Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 1 / 7B (Vectors and Matrices)

From $Hv=\mu v$, $v^{†}Hv=\mu \Vert v{\Vert }^2$ is real, so $\mu \in \mathbb{R}$. For distinct eigenvalues, ${\mu }_i{v}_i^{†}{v}_j=v_i^{†}H{v}_j={\mu }_j{v}_i^{†}{v}_j$, hence orthogonality. Normalising the eigenvectors and taking them as the rows of $U$ gives $H=U^{†}\Lambda U$.

${\chi }_H(\lambda )=\det (H-\lambda I)={\prod }_i({\mu }_i-\lambda )$ for this sign convention.

The subspace $y^{\perp }$ is fixed by $M=I-x{y}^{†}$, while the remaining determinant factor is $1-y^{†}x$. Thus $\det M=1-y^{\ast }\cdot x$.

Apply the matrix determinant lemma to $HP-\lambda I=(H-\lambda I)-Hu,u^T$: its determinant is ${\chi }_H(\lambda )[1-u^T(H-\lambda I{)}^{-1}Hu]$.

$Pu=0$, 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 ${\mu }_1\le {\sigma }_1\le {\mu }_2\le \cdots \le {\sigma }_{n-1}\le {\mu }_n$.