In the book I'm reading there is a question,
Prove that given a rotation $R(\phi \hat{n})$ and a reflection $\sigma$, they commute iff $\sigma \perp \hat{n}$ and $\phi=\pi$. (Binary rotation normal to the reflection plane.)
The answer in the back of the book is,
Take $R(\phi z)$. For $\phi=\pi$, the eigenvectors, with eigenvalues in brackets, are: $\vec{z}(+1), \vec{x}(-1), \vec{y}(-1)$. If $\sigma$ is the $\vec{x}\vec{y}$ plane its eigenvectors are $\vec{x}(+1), \vec{y}(+1), \vec{z}(-1)$ and coincide with those of $R(\pi\vec{z})$.
My question:
I thought an eigenvector of a symmetry operation is one that after the symmetry operation $g\vec{r}$ leaves r changed only by the eigenvalues +1 or -1. In my mind, this means that the eigenvectors of a reflection operator $\sigma$ would be all vectors on the reflection plane and normal to the plane. Not just the two particular vectors that span the plane.
Eigenvectors are not unique in general and can be with respect to any eigenvalue. Generally what one does instead is pick a basis for a given eigenspace (the space of eigenvectors with respect to a particular eigenvalue) in the cases where such a basis exists. In this case the eigenspace associated to eigenvalue $1$ is the entire reflection plane, and any basis of this works.
However, in the generic case eigenspaces have dimension $1$ and this issue doesn't occur.