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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?

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  • $\begingroup$ I believe $H(u)=u-(2u\cdot v^T)v$ is what you meant. This is a reflection. The product of two reflections is a rotation. $\endgroup$
    – David Hill
    Jul 8, 2015 at 0:01
  • $\begingroup$ Place an arrow on a table. Rotate it through π round the vertical axis. It's now pointing the other way. This is trivial. $\endgroup$
    – Dan Piponi
    Jul 8, 2015 at 0:59

1 Answer 1

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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 $SO(3)$.

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  • $\begingroup$ Yes, I was aware of the fact but failed to make the connection. Thank you for your answer $\endgroup$
    – Ricardo
    Jul 8, 2015 at 17:45

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