Can anyone give a reference or a proof to show the generalized eigenspace corresponding to a non zero eigenvalue of a compact operator defined on a Banach space is finite dimensional? I already saw the answer when the operator defined on a Hilbert space.


  • $\begingroup$ What do you mean when you write that a generalized eigenspace is finite? There are at least a couple of things that could mean. $\endgroup$ – DisintegratingByParts Jun 17 '17 at 15:44
  • $\begingroup$ I'm sorry. It should be finite dimensional. thanks. $\endgroup$ – Mathsira Jun 17 '17 at 15:47

I learned this fact in the lecutre course "Functional Analysis 1", held at ETH Zürich in Fall 2015. You can find lecture notes HERE. These might turn into a book once. You will find Theorem 5.28, part (i) on page 226 helpful. The proof uses Fredholm Theory.

  • $\begingroup$ Thank you. I'll read it. $\endgroup$ – Mathsira Jun 18 '17 at 15:50

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