 Solution

The functional $\ell (u)=⟨u,v⟩$ is linear and nonzero because $\ell (v)=⟨v,v⟩>0$. It therefore has rank one, and rank-nullity theorem givesNo Gram–Schmidt argument is needed. Every vector decomposes uniquely as $u=w+av$, where $w\perp v$ and $a=⟨u,v⟩/⟨v,v⟩$. Substitution gives $S_v(w+av)=w-av$. Thus the eigenspace for $-1$ is $\mathbb{R}v$, and that for $+1$ is $v^{\perp }$. The latter is an actual eigenvalue when $n\ge 2$; when $n=1$ its eigenspace is zero and only $-1$ occurs. In the plane the map is reflection in the line perpendicular to $v$.