Let $\mathcal M_n (\mathbb R)$ be the set of all square matrices of order $n$. Show that $\|AB\| \le \|A\| \|B\|$

Question

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,

$$\begin{align} \|AB\|^2 &=\sum_{i=1}^{n} \sum_{j=1}^{n}\left(\sum_{k=1}^n a_{ik} b_{kj} \right)^2 &&\le \sum_{i=1}^{n} \sum_{j=1}^{n} \left [ \left(\sum_{k=1}^n a_{ik}^2 \right ) \left (\sum_{k=1}^n b_{kj}^2 \right) \right ] \\ &=\sum_{i=1}^{n} \left[ \left ( \sum_{k=1}^n a_{ik}^2 \right ) \cdot \left (\sum_{j=1}^{n} \sum_{k=1}^n b_{kj}^2 \right ) \right ] &&= \left (\sum_{j=1}^{n} \sum_{k=1}^n b_{kj}^2 \right ) \cdot \left( \sum\limits_{i=1}^{n} \sum_{k=1}^n a_{ik}^2 \right ) \\ &=\|B\|^2 \|A\|^2 &&=\|A\|^2 \|B\|^2 \end{align}$$

Answer 1

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 $\|AB\|_F^2 = \sum_k \|ABv_k\|_2^2 \le \sum_k \|A\|_2^2 \|Bv_k\|_2^2 \le \|A\|_F^2 \sum_k \|Bv_k\|_2^2 = \|A\|_F^2 \|B\|_F^2$.

Answer 2

The norm induced by your scalar product is known as the Frobenius norm. Indeed, your proof is flawless (see this question).