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- From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
- Newsgroups: sci.math
- Subject: Re: Help - non-integral power of a matrix?
- Keywords: matrix
- Message-ID: <a_rubin.713653963@dn66>
- Date: 12 Aug 92 21:12:43 GMT
- References: <Aug.10.15.45.34.1992.26563@clam.rutgers.edu> <carle.713599691@vex>
- Lines: 32
- Nntp-Posting-Host: dn66.dse.beckman.com
-
- In <carle.713599691@vex> carle@vex.ugcs.caltech.edu (Matthew Thomas Carle) writes:
-
- >gonzalez@clam.rutgers.edu (Ralph Gonzalez) writes:
-
-
- >>Hi. Does anyone know of an algorithm to find a non-integral
- >>power of a matrix, e.g. A^.5 or A^1.3? Thus, A^2 is the same
- >>as AxA and A^0 is the identity.
-
- >>I imagine if such a thing is defined, then there are conditions
- >>on A...
-
- >(I've never actually seen this applied, but I don't know why it wouldn't work)
- >Well, let's look at how we implement a^b, with real a and b
- >c=a^b -> log c = b log a -> c=exp(b*log(a))
- >For real b, but a matrix A, C=A^b=exp(b*log(A))
- >log(I+A) = A + (A^2)/2 + (A^3)/3 + ... +(A^n)/n + ...
- >b*log(A) = b * [(A-I) + ((A-I)^2)/2 + ((A-I)^3)/3 + ... + ((A-I)^n)/n + ...]
- >where the powers of A are defined as usual; then stick that into the
- >power series expansion of the exponential of a matrix . . .
- >C = I + [b*log(A)] + ([b*log(A)]^2)/2 + ... + ([b*log(A)]^n)/(n!) + ...
- >I dunno know, looks kinda freaky to me.
-
- >I would guess exp(A) converges for all A, but I don't know about the log.
-
- log "obviously" converges if all eigenvalues of A are strickly within 1 of 1,
- just like the complex case.
- --
- Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
- 216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal)
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