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Math Help - help with eigenvalue proof

  1. #1
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    help with eigenvalue proof

    Show that if 0 < θ < π then the matrix B=\left(\begin{array}{cc}cos\theta&-sin\theta\\sin\theta&cos\theta\end{array}\right)
    has no real eigenvalues and consequently no eigenvectors.
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  2. #2
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    Quote Originally Posted by chuckienz View Post
    Show that if 0 < θ < π then the matrix B=\left(\begin{array}{cc}cos\theta&-sin\theta\\sin\theta&cos\theta\end{array}\right)
    has no real eigenvalues and consequently no eigenvectors.
    \left(\begin{array}{cc}cos\theta -\lambda &-sin\theta\\sin\theta&cos\theta -\lambda \end{array}\right)

    So we take the determinate of the above matrix to get

    (\cos(\theta)-\lambda)^2+\sin^2(\theta)=0

    \cos^2(\theta)-2\lambda \cos(\theta)+\lambda^2+\sin^2(\theta)=0

    \lambda^2-2\cos(\theta) \lambda+1=0

    This will only have real solutions if the discriminate is posative

    So the discriminate of this quadratic is

    (-2\cos(\theta))^2-4(1)(1)=4\cos^{2}(\theta)-4=4(\cos^2(\theta)-1)

    This is negative for 0< \theta < \pi

    So the equation have no Real solutions
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  3. #3
    Moo
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    FYI, its eigenvalues are e^{i\pi}~,~ e^{-i\pi}
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  4. #4
    Super Member Gamma's Avatar
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    Quote Originally Posted by Moo View Post
    FYI, its eigenvalues are e^{i\pi}~,~ e^{-i\pi}
    e^{i\pi}=-1=e^{-i\pi} \in \mathbb{R}
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  5. #5
    Moo
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    Quote Originally Posted by Gamma View Post
    e^{i\pi}=-1=e^{-i\pi} \in \mathbb{R}
    I meant theta

    e^{i\theta}~,~ e^{-i\theta}
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  6. #6
    Super Member Gamma's Avatar
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    ahhh! Okay, I should have realized that is what you meant. Sorry
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