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Solve for unknown matrix

  1. Aug 2, 2007 #1
    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;
     
  2. jcsd
  3. Aug 2, 2007 #2

    D H

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    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.
     
  4. Aug 2, 2007 #3

    HallsofIvy

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    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.
     
  5. Aug 2, 2007 #4
    Er, B and A are not square matrices, so they do not have inverses, no?
     
  6. Aug 2, 2007 #5

    HallsofIvy

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    I really need to learn to read these posts! Of course, they might have "generalized inverses" in which case the same comments apply.
     
  7. Aug 2, 2007 #6

    radou

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    Just curious, how would these look like?
     
  8. Aug 2, 2007 #7

    D H

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    One way to form the pseudoinverse (the Moore-Penrose psuedoinverse) of a matrix A is to decompose the matrix into a form [tex]A=UVW^T[/tex] 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 [tex]V^\dagger[/tex] of [tex]V[/tex] by taking the inverse of each non-zero element of [tex]V[/tex]. Then the pseudoinverse of [tex]A[/tex] is [tex]A^{\dagger}=WV^{\dagger}U^T[/tex]. Note that the pseudoinverse is the inverse if A is not singular.
     
  9. Aug 3, 2007 #8
    thanx all for the suggestions
     
  10. Aug 3, 2007 #9
    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
     
  11. Nov 4, 2010 #10
    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.
     
  12. Nov 7, 2010 #11

    hotvette

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    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 AT is decomposed into QR (not the same Q as the original problem), Z can be solved as follows:

    Solve RTy = C for y. Then Z = Qy ==> least norm solution

    You then have QB = Z. Taking transpose of both sides, you get BTQT=ZT which can be solved for QT 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.
     
  13. Nov 9, 2010 #12

    hotvette

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    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 AAT isn't singular) is:

    xmin norm = AT(AAT)-1b
     
    Last edited: Nov 9, 2010
  14. Nov 9, 2010 #13

    D H

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