# Showing that Frobenius Norm is Submultiplicative

Considering the Frobenius norm $$\|.\|_{F}$$, how can we prove that it's submultiplicative?

N.b: I noticed there is a user who asked a similar question but proof looked like directing away from the question. I hope anyone can help me.

• Welcome to MSE! next time at least provide an attempt of your own if you find other users' answers to be not what you are looking for. Dec 30 '20 at 3:07
• Well noted, thank you Dec 30 '20 at 3:27
• Alright let's do it! @vitamind Jun 8 at 17:28
• I fixed it and asked a more mathematical question with some quest for reference on this matter @vitamind thank you Jun 8 at 18:01

\begin{align*}\|AB\|_{F}^{2}= \left\|\left(\begin{array}{c} \widehat{a}_{0}^{H} \\ \widehat{a}_{1}^{H} \\ \vdots \\ \widehat{a}_{m-1}^{H} \end{array}\right)\left(\begin{array}{llll} b_{0} & b_{1} & \cdots & b_{n-1} \end{array}\right)\right\|_{F}^{2} &=\left\|\left(\begin{array}{c|c|c|c} \widehat{a}_{0}^{H} b_{0} & \widehat{a}_{0}^{H} b_{1} & \cdots & \widehat{a}_{0}^{H} b_{n-1} \\ \hline \hat{a}_{0}^{H} b_{0} & \widehat{a}_{0}^{H} b_{1} & \cdots & \widehat{a}_{0}^{H} b_{n-1} \\ \vdots & \vdots & & \vdots \\ \hline \widehat{a}_{m-1}^{H} b_{0} & \widehat{a}_{m-1}^{H} b_{1} & \cdots & \widehat{a}_{m-1}^{H} b_{n-1} \end{array}\right)\right\|_{F}^{2}\\ \\&=\sum_{i} \sum_{j}\left|\widehat{a}_{i}^{H} b_{j}\right|^{2} \\ &\leq \sum_{i} \sum_{j}\left\|\hat{a}_{i}^{H}\right\|_{2}^{2}\left\|b_{j}\right\|^{2} \quad \text { (Cauchy-Schwartz) } \\ &=\left(\sum_{i}\left\|\widehat{a}_{i}\right\|_{2}^{2}\right)\left(\sum_{j}\left\|b_{j}\right\|^{2}\right)\\ &=\left(\sum_{i} \widehat{a}_{i}^{H} \widehat{a}_{i}\right)\left(\sum_{j} b_{j}^{H} b_{j}\right) \\ &\leq\left(\sum_{i} \sum_{j}\left|\widehat{a}_{i}^{H} \widehat{a}_{j}\right|\right)\left(\sum_{i} \sum_{j}\left|b_{i}^{H} b_{j}\right|\right)\\&=\|A\|_{F}^{2}\|B\|_{F}^{2} \end{align*} Now, to conclude this proof, we shall take the square root of both sides of the inequality :$$\|AB\|_{F}^{2}\leq\|A\|_{F}^{2}\|B\|_{F}^{2}$$ And hence, this proves that Frobenius norm is submultiplicative.