 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-3/40a/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

\nabla f (x) = A x - b = A (x - x^{*})

f (x) - f (x^{*}) = \frac{1}{2} (x - x^{*})^{T} A (x - x^{*}) > 0

for

x \neq = x^{*}

, so

x^{*}

is the unique global minimizer. Since

x - x^{*} = A^{- 1} \nabla f (x)

f (x) - f (x^{*}) = \frac{1}{2} \nabla f (x)^{T} A^{- 1} \nabla f (x) .

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