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Thread: Linear Algebra Proof Help

  1. #1
    Senior Member Danneedshelp's Avatar
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    Linear Algebra Proof Help

    Hello, I need help with a proof again.

    Q: prove that if $\displaystyle A^{2}=A$, then $\displaystyle I-2A=(I-2A)^{-1}$.

    So, from that I gather...

    Givens: $\displaystyle A^{2}=A \rightarrow A$ is square.

    Goal: $\displaystyle I-2A=(I-2A)^{-1}$

    So, I don't have very much ha.

    Should I just approach this multiplying the the inverse quantity to both sides of the quantity $\displaystyle I-2A$ and hope to get the Identity $\displaystyle I$ back?

    Not sure how to use the first fact $\displaystyle A^{2}=A$.

    Thanks,

    I'll keep plugging away at it.
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  2. #2
    MHF Contributor

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    $\displaystyle \left( {I - 2A} \right)\left( {I - 2A} \right) = I^2 - 2IA - 2AI + 4A^2 = I - 4A + 4A = I$
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Plato View Post
    $\displaystyle \left( {I - 2A} \right)\left( {I - 2A} \right) = I^2 - 2IA - 2AI + 4A^2 = I - 4A + 4A = I$
    Is that all that has to be shown?
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  4. #4
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    Quote Originally Posted by Danneedshelp View Post
    Is that all that has to be shown?
    If each of $\displaystyle A~\&~B$ is a matrix, how does one show that $\displaystyle A=B^{-1}?$
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  5. #5
    Senior Member Danneedshelp's Avatar
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    oh, or is it
    $\displaystyle AB=I=BA$
    Last edited by Danneedshelp; Jun 25th 2009 at 01:54 PM.
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  6. #6
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    Quote Originally Posted by Danneedshelp View Post
    $\displaystyle AB^{-1}=I=B^{-1}A$
    correct?
    No it is not correct? Do you have a textbook?

    $\displaystyle AB=BA=I$
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  7. #7
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Plato View Post
    No it is not correct? Do you have a textbook?

    $\displaystyle AB=BA=I$

    Yeah, I see that now. I just forgot to delete first part of my last post when I made the edit. I put $\displaystyle AB=BA=I$[/quote] at the bottom. My bad.

    Thanks for the help.
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