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

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

    I suppose that if we know $\displaystyle det(A - \lambda I_n) = 0$, when the rank of $\displaystyle A - \lambda I_n$ is not full (i.e. - less than $\displaystyle n$). And then we can try to prove that because of this, the rank of $\displaystyle A^{-1} - \lambda^{-1} I_n$ is also less than $\displaystyle n$...?
    Last edited by Last_Singularity; Oct 29th 2009 at 04:38 PM.
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  2. #2
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    If $\displaystyle \lambda$ is an eigenvalue of $\displaystyle T$ then there exists $\displaystyle v\in V$ such that $\displaystyle T(v)=\lambda v$ which happens iff $\displaystyle v= \lambda T ^{-1}(v)$ (since both $\displaystyle T$ and $\displaystyle T^{-1}$ are linear) and since $\displaystyle T$ is invertible it implies that $\displaystyle 0$ can't be an eigenvalue so $\displaystyle \lambda ^{-1} v = T^{-1} (v)$
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