Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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 $(1 + A + \cdots + A^{n-1}) \sum_{i=0}^\infty A^{ni}$, which surely does converge in norm.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.