Expert: Paul Klarreich - 10/1/2005

Question
I have three matrices M, A and B in an equation M * A = B
M is 5x5
A and B are both 5 row by 1 column.

A and B are known, M isnt. I want to solve for M.


I know M = B * A inverse, but that doesnt help since A isnt square. What other lin alg trick am I missing? I also know that I could diagonalize the M matrix, meaning I could set the 5 diagonal elements to the 5 corresponding B/A elements, and all the other elements would be zero. Is there a more general solution? Thanks alot.


Answer
Hi, Ryan,
I'm not sure I can give you much more than this:  You are trying to find the 25 elements of a matrix, given five equations.  So that would imply that 20 elements can be assigned arbitrarily.

In two dimensions, (I can't type out five dimensions in this space, but the idea is the same.) you are trying to find solutions to: (USE A FIXED-SIZE FONT, LIKE COURIER)
==== M ====    =A=     =B=
m11   m12   a1   b1
m21   m22   a2   b2

or:

m11a1 + m12a2 = b1
m21a1 + m22a2 = b2

If you are solving for the m's, you clearly have four variables and two equations, so there would be multiple solutions:

Assuming the ai's are not zero, we can make m12 and m21 arbitrary, then

m11 = (b1 - m12a2)/a1
m22 = (b2 - m21a1)/a2

So, for example:

m11   m12   2   6
m21   m22   1   4

Obviously the matrix:
3   0   2   6
0   4   1   4
is a solution, but so is:
0   6   2   6
2   0   1   4
and

1   4   2   6
1   2   1   4

and, of course, many others.

There may be some deep property of the matrix M here, but I don't know it.