# 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?

Deveno
it don't see why not. the existence of real eigenvalues will depend on the angle of rotation (most angles will give complex eigenvalues).

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.

Deveno
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.

AlephZero
Homework Helper
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"...

chiro
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?