Operator Norm Inequality Used in The proof of Neumann Series Theorem

Question

Somewhere in the proof presented for Neumann Series Theorem, the following inequality is implicitly assumed. $$||A^k|| \leq ||A||^k.$$ The $||.||$ denotes the operator norm. How can I prove this inequality? Any hint where to begin?

To start, Prove the lemma $||AB||\le ||A|| ||B||$ – Nick Aug 26 '19 at 05:55 — Nick, Aug 26 '19 at 05:55

score 2 · Accepted Answer · answered Aug 26 '19 at 05:55

2

If $\|Ax|| \leq a\|x\|$ and $\|Bx|| \leq b\|x\|$ for all $x$ then $\|ABx\| \leq a\|Bx\| \leq ab\|x\|$. Hence $\|AB\| \leq \|A\|\|B\|$. Now use induction to show that $\|A^{k}\| \leq \|A\|^{k}$.

answered Aug 26 '19 at 05:55

Kavi Rama Murthy

311,013

score 1 · Answer 2 · answered Aug 26 '19 at 06:00

1

In the proof it is assumed that $\|\cdot\|$ is a matrix norm which is induced by a vector norm. Hence the matrix norm $\|\cdot\|$ is submultiplicative:

$$\|AB\| \le \|A\| \cdot \|B\|.$$

answered Aug 26 '19 at 06:00

Fred

77,394

Operator Norm Inequality Used in The proof of Neumann Series Theorem

2 Answers2