Rotation matrix vs regular matrix

[dmoney123]

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]

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?