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In article <3784@news.cerf.net> heimbold@nic.cerf.net (John Heimbold) writes:
>In article <6611@dciem.dciem.dnd.ca> sjm@dciem.dciem.dnd.ca (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