Prove ABC^TA^-1=CB

  • #1

Main Question or Discussion Point

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
 

Answers and Replies

  • #2
34,370
10,446


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].
 
  • #3


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
 
  • #4
34,370
10,446


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.
 
  • #5


Thanks for the quick response, Apologies however A and C are symmetric B is not does this chaneg anything
 
  • #6


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:
  • #7


14.63 to 14.64, should be an easy substitution but I can't get there
 
  • #8


no one ??
 

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