Solution
= Solution
Choose $c=-\operatorname{sgn}_+(x_1)\|x\|$, with $\operatorname{sgn}_+(0)=1$, and put $v=x-ce_1$. The opposite sign choice guarantees $v\ne0$, even if $x$ is already parallel to $e_1$. Since $c^2=\|x\|^2$,
$$
v^Tv=2(\|x\|^2-cx_1)=2v^Tx.
$$
Consequently $H_vx=x-2v(v^Tx)/(v^Tv)=x-v$, yielding
$$
\boxed{H_vx=ce_1,\qquad c\ne0.}
$$
The sign choice also avoids cancellation in constructing $v$ numerically.