Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 6 / b / Solution

Let $A{v}_i={\lambda }_i{v}_i$ and $A{v}_j={\lambda }_j{v}_j$. The skew-Hermitian matrix property giveswhere the last equality uses part (a). Distinct eigenvalues therefore imply $\boxed{v_i^{†}{v}_j=0\text{ for }i\ne j}$. Normalize each eigenvector to length one. Since there are $n$ distinct eigenvalues, these vectors form an orthonormal basis.

Put them in the columns of $U$. Then $U$ is a unitary matrix and $AU=UD$, so