= Solution
Because $A$ is a real <symmetric matrix>, differentiating the <quadratic form> gives
$$
\nabla_v a=(A-\alpha I)v.
$$
Thus the stationary vectors are precisely $\ker(A-\alpha I)$, and at every one of them
$$
\boxed{a=0.}
$$
The zero vector is always stationary. Nonzero stationary vectors exist exactly when $\alpha$ is an <eigenvalue> of $A$, and are exactly its nonzero <eigenvectors> with that <eigenvalue>. There is no norm constraint here; introducing one would change the stationary-value problem.
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