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Similarity is transitive

  1. Nov 22, 2008 #1
    1. The problem statement, all variables and given/known data

    In linear algebra, how to prove that the Similarity is transitive?

    2. Relevant equations

    [T]C = P-1 [T]B P

    3. The attempt at a solution
    My logic: if a is similar to b and b is similar to c, then a is similar to c. But how do you implement that in case of similarity ?
    Last edited: Nov 22, 2008
  2. jcsd
  3. Nov 22, 2008 #2


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    If A is similar to B, A=PBP^(-1). If B is similar to C, B=QCQ^(-1). Where P and Q are invertible. Put the second equation into the first equation.
  4. Nov 22, 2008 #3


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    If A = P B Pinv and B = K C Kinv, where Pinv is the inverse of P, and mutatis mutandis for K, use the property that (PK)inv is (Kinv)(Pinv). and hence A = (PK) C (PK)inv.
  5. Nov 22, 2008 #4
    how do PQ and P-1Q-1 behave ?
  6. Nov 22, 2008 #5


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    You should have asked how do PQ and Q^(-1)P^(-1) behave. Whatever 'behave' means. They don't do anything, they are just invertible matrices. And one is the inverse of the other, see ptr's post.
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