= Solution
We prove the finite-dimensional <spectral theorem for normal operators> by induction on dimension. A complex <linear operator> has an <eigenvector> $v$ with eigenvalue $\lambda$, by the <fundamental theorem of algebra>. For a <normal operator> $T$, the operator $T-\lambda I$ is also normal, and
$$
\|(T-\lambda I)w\|^2=\|(T^*-\overline\lambda I)w\|^2
$$
for every $w$. Applying this to $v$ shows $T^*v=\overline\lambda v$.
Normalize $v$. Its <orthogonal complement> $W=v^\perp$ is invariant under both $T$ and $T^*$: the <adjoint operator> identities pair $Tw$ with $v$ through $T^*v$, and $T^*w$ with $v$ through $Tv$. The restricted operator on $W$ remains normal, and its adjoint is the restriction of $T^*$. Induction supplies an <orthonormal eigenbasis> of $W$, which together with $v$ gives one for $V$.
Conversely, in an <orthonormal basis> of eigenvectors, $T$ is diagonal and $T^*$ is the diagonal matrix of the conjugate eigenvalues. The two diagonal operators commute. Thus \b[normality is equivalent to an orthonormal eigenbasis]. The zero-dimensional case is immediate.
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