Inner product is invariant under reflections in the real hyperplane?

Question

Let $V$ be an (even) finite-dimensional vector space. I’m trying to prove that $$d(r_v(u), r_v(w))=d(u,w),$$ where $d:V \times V \to \mathbb{R}$ is a positive definite symmetric bilinear form (i.e. a real scalar product on $V$, $v$ is a unit vector (that is, that $d(v,v)=1$) and $$r_v(u)=2d(u,v)v-u.$$ Notice that $-r_v$ is the reflection in the real hyperplane orthogonal to $v$. In particular, $r_v^2=1$.

The problem is that when I compute $d(r_v(u),r_v(w))$ I get $$d(r_v(u),2d(v,w)v)+d(2d((v,w)v,-u)+d(u,w)=d(2d(u,v)v,r_v(w)+d(2d(v,w)v,-w)+d(u,w).$$ I fill like I’m close, but, how do I prove that $$d(r_v(u),2d(v,w)v)+d(2d((v,w)v,-u))=0= d(2d(u,v)v,r_v(w)+d(2d(v,w)v,-w)?$$ Any ideas?? :(

Answer 1

Every vector $x$ can be decomposed into the sum of two components, $x_v=d(x,v)v$ and $x_{v^\perp}=x-d(x,v)v$, where $x_v$ is parallel to $v$, and $x_{v^\perp}$ is $d$-orthogonal to $v$ because $$ d(v,x_{v^\perp}) =d\big(v,x-d(x,v)v\big) =d(v,x)-d(x,v)d(v,v)=0. $$ It follows that $d(u_v,w_{v^\perp})=d(u_{v^\perp},w_v)=0$. Now note that $r_v(x)=x_v-x_{v^\perp}$. Therefore $$ \begin{aligned} d\big(r_v(u),r_v(w)\big) &=d\big(u_v-u_{v^\perp},\,w_v-w_{v^\perp}\big)\\ &=d(u_v,w_v) -d(u_v,w_{v^\perp}) -d(u_{v^\perp},w_v) +d(u_{v^\perp},w_{v^\perp})\\ &=d(u_v,w_v) +d(u_v,w_{v^\perp}) +d(u_{v^\perp},w_v) +d(u_{v^\perp},w_{v^\perp})\\ &=d\big(u_v+u_{v^\perp},\,w_v+w_{v^\perp}\big)\\ &=d(u,w). \end{aligned} $$

Answer 2

$r_v(u) = 2d(u,v)v-u$, $$d(r_v(u),r_v(w)) = d(2d(u,v)v-u,2d(w,v)v-w)$$ $$ = -1 \times d(u,2d(w,v)v-w) + 2d(u,v) \times d(v,2d(w,v)v-w)$$ $$ = -1 \times (2d(u,v)\times d(w,v) - d(u,w)) + 2d(u,v) \times (2d(w,v)-d(v,w))$$ $$ = -2d(u,v)\times d(w,v) + d(u,w) + 2d(u,v) \times 2d(w,v)- 2d(u,v) \times d(v,w)$$ $$ = d(u,v)$$ The last step used symmetry of $d$ and third step we used $d(v,v)=1$.