Re: Non-square square root of a matrix
- From: Helmut Jarausch <jarausch@xxxxxxxxx>
- Date: Tue, 31 Jul 2007 22:38:16 +0200
g.flemin@xxxxxxxxx wrote:
Hello,
I am trying to find S, a non-square square root of a matrix G, s.t. G
= S^T S
G is a n x n symmetric matrix; rank(G) = p with p < n.
S should be a p x n matrix.
It seems this problem has a solution but I cannot find it :-(
Btw, G should be positive definite but, due to some previous
computations, chol(G) fails..
Any ideas or references?
Take an svd or eigendecomposition like
G= X S X^T
where S = diag(s_1,s_2,....)
Assume |s_1| >= |s_2| >= ... >= |s_p| >= ... >= |s_n|
(that's the case for the SVD and can be arranged for the
eigendecomposition by permuting columns of X)
Since rank(G) = p, we have s_{p+1}=...=s_n=0
Therefore X(:,1:p)*diag(s_1,..,s_p)*X(:,1:p)^T
comes close to what you want, unless some of
s_1,..,s_p are negative. Then I doubt there is a real
solution of your problem. For positive definite G
this gives a solution.
--
Helmut Jarausch
Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany
.
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- Non-square square root of a matrix
- From: g . flemin
- Non-square square root of a matrix
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