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Prove ABC^TA^-1=CB

  1. Feb 22, 2013 #1
    Hi all,

    I have a suspicion this may be obvious but have lookd and can't seem to obtain the correct answer,

    Can somone please explain the steps required to prove

    ABC^TA^-1=CB

    where C^T is the transpose of C and A^-1 the inverse of A. Matrices B and A are covariance matrices and thus may be considered symmetric if that helps

    Thanks in advance for some direction
     
  2. jcsd
  3. Feb 22, 2013 #2

    mfb

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    Staff: Mentor

    Re: Multiplication

    I don't think your requirements are sufficient. Consider A=1, B=[1,2;2,1], C=[0,0;1,0], for example. BC^T = [0,1;0,2], but CB = [0,0;1,2].
     
  4. Feb 22, 2013 #3
    Re: Multiplication

    Thanks for the reply, on a second look I now have:

    AB^TC^-1BA=AB^TC^-1CAB^TC^-1

    Which I can break down to:

    AB^TC^-1BA=AB^TAB^TC^-1

    is there anyway I can re-order the RHS to equal the left?

    Thanks in advance
     
  5. Feb 22, 2013 #4

    mfb

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    Staff: Mentor

    Re: Multiplication

    Assuming A and B are invertible, this can be simplified to
    C^-1BA=AB^TC^-1

    As both A and B are symmetric, B^T=B and AB=BA (you can check this with the definition of matrix multiplication). Define D=AB and E=C^(-1). D is symmetric as well.

    Therefore, your equation is equivalent to ED=DE for symmetric D and invertible E. But this is wrong, for example for D=[1,2;2,1] and E=[1,1;0,1].

    Or, with the original matrices:
    A=1, B=[1,2;2,1], C=[1,-1;0,1] violates the equation.
     
  6. Feb 22, 2013 #5
    Re: Multiplication

    Thanks for the quick response, Apologies however A and C are symmetric B is not does this chaneg anything
     
  7. Feb 22, 2013 #6
    Re: Multiplication

    I am attempting to work through a derivation and the step attached requires the above mentioned to hold, any suggestions appreciated
     
    Last edited by a moderator: Feb 22, 2013
  8. Feb 22, 2013 #7
    Re: Multiplication

    14.63 to 14.64, should be an easy substitution but I can't get there
     
  9. Feb 22, 2013 #8
    Re: Multiplication

    no one ??
     
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