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How do I show that the rotation by a non zero angle $\theta$ in $\mathbb{R}^2 $ does not have any real eigenvalues. I know the matrix of a rotation but I don't how to show the above proposition.

Thank you

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3 Answers 3

up vote 2 down vote accepted

The characteristical polynomial is $$x^2-2\cos(\theta) x+1$$ and $x^2\geq 0$ and $1>0$ and $|\cos(\theta)|\leq 1$ the polynomial can only have a zero when $|\cos(\theta)|=1$.
As $x^2-2x+1=(x-1)^2$ and $x^2+2x+1=(x+1)^2$

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2  
That's not the characteristic polynomial. –  Jim Mar 9 '13 at 19:59
    
@Jim thanks fixed it –  Dominic Michaelis Mar 9 '13 at 20:02

The answer depends on $\theta$. For example if $\theta = 0$ then $1$ is an eigenvalue and if $\theta = \pi$ then $-1$ is an eigenvalue. In general the characteristic polynomial of the rotation matrix is $$x^2 - 2\cos(\theta)x + 1$$ To find out if this polynomial has real roots we check the discriminant: $$4(\cos^2(\theta) - 1)$$ This needs to be non-negative and you should be able to find out for exactly which values of $\theta$ this is the case.

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Let's do't without calculating the characteristic polynomial!

Remember that a putative eigenvector cannot be the zero vector, and that an eigenvector belonging to a real eigenvalue can be scaled to have real components.

But if you rotate a vector in the real plane by an angle that is not an integer multiple of 180 degrees, the rotated version will not be parallel to the unrotated version. Therefore no such eigenvector exists (unless $\theta=n\cdot\pi$).

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