 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2023/discrete-mathematics-probability-and-continuous-mathematics/5/b/ii/solution

The origin is feasible and has objective zero, while a Euclidean norm is nonnegative. Thus the minimum is $0$ , and the complete argmin is

ker A \cap {x : ∥ x ∥ \leq 1} .

A

is injective, this set contains only the origin. If it has a nontrivial kernel, every feasible kernel vector is also a minimizer. No Lagrange multipliers are needed.

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