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I have a question about the convergence of the Neumann series:

Let $A$ be a matrix with spectral radius $\rho(A)<1$, i.e., all eigenvalues of $A$ are strictly less than $1$. Does that imply that the series \begin{equation} \sum_{i=0}^{\infty}A^i \end{equation} converges (in the operator norm)? I know how to prove it if the operator norm of $A$ is strictly less than $1$, but I don't know how to prove it if I only know that the spectral radius is less than $1$.

Many thanks for any help!

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  • $\begingroup$ Are you familiar with Gelfand's formula relating operator norm to spectral radius? $\endgroup$ – Erick Wong Nov 22 '12 at 4:34
  • $\begingroup$ I know the formula, but that's it. in particular, I would not know how to apply it here... $\endgroup$ – s_2 Nov 22 '12 at 4:39
  • $\begingroup$ If $\|A^n\|^{1/n} \to c < 1$ then for some $n$ large enough, $\|A^n\| < 1$. $\endgroup$ – Erick Wong Nov 22 '12 at 4:53
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    $\begingroup$ You said you know how to show convergence given the operator norm is $<1$...How does the proof go? $\endgroup$ – Erick Wong Nov 22 '12 at 4:58
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    $\begingroup$ ok, there is the answer - I was just not able to see it! sorry for taking your time. and many thanks Erick and Jonas!!! I really appreciate your help. $\endgroup$ – s_2 Nov 22 '12 at 5:02
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Gelfand's formula shows that if $\rho(A) < 1$, then for some $n$, $\|A^n\| < 1$. One can then rewrite the series as $(1 + A + \cdots + A^{n-1}) +\sum_{i=0}^\infty A^{n+i}$, which surely does converge in norm.

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  • $\begingroup$ Also note that $\rho(A^{n+i})=\bigl(\rho(A^{n})\bigr)^{i}$. $\endgroup$ – bkarpuz Jul 20 '18 at 5:23

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