= Solution
With a fixed <unit normal> $n$, the <Householder reflection> is $P=I-2nn^T$. It is a <linear map> because both the identity and the map $v\mapsto n(n^Tv)$ are linear. Decompose $v=v_\parallel+v_\perp$, where $v_\perp=(n\cdot v)n$ and $v_\parallel\cdot n=0$. Then
$$
Pv=v_\parallel-v_\perp.
$$
Thus the tangential component is unchanged and the normal component reverses. Its norm and absolute normal component are preserved, proving equality of the physical incidence and reflection angles measured against the appropriate normal. Angles of the incident and reflected \i[directed vectors] to one fixed oriented normal are instead supplementary; equality uses the conventional acute angle to the normal line, or reverses the incident direction.
Since $(nn^T)^2=nn^T$, direct multiplication gives $P^2=I$. Every <eigenvalue> therefore satisfies $\lambda^2=1$. More specifically, the tangent plane is the $+1$ <eigenspace> and the normal line the $-1$ <eigenspace>. In the stated <orthonormal basis>, with the first two vectors tangent to the plane,
$$
\boxed{[P]_{\{e_1,e_2,n\}}=\operatorname{diag}(1,1,-1).}
$$
This also proves both signs occur, rather than merely showing that any eigenvalue must belong to $\{+1,-1\}$.
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