Eigenvalues of a rotation matrix

1. Jan 8, 2012

Lucy Yeats

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. Jan 8, 2012

I like Serena

What do you get if you write down R-λI?

3. Jan 8, 2012

Lucy Yeats

I think you get:

cosθ-λ -sinθ
sinθ cosθ-λ

4. Jan 8, 2012

I like Serena

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...

5. Jan 8, 2012

Lucy Yeats

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

6. Jan 8, 2012

I like Serena

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.

7. Jan 8, 2012

Lucy Yeats

Ah, I can't believed I missed such an obvious mistake! Thanks for pointing it out.

8. Jan 8, 2012

I like Serena

Okay... so you found your eigenvectors?

9. Jan 8, 2012

Lucy Yeats

Yes, are they (ia, a) and (-ia, a) for all a?
If so, how do you normalize imaginary eigenvectors?

10. Jan 8, 2012

I like Serena

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. Jan 8, 2012

Lucy Yeats

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. Jan 8, 2012

I like Serena

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. Jan 8, 2012

Lucy Yeats

So are they (i√0.5, √0.5) and (-i√0.5, √0.5)?

14. Jan 8, 2012

I like Serena

Yep!

15. Jan 8, 2012

Lucy Yeats

Thank-you! :-)