Consider $\mathcal{A}:\mathbb{R}^{n\times m}\to \mathbb{R}^{p\times q}$ to be a linear operator. I know that by considering the trace norm and using the submultiplicativity of the operator norm we have
$\|\mathcal{A}(X)\|_F \leq \|\mathcal{A}\|_{op} \|X\|_F$
where $\|.\|_{op}$ denotes the operator norm and $\|.\|_F$ is the Frobenius norm. The question is whether we can also claim that
$\|\mathcal{A}(X)\| \leq \|\mathcal{A}\|_{op} \|X\| $
where $\|.\|$ is the matrix spectral norm (the largest singular value of $X$)? It is certainly true that
$\|\mathcal{A}(X)\| \leq \sqrt{r}\|\mathcal{A}\|_{op} \|X\| $
where $r = rank(X)$, but I am not sure about the one without $\sqrt{r}$. I would appreciate any thoughts or even references.