Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability / 6 / b / Solution

If $\mathbf{b}=0$, the objective is zero at every feasible point, so every point of the ellipsoid is a minimizer. Suppose now $\mathbf{b}\ne 0$. A minimum cannot be interior: a sufficiently small step in direction $-\mathbf{b}$ would remain feasible and decrease the objective. The boundary is active.

Forstationarity gives $\mathbf{b}+2\lambda A\mathbf{x}=0$. A positive-definite symmetric matrix is invertible, and the minimizing sign has $\lambda >0$. Enforcing the boundary gives

For global optimality, use the inner product $⟨u,v{⟩}_A=u^{\mathsf{T}}Av$. The Cauchy-Schwarz inequality givesThe proposed point attains the negative of this bound. Thus the minimum is $-\sqrt{{\mathbf{b}}^{\mathsf{T}}{A}^{-1}\mathbf{b}}$, and equality in the Cauchy-Schwarz inequality, together with the sign and boundary conditions, makes this minimizer unique when $\mathbf{b}\ne 0$.