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.