 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2012/p/p3/b/solution

∥ Q x ∥^{2} = (Q x)^{T} Q x = x^{T} Q^{T} Q x = x^{T} x = ∥ x ∥^{2} .

Both norms are nonnegative, so taking square roots proves

∥ Q x ∥ = ∥ x ∥.

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