There is at most one orthogonal reflection for a given reflection line

Question

$ \def\char{\operatorname{char}} \def\id{\operatorname{id}} \def\rank{\operatorname{rank}} \def\Hom{\operatorname{Hom}} $ Let $k$ be a field with $\char k=0$ and let $V$ be a finite-dimensional $k$-vector space. A reflection is a linear automorphism $s:V\to V$ such that $s^2=\id_V$ and $\rank(s-\id_V)=1$. So there is a hyperplane $H_s\subset V$ such that $s(h)=h$ for all $h\in H_s$.

It can be proven that a linear endomorphism $s:V\to V$ is a reflection if and only if there is $\alpha\in V\setminus\{0\}$ and $\alpha^\vee\in V^*\setminus\{0\}$ (where $V^*=\Hom_k(V,k)$ is the dual of $V$) such that $\alpha^\vee(\alpha)=2$ and $s(v)=v-\alpha^\vee(v)\alpha$ for every $v\in V$. Note that on this case $s(\alpha)=-\alpha$ and therefore $s$ is diagonalizable, $V=k\alpha\oplus H_s$.

Now let $(-,-)$ be a symmetric non-degenerate $k$-bilinear form over $V$. An orthogonal reflection on $V$ is a reflection $s$ on $V$ which is an isometry with respect to $(-,-)$, i.e., we have $(s(v),s(w))=(v,w)$ for all $v,w\in V$.

I was wondering whether there is at most one orthogonal reflection for a given vector. That is, if $s,s'$ are orthogonal reflections on $V$ which share the same eigenspace of eigenvalue $-1$, then $s=s'$.

If the reflections are not orthogonal, then there are easy counterexamples to the result in $\mathbb{R}^n$ with the standard inner product, but I don't know what happens if we ask for the reflections to be orthogonal.

Answer 1

$\def\dim{\operatorname{dim}}$ I arrived to the following argument from the ideas of Joppy in the comments.

Let $s:V\to V$ be an orthogonal reflection. Call $H$ and $L$ to the $1$-eigenspace and ($-1$)-eigenspace of $s$, respectively. Then, for $v\in H$ and $w\in L$, we have $(v,w)=(s(v),s(w))=(v,-w)=-(v,w)$, so $(v,w)=0$. In particular, $L\subset H^\perp$ and $H\subset L^\perp$. By the formula $\dim W+\dim W^\perp=\dim V$, valid as $(-,-)$ is non-degenerate (see Theorem 3.12.(2) here), we deduce $L=H^\perp$ and $H=L^\perp$. Therefore, the $1$-eigenspace determines the ($-1$)-eigenspace of an orthogonal projection and viceversa. Thus, two orthogonal reflections are the same iff they share the same reflection line or the same reflection hyperplane.