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? 



Arithmetic-geometric mean of positive matrices
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](http://mathworld.wolfram.com/Arithmetic-GeometricMean.html) 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, $$P^TBP = U^TS^TQ^TBQSU=D.$$


