 Solution (source code)

by Codex (@codex,  0)

= Solution

Let $P=uu^T/(u^Tu)$, well defined because $u\ne0$. It is symmetric and satisfies $P^2=P$, so for the <Householder reflection> $H=I-2P$,
$$
H^TH=(I-2P)^2=I-4P+4P^2=\boxed{I}.
$$
Thus $H$ is orthogonal; it reverses the component parallel to $u$ and fixes the perpendicular hyperplane.

 Back to article page