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I am trying to undestand a proof and there is one part that's holding me back. By assumption we have that spectral radius $\rho(A) < 1$. Hence, following inequality should hold

$$\|A^k\| < C \mu^k,$$

where $C>0$ and $\mu \in (0,1)$. I tried to understand it through Gelfand's formula saying $$\rho(A) = \lim_{k\to \infty} \|A^k\|^{\frac{1}{k}}$$

for any matrix norm $\|.\|$, but I still do not undestand why $C>0$ is uniform for any $k$.

Any help please?

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  • $\begingroup$ You seem to be convinced that we can find $0 < \mu < 1$ and $C$ which work for $k >K$, where $K$ is some large positive integer. If necessary, replace $C$ by ${\rm max}\{C, \frac{\|A^{k}\|}{\mu^{k}} : k \leq K \}.$ $\endgroup$ – Geoff Robinson Jun 27 '15 at 12:49
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Since $\rho(A)<1$, there is an induced norm $N(.)$ s.t. $N(A)<1$. Let $\mu=N(A)$; thus $N(A^k)\leq \mu^k$. Since the norms are equivalent, there is a fixed $C$ s.t. $||.||\leq CN(.)$. Finally $||A^k||\leq CN(A^k)\leq C\mu^k$.

EDIT. If $E$ is a Banach space and $A\in L(E)$ is bounded, then one has the same result. According to Gelfand, there is $\epsilon>0,k_0$ s.t. for $k\geq k_0$, $||A^k||^{1/k}\leq (1-\epsilon)$, that is $||A^k||\leq(1-\epsilon)^k$. Putting $D=\sup_{1\leq k\leq k_0}\dfrac{||A^k||}{(1-\epsilon)^k}$ and $C=\sup(D,1)$, we are done.

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  • $\begingroup$ Thank you. I see, that this works fine for finite dimensional spaces as we are using equivalence of norms. But what to do when I am in infinite dimensional space? $\endgroup$ – MathFan Sep 1 '15 at 19:21

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