How to solve this Matrix Equation?

All matrices are 2*2. A,B,C,D are known, need to solve for X.

A*X*B = diagonal matrix

C*X*D = diagonal matrix

Will this give a unique solution of X? How to solve this?

Many thanks,

Ellen

[Walter Roberson]

To check: is it the *same* diagonal matrix to be reached, or potential *different* diagonal matrices?

I'm presuming here you aren't looking for the trivial X = eye(2)

Answer 1

[Walter Roberson]

The solution is definitely not unique.

Let the two result matrices be [rab11,0;0,rab22] and [rcd11;0;0;rcd22]

Set rcd22 to any non-zero non-infinite value. Then, except for possible singularities,

rab11 = rcd22*(((c12*c21*d12*d21-c11*c22*d11*d22)*b11+b21*d12*d11*(c11*c22-c12*c21))*a12+a11*b11*c12*c22*(d11*d22-d12*d21))*(a11*a22-a12*a21)*(b11*b22-b12*b21)/((((d12*c21*b22*a22-b12*a21*d22*c22)*b11+d12*b12*b21*(-c21*a22+a21*c22))*a12+a11*a22*b11*c22*(-d12*b22+b12*d22))*(c11*c22-c12*c21)*(d11*d22-d12*d21))

rab22 = -rcd22*(((c12*c21*d12*d21-c11*c22*d11*d22)*b12+b22*d12*d11*(c11*c22-c12*c21))*a22+a21*b12*c12*c22*(d11*d22-d12*d21))*(a11*a22-a12*a21)*(b11*b22-b12*b21)/((((b11*d22*c22*a11-d12*c21*a12*b21)*b12-d12*b11*b22*(c22*a11-a12*c21))*a22-a12*a21*c22*b12*(b11*d22-b21*d12))*(c11*c22-c12*c21)*(d11*d22-d12*d21))

rcd11 = -rcd22*((d11*c11*(b11*b22-b12*b21)*a22-a21*c12*b12*(b11*d21-b21*d11))*a12+a11*a22*b11*c12*(b12*d21-d11*b22))/((c21*d12*(b11*b22-b12*b21)*a22-a21*c22*b12*(b11*d22-b21*d12))*a12+a11*a22*b11*c22*(-d12*b22+b12*d22))

You can see in this that rcd22 acts as an arbitrary scale factor for the other coefficients.

The X matrix then becomes

X(1,1) = (b22*a22*rcd22*(((c12*c21*d12*d21-c11*c22*d11*d22)*b11+b21*d12*d11*(c11*c22-c12*c21))*a12+a11*b11*c12*c22*(d11*d22-d12*d21))*(a11*a22-a12*a21)*(b11*b22-b12*b21)/((((d12*c21*b22*a22-b12*a21*d22*c22)*b11+d12*b12*b21*(-c21*a22+a21*c22))*a12+a11*a22*b11*c22*(-d12*b22+b12*d22))*(c11*c22-c12*c21)*(d11*d22-d12*d21))-b22*a12*rab21-b21*a22*rab12-b21*a12*rcd22*(((c12*c21*d12*d21-c11*c22*d11*d22)*b12+b22*d12*d11*(c11*c22-c12*c21))*a22+a21*b12*c12*c22*(d11*d22-d12*d21))*(a11*a22-a12*a21)*(b11*b22-b12*b21)/((((b11*d22*c22*a11-d12*c21*a12*b21)*b12-d12*b11*b22*(c22*a11-a12*c21))*a22-a12*a21*c22*b12*(b11*d22-b21*d12))*(c11*c22-c12*c21)*(d11*d22-d12*d21)))/((a11*a22-a12*a21)*(b11*b22-b12*b21))

X(1,2) = -(b12*a22*rcd22*(((c12*c21*d12*d21-c11*c22*d11*d22)*b11+b21*d12*d11*(c11*c22-c12*c21))*a12+a11*b11*c12*c22*(d11*d22-d12*d21))*(a11*a22-a12*a21)*(b11*b22-b12*b21)/((((d12*c21*b22*a22-b12*a21*d22*c22)*b11+d12*b12*b21*(-c21*a22+a21*c22))*a12+a11*a22*b11*c22*(-d12*b22+b12*d22))*(c11*c22-c12*c21)*(d11*d22-d12*d21))-b12*a12*rab21-b11*a22*rab12-b11*a12*rcd22*(((c12*c21*d12*d21-c11*c22*d11*d22)*b12+b22*d12*d11*(c11*c22-c12*c21))*a22+a21*b12*c12*c22*(d11*d22-d12*d21))*(a11*a22-a12*a21)*(b11*b22-b12*b21)/((((b11*d22*c22*a11-d12*c21*a12*b21)*b12-d12*b11*b22*(c22*a11-a12*c21))*a22-a12*a21*c22*b12*(b11*d22-b21*d12))*(c11*c22-c12*c21)*(d11*d22-d12*d21)))/((a11*a22-a12*a21)*(b11*b22-b12*b21))

