounded" condition.



Spectral norm of matrices of bounded random variables
Your chain of equality is slightly off, the idea is there however. I think that the following one is correct :
\begin{align\*}
\| A \|\_2^2 &= \sup\_{\| v \|\_2=1} v^TA^TAv\\
&= \sup\_{\|v \|\_2 = 1} \left| \sum\_{i,j} v\_i v\_j \left( \sum\_{k} A\_{ki} A\_{kj} \right) \right|\\
&\leq \sup\_{\|v \|\_2 = 1} \sum\_{i,j} |v\_i|\cdot |v\_j|\cdot \left| \sum\_{k} A\_{ki} A\_{kj} \right|\\
&\leq n \cdot \max(a^2,b^2)\sup\_{\|v \|\_2 = 1} \|v\|\_1^2\\
&=n^2\cdot \max(a^2, b^2)
\end{align\*}
Note that all inequalities can be achieved ($A\_{i,j}=\max(a,b)$ for all $i,j$ and $v\_i=\frac{1}{\sqrt{n}}$ for all $i$) and therefore your bound cannot really be correct.


Now this yields that $0\leq \| A \|\_2 \leq n\cdot\max(|a|,|b|)$, and therefore $\|A \|\_2$ is a sub-Gaussian.


