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Math Help - relationship between eigenvalue and inverse

  1. #1
    Member Jskid's Avatar
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    [SOLVED]relationship between eigenvalue and inverse

    Let \lambda be an eigenvalue of the nonsingular matrix A with associated eigenvector \vec x. Show that \frac{1}{\lambda} is an eigenvalue of A^{-1} with associated eigenvector \vec x.

    Things I thought about: roots of det(\lambda I_n-A) are eigenvalues, inverse is calculated by rref[A|In]
    Last edited by Jskid; March 29th 2011 at 12:19 PM. Reason: solved
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Jskid View Post
    Let \lambda be an eigenvalue of the nonsingular matrix A with associated eigenvector \vec x. Show that \frac{1}{\lambda} is an eigenvalue of A^{-1} with associated eigenvector \vec x.

    Things I thought about: roots of det(\lambda I_n-A) are eigenvalues, inverse is calculated by rref[A|In]
    you know that there exists some x such that Ax=\lambda x so that x=A^{-1}\lambda x or \frac{1}{\lambda}x=A^{-1}x (where we used the fact that A is invertible thus \lambda\in R^\times)
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  3. #3
    Member Jskid's Avatar
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    Quote Originally Posted by Drexel28 View Post
    x=A^{-1}\lambda x or \frac{1}{\lambda}x=A^{-1}x
    But you can't divide by a vector or matrices, in this case \lambda? Are you allowed to multiply by its reciprocal?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Jskid View Post
    But you can't divide by a vector or matrices, in this case \lambda? Are you allowed to multiply by its reciprocal?
    Huh? I'm sorry I must be misuderstanding you. Isn't \lambda a scalar?
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  5. #5
    Member Jskid's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Huh? I'm sorry I must be misuderstanding you. Isn't \lambda a scalar?
    I thought it was a vector but it turned out I was confused
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