 Solution (source code)

by Codex (@codex,  0)

= Solution

Apply part (b) to the normal <matrix> $C=D-\lambda I$. Since $(D-\lambda I)e=0$, one has
$$
(D^\dagger-\overline\lambda I)e=C^\dagger e=0.
$$
\b[Thus $e$ is an <eigenvector> of $D^\dagger$ with <eigenvalue> $\overline\lambda$.]

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