most recent 30 from http://mathoverflow.net 2013-05-23T20:16:32Z http://mathoverflow.net/feeds/question/50120 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/50120/eigenvalues-of-matrix-product Maria Kinget 2010-12-22T02:48:10Z 2010-12-22T20:26:44Z

<p>Hi Everyone,</p> <p>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?</p> <p>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! :-)</p>

http://mathoverflow.net/questions/50120/eigenvalues-of-matrix-product/50125#50125 Michael Renardy 2010-12-22T03:32:45Z 2010-12-22T03:32:45Z

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

http://mathoverflow.net/questions/50120/eigenvalues-of-matrix-product/50138#50138 Denis Serre 2010-12-22T08:14:49Z 2010-12-22T08:14:49Z

<p>The answer is <strong>Yes</strong>. 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.</p>

http://mathoverflow.net/questions/50120/eigenvalues-of-matrix-product/50184#50184 D H 2010-12-22T20:26:44Z 2010-12-22T20:26:44Z

<p>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?</p>