Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 1 / 7B / e / Solution

$Pu=0$, hence $HPu=0$. In the diagonal basis, $f(\lambda )=1-{\sum }_j{u}_j^2{\mu }_j/({\mu }_j-\lambda )=-\lambda {\sum }_j{u}_j^2/({\mu }_j-\lambda )$. Between consecutive ${\mu }_j$ its nonzero factor is strictly monotone from one infinite sign to the other, giving one root; zero weights give equality at endpoints. Hence ${\mu }_1\le {\sigma }_1\le {\mu }_2\le \cdots \le {\sigma }_{n-1}\le {\mu }_n$.