Different version of Cauchy-Schwarz Inequality?

Question

Show that $$\|AC\|_2 \leq \|A\|_2 \|C\|_2,$$

where $$A, C$$ are square matrices of the same size.

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I understand this probably needs to use Cauchy-Schwarz Inequality, but I don't know how should I start the problem, any help?

Answer 1

Observe that $$ \|AC\|=\sup_{\|x\|=1}\|ACx\|\le \sup_{\|x\|=1}\|A\|\|Cx\|=\|A\|\sup_{\|x\|=1}\|Cx\|=\|A\|\|C\|. $$ Here we only used that $$ \|ACx\|\le \|A\|\|Cx\| $$ which is an implication of the definition of $\|A\|$.

Note. This inequality HAS NOTHING TO DO with Cauchy-Schwarz, although it looks the same!

Answer 2

The answer will depend on your definition of $\|\cdot \|_2$.

In the case of the induced $2$-norm. The answer is given by Yiorgos S. Smyrlis.

In the case of the "entrywise" $2$-norm, we see that \begin{align} \|AC\|_2^2 =&\ \sum^n_{j=1}\sum^n_{i=1}|(AC)_{ij}|^2 =\sum^n_{j=1}\sum^n_{i=1}\left|\sum_{k=1}^n a_{ik}c_{kj} \right|^2\\ \leq&\ \sum^n_{j=1}\sum^n_{i=1} \left(\sum^n_{k=1}|a_{ik}|^2 \right)\left(\sum^n_{k=1}|c_{kj}|^2 \right)\\ =&\ \left(\sum^n_{i=1}\sum^n_{k=1}|a_{ik}|^2\right)\left(\sum^n_{j=1}\sum^n_{k=1}|c_{kj}|^2\right) = \|A\|_2^2\|C\|_2^2 \end{align} which we used Cauchy-Schwarz.