How do I solve for the unknown matrix Q in this equation?

[tapashettisr]

I have an equation as follows:

A*Q*B=C
Here A is 2X4 known matrix
Q is a 4X4 unknown matrix
B is a 4X2 known matrix
C is a 2X2 known matrix
How can I solve for unknown matrix Q;

 

[D H]

You have 4 equations in 16 unknowns. Whatever Q you find will not be unique. Decompose A and B via singular value decomposition to find the pseudoinverses of A and B.

 

[HallsofIvy]

Or, treating it purely as a matrix problem, multiply on the right on both sides of the equation by B^-1 and on the left by A^-1, assuming those inverses exist:

A^-1*A*Q*B*B^-1= A^-1CB^-1
Q= A^-1CB^-1

Of course, if either A or B does not have an inverse, there may be no solution or there may be an infinite number of solutions.

 

[nicktacik]

Er, B and A are not square matrices, so they do not have inverses, no?

 

[HallsofIvy]

I really need to learn to read these posts! Of course, they might have "generalized inverses" in which case the same comments apply.

 

[radou]

I really need to learn to read these posts! Of course, they might have "generalized inverses" in which case the same comments apply.

Just curious, how would these look like?

 

[D H]

One way to form the pseudoinverse (the Moore-Penrose psuedoinverse) of a matrix A is to decompose the matrix into a form $$A=UVW^T$$ where U and W are orthonormal matrices and V is a diagonal matrix. (This is the singular value decomposition of the matrix.) Form the pseudoinverse $$V^\dagger$$ of $$V$$ by taking the inverse of each non-zero element of $$V$$. Then the pseudoinverse of $$A$$ is $$A^{\dagger}=WV^{\dagger}U^T$$. Note that the pseudoinverse is the inverse if A is not singular.

 

[tapashettisr]

thanx all for the suggestions

 

[tapashettisr]

Unlike in case of square matrix, in case of rectangular matrix inv(A)*A is not equal to A*inv(A). If we find the pseudoinverse of A then A*inv(A)=identity matrix but inv(A)*A does not yield identity matrix. Hence, I am not able to understand how the suggested solution given below would work?

Inv(A)*A*Q*B*inv(B)=iinv(A)*C*inv(B)=Q

 

[tayyaba aftab]

If i have a unknown 4*4 matrix A
and i have an equation like AQA(dagger)=block diagnol matrix
all the matrices are 4*4
how can i evaluate 16 unknowns from these 16 equations using mathematica,as terms involved are much complicated.

 

[hotvette]

I have an equation as follows:

A*Q*B=C
Here A is 2X4 known matrix
Q is a 4X4 unknown matrix
B is a 4X2 known matrix
C is a 2X2 known matrix
How can I solve for unknown matrix Q;

This problem can be treated as a succession of two under determined problems that can be solved for least norm solutions. If you let Z = QB, the result is AZ=C, which can be solved as a mininum norm problem using QR factorization. If A^T is decomposed into QR (not the same Q as the original problem), Z can be solved as follows:

Solve R^Ty = C for y. Then Z = Qy ==> least norm solution

You then have QB = Z. Taking transpose of both sides, you get B^TQ^T=Z^T which can be solved for Q^T in exactly the same fashion. It does work. I tried it on a made up problem. The solution isn't unique, there are multiple possible values for Q, but the above does give you a way to find a solution.

 

[hotvette]

I think my last reply was too convoluted. Simply put, if Ax=b is an underdetermined system (where A has more columns than rows), it is easy to show that the minimum norm solution (assuming AA^T isn't singular) is:

x_{min norm} = A^T(AA^T)^-1b

 

[D H]

That works. That is a special case of the Moore-Penrose psuedoinverse cited in post #7.