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?
Best Answer
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 $$ \|AB\| = \sup_{x\ne0} \frac{|ABx|}{|x|}=\sup_{Bx\ne0}\left( \frac{|ABx|}{|Bx|}\frac{|Bx|}{|x|}\right)\le\sup_{Bx\ne0}\frac{|ABx|}{|Bx|}\sup_{x\ne0}\frac{|Bx|}{|x|}\le\|A\|\cdot\|B\|. $$