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Eigenvalues of a rotation matrix

  1. Jan 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the eigenvalues and normalized eigenvectors of the rotation matrix
    cosθ -sinθ
    sinθ cosθ


    2. Relevant equations



    3. The attempt at a solution

    c is short for cosθ, s is short for sinθ
    I tried to solve the characteristic polynomial (c-λ)(c-λ)+s^2=0, and got λ=cosθ±sinθ.
    I then tried to find the eigenvectors, but only got (0, 0) and couldn't find any others. Where did I go wrong?
     
  2. jcsd
  3. Jan 8, 2012 #2

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    What do you get if you write down R-λI?
     
  4. Jan 8, 2012 #3
    I think you get:

    cosθ-λ -sinθ
    sinθ cosθ-λ
     
  5. Jan 8, 2012 #4

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    Oh, I think I see your problem.
    Your eigenvalues should be λ=cosθ ± i sinθ.

    And I meant for you to write down R-λI with λ=cosθ + i sinθ...
     
  6. Jan 8, 2012 #5
    I'm not sure where the i comes from.

    I did this to get the eigenvalues:
    (c-λ)(c-λ)+s^2=0
    c^2-2λc+λ^2+s^2=0
    as C^2+s^2=1,
    λ^2-2cλ+1=0
    λ=[itex]\frac{2c±√(4c^2-4)}{2}[/itex]
    =c±√(c^2-1)=c±s=cosθ±sinθ
     
  7. Jan 8, 2012 #6

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    You have √(c^2-1) = √(cos2θ-1)
    Since cosθ≤1, this means that you're taking the square root of a negative number.
    Hence the "i".


    I was assuming you are supposed to solve this with imaginary numbers.
    If you are supposed to only find real eigenvalues, then there are none.
     
  8. Jan 8, 2012 #7
    Ah, I can't believed I missed such an obvious mistake! Thanks for pointing it out.
     
  9. Jan 8, 2012 #8

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    Okay... so you found your eigenvectors?
     
  10. Jan 8, 2012 #9
    Yes, are they (ia, a) and (-ia, a) for all a?
    If so, how do you normalize imaginary eigenvectors?
     
  11. Jan 8, 2012 #10

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    Good!

    The length of a vector (a,b) in ##\mathbb{C}^2## is given by ##\sqrt{|a|^2 + |b|^2}##.
    Normalization works the same as with real vectors.
     
  12. Jan 8, 2012 #11
    So you normalize a vector by writing √(x^2+y^2)=1?
    But doesn't (ia)^2+a^2=a^2-a^2=0?
     
  13. Jan 8, 2012 #12

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    Not quite, |ia| = |a|.
    The absolute value of a complex number x+iy is given by |x+iy|=√(x^2+y^2).
    For the length of a complex vector, you need to use these absolute values.
     
  14. Jan 8, 2012 #13
    So are they (i√0.5, √0.5) and (-i√0.5, √0.5)?
     
  15. Jan 8, 2012 #14

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  16. Jan 8, 2012 #15
    Thank-you! :-)
     
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