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Thread: Diagonalizablilty!! Urgent!! (2)

  1. #1
    Sep 2008

    Diagonalizablilty!! Urgent!! (2)

    Let T be an invertible linear operator on a finite-dimensional vector space V.

    a) Recall that for any eigenvalue λ of T, λ^-1 is an eigenvalue of T^-1. Prove that the eigenspace of T corresponding to λ is the same as the eigenspace of T^-1 corresponding to λ^-1.

    b) Prove that if T is diagonalizable, then T^-1 is diagonalizable
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  2. #2
    MHF Contributor

    Apr 2005
    If $\displaystyle \lambda$ is an eigenvalue of A, then $\displaystyle Ax= \lambda x$ for any x in the eigenspace corresponding to $\displaystyle \lambda$

    Now apply $\displaystyle T^{-1}$ to both sides of that equation.

    (b) is trivial if you think about what "diagonaizable" means.
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