 Solution

Write the constraint as $c(x)=1-x^{T}x\ge 0$ and its multiplier as $y\ge 0$. The KKT conditions becomeIf $A$ is positive definite, $A+yI$ is positive definite for every $y\ge 0$, so $\boxed{x=0,y=0}$ is the only critical point.

For indefinite $A$, the origin remains a critical point, contrary to an unqualified reading of the question. Every nonzero critical point is an eigenvector with $\boxed{Ax=-yx}$, so its eigenvalue is nonpositive. A negative-eigenvalue vector must lie on the unit sphere, with multiplier equal to the negative of that eigenvalue. A nonzero vector in $\ker A$ may have any norm at most one and has multiplier zero. Positive-eigenvalue directions do not satisfy the multiplier sign condition. These first-order critical points include saddles; the global minimum selects a unit vector in the lowest-eigenvalue eigenspace.