Eigenvalues of product of two symmetric matrices - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T17:21:17Z http://mathoverflow.net/feeds/question/106191 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106191/eigenvalues-of-product-of-two-symmetric-matrices Eigenvalues of product of two symmetric matrices Kjetil B Halvorsen 2012-09-02T17:25:49Z 2012-09-02T21:25:39Z <p>This is mostly a reference request, as this must be well known!</p> <p>Let $A$ and $B$ be two real symmetric matrices, one of which is positive definite. Then it is easy to see that the product $AB$ (or $BA$, which has the same eigenvalues) is similar to a symmetric matrix, so has real eigenvalues. Now take the vectors of eigenvalues of $A$ and of $B$, sorted in decreasing order, and let their componentwise product be $ab$. Then the question is: what is known about the relationship (for instance, inequalities) between $ab$ and the vector of eigenvalues of the product $AB$ (also taken in decreasing order). Some experimentation gives the conjecture that there is a majorization order between them, for instance. This must be well-known!</p> http://mathoverflow.net/questions/106191/eigenvalues-of-product-of-two-symmetric-matrices/106199#106199 Answer by S. Sra for Eigenvalues of product of two symmetric matrices S. Sra 2012-09-02T21:25:39Z 2012-09-02T21:25:39Z <p>Here are the results that you are probably looking for.</p> <p>The first one is for positive definite matrices only.</p> <blockquote> <p><strong>Theorem</strong> (Prob.III.6.14; Matrix Analysis, Bhatia 1997). Let $A$ and $B$ be Hermitian positive definite matrices. Let $\lambda^\downarrow(X)$ denote the vector of eigenvalues of $X$ in decreasing order; define $\lambda^\uparrow(X)$ likewise. Then, \begin{equation*} \lambda^\downarrow(A) \cdot \lambda^\uparrow(B) \prec \lambda(AB) \prec \lambda^\downarrow(A) \cdot \lambda^\downarrow(B) \end{equation*}</p> </blockquote> <p>However, when dealing with matrix products, it is more natural to consider singular values rather than eigenvalues. </p> <blockquote> <p>Therefore, the relation that you might be looking for is the <em>log-majorization</em> \begin{equation*} \log \sigma^\downarrow(A) + \log\sigma^\uparrow(B) \prec \log\sigma(AB) \prec \log\sigma^\downarrow(A) + \log\sigma^\downarrow(B), \end{equation*} where $A$ and $B$ are arbitrary matrices, and $\sigma(\cdot)$ denotes the singular value map.</p> </blockquote> <p><strong>Reference</strong></p> <ol> <li>R. Bhatia. <em>Matrix Analysis</em>. Springer, GTM 169. 1997.</li> </ol>