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!

• Are you familiar with Gelfand's formula relating operator norm to spectral radius? – Erick Wong Nov 22 '12 at 4:34
• I know the formula, but that's it. in particular, I would not know how to apply it here... – s_2 Nov 22 '12 at 4:39
• If $\|A^n\|^{1/n} \to c < 1$ then for some $n$ large enough, $\|A^n\| < 1$. – Erick Wong Nov 22 '12 at 4:53
• You said you know how to show convergence given the operator norm is $<1$...How does the proof go? – Erick Wong Nov 22 '12 at 4:58
• 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. – s_2 Nov 22 '12 at 5:02

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