For a <Hermitian matrix>, $A^\dagger=A$, so $A^\dagger A=A^2=AA^\dagger$ and $A$ is a <normal matrix>. If $Ax=\lambda x$ with $x\ne0$, part (a)(iii) gives $A^\dagger x=\overline\lambda x$. Since also $A^\dagger x=Ax=\lambda x$, we obtain $\lambda=\overline\lambda$. Thus \b[every eigenvalue is real].

Solution

Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 1 / 8A / b / Solution

 Solution (source code)