Every reflection is an isometry proof

Question

The theorem is that every reflection $R_{S}$ in an affine subspace $S$ of $\mathbb{E}^{n}$ is an isometry:

$R_S:\ \mathbb{E}^{n} \rightarrow \mathbb{E}^{n}:\ x \mapsto R_{S}(x) = x + 2 \overrightarrow{x\pi_{S}(x)}$

I'm horrendously stuck with the proof. I get that I'm trying to prove that $R_{S}$ preserves the distance between the two points, but I'm lost as to how.

EDIT:

$\pi_{S}(x)$ is defined as the intersection of $S$ and the euclidean subspace $T_x$ through $x$, perpendicular to $S$.

Answer 1

Let $\{u_1,\dotsc,u_k\}$ be an orthonormal basis for $T_x$. Now pick a point $p$ on $S$. $\pi_S(x)$ can then be written as follows: $$ \pi_S(x)$ = x + \sum_{i=1}^k ((p-x) \cdot u_i)u_i$$

For all points $x$ and $y$ of $\mathbb{E}^n$, watch what happens to the distance:

First (for simplicity later), I will work out the vector between the reflection of $x$ and the reflection of $y$.

$$ \begin{array}{rll} R_{S}(x) - R_{S}(y) &= (x + 2\overrightarrow{x\pi_{S}(x)}) - (y + 2\overrightarrow{y\pi_{S}(y)}) &\\ &= (x + 2(\pi_{S}(x) - x)) - (y + 2(\pi_{S}(y) - y)) &\\ &= (-x + 2\pi_{S}(x)) - (-y + 2\pi_{S}(y)) &\\ &= (-x + 2(x + \sum_{i=1}^{k}((p-x)\cdot u_{i})u_{i})) - (-y + 2(y + \sum_{i=1}^{k}((p-y)\cdot u_{i})u_{i})) &\\ &= (x + 2\sum_{i=1}^{k}((p-x)\cdot u_{i})u_{i}) - (y + 2\sum_{i=1}^{k}((p-y)\cdot u_{i})u_{i}) &\\ &= (x-y) + 2\left(\sum_{i=1}^{k}((p-x)\cdot u_{i})u_{i} - \sum_{i=1}^{k}((p-y)\cdot u_{i})u_{i}\right) &\\ &= (x-y) + 2\left(\sum_{i=1}^{k}\left(((p-x)\cdot u_{i})u_{i} - ((p-y)\cdot u_{i})u_{i}\right)\right) &\\ &= (x-y) + 2\left(\sum_{i=1}^{k}\left((((p-x)\cdot u_{i})- ((p-y)\cdot u_{i}))u_{i}\right)\right) &\\ &= (x-y) + 2\left(\sum_{i=1}^{k}(((p-x)-(p-y))\cdot u_{i} )u_{i}\right) &\\ &= (x-y) + 2\left(\sum_{i=1}^{k}((y-x)\cdot u_{i} )u_{i}\right) &\\ &= (x-y) - 2\left(\sum_{i=1}^{k}((x-y)\cdot u_{i} )u_{i}\right) &\\ \end{array}$$ Now we will look at the quadrate of the distance betwee the reflection of the points: $$ \begin{array}{rll} d(R_{S}(x),R_{S}(y))^2 &= \left\| R_{S}(x) - R_{S}(y)\right\|^{2} & \\ &= (R_{S}(x) - R_{S}(y))^{2} &\\ &= \left((x-y) - 2\left(\sum_{i=1}^{k}((x-y)\cdot u_{i} )u_{i}\right)\right)^{2} &\\ &= (x-y)^{2} -4(x-y)\left(\sum_{i=1}^{k}((x-y)\cdot u_{i} )u_{i}\right) + 4\left(\sum_{i=1}^{k}((x-y)\cdot u_{i} )u_{i}\right)^{2} &\\ &= (x-y)^{2} -4(x-y)\left(\sum_{i=1}^{k}u_{i}((x-y)\cdot u_{i} )\right) + 4\left(\sum_{i=1}^{k}((x-y)\cdot u_{i} )u_{i}\right)^{2} &\\ &= (x-y)^{2} -4\left(\sum_{i=1}^{k}((x-y) \cdot u_{i})\cdot((x-y)\cdot u_{i} )\right) + 4\left(\sum_{i=1}^{k}((x-y)\cdot u_{i} )u_{i}\right)^{2} &\\ &= (x-y)^{2} -4\left(\sum_{i=1}^{k}((x-y)\cdot u_{i} )^{2}\right) + 4\left(\sum_{i=1}^{k}((x-y)\cdot u_{i} )^{2}\right) &\\ &= (x-y)^{2} &= d(x,y)^2 \end{array}$$

Notice that the third to last equality only holds because the $u_i$ form an orthonormal basis.

Answer 2

Here's a partial answer which is suitable for Euclidean space (not quite as general as an affine space). Given a vector $a$ in Euclidean space $\mathbb{R}^n$, the formula for the reflection in the hyperplane through the origin, orthogonal to $a$, is given by

$$\text{Ref}_a(v) = v - 2\frac{v\cdot a}{a\cdot a}a$$

Recall that the distance between two vectors $v$ and $w$ is $d(v,w)=|v-w|$. So we can show that

$$d(\text{Ref}_a(v),\text{Ref}_a(w))=| v - 2\frac{v\cdot a}{a\cdot a}a - (w - 2\frac{w\cdot a}{a\cdot a}a)| = |v-w+2(wa-va)\frac{a}{a\cdot a}| = $$

$$=|v-w+2(w-v)\frac{a\cdot a}{a\cdot a}|=d(v,w)$$

Answer 3

Pick an origin for $\Bbb E^n$, and call the resulting vector space $\Bbb R^n$. To save notation, let's put the origin in $S$, so that $S$ now becomes a (vector) subspace. Write $\Bbb R^n=S\oplus S^\perp$. Now, identifying points $x\in\Bbb E^n$ with the corresponding vectors in $\Bbb R^n$, write $x=x_1+x_2$, where $x_1\in S$ and $x_2\in S^\perp$. Then $\pi_S(x)=x_1$ and $x-\pi_S(x) = x_2\in S^\perp$.

Here's a lemma you need to prove, using dot products. (It's just a higher-dimensional Pythagorean Theorem.)

Lemma: Writing $x=x_1+x_2$, $x_1\in S$, $x_2\in S^\perp$, we have $\|x\|^2 = \|x_1\|^2 + \|x_2\|^2$.

The vector $\overrightarrow{x\pi_S(x)}$ can now be written as $\pi_S(x)-x=-x_2$, so our formula for $R_S$ becomes $R_S(x) = (x_1+x_2)+2(-x_2)= x_1-x_2$. Now it's clear that for any $x,y\in\Bbb R^n$, we have $$R_S(x)-R_S(y) = (x_1-x_2)-(y_1-y_2) = (x_1-y_1) + (y_2-x_2).$$ Because $x_1-y_1\in S$ and $y_2-x_2\in S^\perp$, we have \begin{align*} \|R_S(x)-R_S(y)\|^2 &= \|x_1-y_1\|^2 + \|y_2-x_2\|^2 = \|x_1-y_1\|^2 + \|x_2-y_2\|^2 \\&= \|(x_1+x_2)-(y_1+y_2)\|^2 = \|x-y\|^2, \end{align*} as desired.