Square Root of Matrix ADate: 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/ |
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