Householder transformation

Remarks

$P$ is Hermitian and unitary, and $P^2 = I$; for $\mathbb{K} = \R$ it is a symmetric orthogonal matrix with determinant $-1$. It fixes every vector orthogonal to $v$ and sends $v$ to $-v$, so it is the reflection in the hyperplane $v^\perp$. If $x, y \in \R^n$ are distinct with $\norm{x} = \norm{y}$, then $v = (x - y)/\norm{x - y}$ gives $P x = y$. Example: in $\R^2$, take $x = (3, 4)$ and $y = (5, 0)$; then $v = (-2, 4)/\sqrt{20}$, and $$P = I - \frac{2}{20} \begin{pmatrix} 4 & -8 \\ -8 & 16 \end{pmatrix} = \frac{1}{5} \begin{pmatrix} 3 & 4 \\ 4 & -3 \end{pmatrix},$$ and indeed $P (3, 4)^T = (5, 0)^T$. Non-example: $v = (2, 0)$ is not a unit vector, and $I - 2 v v^T = \begin{pmatrix} -7 & 0 \\ 0 & 1 \end{pmatrix}$ is not orthogonal. For a nonzero $u$ that is not normalized, one writes $P = I - 2 u u^* / (u^* u)$. In computations, $P$ is never formed: $P A = A - 2 v (v^* A)$ costs $O(n m)$ operations for an $n \times m$ matrix $A$.