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 Post subject: Positive definite Hermitian matrix as a productPosted: Sun, 30 Nov 2008 23:41:49 UTC
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Joined: Sun, 30 Nov 2008 23:30:36 UTC
Posts: 2
I would like if you people could help me out with this cause I don't have a clue and I spent the whole evening searching for it.

(in the following ' stands for the Hermitian transpose)
Matrix F is an n*n matrix which is computed from the row vectors (1*n sized) a and b as:

F=a'*a + b'*b

then I would like to compute (I guess there might be a straightforward expression) the row vector f (1*n sized) that fulfills:

F=f'*f

Analogously, if F is computed from two matrices (n*n) X and Y as:

F=X'*X + Y'*Y

I would like to compute the n*n matrix G that fulfills:

F=G'*G

Maybe there are several solution matrices f or G, but only one is enough for me.

Thank you very much

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 Post subject: Re: Positive definite Hermitian matrix as a productPosted: Mon, 1 Dec 2008 12:38:06 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Sat, 7 Jan 2006 18:29:24 UTC
Posts: 1401
Location: Leeds, UK
juanaco wrote:
(in the following ' stands for the Hermitian transpose)
Matrix F is an n*n matrix which is computed from the row vectors (1*n sized) a and b as:

F=a'*a + b'*b

then I would like to compute (I guess there might be a straightforward expression) the row vector f (1*n sized) that fulfills:

F=f'*f

In general no such vector f will exist. The reason is that any matrix of the form f'*f has rank 1. A sum of two such matrices will normally have rank 2, which means that a'*a + b'*b cannot be expressed in the form f'*f.

juanaco wrote:
Analogously, if F is computed from two matrices (n*n) X and Y as:

F=X'*X + Y'*Y

I would like to compute the n*n matrix G that fulfills:

F=G'*G

Maybe there are several solution matrices f or G, but only one is enough for me.

For this problem, a solution does exist, but there is no simple formula for G in terms of X and Y.

A matrix of the form X'*X is positive semi-definite. The sum of two such matrices is also positive semi-definite, and therefore possesses a (unique) positive semi-definite square root.

The way to calculate it is like this. The matrix F=X'*X + Y'*Y is positive semi-definite, so it can be diagonalised. That means that there is an orthogonal matrix P such that P*F*P' = D, a diagonal matrix with non-negative entries on the diagonal. Let S be the diagonal matrix whose diagonal entries are the square roots of those of D. Then S is the unique positive semi-definite matrix whose square is D. Let G = P'*S*P. Then G is positive semi-definite, and G'*G = P'*D*P = F.

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 Post subject: Posted: Mon, 1 Dec 2008 13:03:07 UTC
 S.O.S. Newbie

Joined: Sun, 30 Nov 2008 23:30:36 UTC
Posts: 2
thank you very much!

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