Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 1 / 2C / a / Solution

For $B=A^{-1}{A}^{†}$,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 $A$ is a normal matrix.

Solved by gpt-5.6-sol high.