Geometric proof for Composition bound property of operator norms?

Question

This is just a curiosity. For linear transformations $A$ and $B$, $||AB|| \le ||A|| \cdot ||B||$ where$||\cdot||$ denotes the operator norm (Of course provided $AB$ exists.) This fact has a proof, but I was wondering if there was any proof for it for vectors in $\mathbb{R^2}$ or even $\mathbb{R^3}$. I know this wouldn't come close to proving this general result, but I was essentially looking for some geometric intuition if there was one.

Answer 1

For vectors on $\mathbb{R}^2$ for example, if we take $A=\begin{pmatrix} a\\b\end{pmatrix}$ and $B=\begin{pmatrix} c\\d\end{pmatrix}$, with the usual scalar product we have $AB=ac+bd$, and then the inequality is just the classical Cauchy-Schwarz inequality $$|ac+bd|\leq \sqrt{a^2 +b^2} \sqrt{c^2 +d^2}.$$ Of course it's an obvious inequality with the definition of scalar product.

Answer 2

Informally, I would describe the situation as follows: $\| A\|$ is the largest possible factor by which the length of a vector $x$ can grow under the action of $A$ (this is just: $\|A \, x\| \le \|A\| \, \|x\|$.)

Since $AB$ is the concatenation of $A$ and $B$, the inequality is now obvious.

However, I am not sure whether this counts as 'geometric'.