Solution
= Solution
Let $P=vv^T/(v^Tv)$, which is a nonzero symmetric <orthogonal projection>: $P^T=P$ and $P^2=P$. Then $H_{v,\alpha}=I-\alpha P$ is symmetric and
$$
H_{v,\alpha}^TH_{v,\alpha}=(I-\alpha P)^2=I+(\alpha^2-2\alpha)P.
$$
Since $P\ne0$, this equals the identity exactly when $\alpha(\alpha-2)=0$. Thus
$$
\boxed{H_{v,\alpha}\text{ is orthogonal}\iff\alpha\in\{0,2\}}.
$$