Put . Differentiating the barrier gradient gives
At a nonzero stationary point on the branch , . Choose an orthonormal eigenbasis of the symmetric matrix whose first vector is . On each of the remaining eigenvectors the rank-one term vanishes, so the corresponding Hessian eigenvalues are
Along , the first two terms cancel and the remaining eigenvalue is
If repeats, a transverse eigenvalue can be zero. If a lower eigenvalue than exists, a transverse eigenvalue is negative, identifying this stationary branch as a saddle rather than a minimum.

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