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Hello all,

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 $(A,B)$.

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 $B$ do no necessarily commute. 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?

Thank you very much in advance!

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What do you mean by "jointly concave" for a map that does not take value in the self-adjoint matrices? – Mikael de la Salle Nov 19 '12 at 13:32
Sorry for the ignorance but why is it necessary that the operator maps to an Hermitian space in order to define joint concavity? – Ziv Goldfeld Nov 19 '12 at 14:04
Not areal reply, however you may find some hints in the following book "Positive definite matrices" by Rajendra Bhatia (MR2284176). You may find some hints there (I do not work on that topic,and do not have a copy with me. However, I remember that there is a chapter devoted to matrix means). – RSG Nov 19 '12 at 14:35
@ZivGoldfeld: you need an order relation $\leq$ to be defined on the map image in order to speak about convexity/concavity. You have to specify what $M\leq N$ means if $M$ and $N$ are not self-adjoint. – Federico Poloni Nov 19 '12 at 14:48
No, real vs complex is not the problem here. The problem is that the positive definite semiordering is well defined only for Hermitian matrices (or symmetric matrices if you want to restrict to the reals). – Federico Poloni Nov 19 '12 at 15:36

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