 Solution

Because $A$ is a real symmetric matrix, differentiating the quadratic form givesThus the stationary vectors are precisely $\ker (A-\alpha I)$, and at every one of themThe 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.