NetNews Usenet Archive 1992 #27/NN_1992_27.iso/spool/sci/engr/control/302

Internet Message Format | 1992-11-22 | 1.8 KB

Path: sparky!uunet!psinntp!kepler1!andrew
From: [e-mail] (Andrew Mullhaupt)
Newsgroups: sci.engr.control
Subject: Re: Solution of matrix Equation AX=XB
Message-ID: <[e-mail]>
Date: 21 Nov 92 23:21:04 GMT
References: <[e-mail]> <[e-mail]>
Distribution: na
Organization: Renaissance Technologies Corp., Setauket, NY.
Lines: 36

In article <[e-mail]> [e-mail] (John Heimbold) writes:
>In article <[e-mail]> [e-mail] (Stewart Moorehead) writes:
>> calculated. To calculate X, the equation AX=XB must be solved, with A
>> and B being known 4x4 matrices. I have found two methods which claim to
>> solve for X.

To start with, a nonsingular solution X to AX=XB means that A=XBX^-1, and
this means that A and B must be similar. X is available by computing the
normal forms of A and B in this case. Computationally, the Shur
triangularization is probably the best. Note that this allows us to consider
the case where A and B are upper triangular without loss...

Now let L be a linear transformation on X defined by

        L[X] = AX - XB

and notice that if A and B^T have a common eigenvalue, say

        Au = cu,    v^T B = cv^T

then the matrix uv^T is an eigenvector of L with eigenvalue (c - d).

    L[uv^T] = Auv^T - uv^TB = cuv^T - duv^T = (c-d) uv^T.

This means that any common eigenvalue generates a rank one solution,
and any linear combination of these rank one solutions is also a solution.

There will be lots of these, even for A and B which are similar, so it only
makes sense to worry about uniqueness of solutions X of maximal rank. If
A and B are similar, the maximal rank solution X is unique.

More on this equation can be found in Gantmacher's _Matrix Theory_ and
Horn and Johnson's _Matrix Analysis_.

Later,
Andrew Mullhaupt