Past exam of the mathematical methods I course of the University of Oxford / 2018 / 5 / c / Solution

For Hermitian matrices, $A^{†}=A$ and $B^{†}=B$, soThe commutator is an anti-Hermitian matrix. Consequentlymakes $C^{†}=C$. More generally, any nonzero purely imaginary $\alpha$ works. For a nonzero commutator this is also necessary if $C$ is to be Hermitian: comparing the Hermitian adjoint equations gives $(\overline{\alpha }+\alpha )C=0$.