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- Path: sparky!uunet!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.92Aug17165135@minerva.harvard.edu>
- Date: 17 Aug 92 21:51:35 GMT
- Article-I.D.: minerva.KOSOWSKY.92Aug17165135
- References: <Aug.10.15.45.34.1992.26563@clam.rutgers.edu>
- <56923@mentor.cc.purdue.edu>
- Sender: usenet@das.harvard.edu (Network News)
- Organization: Harvard Robotics Lab, Harvard University
- Lines: 42
- In-Reply-To: hrubin@pop.stat.purdue.edu's message of 15 Aug 92 15:30:45 GMT
-
- In article <56923@mentor.cc.purdue.edu> hrubin@pop.stat.purdue.edu (Herman Rubin) writes:
-
- > In article <Aug.10.15.45.34.1992.26563@clam.rutgers.edu> 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.
- >
- > There have been N postings in reply to this, which give partial
- > results, but none which I would consider sufficiently complete.
- >
- > There are conditions, and it even depends on exactly what is
- > wanted. The simplest way to look at this is by considering a
- > reduction to one of the canonical forms. A singular component
- > can be considered as having 0 down the diagonal and possibly
- > something above the diagonal; if this block is nxn, all powers
- > greater than n can be considered 0, but below n, it has to be
- > looked at carefully.
- >
- > If one takes a block with a positive constant diagonal element h,
- > the Taylor series expansion around hI gives h^k*I plus a FINITE
- > series in (A-hI).
- >
- > If a characteristic root is negative or complex, the same can be
- > done, but if a real solution is wanted, this may or may not be
- > possible. For example, the matrix
- >
- > cos x -sin x
- > sin x cos x
- >
- > is essentially exp(ix). So powers of orthogonal matrices with
- > determinant 1 can be defined non-uniquely as real matrices.
- > However, if the determinant is negative, there clearly are problems.
- > --
- > Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
- > Phone: (317)494-6054
- > hrubin@pop.stat.purdue.edu (Internet, bitnet)
- > {purdue,pur-ee}!pop.stat!hrubin(UUCP)
- >
-
- Is this result really the definitive "sufficiently complete" answer
- that we have all been waiting for?
-