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Thread: Induction matrix proof.

  1. #1
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    Induction matrix proof.

    Apparently this is simple, but I'm not seeing it.

    P, A are matrices, n is a real.

    Use induction to prove that $\displaystyle (P^{-1}AP)^n = P^{-1}A^nP$.

    Presumably the first step is:

    $\displaystyle (P^{-1}AP)^n = P^{-n}A^nP^n$

    I'm not sure what's next, or if that is correct.
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  2. #2
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    Quote Originally Posted by scorpion007 View Post
    Apparently this is simple, but I'm not seeing it.

    P, A are matrices, n is a real.

    Use induction to prove that $\displaystyle (P^{-1}AP)^n = P^{-1}A^nP$.

    Presumably the first step is:

    $\displaystyle (P^{-1}AP)^n = P^{-n}A^nP^n$

    I'm not sure what's next, or if that is correct.
    In brief:

    Step 1: True for n = 1.

    Step 2: Assume true for n = k. That is, assume $\displaystyle (P^{-1}AP)^k = P^{-1} A^k P$ is true.

    Step 3: Show true for n = k + 1. That is, assuming $\displaystyle (P^{-1}AP)^k = P^{-1} A^k P$ is true, show that $\displaystyle (P^{-1}AP)^{k+1} = P^{-1} A^{k+1} P$.

    To do this, note that $\displaystyle (P^{-1}AP)^{k+1} = (P^{-1}AP)^{k} \cdot (P^{-1}AP)$ and substitute $\displaystyle (P^{-1}AP)^k = P^{-1} A^k P$.
    Last edited by mr fantastic; Mar 28th 2009 at 11:23 PM.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    In brief:

    Step 1: True for n = 1.

    Step 2: Assume true for n = k. That is, assume $\displaystyle (P^{-1}AP)^k = P A^k P$ is true.
    Did you mean $\displaystyle P^{-1} A^k P$ in the RHS?
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  4. #4
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    Quote Originally Posted by scorpion007 View Post
    Did you mean $\displaystyle P^{-1} A^k P$ in the RHS?
    Yes. I have made the necessary edit.
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  5. #5
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    Thanks, I see it now:

    Expanding the last step would be:

    $\displaystyle (P^{-1}AP)^{k+1} = (P^{-1}AP)^{k} \cdot (P^{-1}AP) =$
    $\displaystyle (P^{-1} A^k P) \cdot (P^{-1}AP) = P^{-1}A^k\underbrace{PP^{-1}}_{I}AP=P^{-1}A^k\cdot AP=\color{red}P^{-1}A^{k+1}P$

    Correct? Too verbose?
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  6. #6
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    Quote Originally Posted by scorpion007 View Post
    Thanks, I see it now:

    Expanding the last step would be:

    $\displaystyle (P^{-1}AP)^{k+1} = (P^{-1}AP)^{k} \cdot (P^{-1}AP) =$
    $\displaystyle (P^{-1} A^k P) \cdot (P^{-1}AP) = P^{-1}A^k\underbrace{PP^{-1}}_{I}AP=P^{-1}A^k\cdot AP=\color{red}P^{-1}A^{k+1}P$

    Correct? Too verbose?
    It's fine.

    Now you need to write out the complete proof by induction, dotting the i's and crossing the t's.
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