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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 $T$ ?

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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

1 Answer 1

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.

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