Solution
= Solution
Set $P=vv^T/(v^Tv)$. Since $v\ne0$, it is well defined, symmetric and idempotent: $P^T=P$, $P^2=P$. Therefore $H_{v,\alpha}=I-\alpha P$ is symmetric, and
$$
H_{v,\alpha}^TH_{v,\alpha}=I+(\alpha^2-2\alpha)P.
$$
Because $P\ne0$, this equals $I$ exactly when $\alpha(\alpha-2)=0$. Hence $\boxed{H_{v,\alpha}\text{ orthogonal}\iff\alpha\in\{0,2\}}$.