Shel (sjk@kaphan.org) writes << ...Apropos your response to math-fun, what relationship does Log(A) have to the eigenvector of A, if any?...

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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