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Similar matrices please check my proof.

  1. Dec 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove that every periodic matrix is similar to a diagonal matrix.

    2. Relevant equations

    I use ~ to denote A similar to B as A~B

    3. The attempt at a solution

    Let A be periodic, and let A^m = I.

    If A^m = I this implies A^m ~ I.

    Claim: if X~Y implies X^n ~ Y^n for n a positive integer.
    By induction.

    Base case n = 2.
    If X~Y then there exists a Z such that X = ZY(Z^-1).
    X^2 = X*X = ZY(Z^-1)ZY(Z^-1) = Z*(Y^2)*(Z^-1).

    Inductive step Assume true for some n.

    X^(n+1) = (X^n) * X = Z(Y^n)(Z^-1)* ZY(Z^-1) = Z (Y^(n+1))(Z^-1) and we are done.

    So since A^m ~ I implies A^(m+1) ~ I ^2 implies A^(m+1) ~ I, but since A^(m+1) = A we have A~I and since I is a diagonal matrix we are done.
  2. jcsd
  3. Dec 11, 2008 #2


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    No, that doesn't work. You've shown X~Y implies X^n~Y^n. That's fine. But applying that to A^m~I only gives you e.g. (A^m)^2~I^2 or A^(2m)~I. That doesn't help. Think about what the blocks must look like in the Jordan normal form of A.
    Last edited: Dec 11, 2008
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