 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2013/p4/a/solution

Q^{T} Q = I

, equivalently when its columns form an orthonormal basis. Taking determinants gives

(det Q)^{2} = 1

, so

det Q \in {1, - 1} .

In particular it is invertible with inverse

Q^{T}

and preserves the Euclidean norm.

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