From: Eric Rudd <rudd@cyberoptics.com>
Subject: Re: rectangular matrix logarithms
Date: Tue, 15 Aug 2000 16:21:57 -0500
Newsgroups: sci.math.num-analysis
Summary: [missing]

"Andrew M. Ross" wrote:

> Where did you learn that exp(A)*exp(B) != exp(A+B) -- none of the
> linear algebra books I've looked through mention it.  Is there a good
> reference for this kind of thing?

A good place to start would be

     Moler, Cleve and Charles van Loan, "Nineteen Dubious Ways to
     Compute the Exponential of a Matrix," SIAM Review, vol 20
     (1978) pp. 801-836.

There is an outstanding review of the theory of matrix exponentials
available over the Web at
ftp://das-ftp.harvard.edu/techreports/tr-33-94.ps.gz .  I highly
recommend it.

Golub and Van Loan also have a good section on the matrix exponential.

You don't have to look very hard to find matrices where exp(A)*exp(B) !=
exp(A+B); try A = [1 1;-1 -1], B=A^T.  However, if AB = BA, the identity
*does* hold.

-Eric Rudd
rudd@cyberoptics.com
==============================================================================

From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: rectangular matrix logarithms
Date: 15 Aug 2000 19:21:13 -0400
Newsgroups: sci.math.num-analysis

In article <3999B475.5F07B825@cyberoptics.com>,
Eric Rudd  <rudd@cyberoptics.com> wrote:
[quote of previous message deleted --djr]

(The proof of the latter uses the familiar re-arrangement of power series,
which is legitimate because of absolute convergence.)

A small additional example (a standard exercise): if you calculate

  f(t) = exp(t*A) * exp(t*B) - exp(t*A + t*B)

using power series, you will find that f(0)=0, f'(0)=0 but

  f''(0) = A*B - B*A

illustrating the importance of the relation A*B=B*A.

Cheers, ZVK(Slavek)