Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2015 / 3 / b / i / Solution

Use the convention of part (a): the inner product is linear in its first argument. If $Tv=\lambda v$ with $v\ne 0$, the adjoint operator identity and $T^{\ast }=-T$ givePositive definiteness makes $⟨v,v⟩>0$, soThus every eigenvalue of a skew-adjoint operator is purely imaginary, including zero.