Eigenvalues of reflection Why are the eigenvalues of a reflection $Rx=\rho x$ in a $n$-dimensional vector space just $\lambda=-1,1$? I can't seem to convince myself of this.
 A: Suppose that $R$ reflects through an $(n-1)$-dimensional hyperplane $H$ in $\mathbb{R}^n$.  Consider a basis of $\mathbb{R}^n$ consisting of $n-1$ linearly independent vectors in $H$ and a single vector that is orthogonal to $H$.  (Can you prove that these $n$ vectors do actually form a basis of $\mathbb{R}^n$?)  By considering the effect of $R$ on this basis, you will see that it is a basis of eigenvectors and you will be able to determine the eigenvalues (along with their geometric multiplicities).
A: $R\circ R$ is the identity, hence the square of an eigenvalue of $R$ must be an eigenvalue of the identity.
A: One of the most basic things to know about eigenvalues of linear operators (such as $R$ in the question) is that if the operator satisfies a polynomial equation, then its eigenvalues (if any) must satisfy the same polynomial equation. Here that equation is $R^2=I$, so any eigenvalue$~\lambda$ must satisfy $\lambda^2=1$.

This is of course just repeating the answer by Hagen van Eitzen, but I wanted to stress the general principle going on here, which has nothing to do with reflections in particular. By the same argument eigenvalues of a projection$~P$ (which satisfies $P^2=P$) must be roots of $X^2-X$, and eigenvalues of an operator$~\phi$ of order$~n$ (i.e., with $\phi^n=I$) can only be $n$-th roots of unity. In general this allows finding eigenvalues of "special" linear operators (satisfying some given polynomial relation) without having to bother about their characteristic polynomials. There is no guarantee however that all roots allowed by the relation will in fact be eigenvalues (though this is the case for reflections in dimension $n\geq2$).
A: Reflection preserves length, and any other eigenvalue would mean the reflection is shrinking or stretching a vector.
