Solution
= Solution
Let $r=\|u\|_2>0$ and choose $s=1$ when $u_1\geq0$, otherwise $s=-1$. Set $w=u+sr e_1$, which is nonzero because $r+su_1>0$. Compute
$$
w^Tu=r(r+su_1),\qquad w^Tw=2r(r+su_1).
$$
Thus the reflecting term in $H(w)u$ is exactly $w$, giving
$$
\boxed{H(w)u=u-w=-sr e_1.}
$$
Its first coordinate is nonzero and all remaining coordinates vanish. The choice of sign also avoids subtracting nearly equal vectors when the first component is large.