I know that in 2-dimension, a rotation matrix has no non-zero eigenvector.
But in 3-d space, I can only imagine the rotation axis to be an eigenvector, and it looks like that the rotation angle can't be represent through eigenvectors or eigenvalues. So I guess in 3-dimension, any rotation matrix has and only has one eigenvector. Am I right?
[Math] Does any rotation matrix in 3-d space have only one non-zero eigenvector
linear algebramatrices
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Best Answer
In two dimension the result you quoted is false as stated: the matrix $$ \begin{pmatrix} -1 & 0 \\ 0 & -1\end{pmatrix} = \begin{pmatrix} \cos \pi & \sin \pi \\ -\sin \pi & \cos \pi \end{pmatrix} $$ is a rotation matrix. And every vector is an eigenvector.
It is true, however, if you explicitly disallow this particular case.
In three dimensions, note that since rotations preserves vector norms, you have that if $v$ is an eigenvector of a rotation $A$ you must have $Av = \pm v$. Now supposing you have two linearly independent eigenvectors $v$ and $w$. Let $u$ be the unique vector orthogonal to $v$ and $w$. Then you have $$ (Au)^Tv = u^T A v = u^T (\pm v) = 0 $$ and similarly $$ (Au)^Tw = 0 $$ and hence you get that $Au$ is proportional to $u$, and hence you must have $u$ is an eigenvector also.
This means that if a rotation matrix has more than 1 eigendirections, it must have a set of three linearly independent eigendirections.