Suppose $x\in SG(\sigma^2)$ is a sub-Gaussian random vector, i.e.
$\left<u,x\right>\quad \forall u\in \mathbb{S}^{n-1}$ is a sub-Gaussian random variable.
My question is : what condition on the random matrix $A$ can guarantee that $Ax$ is again a sub-Gaussian random vector?
I know that $\|A\|\in L^{\infty}$ is one of the conditions. But this one is too strong. Is there any weaker condition?
If you only have the hypothesis of sub-Gaussianity, this is the best you can do. Work in dimension $n=1$ for simplicity, let $X\sim N(0,1)$, and let $A$ be independent of $X$. If $AX$ is to be sub-Gaussian, the Laplace transform condition will demand
$$ \mathbb{E}[e^{\lambda A X}]= \mathbb{E}[e^{\lambda^2 A^2 /2}] \leq B e^{\lambda^2 b} $$
for some constants $B,b>0$ and all $\lambda$. But, this tells us something about the distribution of $A$. Namely, by a Chernoff bound, we have the concentration estimate
$$ P\{|A|^2/2 > t^2\} \leq e^{-\lambda^2 t^2} \mathbb{E}[e^{\lambda^2 A^2 /2}] \leq B e^{-\lambda^2 t^2} e^{\lambda^2 b}, ~~~\lambda\in \mathbb{R}. $$
Hence, $P\{|A|^2 > 2 b\} = 0$, and $A$ is bounded a.s.
