Past exam of the mathematical methods I course of the University of Oxford / 2019 resit / 11 / a / iv / Solution

One has $(C^{†}C{)}^{†}=C^{†}C$, with no requirement that $C$ itself be Hermitian. Moreover,Thus $C^{†}C$ is Hermitian and positive semidefinite. The positive semidefinite property is stronger than the Hermitian conclusion alone.