Talk:Compact operator

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I have corrected the statement of the spectral theorem. It read

The spectral theory for compact operators in the abstract was worked out by Frigyes Riesz. It shows that a compact operator K has a discrete spectrum, with finite multiplicities (so that K − λI has a finite-dimensional kernel for all complex λ).

Which is close, but case where the spectrum has 0 as a limit point is not a discrete subset of C. Moreover, 0 need not be an eigenvector even though it is always in the spectrum (e.g. Volterra operator) and if 0 is an eigenvector it may have infinite multiplicity (e.g. 0 operator)

It might be worth expanding on volterra operator, either here or in a new page, but I don't have time now. --AndrewKepert 07:57, 7 Apr 2004 (UTC)

OK, fine. 'Discrete spectrum' as opposed to 'continuous spectrum' is sort of lax terminology, I guess.

Charles Matthews 08:23, 7 Apr 2004 (UTC)

[edit] Some properties of compact operators

In the following, X,Y,Z,W are Banach spaces, B(X,Y) is space of bounded operators from X to Y. K(X,Y) is space of compact operators from X to Y. B(X)=B(X,X), K(X)=K(X,X). B_X is the unit ball in X.

• A bounded operator is compact if and only if any of the following is true
  • there exists a neighbourhood of 0, , and compact set such that .
  • T(B_X) is relatively compact
  • Image of any bounded set under T is relatively compact
  • Image of any bounded set under T is totally bounded in Y.
  • For any sequence from the unit ball, B_X, the sequence contains a Cauchy subsequence.
• K(X,Y) is closed subspace of B(X,Y)
• This is a generalization of the statement that K(X) forms a two-sided operator ideal in B(X)
• id_X is compact if and only if X has finite dimension
• For any , is closed.

this property immediately above somehow didn't get added. i will do so shortly. Mct mht 08:45, 3 April 2007 (UTC)

It was not missed, it follows trivially from the properties of Fredholm operators. ((Igny 13:34, 3 April 2007 (UTC))—the preceding unsigned comment is by Igny (talk • contribs)

ok, Fredholm operators have closed range and Fredholm-ness is preserved by homotopy, in the set of Fredholm operators. seems to me that the latter fact is not entirely trivial. just wanted to note that the property can also be shown directly. Mct mht 03:51, 5 April 2007 (UTC)

You are welcome to add this to the article, as it fits nicely overall in the article. Oleg Alexandrov (talk) 20:21, 7 December 2005 (UTC)

Done. (Igny 21:44, 7 December 2005 (UTC))

Thanks! Oleg Alexandrov (talk) 01:08, 8 December 2005 (UTC)