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Proof Similarity

  1. Dec 10, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove B^T ~ A^T (tranpose)
    Given: B ~ A
    Can anyone check if my proof is correct? Towards the end I'm not quite sure if I can do it like that. Do I have to say what P is exactly? The only matrix I can think that would satisfy that is the identity matrix.

    2. Relevant equations

    3. The attempt at a solution

    B ~ A
    B = P^-1 * A * P
    B^T = P^T * A^T * (P^-1)^T
    so P is a matrix that where P^T = P^-1 therefore (P^T)^-1 = P
    B^T = P^-1 * A^T * P
    B^T ~ A^T
    Last edited: Dec 10, 2007
  2. jcsd
  3. Dec 10, 2007 #2


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    Science Advisor
    Homework Helper

    You don't need to assume P^(-1)=P^T. That means P is a special kind of matrix. It's enough that (P^T)^(-1)=(P^(-1))^T, which is true for any invertible matrix.
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