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Thread: [SOLVED] eigenvalue help!

  1. #1
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    [SOLVED] eigenvalue help!

    If a is an eigenvalue of A. show that a^(-1) is an eigenvalue of A^-1. I don't know where to start.
    Last edited by Jhevon; Mar 30th 2009 at 09:13 PM. Reason: Restored problem
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  2. #2
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    Quote Originally Posted by lord12 View Post
    If x is an eigenvalue of A. show that a^(-1) is an eigenvalue of A^-1. I don't know where to start.
    Do you mean that if $\displaystyle a$ is an eigenvalue of $\displaystyle A$, show that $\displaystyle a^{-1}$ is an eigenvalue of $\displaystyle A^{-1}$?
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    yes!
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    Quote Originally Posted by lord12 View Post
    yes!
    If $\displaystyle a$ is an eigenvalue of $\displaystyle A$, then

    $\displaystyle A\mathbf{x} = a\mathbf{x}$ for some eigenvector $\displaystyle \mathbf{x}$.

    Notice that $\displaystyle a\mathbf{x} = aI\mathbf{x}$, where $\displaystyle I$ is the Identity matrix.

    Therefore

    $\displaystyle A\mathbf{x} - aI\mathbf{x} = \mathbf{0}$

    $\displaystyle (A - aI)\mathbf{x} = \mathbf{0}$.

    For a nonzero $\displaystyle \mathbf{x}, A - aI = \mathbf{0}$.


    Therefore $\displaystyle A = aI$.

    If $\displaystyle A$ is nonsingular, then

    $\displaystyle A^{-1} = (aI)^{-1}$.


    By definition of inverses

    $\displaystyle AA^{-1} = aI(aI)^{-1} = I$

    $\displaystyle aI(aI)^{-1} = I$

    $\displaystyle I(aI)^{-1} = a^{-1}I$

    $\displaystyle (aI)^{-1} = a^{-1}I$

    $\displaystyle A^{-1} = a^{-1}I$.


    Therefore $\displaystyle a^{-1}$ is an eigenvalue of $\displaystyle A^{-1}$.
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