Let $\mathcal M_n (\mathbb R)$ be the set of all square matrices of order $n$. Show that $\|AB\| \le \|A\| \|B\|$ Good night, I'm trying to solve this exercise:

Let $\mathcal M_n (\mathbb R)$ be the set of all square matrices of order $n$. Show that $\langle A,B \rangle := \operatorname {tr} (A^T B)$ is an inner product on $\mathcal M_n (\mathbb R)$ such that $\|AB\| \le \|A\| \|B\|$.

Could you please verify if my proof of $\|AB\| \le \|A\| \|B\|$ looks fine or contains logical gaps/errors? Thank you so much for your help!

My attempt:
We have $$\|A\|^2 = \operatorname {tr} (A^T A) =\sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij}^2$$
As such,
$$\begin{align} 
\|AB\|^2 &=\sum_{i=1}^{n} \sum_{j=1}^{n}\left(\sum_{k=1}^n a_{ik} b_{kj} \right)^2
&&\le \sum_{i=1}^{n} \sum_{j=1}^{n} \left [ \left(\sum_{k=1}^n a_{ik}^2 \right ) \left (\sum_{k=1}^n b_{kj}^2 \right) \right ]
\\
&=\sum_{i=1}^{n} \left[ \left ( \sum_{k=1}^n a_{ik}^2 \right ) \cdot \left (\sum_{j=1}^{n}  \sum_{k=1}^n b_{kj}^2 \right ) \right ] 
&&= \left (\sum_{j=1}^{n}  \sum_{k=1}^n b_{kj}^2 \right ) \cdot \left( \sum\limits_{i=1}^{n} \sum_{k=1}^n a_{ik}^2 \right )
\\
&=\|B\|^2 \|A\|^2 &&=\|A\|^2 \|B\|^2
\end{align}$$
 A: Showing that $\langle \cdot, \cdot \rangle$ is an inner product is straightforward, the
only marginally tricky part is to observe that all the eigenvalues (and hence the trace)
of a positive semi definite matrix are non negative.
Your proof that $\|\cdot\|_F$ is sub multiplicative is correct. Here is a more complex alternative
approach that illustrates some other useful facts about the Frobenius norm and avoids all that nasty summation:
Note that if $V$ is an orthogonal matrix, $\|A\|_F^2 = \operatorname{tr} (A^T A) = \operatorname{tr} (A^T AV V^T) = \operatorname{tr} (V^TA^T AV) = \|AV\|_F^2$.
In particular, if $v_k$ form an orthonormal basis, then  $\|A\|_F^2 = \sum_k \|Av_k\|_2^2 $. It is straightforward to show from this that $\|A\|_2 \le \|A\|_F$.
Then we have 
$\|AB\|_F^2 = \sum_k \|ABv_k\|_2^2 \le \sum_k \|A\|_2^2 \|Bv_k\|_2^2 \le \|A\|_F^2 \sum_k \|Bv_k\|_2^2 = \|A\|_F^2 \|B\|_F^2$.
A: The norm induced by your scalar product is known as the Frobenius norm. Indeed, your proof is flawless (see this question).
