Here is just one of many possible examples of [exp(A), exp(B)] = 0 for [A, B] != 0, using commutator notation. The Pauli spin matrices are σ1 = [[0,1],[1,0]], σ2 = [[0,-i],[i,0]], σ3 = [[1,0],[0,-1]]. They satisfy σ1 σ2 = -σ2 σ1 = i σ3, and the rest by cyclic permutation. Let u be a real unit vector, u1^2 + u2^2 + u3^2 = 1. Let U = i(u1 σ1 + u2 σ2 + u3 σ3). It's easy to see that the eigenvalues of U are +i and -i. Then it follows from the series expansion, and substituting U^2 = -1, that exp(tU) = cos t + U sin t. Thus if t is a nonzero multiple of π, and A = tU, exp(A) is plus or minus the identity, and so commutes with every matrix, while A itself does not. The second part of the problem asks for exp(A) exp(B) = exp(A+B) = exp(B) exp(A), where by the symmetry of the middle expression, one equality implies the other. Let v be another real unit vector, orthogonal to u, and similarly construct V. Let A = 3πU, B = 4πV, so that A+B has "length" 5π. Then exp(A) exp(B) = (-1)^3 (-1)^4 = (-1)^5 = exp(A+B). -- Gene
________________________________ From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Monday, August 4, 2014 6:17 PM Subject: [math-fun] exp(A)exp(B) = exp(A+B) = exp(B)exp(A)
It's well known and easy to prove that if NxN matrices A and B commute, then
(*) exp(A)exp(B) = exp(A+B) = exp(B)exp(A)
.
But AB = BA is not a necessary condition for (*), as googling will readily reveal.
Does anyone know necessary and sufficient conditions on A and B for (*) to hold?
If not, how about a large class of A and B for which (*) holds despite AB and BA being unequal ?
I think such examples of complex matrices are easier to come by than of real ones, so I'm particularly interested in the case where A and B are real.
--Dan