 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-3/8d/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Q \in O (2)

, then

Q^{T} Q = I

, so

(det Q)^{2} = 1

and

det Q = \pm 1

. If

det Q = 1

, its first column is

(cos θ, sin θ)^{T}

for some

θ

Q = (cos θ sin θ - sin θ cos θ) .

Thus every element of

SO (2)

Let

J = diag (1, - 1)

. If

det Q = - 1

, then

Q J \in SO (2)

and

Q = (Q J) J \in SO (2) J

. Conversely every matrix in

SO (2) J

- 1

. Therefore the union is disjoint and

O (2) = SO (2) ⊔ SO (2) J .

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