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I have corrected the statement of the spectral theorem. It read
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)
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). is the unit ball in X.
Article said:
This is wrong. A compact operator may have spectral radius 0, hence finite spectrum without being of finite rank. Consider for example the integration operator
on -- Bdmy ( talk) 22:31, 28 February 2009 (UTC)
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)
-- 195.113.30.252 ( talk) 15:05, 25 January 2017 (UTC)
Under the section Important properties, the third bullet point reads as follows:
" In particular, K(X) forms a two-sided ideal in B(X)."
But no explanation is given for the circle symbol .
Can someone who knows what this means please include an explanation? 2600:1700:E1C0:F340:7967:7502:8C03:5EC0 ( talk) 18:31, 30 September 2018 (UTC)
This article is rated C-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||
|
I have corrected the statement of the spectral theorem. It read
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)
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). is the unit ball in X.
Article said:
This is wrong. A compact operator may have spectral radius 0, hence finite spectrum without being of finite rank. Consider for example the integration operator
on -- Bdmy ( talk) 22:31, 28 February 2009 (UTC)
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)
-- 195.113.30.252 ( talk) 15:05, 25 January 2017 (UTC)
Under the section Important properties, the third bullet point reads as follows:
" In particular, K(X) forms a two-sided ideal in B(X)."
But no explanation is given for the circle symbol .
Can someone who knows what this means please include an explanation? 2600:1700:E1C0:F340:7967:7502:8C03:5EC0 ( talk) 18:31, 30 September 2018 (UTC)