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Math Help - Lambda is eigenvalue of T if and only if inverse lambda is eigenvalue of inverse T

  1. #1
    Member Last_Singularity's Avatar
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    Lambda is eigenvalue of T if and only if inverse lambda is eigenvalue of inverse T

    Question: Suppose that T is an invertible linear operator. Prove that \lambda is an eigenvalue of T if and only if \lambda^{-1} is an eigenvalue of T^{-1}.

    Let A be defined as an n \times n matrix such that T(x) = Ax. Then I tried approaching this by demonstrating that
    det(A - \lambda I_n) = det(A^{-1} - \lambda^{-1} I_n) = 0
    But the only information that I have is AA^{-1}=A^{-1} A = I_n - how does that get me to the desired result?

    I suppose that if we know det(A - \lambda I_n) = 0, when the rank of A - \lambda I_n is not full (i.e. - less than n). And then we can try to prove that because of this, the rank of A^{-1} - \lambda^{-1} I_n is also less than n...?
    Last edited by Last_Singularity; October 29th 2009 at 05:38 PM.
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  2. #2
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    If \lambda is an eigenvalue of T then there exists v\in V such that T(v)=\lambda v which happens iff v= \lambda T ^{-1}(v) (since both T and T^{-1} are linear) and since T is invertible it implies that 0 can't be an eigenvalue so \lambda ^{-1} v = T^{-1} (v)
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