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Thread: invertible operator

  1. #1
    jax
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    invertible operator

    I'm stuck on this h/w problem...any help would be appreciated!!! thank u

    Let V be a vector space of dimension n, and T: V --> V an invertible operator. Prove that if U is a T-invarient subspace of V, then U is also invariant under T^-1. Justify each step.
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  2. #2
    A Plied Mathematician
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    What ideas have you had so far?
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  3. #3
    jax
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    Well I don't know too much about T-invarient subspaces, but I figured it means eigenvalues. So here is my prrof so far-

    First suppose that U is an eigenvalue of T. Thus there exists a nonzero vector v exists in V such that: Tv=Uv
    Appyling T^-1 to both sides of the equation above and we get v=UT^-1v, which is equivantly to the equation T^-1v=(1/U)v. Thus (1/U) is an eigenvalue of T^-1. Therefore proving that if U is a T-invarient subspace of V, then U is also invariant under T^-1.

    What do u think??
    Thank you.
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  4. #4
    A Plied Mathematician
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    No, I don't think you understand what's going on. A $\displaystyle T$-invariant subspace $\displaystyle U$ is a subspace of $\displaystyle V$ that remains the same under action from $\displaystyle T$. That is, $\displaystyle T(U) \subseteq U$. This relation says that $\displaystyle \forall\,u\in U,$ it is the case that $\displaystyle T(u)\in U.$

    Can you write down, now, a mathematical statement that represents what you need to prove?
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  5. #5
    jax
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    ok that helps thank you!! I will play around with the proof now.
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  6. #6
    A Plied Mathematician
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    Ok, let me know how it goes.
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