 Solution

ID: past-exam-of-the-mathematical-and-computational-biology-course-of-the-university-of-cambridge/2024/paper-1/b3/a/solution

W = diag (w_{1}, \dots, w_{m})

. Multiplying a vector by

W

scales its

i

th component by

w_{i}

. Therefore

∥ Wb - W A x ∥^{2} = i = 1 \sum m w_{i}^{2} (b_{i} - (A x)_{i})^{2} .

(W A, Wb)

is exactly the weighted least squares objective, so the sets of minimizing vectors coincide. Zero weights simply discard the corresponding residuals.

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