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Thread: Eigenvalues involving trig functions and imaginary/complex numbers

  1. #1
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    Eigenvalues involving trig functions and imaginary/complex numbers

    Hi overlords.

    I have a pretty grand request if someone has the time. In my revision for an upcoming exam there is a fairly advanced (for intro level LA) question regarding eigenvectors. Now, I'm very familiar with what eigenvectors are and how to obtain them, however because of my poor math background I am really unfamiliar with factorising trig functions and have literally no idea how to operate with imaginary/complex numbers. Could someone possibly give a 10 seconds lesson on how to approach this question after the determinant has been found? Everything above the blue line I understand, everything else I need some help.

    I understand this is quite a task, but I don't have the time to fully learn all of the above, as well as revise for everything else in time for exams, so I figured I'd ask here as a last resort. No feelings hurt if no takers
    Eigenvalues involving trig functions and imaginary/complex numbers-q12.pngEigenvalues involving trig functions and imaginary/complex numbers-a12.png
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  2. #2
    Senior Member zzephod's Avatar
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    Re: Eigenvalues involving trig functions and imaginary/complex numbers

    The first line under you blue line is obtained by expanding the term $(\cos \theta-\lambda)^2=\cos^2 \theta -2\lambda \cos \theta+\lambda^2$ on the line above.

    Then you need to use the trig-identity corresponding to Pythagoras' theorem $\cos^2 \theta +\sin^2 \theta =1$ to eliminate the squared trig functions:

    $$\begin{array}{ll} (\cos \theta-\lambda)^2+\sin^2 \theta &=\cos^2 \theta -2\lambda \cos \theta+\lambda^2+\sin^2 \theta\\
    &=1-2\lambda \cos \theta+\lambda^2
    \end{array}$$

    To find the roots of $1-2\lambda \cos \theta+\lambda^2$ you use the quadratic formula to get:
    $$
    \begin{array}{ll}
    \lambda &=\frac{2\cos(\theta \pm \sqrt{4 \cos^2 \theta -4}}{2}\\
    &=\frac{2\cos\theta \pm 2\sqrt{ \cos^2 \theta -1}}{2}\\
    &=\cos\theta \pm \sqrt{ \cos^2 \theta -1}
    \end{array}
    $$

    Now we use the same trig-identity we used before $\cos^2 \theta+\sin \theta=1$, so $\cos^2 \theta -1=-\sin \theta$, so we have:
    $$
    \lambda=\cos\theta \pm \sqrt{-\sin^2 \theta}=\cos \theta \pm \rm{i} \sin \theta
    $$
    Then that $\cos$ is an even function and $\sin$ and odd function means that:
    $$
    \lambda=\cos \theta + \rm{i} \sin \pm \theta
    $$
    Last edited by zzephod; Jun 12th 2017 at 06:07 AM.
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