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Square Root of Matrix A

Date: 06/29/2003 at 14:08:09
From: Alfredo Alvarez
Subject: How can we find the square root of matrix A... 

How can we find the square root of matrix A if 
      
       2   -1
A =         
       -1   2  

Thank you

I understand that A = M x M
and
    a  b
M =     
    c  d
           [ a  b ]   [ a  b ]   [ a^2+bc  ab+bd ]
==>  M^2 = [      ] x [      ] = [               ]
           [ c  d ]   [ c  d ]   [ ca+dc   cb+d^2]

then a^2+bc = 2, ab+bd = -1, ca+dc = 2 and cb+d^2 = -1


Date: 06/29/2003 at 14:50:42
From: Doctor Rob
Subject: Re: How can we find the square root of matrix A...

Thanks for writing to Ask Dr. Math, Alfredo.

The usual way of finding the square root of a real symmetric matrix
involves diagonalizing it. Find an invertible matrix U such that

                [d1  0]
   U*M*U^(-1) = [     ] = D.
                [0  d2]

d1 and d2 are the eigenvalues of M. U is not unique, but any one that 
works in the above equation will do. In the case at hand, it turns out 
that d1 = 1 and d2 = 3, or vice versa. Then there are four square 
roots of D:

   [sqrt(d1}     0   ]   [sqrt(d1}     0   ]
   [                 ],  [                 ],
   [   0     sqrt(d2)]   [   0    -sqrt(d2)]

   [-sqrt(d1}    0   ]   [-sqrt(d1}    0   ]
   [                 ],  [                 ].
   [   0     sqrt(d2)]   [   0    -sqrt(d2)]

(Be sure you understand why each of the four, when squared, gives D as 
the answer.) Let R be any one of them. Then

   U*M*U^(-1) = D = R^2,
   M = U^(-1)*R^2*U = [U^(-1)*R*U]*[U^(-1)*R*U],

so U^(-1)*R*U is one of the four square roots of M.

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Linear Algebra

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