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How to show $||A^k||$ $\leq$ $||A||^k$ where $A\in \mathbb R^{n\times n}$ and $k \in \Bbb Z^+$?

I want to start with $||A^k||=\max||A^kx||$ and $||A||^k=(\max||Ax||)^k$, and then multiply by the $\frac{1}{k}$-th power on both side, but I don't know what to do next.

Sebastiano
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sensationti
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1 Answers1

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Recall that the operator norm on a matrix $A$ is defined to be $$\lVert A\rVert := \text{sup}_{x}\frac{\lVert Ax\rVert_2}{\lVert x\rVert_2}.$$

For $k > 1$ and a nonzero vector $x$ we get $$\lVert A^k x\rVert_2 = \lVert AA^{k-1} x\rVert_2 \le \lVert A\rVert \lVert A^{k-1}x\rVert_2.$$

Can you take it from there?

Pirate Prentice
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  • Thanks, I think I got it! So, just repeating the above process and we will end up with $||A||$ $||A||$...$||A||$ on the right hand side, which is the same as $||A||^k$ – sensationti Sep 20 '21 at 03:58
  • You'll end up with $\lVert A^kx \rVert_2 \le \lVert A\rVert ^k\lVert x\rVert_2$, which means $$\frac{\lVert A^kx \rVert_2}{\lVert x\rVert_2} \le \lVert A\rVert^k.$$ Now take the supremum of the LHS. – Pirate Prentice Sep 20 '21 at 04:34