Let $u$ be a nonzero vector in $\mathbb{R}^n$, and define $\gamma=\frac{2}{||u||_2^2}$ and $Q=I-\gamma uu^T$.
Prove Q is a reflector satisfying
A) $ Qu=-u$
B) $Qv=v$ if $<u,v>=0$
My approach: I'm letting some $$\hat u=\frac{u}{||u||_2}$$ so that $||\hat u||=1$.
I don't know where this is getting me though I'm missing a few details to prove A and B. Any help would be appreciated. Thanks guys.
$Qu = u - \frac{2}{||u||_2^2}uu^Tu = u - \frac{2}{||u||_2^2}u||u||_2^2 = u-2u = -u$
$Qv = v-\gamma uu^Tv = v-\gamma u \langle u,v \rangle = v$, if $\langle u,v \rangle = 0$
and because by definition $||u||_2^2 = \langle u,u \rangle = u^Tu$.
Let $x = \alpha u + v$, where $v \in \operatorname{sp} \{u\}^\bot$. Then $Q x = (I-\gamma u u^T) (\alpha u + v) = \alpha (1-\gamma \|u\|^2)u + v = -\alpha u +v$.
Hence if $v=0$, $\alpha=1$, we have A), that is, $Q u = -u$, and if $\alpha = 0$, (and $v^Tu = 0$, of course) we have B), that is, $Qv = v$.
