Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi Everyone,

Assume that we have a real symmetric matrix $H$, which can be written in the form $H=D \cdot B$, where $D$ is a positive diagonal matrix, and $B$ is a diagonally dominant matrix. All elements of all matrices are positive real numbers. We know that real symmetric matrices have real eigenvalues, and that diagonally dominant matrices have (potentially complex) eigenvalues with positive real parts. Could we infer from the above that $H$ is positive definite?

More generally, if $D$ is a diagonal matrix and $B$ is a positive definite matrix, could we infer that the product $D \cdot B$ is positive definite? My feeling is that this problem should have long been solved, I would really appreciate any pointers to books/research articles that talk about this problem. Thanks! :-)

share|improve this question
1  
Erm... Multiplying by a diagonal matrix on the left means multiplying rows, so $H$ is still diagonally dominant. Did you really mean $BD$? –  fedja Dec 22 '10 at 3:33
    
BD and DB are similar matrices, so they have the same eigenvalues. –  Michael Renardy Dec 22 '10 at 3:52
    
Presumably fedja's comment should really be interpreted as: BD is a diagonally dominant symmetric matrix, hence positive definite, hence the answer to the OP's question is: YES, while Michael's comment is: and so is BD. –  Igor Rivin Dec 22 '10 at 4:05
    
For the "more generally" part: if $\mathbf D$ is the negative of the identity matrix, $\mathbf D\mathbf B$ is negative definite; do you not have any restrictions on the entries of $\mathbf D$, just like in the first part? –  J. M. Dec 22 '10 at 5:09
add comment

3 Answers

up vote 2 down vote accepted

The answer is Yes. Write $B=D^{-1}H$. Thus $B$ is the product of two Hermitian matrices, ones of which ($D$) being positive definite. It is a classical fact (see my book on Matrices, 2nd edition, Prop. 6.1) that this product is diagonalisable with real eigenvalues of the same signs as those of $H$. The $B$ has real eigenvalues, which are positive because you already knew that their real parts are positive. This implies that the eigenvalues of $H$ are positive. Hence $H$ is positive definite.

share|improve this answer
add comment

I think D was supposed to have positive entries. If B is positive definite (meaning that the associated quadratic form is positive definite), then so is $D^{1/2}BD^{1/2}$. This matrix is similar to $DB$, hence it has the same eigenvalues. So if $DB$ is symmetric, it is positive definite.

I note, however, that a diagonally dominant matrix is not necessarily positive definite, although it has eigenvalues of positive real part.

share|improve this answer
    
I think you have to wake up pretty early in the morning to find interesting examples of H = D B with both H and B symmetric... –  Igor Rivin Dec 22 '10 at 3:59
1  
The problem as I understood it did not say B was symmetric, only that H was. –  Michael Renardy Dec 22 '10 at 4:02
add comment

This might be a trivial question, but at the end of the proof above, it is assumed that $H=D*B$ is positive definite as a product of a positive diagonal matrix and a matrix with positive real eigenvalues. Why is this always true?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.