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!
Inequality from a matrix norm
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linear-algebra
real-analysis
matrices
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0You'll have to be more specific about what you mean by $\|A^t\|$. Which norm? Spectral norm, Frobenius norm, ...? – 2012-12-17
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0Any matrix norm! could this work with any matrix norm? aren't that norms are equivalent in finite dimensional space? – 2012-12-17
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0What 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. – 2012-12-17
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0$\eta=100000000$ trivially satisfies the problem with $c=1$ otherwise. – 2012-12-17
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0@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 – 2012-12-17
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0I got a result that says that if $ \rho(A) < 1,$ there is a matrix norm $ \| \cdot\|$ such that $ \|A\| < 1.$ I don't know how would this help me? – 2012-12-17
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0I still don't understand what you mean when you write a superscript $t$ above the real numbers $\eta$ and $\rho(A)$. Is $\eta^t=\eta$? – 2012-12-17
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0$ \eta$ is a real number and $t$ is non-negative integer. – 2012-12-17
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0Not fit for an answer since I am not sure: If $A$ is diagonalizable, then diagonalize it to $P^{-1}DP$. $$$$ \begin{align} A^t&\equiv P^{-1}D^tP\\ ||A^t||&=||P^{-1}||\cdot||D^t||\cdot||P||\\ \text{If } ||A^t||&>c\eta^t\quad \forall c,t\\ \implies ||P^{-1}||\cdot||D^t||\cdot||P||&>c\eta^t\quad \forall c,t\\ \implies ||D^t||&>c'\eta^t\quad \forall c',t\\ \end{align} But I predict there is a contradiction here. – 2012-12-17
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0Oh. Silly me, it looked like a transpose :-) – 2012-12-17
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0@Inquest You're right! the contradiction could come from $c'$ which is arbitrary . Because, since $ \rho(A) < 1$ then, $\|D^t\| \rightarrow 0$ as $t \rightarrow \infty,$ hence $c' \eta^t$ must tend $0$ as $ t \rightarrow \infty$ but if we choose $c' = \frac{1}{ \eta^t}$ then $ c^ \eta^t$ tends to $1$ instead which is a contradiction. We could may be think of more regiorous way to say this or a shorter statement – 2012-12-17
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0You cannot let $c'=\dfrac{1}{\eta^t}$ since $c'$ is a constant. – 2012-12-17
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0@Inquest I agree, thinking of a better thing to say.. – 2012-12-17
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0It looks like this is closely related to Gelfand's formula http://en.wikipedia.org/wiki/Spectral_radius – 2012-12-17
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0@user108903 yes, its in some sense! Gelfand's formula said for any $ \epsilon> 0,$ there is an integer $T$ such that for all $t>T,$ we have $\|A^t\| < (\rho(A) + \epsilon)^t.$ – 2012-12-17
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0@Inquest Just one thing!! The same thing could still be done for general matrix $A$ ( weather its digonlizable or not) from Jordan forms. – 2012-12-17
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0@Inquest L'Hôpital's rule would do the trick. By the way, how could you've $\|A^t\| = \|P\|\|D\|\|P^{-1}\|.$ Isn't $\|A^t\| \leq \|P\|\|D\|\|P^{-1}\|.$ If its the case I don't think the next will follow. – 2012-12-18
1 Answers
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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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0That's what I am talking about.. Nice proof. – 2012-12-18