 Solution

ID: past-exam-of-the-numerical-linear-algebra-and-continuous-optimisation-course-of-the-university-of-oxford/2007/b/8/iii/solution

Ψ (x, μ) = x^{T} A x - μ lo g (1 - x^{T} x), ∥ x ∥_{2} < 1, μ > 0.

2 A x + 2 μx / (1 - x^{T} x)

. Thus every nonzero stationary point satisfies

A x = λ_{1} x, λ_{1} = - \frac{μ}{1 - ∥ x ∥ _{2}^{2}} < 0, ∥ x ∥_{2}^{2} = 1 + μ / λ_{1} .

A fixed negative-eigenvalue branch exists for

0 < μ < - λ_{1}

. The zero point is stationary for every

μ

as well, and it is not an eigenvector; at zero the displayed scalar

- μ

need not be an eigenvalue of

A

. Thus the claimed eigenvalue conclusion requires the nonzero-point qualification. This distinction is essential for the following Hessian analysis.

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