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Math Help - Orthogonal Matrix

  1. #1
    Senior Member slevvio's Avatar
    Joined
    Oct 2007
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    Orthogonal Matrix

    Let \alpha be an imaginary eigenvalue of an orthogonal matrix Q adn let X be a corresponding eigenvector. Prove that  X^T X = 0

    Ok here is what I did:

     X^TQ^{-1}QX = X^T Q^{-1} \alpha X = \alpha X^T Q^{-1}X = \alpha X^T Q^T X = \alpha (QX)^T X

    = \alpha ( \alpha X)^T X = a X \overline{\alpha} X = \alpha \overline{\alpha} X^T X = {|\alpha|}^2 X^T X

    Also  X^TQ^{-1}QX = X^T X

    So  X^T X = {|\alpha|}^2 X^T X
    i.e.  (1-{|\alpha|}^2) X^T X = 0

    i.e.  X^T X = 0 or (1- {|\alpha|}^2) = 0

    Why can't (1- {|\alpha|}^2) = 0 ?
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  2. #2
    Junior Member
    Joined
    Jan 2009
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    Hi, it should be alpha^2 and not |alpha|^2 cause you don't take complex conjugate but rather the transpose only, and because alpha=ai a is real, alpha^2=-a^2 so it cannot be equal 1.
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