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Math Help - Eigenvalue related question.

  1. #1
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    Eigenvalue related question.

    if T: V -> V has the property that T^2 has a non negative eigenvalue (lambda)^2, prove that at least one of (lambda) or -(lambda) is an eigenvalue for T.

    The Hint that it gives is the equality:

    T^2 - (lambda)^2 * I = (T + lambda * I)*(T - lambda * I)

    where I is the identity matrix.

    Any help?
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  2. #2
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    Quote Originally Posted by hashshashin715 View Post
    if T: V -> V has the property that T^2 has a non negative eigenvalue (lambda)^2, prove that at least one of (lambda) or -(lambda) is an eigenvalue for T.

    The Hint that it gives is the equality:

    T^2 - (lambda)^2 * I = (T + lambda * I)*(T - lambda * I)

    where I is the identity matrix.

    Any help?
    What is T?
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  3. #3
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    T is the linear transformation. Like as in T(x) = (lambda)x. But we are given T^2(x), which I think is T(T(x)) = (lambda)^2 * x
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by hashshashin715 View Post
    if T: V -> V has the property that T^2 has a non negative eigenvalue (lambda)^2, prove that at least one of (lambda) or -(lambda) is an eigenvalue for T.

    The Hint that it gives is the equality:

    T^2 - (lambda)^2 * I = (T + lambda * I)*(T - lambda * I)
    By hypothesis,

    \exists x\in V\;(x\neq 0):T^2(x)=\lambda^2 x

    equivalently

    T^2-\lambda^2 I)(x)=0" alt="\exists x\in V\;(x\neq 0)T^2-\lambda^2 I)(x)=0" />.

    Now, use:

    (T+\lambda I)[ (T-\lambda I)(x)]=0

    and analyze what would happen if

    (T-\lambda I)(x)\neq 0 .



    Fernando Revilla
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