# 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)

## 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). ${\displaystyle B_{X}}$ is the unit ball in X.

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)
A question: The article says "A bounded operator ${\displaystyle T\in B(X,Y)}$ is compact if and only if any of the following is true". This seems to suggest boundedness is necessary for the following conditions to be equivalent to compactness. Is this true? —Preceding unsigned comment added by 130.207.197.164 (talk) 17:40, 25 October 2010 (UTC)

## Finite spectrum

Article said:

It shows that a compact operator K on an infinite-dimensional Banach space has spectrum that is either a finite subset of C which includes 0 (in that case, the operator has finite rank)

This is wrong. A compact operator may have spectral radius 0, hence finite spectrum ${\displaystyle \{0\}}$ without being of finite rank. Consider for example the integration operator

${\displaystyle f\rightarrow \{x\in [0,1]\rightarrow \int _{0}^{x}f(t)\,dt\}}$

on ${\displaystyle L^{2}([0,1]).}$ --Bdmy (talk) 22:31, 28 February 2009 (UTC)

This is the Volterra operator, mentioned previously on this talk page. linas (talk) 03:46, 15 November 2010 (UTC)

## Compact operator on Hilbert spaces

There is something wrong here. As it stands it would imply the identity is compact. Surely it needs to say the singular values tend to zero. I will look for a good reference then fix it.Billlion (talk) 08:06, 2 May 2013 (UTC)

No, the text is OK, even if not formulated in the clearest way. It's OK because you need infinitely many orthonormal vectors to represent the identity in an infinite dimensional Hilbert space, and then the ${\displaystyle \lambda _{n}}$ will accumulate at a non zero value, which has been excluded in the text. Bdmy (talk) 08:19, 2 May 2013 (UTC)