 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-1/2c/a/solution

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

For

B = A^{- 1} A^{†}

B^{†} B = A (A^{†})^{- 1} A^{- 1} A^{†} .

Multiplying

B^{†} B = I

on the left by

A^{- 1}

and on the right by

(A^{†})^{- 1}

shows that it is equivalent to

(A^{†})^{- 1} A^{- 1} = A^{- 1} (A^{†})^{- 1}

. Taking inverses gives

A^{†} A = A A^{†}

. Thus

B

is unitary exactly when

A

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