## Arithmetic-geometric mean of positive matrices

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?

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 I computed the geometric mean of $A=[1,-1/2;-1/2,1]$ and $B=[1,1/2;1/2,1]$ (both matrices are symmetric positive definite) using Eq.(1) from the link. The result is a matrix whose diagonal entries are not real (in particular, it isn't Hermitian). You have to find a proper set of matrices (or a better geometric mean) to have your question well defined. – Wadim Zudilin Aug 1 2011 at 11:56 @Wadim: Are you sure with your calculation? $A\natural B$ must be Hermitian provided it is defined (i.e., $A,B$ are psd). – Sunni Aug 1 2011 at 13:25 @Wadim: Did you use the definition? $a\sharp b := a^{1/2}(a^{-1/2}ba^{-1/2})^{1/2}a^{1/2}$ – S. Sra Aug 1 2011 at 17:09 Sounds fishy, since that matrix is clearly Hermitian from its definition. Is the loss of Hermitianity small? It could be due to numerical errors and/or a poor sqrtm() implementation. (Incidentally, you're not using Matlab's sqrt() by chance, are you?) – Federico Poloni Aug 1 2011 at 19:29 @Wadim: for that $A$ and $B$ I get geometric mean with both diagonal entries $$\frac{\sqrt{2} \sqrt{3}}{12} + \frac{3 \sqrt{2}}{4}$$ and both off-diagonal entries $$- \frac{\sqrt{2} \sqrt{3}}{12} + \frac{3 \sqrt{2}}{4}$$ – Gerald Edgar Aug 1 2011 at 20:46
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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, $$P^TBP = U^TS^TQ^TBQSU=D.$$

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I don't understand the "no loss of generality" here. Of course if $A,B$ commute, then we can reduce to the diagonal case. – Gerald Edgar Aug 1 2011 at 1:13
@Gerald: No commutability was assumed; I had first put a comment, but now I have edited it into the answer. – S. Sra Aug 1 2011 at 17:22
Sorry for the sloppiness, by "invariant" I mean that $S(A\sharp B)S^*=(SAS^*) \sharp (SBS^*)$. – Federico Poloni Aug 1 2011 at 19:32
@Gerald: this result is quite well-known; it is easier to prove using other characterizations of the geometric mean. For instance, using the fact that $Af(BA)=f(AB)A$ for all holomorphic $f$ (and in particular for the square root function), you can rewrite the definition as $A \sharp B = A(A^{-1}B)^{1/2}$, and from this form congruence invariance follows easily. – Federico Poloni Aug 2 2011 at 7:34
Maybe the easiest way to prove this is to note that the geometric mean is the unique positive definite solution to the Riccati equation: $XA^{-1}X=B$. – S. Sra Aug 2 2011 at 18:19
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