Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 2 / 17F / iii / Solution

Part (ii) shows that ${\alpha }^{\ast }v=\bar{\lambda }v$ for every $v\in E_{\lambda }$, so $E_{\lambda }$ is invariant under ${\alpha }^{\ast }$. Apply the adjoint criterion for an invariant orthogonal complement with ${\alpha }^{\ast }$ in place of $\alpha$, using $({\alpha }^{\ast }{)}^{\ast }=\alpha$. It follows thatThe same argument interchanging $\alpha$ and ${\alpha }^{\ast }$ shows that $E_{\lambda }^{\perp }$ is invariant under both operators, a fact needed for the induction in part (iv).