 Solution

For $Av=\lambda v$, $v^{\ast }Av=\lambda v^{\ast }v$ is real, so $\lambda$ is real. The spectral theorem writes positive definite $C=U\mathrm{diag}({\lambda }_i)U^{\ast }$; taking positive square roots gives $B$. For $H$, eigenvalues are $1,3$ with normalized eigenvectors $(-i,1)/\sqrt{2}$ and $(i,1)/\sqrt{2}$, so these columns form a suitable $U$. Finally expand $v$ in an orthonormal eigenbasis: the Rayleigh quotient is a weighted average of eigenvalues, bounded above by ${\lambda }_{max}$ and attaining it on a top eigenvector.