Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / P / P4 / c / Solution

Choose $c=-{\mathrm{sgn}}_{+}(x_1)\Vert x\Vert$, with ${\mathrm{sgn}}_{+}(0)=1$, and put $v=x-c{e}_1$. The opposite sign choice guarantees $v\ne 0$, even if $x$ is already parallel to $e_1$. Since $c^2=\Vert x{\Vert }^2$,Consequently $H_{v}x=x-2v(v^{T}x)/(v^{T}v)=x-v$, yieldingThe sign choice also avoids cancellation in constructing $v$ numerically.