Section 5.2 Hyperplanes and reflections
Definition 5.2.
For each unit vector \(e \in \R^N\) and each \(\lambda \in \R\text{,}\) we define the hyperplane
\begin{equation*}
T_\lambda(e):= \{ \xi \in \R^N : \xi \cdot e =\lambda \}
\end{equation*}
The reflection in \(T_\lambda(e)\) of any point \(x \in \R^N\) is
\begin{equation*}
x^{\lambda,e}:= x+ 2 (\lambda -x \cdot e) e.
\end{equation*}
Lemma 5.3.
If \(v \maps B \to \R\) satisfies \(v(x^{0,e}) = v(x)\) for every unit vector \(e\) and \(x \in B\text{,}\) then \(v\) is radially symmetric.
Proof.
It suffices to show that, for any \(y,z \in B\) with \(\abs y = \abs z\) and \(y \ne z\text{,}\) we have \(v(y) = v(z)\text{.}\) This in turn will follow if we can show that \(z=y^{0,e}\) for some unit vector \(e\text{.}\) We claim that this holds with
\begin{equation*}
e = \frac{z-y}{|z-y|}.
\end{equation*}
To check this, we first calculate
\begin{align*}
y^{0,e} \amp= y + 2(0-y \cdot e )e\\
\amp= y - 2\left(y\cdot \frac{z-y}{|z-y|} \right)
\frac{z-y}{|z-y|}.
\end{align*}
Subtracting \(z\text{,}\) we find
\begin{align*}
y^{0,e} - z \amp= y - z + 2(y-z)y \cdot \frac{z-y}{|z-y|^2}\\
\amp= \frac{y - z}{|z-y|^2}|z-y|^2+ 2(y-z)y \cdot \frac{z-y}{|z-y|^2}\\
\amp= \frac{y-z}{|z-y|^2} \left\{ (|z|^2 - 2y\cdot z + |y|^2) + 2(y \cdot z - |y|^2) \right\}\\
\amp= 0,
\end{align*}
where in the last step we have used \(|y| = |z|\text{.}\)
Exercises Exercises
1. (PS9) Reflections are idempotent.
Let \(\lambda \in \R\) and let \(e \in \R^N\) be a unit vector. The reflection through the plane \(T_\lambda(e) = \{ x : x \cdot e = \lambda \}\) is defined by \(x^{\lambda,e} = x + 2(\lambda - x\cdot e)e\text{.}\) Verify that \((x^{\lambda,e})^{\lambda,e} = x\text{.}\)
Solution.
We calculate
\begin{align*}
(x^{\lambda,e})^{\lambda,e}
\amp= x^{\lambda,e} + 2(\lambda - x^{\lambda,e}\cdot e)e\\
\amp= x + 2(\lambda - x\cdot e)e + 2\Big(\lambda - \big(x + 2(\lambda - x\cdot e)e\big) \cdot e\Big)e\\
\amp= x + 2(\lambda - x\cdot e)e + 2\Big(\lambda - x \cdot e - 2(\lambda - x\cdot e)\Big)e\\
\amp= x + \Big[2(\lambda - x\cdot e) - 2(\lambda - x \cdot e)\Big]e\\
\amp= x\text{.}
\end{align*}
