 Solution

On the constraint set, $\Vert x\Vert =1$, so $\nabla f(x)=2x\ne 0$. Part (a)(iv) and the Lagrange multiplier theorem give $2A^{T}Ax=2\lambda x$, henceSince $x$ is nonzero, it is an eigenvector. Moreover $\lambda =x^T{A}^{T}Ax=\Vert Ax{\Vert }^2\ge 0$, so the associated eigenvalue is real and nonnegative.