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Proving Similar Matrices

  1. Mar 18, 2006 #1
    Hey,
    i'm having trouble proving how two matrices are similar using:
    A = PBP^-1

    Given:
    A is similar to B
    And B is similar to C

    Prove that A is Similar to C?

    A = PBP^-1
    B = PCP^-1

    so i.e. A = PPCP^-1P^-1 ..... ?

    Can anyone help me?
     
  2. jcsd
  3. Mar 18, 2006 #2

    robphy

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    You have the right idea.

    If "A is similar to B" means A = PBP^-1 ,
    does "B is similar to C" mean B = PCP^-1 ? Or is that asking a little too much?

    In light of the last question, what would "A is Similar to C" mean?

    Is there some property of matrix multiplication that would be useful here?
     
  4. Mar 18, 2006 #3
    Are you referring to the determinant?

    So you're saying i can't apply the same formula for B is similar to C to give
    B = PCP^-1?

    I think A is similar to C means the size of both the matrix are the same. The numbers are arranged in such a way that their determinants are the same.
     
  5. Mar 18, 2006 #4

    robphy

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    "A is similar to B" means "there exists an invertible square matrix P such that A = PBP^-1".

    "B is similar to C" means "there exists an invertible square matrix Q such that B = QCQ^-1". (It need not be that P=Q... since this sentence knows nothing about A!)

    What would "A is similar to C" mean (without referencing the previous two statements)?
     
  6. Mar 18, 2006 #5

    HallsofIvy

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    It will be helpfull to know that Q-1P-1= (PQ)-1!
     
  7. Mar 19, 2006 #6
    A similar to C is

    A = RCR^-1 ?

    Where R = PQ, from A = PQCP^-1Q^-1 ?

    and to prove that, all i need to do is find R?
     
  8. Mar 19, 2006 #7

    0rthodontist

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    You need to find R^-1.
     
  9. Mar 19, 2006 #8
    Thanks a lot guys! I got it now =)
     
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