Matrix norm inequality proof - does this use Cauchy-Schwarz?

Question

The matrix norm for $A : \mathbb{R}^n \rightarrow \mathbb{R}^m$ (so $A$ is an $m \times n$ matrix) is given by $$\|A\| = \sup_{X \in \mathbb{R}^n \setminus \{0\}} \frac{|AX|}{|X|}$$ where $| \cdot |$ is the Euclidean norm.

I have been given the properties (without proof) that $|AX| \leq \|A\||X|$ and $\|AB\| \leq \|A\|\|B\|$, where $B$ is some $n \times p$ matrix.

Would a proof of this use the Cauchy-Schwarz ineqaulity?

Answer 1

The first inequality is immediate, since $$ \|A\| = \sup_{x \in \mathbb{R}^n \setminus \{0\}} \frac{|Ax|}{|x|}\ge \frac{|Ax|}{|x|}. $$ For the second note that for $x\ne0$ with $Bx\ne0$, we have $$ \|AB\| = \sup_{x\ne0} \frac{|ABx|}{|x|}=\sup_{Bx\ne0}\left( \frac{|ABx|}{|Bx|}\frac{|Bx|}{|x|}\right)\le\sup_{Bx\ne0}\frac{|ABx|}{|Bx|}\sup_{x\ne0}\frac{|Bx|}{|x|}\le\|A\|\cdot\|B\|. $$

Answer 2

The first inequality is clear by the definition of the matrix norm. The second is clear by the definition and

$\frac{|ABx|}{|x|}=\frac{|ABx|}{|Bx|}\frac{|Bx|}{|x|}$.