Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 4 / 1F / Solution

A Hermitian form is conjugate-linear in its first argument, linear in its second, and satisfies $\varphi (w,v)=\overline{\varphi (v,w)}$. For $H=\varphi -\psi$, the hypothesis says $H(v,v)=0$. The complex polarization identitytherefore gives $H(v,w)=0$ for all $v,w$.

The displayed matrix is Hermitian. By Sylvester's criterion, its leading principal minors areHence the form is