= Solution
Put $d=1-x^Tx>0$. Differentiating the barrier <gradient> gives
$$
\boxed{H=2A+\frac{2\mu}dI+\frac{4\mu}{d^2}xx^T.}
$$
At a nonzero stationary point on the branch $Ax=\lambda_1x$, $\mu/d=-\lambda_1$. Choose an orthonormal eigenbasis of the symmetric matrix $A$ whose first vector is $x/\|x\|_2$. On each of the remaining eigenvectors the rank-one term vanishes, so the corresponding Hessian eigenvalues are
$$
\boxed{2(\lambda_i-\lambda_1),\qquad i=2,\ldots,n.}
$$
Along $x$, the first two terms cancel and the remaining eigenvalue is
$$
\boxed{\frac{4\mu\|x\|_2^2}{d^2}=\frac{4\lambda_1^2\|x\|_2^2}{\mu}.}
$$
If $\lambda_1$ 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.
Back to article page
