 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/discrete-mathematics-continuous-mathematics-and-probability-resit/5/a/ii/solution

The eigenvector identity

A b = \frac{1}{2} b

gives

\nabla f (- b) = - 2 A b + b = 0

. Since

H_{f} = 2 A

is a positive-definite matrix, this is a strict local minimum. A global certificate follows by completing the quadratic form:

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

Positive definiteness makes the first term nonnegative, with equality only at

x = - b

. Thus $- b$ is the unique global minimum, with value

- 1/2

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