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

Let $T$ be bounded linear operator on some complex Banach space, and $\lambda$ an eigenvalue of $T$ which is isolated in its spectrum, and such that $\bigcup_{n\ge 1} N((T- \lambda I)^n)$ is one-dimensionnal.

How can we prove that $\lambda$ is a simple pole of the resolvent of $T$ ?

share|cite|improve this question
Dear Ahriman, Can you explain what you mean by $N((T-\lambda)^n)$? Thanks! Regards, – Matt E Nov 27 '11 at 13:17
This denotes the kernel of the operator $(T-\lambda I)^n$. – Ahriman Nov 27 '11 at 16:25

Have a look at Theorems 3 and 4 in Chapter VIII.8. of K. Yosida's "Functional Analysis" (Springer, 1971) which deal with precisely this issue.

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.