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- Path: sparky!uunet!pipex!warwick!doc.ic.ac.uk!uknet!axion!fmg!jcrw
- From: jcrw@fmg.bt.co.uk (Jeremy Wilson)
- Newsgroups: sci.math
- Subject: Re: Square root of a matrix
- Message-ID: <1992Dec11.095732.9802@fmg.bt.co.uk>
- Date: 11 Dec 92 09:57:32 GMT
- References: <1992Dec8.190620.39091@ns1.cc.lehigh.edu>
- Organization: British Telecom
- Lines: 83
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-
- Frederick W. Chapman (fc03@ns1.cc.lehigh.edu) wrote:
- : In article <1992Dec8.053151.20414@news.eng.convex.com>,
- : dodson@convex.COM (Dave Dodson) writes: >In article
- : <1992Dec7.204135.18257@husc15.harvard.edu> blom@husc15.harvard.edu
- : writes:
- : >>How can one take the square root of a matrix? I have done so for a
- : >>few matrices but I am having trouble generalizing my techniques. It
- : >>seems that there are 4 solutions, in general, which are not
- : >>necessarily unique. These can be paired such that there are two
- : >>solutions with positive determinants and the other two solutions are
- : >>the previous ones times -1. This is much like typical square roots.
- : >
- : >If a matrix is diagonalizable, its square roots can be constructed as follows:
- : >
- : > -1
- : >Let A = X D X, where D is a diagonal matrix. Let E be a matrix such that
- : > 2 -1
- : >E = D. Let B = X E X.
- : >
- : > 2 -1 2 -1 -1 -1
- : >Then B = (X E X) = X E X X E x = X E E X = X D X = A, so B is a square
- : >root of A.
- : >
- : > 2
- : >In general, there will be many E such that E = D.
- :
- : What if A is *not* diagonalizable? For example, the matrix
- :
- : [ r 1 ]
- : A := [ ]
- : [ 0 r ]
- :
- : for any real r is already in Jordan canonical form, and is not similar
- : to a diagonal matrix. How do we compute the square root of the matrix
- : in this case?
- :
- : (One possibility is to try to use the Maclaurin series expansion for
- : sqrt(1+x) and evaluate the result at A - I, if the resulting series of
- : matrices converges; taking into account that A is of the form r I + N,
- : where N is nilpotent of order 2, there should be considerable
- : simplification.)
- :
- : One thing is clear in general: If B^2 = A, then the eigenvalues of A are
- : the squares of the eigenvalues of B, and algebraic multiplicities are
- : preserved. Using this information, we can take a wild guess for my
- : example: considering a matrix of the form B = sqrt(r) I + c N, and solving
- : B^2 = A for the real constant c yields
- :
- :
- : [ 1/2 1 ]
- : [ r ------ ]
- : [ 1/2 ]
- : B := [ 2 r ]
- : [ ]
- : [ 1/2 ]
- : [ 0 r ]
- :
- :
- : where the same value of sqrt(r) is taken for each matrix element. I
- : suspect that a similar approach could be derived for the Jordan canonical
- : form of any matrix.
- :
-
- In fact every matrix with all its eigenvalues equal and non-zero can be written as a product of a scalar matrix and an upper unitriangular matrix.
- It is easy to show by induction on the matrix size that the
- unitriangular matrix has only one square root which is unitriangular
- so we now seem to have enough using Jordan normal form to determine all
- square roots, except those with zero eigenvalues. A matrix with all its
- eigenvalues zero, however, cannot have a square root unless it is
- the zero matrix, as far as I can see.
-
- Any thoughts about what happens for fields other than the complaex numbers?
- In particular what happens for fields with characteristic p?
- +---------------------------------+----------------------------------------+
- ! Jeremy Wilson ! email: jcrw@fmg.bt.co.uk !
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