Shel (sjk@kaphan.org) writes

<<
...Apropos your response to math-fun,  what relationship does Log(A) have to the eigenvector of A, if any?...
>>

Sorry about the blank message again!!!  (I'm having a touchpad hardware problem and single clicks often get interpreted as double -- which sent the blank msg.  The hardware's supposed to be replaced on Monday)

Anyhow, any rotation Q in SO(n,R) has assoc. with it a decomp. of R^n into   orthogonal eigenspaces that are 2-planes P_i that are each carried into themselves by rotation  by, say, theta_i for 0 <= theta_i <= pi, (and if n odd, an extra R^1 that is fixed).  This decomp is unique if the |theta_i|'s are all unequal.

Q has a pair of eigenvalues cos(theta_i) + - sin(theta_i) for each P_i, plus an extra 1 if n is odd.

Choosing an orthonormal basis e_1,...,e_n which all lie in the eigenspaces (and the ith pair having the right orientation on its P_i), the Log can be expressed by the 2x2 block about the diagonal equal to

0        theta_i
                    
-theta_i       0

It's easy to check that exp of such a matrix gets you blocks like

cose(theta_i)    sin(theta_i)

-sin(theta_i)     cos(theta_i)

along the diagonal, and so it performs as advertised.

This much I understand, but what happens to the axes and angles when you add Log(A) and Log(B) (and then take exp) is a mystery to me.

                               *                            *                             *

BTW, I wasn't quite right in what I claimed: The expression Shel asked about,

lim n -> oo (A^(1/n) B^(1/n))^n is equal to exp(Log(A) + Log(B)).  The expression

lim n-> oo (A^(1/2n) B^(1/2n))^n is what's equal to exp( (Log(A) + Log(B)) / 2 ).

--Dan