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If I have a matrix M which can be decomposed as:

$M = DH$

where $D$ is a diagonal matrix and $H$ is another matrix with known positive semi-definite square root $H^{1/2}$, does this give a formula or method for finding $M^{1/2}$?

Some additional details for the specific problem that I'm looking at, both D and H are positive semi-definite, Hermitian and have trace 1. In general they are non-commuting.

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up vote 2 down vote accepted

By "the unique non-negative square root", I presume that you mean a positive semidefinite square root. But if this is the case, your question is not well posed: if $M^{1/2}$ is positive semidefinite, so is $M$. Hence $M$ is Hermitian. Yet if $M=DH$ where $M$ is Hermitian, $D$ is diagonal and $H$ is Hermitian, then $D$ and $H$ must commute.

Edit: At any rate, I don't think there is any easy formula for calculating a square root of $DH$ in general. When $D$ is nonsingular, we can calculate $(DH)^{1/2}$ as $$ (DH)^{1/2} = D^{1/2} (D^{1/2} H D^{1/2})^{1/2} D^{-1/2},, $$ where $(D^{1/2} H D^{1/2})^{1/2}$ is the unique positive semidefinite square root of $D^{1/2} H D^{1/2}$. This $(DH)^{1/2}$ is the unique square root of $DH$ with nonnegative eigenvalues. But to calculate it, you need to find a square root of $D^{1/2} H D^{1/2}$, and to do so, knowing $H^{1/2}$ is not really useful.

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Sorry, you're right, the square root of M is not necessarily positive semidefinite (although H is). Question updated to reflect this. –  Ben Aaronson Jan 28 '13 at 11:37
    
What about the (potentially) easier task of finding the trace of the square root of DH? –  Ben Aaronson Jan 28 '13 at 12:12
    
@Stereotomy Square roots are not unique. Their traces can vary wildly. For example, when $M=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$, it has $2^n$ square roots $\operatorname{diag}(\pm\sqrt{\lambda_1},\ldots,\pm\sqrt{\lambda_n})$ and they have potentially $2^n$ different traces. Even if you fix the choice of the square root, since the eigenvalues of $M$ in general are not functions of the eigenvalues of $D$ and $H$, there simply is no hope for inferring the trace of $M^{1/2}$ from the traces of $D,D^{1/2},H$ or $H^{1/2}$. –  user1551 Jan 28 '13 at 12:40
    
Should have specified, in this case I would be fixing the choice of square root as positive. And I'm aware that the trace of one cannot be inferred from the trace of the others, I was just hoping that the trace might be possible to infer (easily) from the entire matrices. –  Ben Aaronson Jan 28 '13 at 14:16
    
@Stereotomy Ur... now we are back to square one. When we say that $M$ has a "positive/nonnegative" square root, the usual meaning is that $M$ has a positive/nonnegative definite square root. But in you case this means $D$ and $H$ commute. So, what do you mean by "fixing the choice of square root as positive"? If you mean a square root with all its eigenvalues positive/nonnegative, then the one I gave in my answer is the unique one when $D$ is nonsingular. –  user1551 Jan 28 '13 at 22:51
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