# Different version of Cauchy-Schwarz Inequality?

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

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

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

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 $$\|A\|$$.
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.