Eigenvalues of a rotation matrix

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The discussion focuses on finding the eigenvalues and normalized eigenvectors of a rotation matrix defined by cosθ and sinθ. The initial attempt at solving the characteristic polynomial led to incorrect eigenvalues, λ=cosθ±sinθ, due to a misunderstanding of the complex nature of the eigenvalues. The correct eigenvalues are λ=cosθ ± i sinθ, indicating the presence of imaginary numbers. Normalization of the resulting complex eigenvectors is addressed, emphasizing the use of absolute values for complex numbers. The conversation concludes with clarification on the normalization process for these eigenvectors.
Lucy Yeats
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Homework Statement


Find the eigenvalues and normalized eigenvectors of the rotation matrix
cosθ -sinθ
sinθ cosθ


Homework Equations





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?
 
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What do you get if you write down R-λI?
 
I think you get:

cosθ-λ -sinθ
sinθ cosθ-λ
 
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...
 
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=\frac{2c±√(4c^2-4)}{2}
=c±√(c^2-1)=c±s=cosθ±sinθ
 
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.
 
Ah, I can't believed I missed such an obvious mistake! Thanks for pointing it out.
 
Okay... so you found your eigenvectors?
 
Yes, are they (ia, a) and (-ia, a) for all a?
If so, how do you normalize imaginary eigenvectors?
 
  • #10
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.
 
  • #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?
 
  • #12
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.
 
  • #13
So are they (i√0.5, √0.5) and (-i√0.5, √0.5)?
 
  • #14
Yep! :smile:
 
  • #15
Thank-you! :-)
 

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