 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/5/b/iv/solution

Put

D = A - i B

, so

D^{†} = A + i B

A, B

D^{†} D = D D^{†} = A^{2} + B^{2} .

For every nonzero vector

v

v^{†} (A^{2} + B^{2}) v = ∥ A v ∥^{2} + ∥ B v ∥^{2} > 0,

since

A

D

and

D^{†}

are invertible matrices, justifying the inverse in the definition of

U

U = D^{†} D^{- 1}

(D^{†})^{- 1} D

. Since

D

and

D^{†}

commute, their inverses do too, and

U^{†} U = (D^{†})^{- 1} D D^{†} D^{- 1} = I .

Hence

U

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