= Solution
Proceed by induction on $\dim V$. Over $\mathbb C$, $\alpha$ has an <eigenvector> $v$, say $\alpha v=\lambda v$. The normal operator $\alpha-\lambda I$ satisfies the norm equality from part (c), so
$$
\|(\alpha^*-\overline\lambda I)v\|
=\|(\alpha-\lambda I)v\|=0.
$$
Thus $\alpha^*v=\overline\lambda v$. Part (a) now shows that $v^\perp$ is invariant under both $\alpha$ and $\alpha^*$. The restriction of $\alpha$ to $v^\perp$ is normal. By induction it has an orthonormal eigenbasis, and adjoining the normalized <vector> $v$ proves the finite-dimensional <spectral theorem for normal operators>.
Hence
$$
\boxed{V\text{ has an orthonormal basis of eigenvectors of }\alpha}.
$$
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