most recent 30 from http://mathoverflow.net 2013-06-19T21:05:55Z http://mathoverflow.net/feeds/question/113828 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/113828/geometric-mean-of-positive-matrices Ziv Goldfeld 2012-11-19T13:28:22Z 2012-11-19T13:28:22Z

<p>Hello all,</p> <p>My question regards the geometric mean (GM) of two positive matrices. The definition of the GM for two positive matrices $(A,B)$ is given by: $M_0(A,B)=A^{\frac{1}{2}}(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{\frac{1}{2}}A^{\frac{1}{2}}$. Moreover, when $A$ and $B$ commute, this definition reduces to $M_0(A,B)=A^{\frac{1}{2}}B^{\frac{1}{2}}$. The GM is known to be jointly concave in the pair $(A,B)$. </p> <p>My question regards the reduced structure, namely $M_0(A,B)=A^{\frac{1}{2}}B^{\frac{1}{2}}$, for the general case where $A$ and $B$ <strong>do no necessarily commute</strong>. Is $A^{\frac{1}{2}}B^{\frac{1}{2}}$ jointly concave in the pair $(A,B)$ for any two positive matrices? If so, how can one proof this? In not? Which additional conditions on $(A,B)$ one should assume (without assuming commutativity) in order for it to be jointly concave in the pair?</p> <p>Thank you very much in advance!</p>