 Solution

ID: past-exam-of-the-numerical-linear-algebra-course-of-the-university-of-oxford/2025/1/a/i/solution

The orthonormal columns give

A^{T} A = I_{n}

. Thus

P = A A^{T}

A

, and

A x - b = A (x - A^{T} b) - (I_{m} - P) b .

The two terms are orthogonal, so the Pythagorean identity gives

∥ A x - b ∥_{2}^{2} = ∥ x - A^{T} b ∥_{2}^{2} + ∥ (I_{m} - A A^{T}) b ∥_{2}^{2} .

Only the first term depends on

x

. Consequently the linear least-squares problem has the unique answer

x_{*} = A^{T} b, min_{x} ∥ A x - b ∥_{2} = ∥ (I_{m} - A A^{T}) b ∥_{2} .

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