This came up doing some research in quantum information. Let us consider two orthogonal three-dimensional unit vectors $v$ and $w$


$v^T\cdot w=0$,


and the Householder transformation


$H=I\_{3}-2v\cdot v^{T}$,


where $I\_{3}$ is the $3\times3$ identity matrix. Then we have


$Hv=-v$,


$Hw=w$.


If we now multiply $H$ by $-I\_3$, which is the three-dimensional reflection through the origin (parity transformation for physicists) we obtain an orthogonal matrix with unit determinant, hence a member of SO(3)


$\Omega\equiv-I\_{3}H=2v\cdot v^{T}-I\_3$.


But clearly now


$\Omega w=-w$


even though $\Omega\in\text{SO}(3)$. So my questions are: how is this possible? how come that an SO(3) transformation is able to produce a parity transformation? is this trivial or there's a deep mathematical reason? 



SO(3) transformation that produces a reflection
What's the problem?
$\Omega$ only coincides with the parity transformation on the orthogonal complement of $v$; on the other hand $\Omega v = v$.  

If, for example, $v = \pmatrix{1\cr 0\cr 0}$, then $\Omega = \pmatrix{1 & 0 & 0\cr 0 & -1 & 0\cr 0 & 0 & -1\cr}$ 
which is certainly in $SO(3)$.


