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Mar 3, 2019 at 17:06 comment added Fedor Petrov I forgot to add powers of $r$: $\|x\|+r^{-1}\|Ax\|+\dots+r^{1-m}\|A^{m-1}x\|$.
Mar 3, 2019 at 12:19 comment added Jonas Adler Beautiful proof! Thank you.
Mar 3, 2019 at 12:19 vote accept Jonas Adler
Mar 3, 2019 at 11:57 comment added Fedor Petrov Jordan form is too heavy machinery for this question. And the real case is not covered, I am afraid. Instead, we may fix any $r>\rho$, choose $m$ for which $\|A^m\|<r^m$ and define new norm as $\|x\|+\|Ax\|+\dots+\|A^{m-1} x\|$.
Mar 3, 2019 at 11:38 history edited Federico Poloni CC BY-SA 4.0
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Mar 3, 2019 at 11:31 history answered Federico Poloni CC BY-SA 4.0