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- Newsgroups: sci.math
- Path: sparky!uunet!usc!wupost!eclnews!jezebel!mccarthy
- From: mccarthy@jezebel.wustl.edu (John McCarthy)
- Subject: Re: Help - non-integral power of a matrix?
- Message-ID: <1992Aug23.204524.15892@wuecl.wustl.edu>
- Sender: usenet@wuecl.wustl.edu (Usenet Administrator)
- Nntp-Posting-Host: jezebel
- Organization: Washington University, St. Louis Mo.
- References: <KOSOWSKY.92Aug14122403@schottky.harvard.edu> <1992Aug14.204404.23279@galois.mit.edu> <KOSOWSKY.92Aug17163140@minerva.harvard.edu>
- Date: Sun, 23 Aug 1992 20:45:24 GMT
- Lines: 33
-
- In article <KOSOWSKY.92Aug17163140@minerva.harvard.edu> kosowsky@minerva.harvard.edu (Jeffrey J. Kosowsky) writes:
- >
- >Just for clarification, the two mentions of finite-dimensionality above
- >can be extended to compact operators on any Banach space. In the
- >compact infinite dimensional case:
- > a] spectrum = closure of set of eigenvalues
- > = {eigenvalues} U {0}
- > so, it is still sufficient check the size of the eigenvalues
- > plus the limit point 0.
- > In the case of the logarithm, 0 does not lie in the radius of
- > convergence about 1. So, the Riesz functional calculus won't work
- > here. Actually, the Borel functional calculus won't work either
- > since the logarithm is not defined at 0.
- >
- > So, is there any way to define the logarithm of a (compact)
- > operator on an infinite dimensional space that avoids this
- > problem of 0 being in the spectrum?
- >
- > b]It is still true that normal iff diagonalizable
- >
- >
- >
- >Jeff Kosowsky
-
- No. If log(K) = A, where A is bounded, then
- spectrum(K) = spectrum(exp(A)) = exp(spectrum(A).
- So if 0 is in the spectrum of K (always true if K is compact on
- an infinite dimensional space), - infinity must be in the spectrum of A,
- contradicting its boundedness.
-
- John E. McCarthy
-
-
-