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- Path: sparky!uunet!wupost!uwm.edu!ogicse!das-news.harvard.edu!das-news!kosowsky
- From: kosowsky@minerva.harvard.edu (Jeffrey J. Kosowsky)
- Newsgroups: sci.math
- Subject: Re: Help - non-integral power of a matrix?
- Message-ID: <KOSOWSKY.92Aug17163140@minerva.harvard.edu>
- Date: 17 Aug 92 21:31:40 GMT
- Article-I.D.: minerva.KOSOWSKY.92Aug17163140
- References: <a_rubin.713653963@dn66> <1992Aug12.231708.3644@galois.mit.edu>
- <KOSOWSKY.92Aug14122403@schottky.harvard.edu>
- <1992Aug14.204404.23279@galois.mit.edu>
- Sender: usenet@das.harvard.edu (Network News)
- Organization: Harvard Robotics Lab, Harvard University
- Lines: 36
- In-Reply-To: jbaez@zermelo.mit.edu's message of Fri, 14 Aug 92 20:44:04 GMT
-
- In article <1992Aug14.204404.23279@galois.mit.edu> jbaez@zermelo.mit.edu (John C. Baez) writes:
-
- > I'm particularly worried about the non-normal case, of course. You are
- > right about the holomorphic functional calculus so we are free to say
- > that my series
- >
- > ln A = (A - 1) - (A - 1)^2/2 + ....
- >
- > converges when the spectral radius of A - 1 is < 1. If our space is
- > finite-dimensional this just means that all eigenvalues of A have
- > |lambda - 1| < 1. If A is normal the spectral radius equals the norm
- > so we may equivalently demand ||A - 1|| less than 1. (In the
- > finite-dim case "normal" just means "diagonalizable".)
- >
-
- 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
-