From: jcrw@fmg.bt.co.uk (Jeremy Wilson)
Newsgroups: sci.math
Subject: Re: Square root of a matrix
Date: 11 Dec 92 09:57:32 GMT
Organization: British Telecom
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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?
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