From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Exponential functions of matrixes Date: 22 Jun 1998 18:54:33 GMT mart68@geocities.com wrote: > Let A and B be NxN matrixes. > > Is there a general formula for > exp(A)*exp(B)? In article <6mloen$k39$1@basement.replay.com>, Anonymous wrote: >The general formula, exp(A)*exp(B) = exp(C), where > C = A + B + higher terms, > >The higher terms involve commutators, and you can work out as many terms >as you wish. I believe this is called the Poincare-Birkhoff-Witt theorem. Bzzt! Next contestant. "What is the Baker-Campbell-Hausdorff formula?" Lie Groups for $500, please, Alex. And the answer is, "A Lie Algebra embeds in the Lie Algebra of its Enveloping algebra". "The Poincare-Birkhoff-Witt theorem." Phrase it in form of a question please. "The Poincare-Birkhoff-Witt theorem, eh?" PS - Actually just what is called the PBW theorem is I suppose a matter of taste; alternatively it is the theorem that a couple of associated algebras are isomorphic, a result from which the other one follows. PPS - Alex Trebek is Canadian. ============================================================================== From: torquemada@my-dejanews.com Newsgroups: sci.math Subject: Re: Exponential functions of matrixes Date: Mon, 22 Jun 1998 18:41:07 GMT In article <6mlid9$4fd$1@nnrp1.dejanews.com>, mart68@geocities.com wrote: > Let A and B be NxN matrixes. > > Is there a general formula for > exp(A)*exp(B)? Not really. You can try using the well known Baker-Campbell-Hausdorff formula if you have some extra information about [A,B]=AB-BA. For example if A and B commute then exp(A)*exp(B)=exp(A+B). If, on the other hand [A,B] commutes with A and B you can use exp(A)*exp(B)=exp([A,B]/2)*exp(A+B). This result generalises to an infinite product involving more and more complex commutator terms like [A,[A,B]], [A,[A,[A,B]]] but it's only really useful if you know these eventually becomes zero. -- Torque http://www.grin.net/~tanelorn -----== Posted via Deja News, The Leader in Internet Discussion ==----- http://www.dejanews.com/ Now offering spam-free web-based newsreading