 Solution

Put $d=1-x^{T}x>0$. Differentiating the barrier gradient givesAt a nonzero stationary point on the branch $Ax={\lambda }_{1}x$, $\mu /d=-{\lambda }_1$. Choose an orthonormal eigenbasis of the symmetric matrix $A$ whose first vector is $x/\Vert x{\Vert }_2$. On each of the remaining eigenvectors the rank-one term vanishes, so the corresponding Hessian eigenvalues areAlong $x$, the first two terms cancel and the remaining eigenvalue isIf ${\lambda }_1$ repeats, a transverse eigenvalue can be zero. If a lower eigenvalue than ${\lambda }_1$ exists, a transverse eigenvalue is negative, identifying this stationary branch as a saddle rather than a minimum.