Let $A,B \in \mathcal{M}_{n\times m}(\mathbb{R})$. Show that $\|A^TB \|_F \leq \|A \|_F \|B \|_{\text{op}}$

Definition: The Operator norm, denoted $\| \cdot \|_{\text{op}}$ is defined to be $$\| B \|_{\text{op}} = \inf\left\{ c \geq 0 : \|Bv \|\leq c\|v \|\ , \forall v \in V\right\}$$

Definition: The Frobenius norm, denoted $\|\cdot \|_F$ is defined to be $$\| A \|_F = \sqrt{\sum_{i=1}^n \sum_{j=1}^m |a_{ij}|^2}$$

Now, the inequality

\begin{align*} \|A^TB \|_F &\leq \|A \|_F \|B \|_F \end{align*}

is rather straight forward. However, obtaining $\|B \|_F \leq \|B \|_{\text{op}}$ is not. Any help would be appreciated.

• Wait, $A$ and $B$ are both $n \times m$ matrices? Did you mean $AB^\top$ or $A^\top B$? Or are these matrices square? – Theo Bendit Sep 19 '18 at 14:07
• Yes, sorry $A^TB$. I will edit. Thank you. – Dragonite Sep 19 '18 at 14:18
• Why do you write $\|\ A\|$ instead of $\|A\|$? (I actually changed that in every instance.) – Michael Hardy Sep 19 '18 at 17:47

Let $A = (a_1, \dots, a_m)$ where $a_j$ denotes the columns. \begin{align*} \|A^T B\|_F^2 = \|B^T A\|_F^2 = \|( B^Ta_1, \dots, B^Ta_m)\|_F^2 = \sum_{j=1}^m \|B^T a_j\|_2^2 \\\le \sum_{j=1}^m \|B^T\|_{op}^2 \|a_j\|_2^2 = \|B^T\|_{op}^2 \|A\|_F^2. \end{align*} Note $\|B^T\|_{op} = \|B\|_{op}$.
Though not an answer to the question, I would like to mention that the inequality $\|B\|_F \leq \|B\|_{\mathrm{op}}$ is not correct, it should be $\|B\|_{\mathrm{op}}\leq\|B\|_{F}$. A quick way to see this, is to use the equations $$\|B\|^2_{\mathrm{op}} = \lambda_{\max}(B^TB)= \sigma^2_{\max}(B)$$ and $$\|B\|_F^2 = \operatorname{trace}(B^T B) = \sum_{i=1}^{\min\{m,n\}} \sigma^2_i (B),$$ where $\lambda_i$ and $\sigma_i$ stand for an eigenvalue and singular value respectively.