= Solution
Write the constraint as $c(x)=1-x^Tx\geq0$ and its multiplier as $y\geq0$. The <KKT conditions> become
$$
\boxed{(A+yI)x=0,\qquad\|x\|_2\leq1,\qquad y\geq0,\qquad y(1-\|x\|_2^2)=0.}
$$
If $A$ is positive definite, $A+yI$ is positive definite for every $y\geq0$, 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.
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