 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability-resit/6/c/solution

Let

P = z z^{T}

, so

P^{2} = r^{2} P

and

f^{2} = 1 + r^{2}

. Multiplying the proposed inverse matrix by the Hessian gives

(\frac{I}{f} - \frac{P}{f ^{3}}) f (I + P) = I + P - \frac{P + P ^{2}}{f ^{2}} = I .

Hence $H_{f}^{- 1} = f (I + z z^{T}) = f (1 + x^{2} x y x y 1 + y^{2})$ , as required. Its existence also follows from the Hessian being a positive-definite matrix.

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