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From: "Pinpress" <nothing@nothing.edu>
Newsgroups: comp.soft-sys.matlab
Subject: Re: Solving AM = MB
Date: Sat, 22 Nov 2008 20:41:02 +0000 (UTC)
Organization: Dartmouth College
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OK. Let me provide an example:

A =

    0.9932   -0.1156    0.0144   27.0156
    0.1150    0.9926    0.0385    0.5268
   -0.0187   -0.0366    0.9992    4.8570
         0         0         0    1.0000
B =

    0.9932    0.1160    0.0141  -41.1405
   -0.1154    0.9926   -0.0374    5.7466
   -0.0183    0.0355    0.9992   -8.3134
         0         0         0    1.0000

I have followed the algorithm to compute X: X*A = B*X, and obtained:

X =

   -0.5700    0.0488   -0.0716    0.9674
    0.0654    0.5853   -0.0091   -0.0883
   -0.0465    0.0426   -0.4048    1.4363
    0.0000   -0.0000    0.0000    0.3822

But my "real" solution, which is known, is:

T =
   -0.9999    0.0098   -0.0106   44.6588
    0.0098    1.0000    0.0001  127.7536
    0.0106   -0.0000   -0.9999  -30.5660
         0         0         0    1.0000

You can verify that X*A = B*X, as well as T*A = B*T. By the way, I really need a orthogonal matrix as my solution. So, X is NOT what I want.

>> det(X)

ans =

    0.0512

>> det(T)

ans =

    1.0000


"Pinpress" <nothing@nothing.edu> wrote in message <gg9pp2$q6m$1@fred.mathworks.com>...
> Hi all,
> 
> This program is interesting. But would the solution be unique??  I have tried the algorithms presented here, but it does not give me a solution that makes sense (because I have the known solution). However, checking XA and BX, they are indeed equal. So it makes me think that the solution is not unique.  Any inputs here?