Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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


    User Avatar
    2017 Award

    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:


    Which I can break down to:


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

    Thanks in advance
  5. Feb 22, 2013 #4


    User Avatar
    2017 Award

    Staff: Mentor

    Re: Multiplication

    Assuming A and B are invertible, this can be simplified to

    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 ??
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Prove ABC^TA^-1=CB
  1. Prove it (Replies: 6)

  2. Proves that (Replies: 3)

  3. Prove This (Replies: 1)

  4. ABC conjecture solved? (Replies: 2)