# The eigenvectors of a reflection and rotation symmetry operation

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.

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The given statement is false: If $\sigma\perp\hat n$, the rotation angle may be anything. Furthermore there is an extra case: If the rotation axis $\hat n$ lies in the reflection plane $\sigma$ and $\phi=\pi$ then the rotation and the reflection do also commute. Furthermore the "proof in the book" only proves the easy half of the (wrong) statement. – Christian Blatter Feb 15 '11 at 11:55
Well I'm glad I didn't understand the question so I could find out this wasn't a true statement anyway! I drew out a picture and quickly verified what you said is correct. Thank you. – coderdave Feb 15 '11 at 20:46

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.