Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want an explication of the following fact:

If the spectral radius of a bounded operator $A$ on a Banach space is less than one, then $I - A$ is invertible.

share|improve this question

3 Answers 3

If $\rho(A)<1$, then the operator $$\sum_{n=0}^\infty A^n$$ is defined (why?) and is in fact $(I-A)^{-1}$.

share|improve this answer

This is trivial from the definition of spectrum. If the spectral radius is less than one, then in particular $1$ is not in the spectrum, which means $I-A$ is invertible.

Cameron Buie has answered a more interesting question.

share|improve this answer

You can find hint at Neumann series wikipedia article. Furthermore the question is maybe a duplicate, I think you can find the answer at math.se.

share|improve this answer

Your Answer

 
discard

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.