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Math Help - Matrix Spectra with Troublesome Trigonometric Term

  1. #1
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    Matrix Spectra with Troublesome Trigonometric Term

    Hello!

    I’m trying to find the spectra (eigenvalues) of matrix

    A = \[ \left( \begin{array}{ccc}<br />
cos(x) & -sin(x) & 0  \\<br />
sin(x) & cos(x) & 0 \\<br />
0 & 0 & 1 \end{array} \right)\]

    I start by generating its characteristic polynomial which yields:

    (cos(x)-\lambda)(cos(x)-\lambda)(1-\lambda)+(sin(x))(sin(x))(1-\lambda)=0
    (1-\lambda)[(cos(x)-\lambda)(cos(x)-\lambda)+sin(x)sin(x) ]
    (1-\lambda)[cos(x)^2-2\lambdacos(x)+\lambda^2+sin(x)^2]
    Applying the Euler Formula:
    (1-\lambda)[-2\lambda cos(x) + \lambda^2 + 1]
    (1-\lambda)(\lambda^2 - 2\lambda.cos(x) + 1)

    At this point, I want to extract roots, the solutions for eigenvalues \lambda. One, from the LH term, is clearly \lambda = 1. But that -2cos(x) in the middle is stopping me from factorizing it via any way I can work out. I expect, since A was orthogonal, that the other two roots will be a complex conjugate pair but am not 100% on that – in any case, I can’t see how I can legally break that trig term into a complex one.

    Any ideas?
    Cheers!
    Alexandicity
    Last edited by alexandicity; June 25th 2010 at 02:41 AM.
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  2. #2
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    Quote Originally Posted by alexandicity View Post
    Hello!

    I’m trying to find the spectra (eigenvalues) of matrix

    A = \[ \left( \begin{array}{ccc}<br />
cos(x) & -sin(x) & 0  \\<br />
sin(x) & cos(x) & 0 \\<br />
0 & 0 & 1 \end{array} \right)\]

    I start by generating its characteristic polynomial which yields:

    (cos(x)-\lambda)(cos(x)-\lambda)(1-\lambda)+(sin(x))(sin(x))(1-\lambda)=0
    (1-\lambda)[(cos(x)-\lambda)(cos(x)-\lambda)+sin(x)sin(x) ]
    (1-\lambda)[cos(x)^2-2\lambdacos(x)+\lambda^2+sin(x)^2]
    Applying the Euler Formula:
    (1-\lambda)[-2\lambda cos(x) + \lambda^2 + 1]
    (1-\lambda)(\lambda^2 - 2\lambda.cos(x) + 1)

    At this point, I want to extract roots, the solutions for eigenvalues \lambda. One, from the LH term, is clearly \lambda = 1. But that -2cos(x) in the middle is stopping me from factorizing it via any way I can work out. I expect, since A was orthogonal, that the other two roots will be a complex conjugate pair but am not 100% on that – in any case, I can’t see how I can legally break that trig term into a complex one.

    Any ideas?
    Cheers!
    Alexandicity

    You did everything fine, as far as I can see...! Now, the discriminant of t^2-2\cos x\,\,t+1 is

    \Delta=-4\sin^2x=(2i\sin x)^2 , so using the roots formula for a quadratic eq. we

    get t_{1,2}=\frac{2\cos x\pm 2i\sin x}{2}=\cos x\pm i\sin x=e^{\pm ix}, and here you have your conjugate pair of roots.

    Tonio
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  3. #3
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    Hi Tonio

    Thanks for the quick reply!

    I get most of what you're saying and it looks good. However - one think I don't follow. You identify the discriminant as -4sin^2x. However, when I try to determine it:
    a=1
    b=-2cos(x)
    c=1
    \triangle = b^2-4ac

    Now squaring b gives b^2 = (4cos^2(x)), so \triangle = 4cos^2(x)-4 I'm not sure how it converted to that sine function?

    My guess is I got something wrong with my trig manipulations!
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  4. #4
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    Just a few random thoughts: the eigenvalue problem has to do with operators whose action on an eigenvector is equivalent to multiplying the eigenvector by a scalar. Now, the matrix you've exhibited is a rotation about the z axis through an angle x. Therefore, I would expect any ol' vector along the z axis to be an eigenvector with eigenvalue 1. Any other vector could be an eigenvector with eigenvalue 1 if you rotated through x=2\pi k radians, with k\in\mathbb{Z}. In addition, any vector in the xy plane could be an eigenvector with eigenvalue -1, if you rotate through an angle (2k+1)\pi with k\in\mathbb{Z}. I don't think there are any other eigenvector/eigenvalue pairs. At least, none come to mind, because the fact is, that aside from a vector along the z axis, this rotation is going to change the direction of any other vector, unless x=2\pi k with k\in\mathbb{Z}.

    I just say all this by way of providing a double-check mechanism on your calculations, which seem to be in good hands.
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  5. #5
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    Quote Originally Posted by alexandicity View Post
    Hi Tonio

    Thanks for the quick reply!

    I get most of what you're saying and it looks good. However - one think I don't follow. You identify the discriminant as -4sin^2x. However, when I try to determine it:
    a=1
    b=-2cos(x)
    c=1
    \triangle = b^2-4ac

    Now squaring b gives b^2 = (4cos^2(x)), so \triangle = 4cos^2(x)-4 I'm not sure how it converted to that sine function?

    My guess is I got something wrong with my trig manipulations!

    Ah, that trigonometry! \cos^2x+\sin^2x=1\Longrightarrow \cos^2x-1=-\sin^2x ...

    Tonio
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  6. #6
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    Ah - I should have seen that. Thanks a load!

    Ackbeet - cheers for the background. I've not looked at rotations is much depth yet but will be moving on to it soon. I'll ponder your words!
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