 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2/10/a/solution

The upper-left block of

U

is the planar rotation matrix

R (π /2)

, and

U

fixes the third coordinate. Therefore

U^{p} = cos (p π /2) sin (p π /2) 0 - sin (p π /2) cos (p π /2) 0 001, U^{4} = I .

The diagonal entries of

V

square to one, so

V^{2} = I

. Direct conjugation gives

V U V = U^{- 1}

, hence

V U^{p} V = U^{- p}

for every integer

p

. Consequently

U^{p} V = V U^{- p} = V U^{4 - (p mod 4)} .

When

p mod 4 = 0

, the last exponent is four, which is equivalent to zero because

U^{4} = I

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