 Solution

Along a fixed nonzero branch, $\Vert x(\mu ){\Vert }_2^2=1+\mu /{\lambda }_1\to 1$ and the radial Hessian eigenvalue grows like $4{\lambda }_1^2/\mu$. The nonzero transverse eigenvalues $2({\lambda }_i-{\lambda }_1)$ stay fixed. If ${\lambda }_1$ is simple, the ratio of the largest to smallest absolute eigenvalue therefore tends to infinity. If it repeats, the Hessian is already singular because of the zero transverse directions; there is no finite spectral condition number to begin with. These conclusions concern the nonzero branches: at the stationary origin $H=2(A+\mu I)$, and the same blowup is not implied.

Ill-conditioning does not invalidate barrier minimizers as approximations to the constrained solution. Let ${\lambda }_{\min }<0$. At radius $r$, the smallest possible quadratic value is ${\lambda }_{\min }{r}^2$, so global barrier minimization reduces to minimizing ${\lambda }_{\min }t-\mu \log (1-t)$ for $0\le t<1$. For $0<\mu <-{\lambda }_{\min }$ its unique minimizing radius satisfiesThus every limiting unit vector in that eigenspace solves the original problem, with objective ${\lambda }_{\min }$. For $\mu \ge -{\lambda }_{\min }$ the barrier minimizer is the origin instead. The increasing condition number concerns the numerical difficulty of small-parameter subproblems, which can call for scaling or a primal-dual interior-point method; it does not prevent the mathematical convergence just established. This is the quadratic-ball barrier eigenvalue branch.