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
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