[proofplan]
We construct the Householder vector $u = w + \operatorname{sgn}(w_1)\|w\|_2\,e_1$ and verify that the reflection $H = I - 2uu^\top/\|u\|^2$ maps $w$ to a scalar multiple of $e_1$. The calculation reduces to showing that $Hw = w - 2\frac{u^\topw}{\|u\|^2}u$ is proportional to $e_1$, which follows from the explicit evaluation of the inner product $u^\topw$ and the norm $\|u\|^2$.
[/proofplan]
[step:Define the Householder vector and compute the reflection $Hw$]
Define the Householder vector:
\begin{align*}
u := w + \operatorname{sgn}(w_1)\|w\|_2\,e_1,
\end{align*}
where $\operatorname{sgn}(w_1) = +1$ if $w_1 \geq 0$ and $-1$ if $w_1 < 0$. The Householder reflection is:
\begin{align*}
H := I - \frac{2uu^\top}{u^\topu}.
\end{align*}
Applying $H$ to $w$:
\begin{align*}
Hw = w - \frac{2(u^\topw)}{u^\topu}\,u.
\end{align*}
[guided]
The Householder reflection $H = I - 2uu^\top/\|u\|^2$ reflects any vector across the hyperplane orthogonal to $u$. We need to choose $u$ so that reflecting $w$ across this hyperplane lands on a multiple of $e_1$.
Geometrically, reflecting $w$ to $\pm\|w\|_2\,e_1$ means $u$ must be parallel to $w \mp \|w\|_2\,e_1$ (the bisector of the angle between $w$ and the target). We choose the sign $u = w + \operatorname{sgn}(w_1)\|w\|_2\,e_1$ to avoid cancellation when $w_1$ and $\|w\|_2$ have the same sign.
Applying $H$ to $w$:
\begin{align*}
Hw = w - \frac{2(u^\topw)}{u^\topu}\,u.
\end{align*}
For this to equal a multiple of $e_1$, the coefficient $\frac{2(u^\topw)}{u^\topu}$ must equal $1$, so that $Hw = w - u = -\operatorname{sgn}(w_1)\|w\|_2\,e_1$. We verify this in the next step.
[/guided]
[/step]
[step:Evaluate $u^\topw$ and $\|u\|^2$ and verify $Hw = -\operatorname{sgn}(w_1)\|w\|_2\,e_1$]
Write $\sigma := \operatorname{sgn}(w_1)\|w\|_2$ for brevity, so $u = w + \sigmae_1$.
Computing $u^\topw$:
\begin{align*}
u^\topw = (w + \sigmae_1)^\topw = \|w\|_2^2 + \sigma w_1.
\end{align*}
Computing $\|u\|^2 = u^\topu$:
\begin{align*}
\|u\|^2 = (w + \sigmae_1)^\top(w + \sigmae_1) = \|w\|_2^2 + 2\sigma w_1 + \sigma^2 = \|w\|_2^2 + 2\sigma w_1 + \|w\|_2^2 = 2(\|w\|_2^2 + \sigma w_1),
\end{align*}
where we used $\sigma^2 = \|w\|_2^2$. Therefore:
\begin{align*}
\frac{2(u^\topw)}{\|u\|^2} = \frac{2(\|w\|_2^2 + \sigma w_1)}{2(\|w\|_2^2 + \sigma w_1)} = 1.
\end{align*}
Substituting into the reflection formula:
\begin{align*}
Hw = w - 1 \cdot u = w - (w + \sigmae_1) = -\sigmae_1 = -\operatorname{sgn}(w_1)\|w\|_2\,e_1.
\end{align*}
Setting $c := -\operatorname{sgn}(w_1)\|w\|_2 \neq 0$ (since $w$ is an eigenvector and hence nonzero), we obtain $Hw = c\,e_1$.
[guided]
The calculation splits into two inner products. First, $u^\topw = w^\topw + \sigmae_1^\topw = \|w\|_2^2 + \sigma w_1$.
Second, for $\|u\|^2$, we expand:
\begin{align*}
\|u\|^2 &= \|w\|_2^2 + 2\sigma(e_1^\topw) + \sigma^2\|e_1\|^2 = \|w\|_2^2 + 2\sigma w_1 + \sigma^2.
\end{align*}
Since $\sigma = \operatorname{sgn}(w_1)\|w\|_2$, we have $\sigma^2 = \|w\|_2^2$, so $\|u\|^2 = 2\|w\|_2^2 + 2\sigma w_1 = 2(\|w\|_2^2 + \sigma w_1)$.
The ratio simplifies perfectly: $\frac{2(u^\topw)}{\|u\|^2} = \frac{2(\|w\|_2^2 + \sigma w_1)}{2(\|w\|_2^2 + \sigma w_1)} = 1$.
This is the crucial point: the reflection coefficient equals exactly $1$, which means $Hw = w - u = -\sigmae_1$. The sign choice $\sigma = \operatorname{sgn}(w_1)\|w\|_2$ ensures that $\|w\|_2^2 + \sigma w_1 = \|w\|_2^2 + |w_1|\|w\|_2 > 0$, so the denominator $\|u\|^2 = 2(\|w\|_2^2 + \sigma w_1) \neq 0$. Had we chosen the opposite sign, we would have $\|u\|^2 = 2(\|w\|_2^2 - |w_1|\|w\|_2)$, which vanishes when $|w_1| = \|w\|_2$ (i.e., when $w$ is already a multiple of $e_1$). The sign convention avoids this degeneracy and provides numerical stability.
[/guided]
[/step]
