Let $(V, \langle,\rangle) $ be an euclidean Vector space. For $w \in V, w\ne 0$, the map $s_w: V \rightarrow V$, $s_w (v): = v -2 \tfrac{\langle v,w \rangle}{\langle w,w \rangle}$ is defined (the mapping $s_w$ is the reflection at the hyperplane ${v ∈ V | \langle v,w \rangle = 0}$).
I want to proof that every orthogonal Endomorphism is a composition of such reflections.
First I formalized the statement: For every orthogonal Endomorphism $f: V\rightarrow V$, there are $w_1,\dots, w_r \in V, r ≤ n$ with $f = s_{w_1} ◦. . . ◦ s_{w_r}$.
I am not sure but in order to proof this statement, I had in mind that I could find a $w_1 \in V$ such that $\widetilde{f} = s_{w1} ◦ f$ is a straight line through the origin and then consider the restriction of $\widetilde{f}$ on the orthogonal complement of the straight line.
Maybe someone can share some thoughts on this or suggest an easier way. Thanks for any kind of advice!
Proof: For $f = id_V$ the assertion is clear ("composition of 0 reflections"). Let now $v \in V$ with $f(v) \neq v$. Set $w_1 := f(v) - v$. Set $\widetilde{f}:= s_{w_1} ◦ f$. Then $\widetilde{f}(v) = v$ is true.
Inductively, we can conclude that the assertion holds for the orthogonal complement of $v$. (Due to the orthogonality of $f$, $f$ induces an endomorphism on the orthogonal complement of $v$.)
We can thus write $\widetilde{f}|_{L(v)^⊥} = s_{w_2} ◦ \dots ◦ s_{w_r}$. The reflections $s_{w_i}$ on $L(v)^⊥$ can be extended to reflections on $V$ by writing $s_{w_i}(v) = v$ for all $ i \in \{2, . . . , r\}$. With $s_{w_1} = s_{w_1}^{−1}$ we obtain: $$f = s_{w_1} ◦ s_{w_2} ◦ \dots ◦ s_{w_r}$$