= Householder reflection sending a vector to a coordinate axis
{c}
{title2=$Hx=\alpha e_1,\quad |\alpha|=\|x\|_2$}
For a nonzero real <vector> $x\in\mathbb R^n$, choose $\alpha$ with $|\alpha|=\|x\|_2$ and put $v=x-\alpha e_1$. When $v\ne0$, the <Householder reflection>
$$
H=I-\frac{2vv^T}{v^Tv}
$$
satisfies $Hx=\alpha e_1$: every component except the first becomes zero. Here $e_1$ is the first <Cartesian basis> <vector>. The calculation is short:
$$
v^Tv=2(\|x\|_2^2-\alpha x_1),
\qquad
v^Tx=\|x\|_2^2-\alpha x_1,
\qquad
Hx=x-v=\alpha e_1.
$$
If $v=0$, the <vector> is already at the selected target, so the <identity matrix> suffices.
In <floating-point arithmetic>, a customary choice is
$$
\alpha=
\begin{cases}
-\|x\|_2,&x_1\ge0,\\
\|x\|_2,&x_1<0.
\end{cases}
$$
Then $|v_1|=|x_1|+\|x\|_2$, avoiding subtraction of nearly equal positive quantities in the first component. Either sign can be valid in exact arithmetic; the numerical choice protects the construction from <catastrophic cancellation>. https://math.mit.edu/classes/18.085/summer2016/handouts/qr.pdf[Olver, Orthogonal Bases and the QR Algorithm: Householder factorization and sign selection]{ref}
For $x=(3,4)^T$, this choice gives $\alpha=-5$ and $v=(8,4)^T$. Dividing $v$ by its <Euclidean norm> gives $u=(2,1)^T/\sqrt5$, exactly the normal in the geometric example.
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