TITLE: Arithmetic-geometric mean of positive matrices
QUESTION [10 upvotes]: Let $A,B$ be positive definite (Hermitian) matrices. Define the Arithmetic-geometric means of positive matrices by $A_0=A, G_0=B$, $A_{n+1}=\frac{A_n+G_n}{2}, G_{n+1}=A_n\natural G_n$, where $A_n\natural G_n$ means the geometric mean defined in http://www.isid.ac.in/~statmath/eprints/2011/isid201102.pdf
Will $\{A_n\}$ and $\{G_n\}$ converge to the same matrix?

REPLY [13 votes]: Recall that since A and B are (hermitian) positive definite, we can without loss of generality (see below for proof) assume that $A=I$ and $B=D$, where $D$ is some positive diagonal matrix. With this observation, merely recall the convergence theory for the scalar case to conclude that the sequences $\{A_n\}$ and $\{B_n\}$ converge to the arithmetic-geometric mean of $I$ and $D$.

Note: (Added to improve clarity)
For positive definite $A$ and $B$, let $A=Q\Lambda Q^T$, $S=\Lambda^{-1/2}Q$, and let $U$ diagonalize $S^TQ^TBQS$ to $D$. Then, with $P=QSU$, we have 
$$P^TAP = U^TS^TQ^TQ\Lambda Q^TQSU = U^TQ^T\Lambda^{-1/2}\Lambda \Lambda^{-1/2}QU = I,$$
and by construction, $$P^TBP = U^TS^TQ^TBQSU=D.$$