Confusions about the proof of representations of isometries as products of reflections. I am reading a proof of reflection representations of isometries that fix the origin in $\mathbb{R}^n$. It is a simple induction on the dimension $n$. For $f(0)=0$ we have some $v\neq w=f(v)$. The reflection in the hyperplane orthogonal to $u=v-w$, $r_u$, suffices $r_uf(u)=-u$ and therefore $r_uf(v)=v$. How do we know $\mathbb{R}v$ is left pointwise invariant under $r_uf$? Do we need to know that $r_uf$ is affine? *Edit: we could know that any isometry is affine without knowing whether it is a product of reflections.
Now $r_uf$ is the identity on $\mathbb{R}v$. Then it seems that I need to prove $r_uf$ is also an isometry on the hyperplane that goes through $O$ and orthogonal to $\mathbb{R}v$. I find this very difficult to understand. It seems to me that this is a result of the fact that $r_uf$ is the identity on $\mathbb{R}v$. But I could approach this conclusion. Why is $r_uf$ an isometry on the hyperplane?

These questions perplexed me when I am reading Stillwell's Naive Lie Theory. I tried to do some online studying but I couldn't understand the most essential parts of the proofs:

*

*WimC's answer in this thread: Proving that every isometry of $\mathbb{R}^n$ is of the form of a composition of at most $n+1$ reflections

*Hagen von Eitzen's claim that $r_uf$ is an identity on $\mathbb{R}v$: reflection representation of isometry
My linear algebra background is poor, so can somebody explain this to me with some easy concepts and notations? Thanks in advance. Any help will be appreciated.
 A: I assume that you are aware of the fact that the "isometries of $\Bbb R^n$ that fix the origin" are exactly the "linear orthogonal transformation in $\mathrm O(\Bbb R^n)$".

So, besides $u=v-w$, let us define $u'=v+w$, which satisfies
$$\langle u,u'\rangle  = \langle v-w,v+w\rangle = \|v\|^2-\|w\|^2 = 0.$$
We used that $f$ is an isometry via $\|w\|=\|w-0\|=\|f(v)-f(0)\| = \|v-0\|=\|v\|$.
So we see that $u'$ must be contained in the reflection hyperplane of $r_u$, and so we have $r_uu'=u'$. Thus
$$r_u f(v) = r_u w = r_u(-u/2+u'/2) = u/2 + u'/2 = v.$$
So indeed, $r_u f$ fixes $v$, and since $r_u$ and $f$ are linear, it also fixes $\Bbb R v$ pointwise.

The next is a general result from representation theory (of finite groups), but it applies to single transformations as well:
If an orthogonal map, let's call it $r\in\mathrm O(\Bbb R^n)$, fixes a subspace $U\subseteq\Bbb R^n$ setwise, then it also fixes its orthogonal complement $U^\bot$ setwise: for every $u\in U$ and $u'\in U^\bot$, we have $r^{-1}u\in U$ and
$$\langle u, ru'\rangle = \langle r^{-1}u,u'\rangle = 0.$$
Thus $r u'\in U^\bot$.
So since $r_u f$ fixes $\Bbb Rv$ pointwise (but also setwise), it also fixes the orthogonal complement of that (setwise), in particular, it restricts to an orthogonal map on this complement, and is thus an isometry on it.
