Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 1 / 2C (Vectors and Matrices)

The Hermitian conjugate is $A^{†}={\overline{A}}^{,T}$. A matrix is unitary when $A^{†}A=A{A}^{†}=I$, and Hermitian when $A^{†}=A$.

For $B=A^{-1}{A}^{†}$,Multiplying $B^{†}B=I$ on the left by $A^{-1}$ and on the right by $(A^{†}{)}^{-1}$ shows that it is equivalent to $(A^{†}{)}^{-1}{A}^{-1}=A^{-1}(A^{†}{)}^{-1}$. Taking inverses gives $A^{†}A=A{A}^{†}$. Thus $B$ is unitary exactly when $A$ is a normal matrix.

Normality givesTherefore one norm vanishes exactly when the other does.

Apply part (b) to the normal matrix $C=D-\lambda I$. Since $(D-\lambda I)e=0$, one hasThus $e$ is an eigenvector of $D^{†}$ with eigenvalue $\overline{\lambda }$.