Let $ w\in \mathbb R^n$ be vector of length $1$.
$U$ is orthogonal space $w^\perp $
The reflection $r_w $ about $U$ is defined as follows if $v=cw+u$ , $u\in U$ then $r_w(v)=-cw+u$
Let $ u ,v$ be vector of equal length then how to find $w$ such that $r_w(u)=v?$
I completly stuck for this problem .
Please give me some hint so I can understand this problem .
Any help will be appreciated
On
If $u=v$ then we can take $w=u=v$.
So let's assume that $u \ne v$. Then $u$ and $v$ span a $2$ dimensional subspace of $\mathbb{R}^n$. Let's call this subspace $V$.
$(u,v)$ is a basis of $V$. Another possible basis is $(u+v, u-v)$. For any $au+bv \in V$ we have
$au+bv = \frac{a+b}{2}(u+v) + \frac{a-b}{2}(u-v)$
In particular
$u = \frac 1 2 (u+v) + \frac 1 2 (u-v) \\ v = \frac 1 2 (u+v) - \frac 1 2 (u-v)$
Note that $(u+v).(u-v) = u.u - v.v = 0$ since $|u|=|v|$. So $u+v$ and $u-v$ are perpendicular (in geometric terms we have just proved that the two diagonals of a rhombus are perpendicular).
So if we let $w=u+v$ then what is $r_w(u)$ ?
Hint
Let $u=cw+v_1$ with $v_1\in U$ (i.e. $v_1\cdot w=0$) therefore $$u\cdot w=cw\cdot w+v_1\cdot w=cw\cdot w=c|w|^2\implies c={u\cdot w\over |w|^2}$$and we obtain $$u={u\cdot w\over |w|^2}w+v_1$$then the reflection $r_w(v)$ will become$$r_w(u){=-cw+v_1\\=cw+v_1-2cw\\=u-2cw\\=u-2{u\cdot w\over |w|^2}w}$$since we want $r_w(u)=v$ we should have $$2{u\cdot w\over |w|^2}w=u-v$$what does it imply about $w$?