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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.

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  • $\begingroup$ 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. $\endgroup$ Dec 30 '20 at 3:07
  • $\begingroup$ Well noted, thank you $\endgroup$
    – SPARSE
    Dec 30 '20 at 3:27
  • $\begingroup$ Alright let's do it! @vitamind $\endgroup$
    – SPARSE
    Jun 8 at 17:28
  • $\begingroup$ I fixed it and asked a more mathematical question with some quest for reference on this matter @vitamind thank you $\endgroup$
    – SPARSE
    Jun 8 at 18:01
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\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.

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