等距变换表为平面反射的复合

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王长平《几何学》，预印讲义(Web Archive)
命题 1.10 如果一个空间等距变换至少有一个不动点, 则它可以表成至多三个平面反射的复合.
证明 设 $O$ 是等距变换 $\phi$ 的不动点. 如果 $\phi$ 是恒同映射, 则 $\phi=\Sigma \circ \Sigma$. 如果 $\phi$ 不是恒同映射, 则存在 $P$ 使得 $\phi(P) \neq P$. 由于
$$
d(O, P)=d(\phi(O), \phi(P))=d(O, \phi(P)),
$$
所以 $\triangle O P \phi(P)$ 是一个等腰三角形. 令 $\Sigma$ 是与此三角形垂直, 并过 $O$ 点及线段 $\overline{P \phi(P)}$ 中点的平面. 则有 $O$ 和 $P$ 是等距变换 $\Sigma \circ \phi$ 的两个不动点. 由命题 1.8 知, $\Sigma \circ \phi$ 至多可以表成两个平面反射的复合. 故 $\phi$ 可以表成至多三个平面反射的复合.

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知乎

由此可得，所有的等距变换都至多由三个反射复合而成，得到旋转（包括恒等变换）、平移或平移反射（包括反射）。偶数次反射复合得到旋转和平移，保持定向；而奇数次反射复合得到平移反射，改变定向。

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MSE

1. We use one reflection to get an isometry with $f(0)=0$. I will now assume that $f$ has this property and will show that $n$ reflections suffice. edit: I should probably say at this point that the reason for step 1 is to use that $f$ is now norm-preserving (and thereby distance preserving) which simplifies the notation.

2. I will inductively construct reflections $r_i$ such that their composition $r_1\circ\cdots\circ r_n$ is the inverse of $f$. Since reflections are their own inverse this proves the statement.

3. Assume that $f$ fixes a subspace $N$ of dimension $d$ (and this subspace is maximal in the sense that there exists no higher dimensional subspace which is fixed). We may then write $\mathbb R^n=N\bot N^\bot$. I will show that we can find a reflection $r$, such that $r\circ f$ fixes a subspace of one dimension higher.

4. Take $0\neq x\in N^\bot$. Then $x\neq f(x)\in N^\bot$ since $0=\left< x,n\right>=\left< f(x),f(n)\right>=\left< f(x),n\right>$ for all $n\in N$. Moreover $f(x)$ is either linearly independent to $x$ or just $-x$. Indeed assume $\lambda f(x)=x$, then $\left< x,x\right>=\left< f(x),f(x)\right>=\lambda^2\left< x,x\right>$ and $\lambda^2=1$ with $\lambda\neq 1$ by assumption. If $f(x)=-x$, we just reflect $x$ and keep $N$ fixed at the same time and are done. Hence assume $f(x)$ and $x$ are linear independent.

5. Find a basis of $N^\bot$ containing $x$ and $f(x)$ namely ${x,f(x),b_1,...,b_k}$. Choose $u=x+f(x)$ and $v=x-f(x)$. Then ${u,v,b_1,...,b_k}$ is also a basis of $N^\bot$, since $x=\frac12(u+v)$ and $f(x)=\frac12(u-v)$.

6. Take $r$ to be the reflection which takes $v$ to $-v$ and fixes everthing else. Then $r\circ f$ fixes the $d+1$-dimensional space $N\bot \mathbb Rx$.

7. After $n$ steps we have (at most) $n$ reflections such that $r_1\circ...\circ r_n\circ f$ is the identity. We are done.

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Wikipedia

A Householder reflection is constructed from a non-null vector v as $ {\displaystyle Q=I-2{\frac {{\mathbf {v} }{\mathbf {v} }^{\mathrm {T} }}{{\mathbf {v} }^{\mathrm {T} }{\mathbf {v} }}}.} $

Here the numerator is a symmetric matrix while the denominator is a number, the squared magnitude of v. This is a reflection in the hyperplane perpendicular to v (negating any vector component parallel to v). If v is a unit vector, then Q = I − 2vv^T suffices. A Householder reflection is typically used to simultaneously zero the lower part of a column.

Any orthogonal matrix of size n × n can be constructed as a product of at most n such reflections.

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Keith Conrad - Isometry of $\mathbf R^n$
Corollary A.7. Every isometry of $\mathbf{R}^n$ is a composition of at most $n+1$ reflections. An isometry that fixes at least one point is a composition of at most $n$ reflections.

The difference between this corollary and Cartan's theorem is that in the corollary we are not assuming isometries, or in particular reflections, are taken from $\mathrm{O}_n(\mathbf{R})$, i.e., they need not fix $\mathbf{0}$.

Proof. Let $h$ be an isometry of $\mathbf{R}^n$. If $h(\mathbf{0})=\mathbf{0}$, then $h$ belongs to $\mathrm{O}_n(\mathbf{R})$ (Theorem 2.4) and Cartan's theorem implies $h$ is a composition of at most $n$ reflections through hyperplanes containing 0 . If $h(p)=p$ for some $p \in \mathbf{R}^n$, then we can change the coordinate system (using a translation) so that the origin is placed at $p$. Then the previous case shows $h$ is a composition of at most $n$ reflections through hyperplanes containing $p$.

Suppose $h$ has no fixed points. Then in particular, $h(\mathbf{0}) \neq \mathbf{0}$. By Theorem A.3 there is some reflection $s$ across a hyperplane in $\mathbf{R}^n$ such that $s(h(\mathbf{0}))=\mathbf{0}$. Then $s h \in \mathrm{O}_n(\mathbf{R})$, so by Cartan's theorem $s h$ is a composition of at most $n$ reflections, and that implies $h=s(s h)$ is a composition of at most $n+1$ reflections.