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Eigenvalue of 3D rotation matrix

  1. Dec 15, 2009 #1
    It was pretty cool to stumble upon Euler's formula as the eigenvalues of the rotation matrix.

    det(Rot - kI) = (cos t - k)2 + sin2t
    =k2-2(cos t)k + cos2t + sin2t
    =k2-2(cos t)k + 1

    k = {2cos t +/- [tex]\sqrt{4cos^2(t) - 4}[/tex]}/2
    k = cos t +/- [tex]\sqrt{cos^2(t) - 1}[/tex]
    k = cos t +/- [tex]\sqrt{cos^2(t) - cos^2t - sin^2(t)}[/tex]
    k = cos t +/- [tex]\sqrt{-sin^2(t)}[/tex]
    k = cos t +/- i sin t = e(+/-)it

    I was wondering what the eigenvalues are for the rotation matrix in 3D, and if there's a 3D equivalent to Euler's formula.
     
  2. jcsd
  3. Dec 16, 2009 #2

    CompuChip

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    Actually, I suppose it is not that strange when you look at a rotation as complex multiplication.
    If you write a two-dimensional vector v = (x, y) as z = x + iy, then rotation over an angle t can be written either as R(t) v, where R is the 2d rotation matrix. But you can also write it as eit z.

    A 3D rotation matrix, in the appropriate basis, looks like
    [tex]R_3(t) = \begin{pmatrix} R_2(t) & 0 \\ 0 & 1 \end{pmatrix}[/tex]
    where R2(t) is the 2-dimensional rotation matrix and the z-axis is the rotation axis (i.e. rotation in the (x,y) plane). So, by expansion along the last row or column,
    det(R3 - k I) = (1 - k) det(R2 - k I)
    which you can work out in terms of your previous result. (And then add that an arbitrary change of basis does not alter the eigenvalues).
     
  4. Dec 16, 2009 #3
    In addition to the two imaginary eigenvalues, the most intuitive one (for me) is the real valued one -- it's 1, with eigenvector along the axis of rotation. You can see that one just by noting that rotating won't change a vector along the axis of rotation.

    edit:
    Quaternions could be considered an extension of Euler's formula ...
     
    Last edited: Dec 16, 2009
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