## rotation matrix vs regular matrix

Can you calculate eigenvalues and eigenvectors for rotation matrices the same way you would for a regular matrix?

If not, what has to be done differently?
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 Recognitions: Science Advisor it don't see why not. the existence of real eigenvalues will depend on the angle of rotation (most angles will give complex eigenvalues).
 The determinant of a rotation matrix should always be 1 (since it preserves length) so there should always be eigenvalues and eigenvectors that can be calculated given a rotation matrix.

Recognitions:

## rotation matrix vs regular matrix

 Quote by chiro The determinant of a rotation matrix should always be 1 (since it preserves length) so there should always be eigenvalues and eigenvectors that can be calculated given a rotation matrix.
the characteristic polynomial of:

$$\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}$$

is:

$$x^2 - (2\cos\theta)x + 1$$

which has real solutions only when:

$$4\cos^2\theta - 4 \geq 0 \implies \cos\theta = \pm 1$$

for angles that aren't integer multiples of pi, this will lead to complex eigenvalues.

Recognitions:
 Quote by Deveno the characteristic polynomial of: $$\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}$$ is: $$x^2 - (2\cos\theta)x + 1$$
... and the eigenvalues are $\cos\theta \pm i \sin\theta$. Now, I wonder what that fact might have to do with "rotation"...

 Quote by Deveno the characteristic polynomial of: $$\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}$$ is: $$x^2 - (2\cos\theta)x + 1$$ which has real solutions only when: $$4\cos^2\theta - 4 \geq 0 \implies \cos\theta = \pm 1$$ for angles that aren't integer multiples of pi, this will lead to complex eigenvalues.
What has that got to do with what I said?

 Tags calculate, eigenvalue, eigenvector, matrix, rotation