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Let $\rho(A)$ be the spectral radius of $A,$ that is the maximal eigenvalue of $A$ in absolute value. I want to show that for any $ \rho(A) < \eta < 1,$ there is $c>0$ such that $\|A^k\| \leq c \eta^k$ for $k = 0,1, \dots.$ Can somebody show me how to do this? Thanks!

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You'll have to be more specific about what you mean by $\|A^t\|$. Which norm? Spectral norm, Frobenius norm, ...? –  John Moeller Dec 17 '12 at 22:27
    
Any matrix norm! could this work with any matrix norm? aren't that norms are equivalent in finite dimensional space? –  Zizo Dec 17 '12 at 22:41
    
What is $\eta^t$ here? And do you mean "... want to show that there is $c>0$ so that for any $\eta>\rho(A)$ ... " ? That is, your quantifiers appear to be in the wrong order, and I don't see why $\eta<1$ might be important. –  user108903 Dec 17 '12 at 22:45
    
$\eta=100000000$ trivially satisfies the problem with $c=1$ otherwise. –  Inquest Dec 17 '12 at 22:45
    
@user108903 it does not matter, we could show that $ \|A^t\| < \rho(A)^t$ instead! Because this would implies that $\|A^t\| < \rho(A)^t.$ Since there is $c>0$ such that $ \rho(A)^t \leq c \eta^t$ for $t = 1,2, \dots$ then the inequality would follow. So the question would be, is it true that $\|A^t\| < \rho(A)^t$ for a matrix norm. I want the proof to work for any arbitrary matrix norm –  Zizo Dec 17 '12 at 22:54
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1 Answer 1

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Let $B=\eta^{-1}A$. Then $\rho(B)<1$. Hence $B^k\rightarrow0$ as $k\rightarrow\infty$. Therefore the entries of $B^k$ are bounded. Hence the Frobenius norm of $B^k$ is bounded and in turn, $\|B^k\|$ is also bounded because all matrix norms are equivalent. Therefore $\|B^k\|\le c$ for some $c>0$, i.e. $\|A^k\|\le c\eta^k$.

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That's what I am talking about.. Nice proof. –  Zizo Dec 18 '12 at 6:12
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