= Solution
Write $n=\dim V$, which is finite by hypothesis. This supplies the meaning of the $n$ used later in the variant; the proof is otherwise unchanged.
The functional $\ell(u)=\langle u,v\rangle$ is linear and nonzero because $\ell(v)=\langle v,v\rangle>0$. It therefore has rank one, and <rank-nullity theorem> gives
$$
\boxed{v^\perp=\ker\ell,\qquad\dim v^\perp=n-1.}
$$
No Gram–Schmidt argument is needed. Every vector decomposes uniquely as $u=w+av$, where $w\perp v$ and $a=\langle u,v\rangle/\langle v,v\rangle$. Substitution gives $S_v(w+av)=w-av$. Thus the <eigenspace> for $-1$ is $\mathbb Rv$, and that for $+1$ is $v^\perp$. The latter is an actual <eigenvalue> when $n\ge2$; when $n=1$ its <eigenspace> is zero and only $-1$ occurs. In the plane the map is \b[<reflection> in the line perpendicular to $v$].
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