 Solution

ID: 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,

⟨ v, C^{†} C v ⟩ = ∥ C v ∥^{2} \geq 0.

Thus $C^{†} C$ is Hermitian and positive semidefinite. The positive semidefinite property is stronger than the Hermitian conclusion alone.

 New to topics? Read the docs here!