 Solution

For any complex $\lambda$, the shifted linear operator $T=\alpha -\lambda I$ is normal: expansion of $T^{\ast }T-T{T}^{\ast }$ leaves just ${\alpha }^{\ast }\alpha -\alpha {\alpha }^{\ast }=0$. Apply part (i) to this shifted normal operator:Hence a nonzero vector is an eigenvector of $\alpha$ with eigenvalue $\lambda$ if and only if it is an eigenvector of ${\alpha }^{\ast }$ with eigenvalue $\bar{\lambda }$. The eigenspaces coincide, with conjugate eigenvalues.