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