In $V = \mathbb{R}^n$, a reflection through a linear subspace $S \subset V$ of dimension $k \leq n$ can be defined to be the linear map $R : V \rightarrow V$ that satisfies
\begin{equation}
v \in S \implies Rv = v, \quad v \in S^\perp \implies Rv = -v. \qquad (1)
\end{equation}
(This is a generalization of the usual notion of a reflection, in which $k = n – 1$.)
According to the second paragraph of the Wikipedia article on reflections, this definition seems to be equivalent to defining $R$ to be a linear map that is isometric ($\forall v, ||Rv|| = ||v||$) and involutory ($R^2 = 1$). However, I am having trouble coming up with a proof of this statement. The forward direction is easy, since any map that satisfies $(1)$ has a full set of eigenvalues that are $\pm 1$, so the involutory and isometry properties follow trivially. However, how does one prove the converse?
Best Answer
Suppose $R$ is an involutory isometry. Then in particular $R$ is normal (since it is an isometry), so it is diagonalizable over $\mathbb{C}$ and its eigenspaces are orthogonal to each other. But since $R^2=1$, its only eigenvalues can be $1$ and $-1$, so $R$ is in fact diagonalizable over $\mathbb{R}$ and $V$ splits as an orthogonal direct sum of the eigenspaces with eigenvalues $1$ and $-1$. Let $S$ be the eigenspace of $R$ with eigenvalue $1$, so $S^\perp$ is the eigenspace with eigenvalue $-1$. We then have $Rv=v$ for all $v\in S$ and $Rv=-v$ for all $v\in S^\perp$, so $R$ is the reflection through $S$.