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Similar Matrices - Proof

  1. Jul 31, 2009 #1
    1. The problem statement, all variables and given/known data

    Let A and B be similar matrices. Prove that [tex]A^T[/tex] and [tex]B^T[/tex] are also similar.

    2. The attempt at a solution

    We know that, since A and B are similar matrices, there exists an invertible matrix P such that

    [tex]B=P^{-1}AP[/tex]

    so, I thought if we transposed both sides, this could lead to a proof, but this gives

    [tex]B^T=\left(P^{-1}AP\right)^T[/tex]

    [tex]B^T=P^TA^T\left(P^{-1}\right)^T[/tex]

    [tex]B^T=P^TA^T\left(P^T\right)^{-1}[/tex]

    Which feels close, but how do I get it to the form

    [tex]B^T=\left(P^T\right)^{-1}A^TP^T[/tex] ?

    Provided of course, that this is an actually viable route to take (I have virtually zero experience with mathematical proofs).

    Thanks!
    phyz
     
  2. jcsd
  3. Jul 31, 2009 #2
    hi phyz,
    you reached [tex] B^T=P^TA^T(P^T)^{-1} [/tex].
    if you put [tex] Q = (P^T)^{-1}[/tex] , what form would the above statement take?
     
  4. Aug 1, 2009 #3
    Ha! Easy as that is it? :biggrin:

    Thanks winter85!
     
  5. Aug 1, 2009 #4
    yep :)
     
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