X(2,1) = -(b22*a21*rcd22*(((c12*c21*d12*d21-c11*c22*d11*d22)*b11+b21*d12*d11*(c11*c22-c12*c21))*a12+a11*b11*c12*c22*(d11*d22-d12*d21))*(a11*a22-a12*a21)*(b11*b22-b12*b21)/((((d12*c21*b22*a22-b12*a21*d22*c22)*b11+d12*b12*b21*(-c21*a22+a21*c22))*a12+a11*a22*b11*c22*(-d12*b22+b12*d22))*(c11*c22-c12*c21)*(d11*d22-d12*d21))-b22*a11*rab21-b21*a21*rab12-b21*a11*rcd22*(((c12*c21*d12*d21-c11*c22*d11*d22)*b12+b22*d12*d11*(c11*c22-c12*c21))*a22+a21*b12*c12*c22*(d11*d22-d12*d21))*(a11*a22-a12*a21)*(b11*b22-b12*b21)/((((b11*d22*c22*a11-d12*c21*a12*b21)*b12-d12*b11*b22*(c22*a11-a12*c21))*a22-a12*a21*c22*b12*(b11*d22-b21*d12))*(c11*c22-c12*c21)*(d11*d22-d12*d21)))/((a11*a22-a12*a21)*(b11*b22-b12*b21))

X(2,2) = -(b22*a21*rcd22*(((c12*c21*d12*d21-c11*c22*d11*d22)*b11+b21*d12*d11*(c11*c22-c12*c21))*a12+a11*b11*c12*c22*(d11*d22-d12*d21))*(a11*a22-a12*a21)*(b11*b22-b12*b21)/((((d12*c21*b22*a22-b12*a21*d22*c22)*b11+d12*b12*b21*(-c21*a22+a21*c22))*a12+a11*a22*b11*c22*(-d12*b22+b12*d22))*(c11*c22-c12*c21)*(d11*d22-d12*d21))-b22*a11*rab21-b21*a21*rab12-b21*a11*rcd22*(((c12*c21*d12*d21-c11*c22*d11*d22)*b12+b22*d12*d11*(c11*c22-c12*c21))*a22+a21*b12*c12*c22*(d11*d22-d12*d21))*(a11*a22-a12*a21)*(b11*b22-b12*b21)/((((b11*d22*c22*a11-d12*c21*a12*b21)*b12-d12*b11*b22*(c22*a11-a12*c21))*a22-a12*a21*c22*b12*(b11*d22-b21*d12))*(c11*c22-c12*c21)*(d11*d22-d12*d21)))/((a11*a22-a12*a21)*(b11*b22-b12*b21))

This was calculated using straight-forward linear algebra by multiplying appropriate inverses on both sides to get two isolated X equations, and then solve()'ing for the corresponding components to be the same.

[Ellen]

Thanks a lot!!

Answer 2

[Walter Roberson]

No, it will not necessarily give a unique solution for X.

Consider the possibility that A*B gives a diagonal matrix, and C*D gives a diagonal matrix. Then if X is a diagonal matrix with the two diagonal elements the same, X*B would be a constant multiple of B and so A*X*B would be a constant multiple of A*B. The same X would also lead to C*X*D being a constant multiple of C*D. This will occur for any finite non-zero X of that form, so in this set of circumstances no unique X could be found.

There might be circumstances under which the solution is unique, but I do not have a feel for them at the moment.

[Ellen]

Walter,

Thanks a lot for you reply.

Here, A B C D and X are usually not diagonal matrices. And two "diagonal matries" are not necessarily the same.

Any input?

Have a nice weekend,

Ellen

[Walter Roberson]

A, B, C, and D might not be diagonal matrices, but A*B and C*D might be, I think (I'd have to double-check whether that can happen for 2x2.)

Anyhow the point is that you asked if the solution was unique, and I showed at least one case (which you do not rule out) where the solution cannot be unique.