Showing that reflections in hyperplanes does not preserve orientation of isometries

Question

I have as a definition in my geometry notes that an isometry $f:\Bbb R^n \rightarrow \Bbb R^n$ preserves isometry if the determinant of the matrix $[f(e_1)-f(0), ..., f(e_n)-f(0)]$ is greater than $0$.

Now for a generic reflection in a hyperplane $\{ H | x \in \Bbb R^n| a \cdot x=b$} for $a \neq 0, a,b \in \Bbb R^n$, I have the equation $R_H(v) = v + 2 \frac {(b-a\cdot v)}{||a||^2} a$. Suppose $a = \{ a_1,...,a_n \} $. Then I wish to show that reflections in hyperplanes do not preserve orientations.

For the first column of the matrix, I have

$f(e_1)-f(0)=e_1+ 2 \frac {(b-a_1)}{||a||^2} a - (0 + 2 \frac {b}{||a||^2})a = e_1 - \frac {a_1}{||a||^2}a$ . This is where I am not really sure how to proceed. Any hints much appreciated.

Answer 1

It suffices to show it for vector hyperplane symmetries (i.e., for hyperplanes that pass through origin) in $\mathbb{R^n}.$

If such an hyperplane is defined by unit normal vector $U_1$ (identified with a column vector), the matrix $R$ of reflection with respect to it is:

$$\tag{1}R=I_n-2U_1U_1^T$$

(as you are aware). Take $n-1$ vectors $U_2,...U_n$ to be an orthonormal basis of the hyperplane.

Evidently, $(U_1,U_2,\cdots U_n)$ constitute a natural basis of $\mathbb{R^n}.$

Remark: matrix $\Omega:=[U_1,U_2,\cdots U_n]$ is an orthogonal matrix, thus has $$\tag{2}det(\Omega)=\pm1.$$

Let us consider matrix $[RU_1,RU_2,\cdots RU_n]$. It can be written under two forms:

$$\begin{cases}R[U_1,U_2,\cdots U_n] \ \text{(factorization)}\\ [-U_1,U_2,...U_n] \ \text{property of hyperplane symmetry)}\end{cases}$$

We thus have : $det(R[U_1,U_2,\cdots U_n])=det([-U_1,U_2,...U_n])$

Therefore, $det(R)det(\Omega)=-det(\Omega)$

As we can simplify by $det(\Omega) \neq 0$ (see (2)), we can conclude that:

$$det(S)=-1$$