Example of operators which are strictly submultiplicative $\|ST\| \overset{!}{<}\|S\|\|T\|$

Question

Let us consider normed spaces $E, F$ and $G$ over some field $\mathbb K$ and operators $T\in L(E,F)$ and $S\in L(F,G)$. I know that $\|ST\| \leqslant \|S\|\|T\|$, where $\|\cdot\|$ is the operator norm (see here).

What are examples of $S,T$ where the inequality is strict? $$\|ST\| \overset{!}{<}\|S\|\|T\|$$

I was thinking about setting where $E,F,G$ are $\mathbb R$ vector spaces and $S, T$ are matrices with real coefficients but I couldn't come up with an example.

Thank you.

Try finding some matrices whose product is the zero matrix, for example. — Misha Lavrov, Jan 09 '23 at 17:04

score 3 · Answer 1 · answered Jan 09 '23 at 17:32

3

An extreme case is $$S = T = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$$ where $\|S\| = \|T\| = 1$ but $\|ST\|=0$.

answered Jan 09 '23 at 17:32

GEdgar

111,679

1

An extreme lazy version $|S|=|T|\neq 0$ as $S=T\neq 0.$ :) – Ryszard Szwarc Jan 09 '23 at 17:35

score 1 · Accepted Answer · answered Jan 09 '23 at 17:05

Consider the following linear transformations of the plane $$ T: (x,y)\to (2x, y) $$ And $$ S: (x,y)\to (x,2y) $$ So $T$ is a horizontal dilation, $S$ is a vertical one. Clearly $\|T\|=2$, $\|S\|=2$ and, if you apply them sequentially, $\|ST\|=2$ (homothety).

Example of operators which are strictly submultiplicative $\|ST\| \overset{!}{<}\|S\|\|T\|$

2 Answers2