# Operator Norm and Submultiplicativity against the Spectral Norm

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.

• Actually, the notion of operator norm depends of the chosen norms in the domain and the co-domain of the operator. – Dirk Jul 6 '15 at 4:55
• Thanks for your note. In fact by stating the first equation I tried to imply that $\|\mathcal{A}\|_{op} = sup \frac{\|\mathcal{A}(X)\|_F}{\|X\|_F}$. – Ali Jul 6 '15 at 5:06