= Solution
Proceed by induction on $\dim V$. The result is immediate for dimension zero. In positive dimension, the <characteristic polynomial> has a root over the <complex numbers>, so there is a nonzero <eigenvector> $v$ with <eigenvalue> $\lambda$. Normalize it to have norm one. <Self-adjointness> gives
$$
\overline\lambda\langle v,v\rangle
=\langle Tv,v\rangle
=\langle v,Tv\rangle
=\lambda\langle v,v\rangle,
$$
so $\lambda$ is real.
The <orthogonal complement> $v^\perp$ is an <invariant subspace>: if $w\perp v$, then
$$
\langle v,Tw\rangle=\langle Tv,w\rangle
=\overline\lambda\langle v,w\rangle=0.
$$
The restriction of $T$ to $v^\perp$ is again <self-adjoint>, since the same <inner product> identity holds on that subspace. By induction it has an <orthonormal basis> of <eigenvectors>. Adjoining $v$ gives
$$
\boxed{\text{an orthonormal basis of eigenvectors of $T$}.}
$$
The existence of the first <eigenvector> is the essential use of the <complex numbers> in this proof of the finite-dimensional <spectral theorem>.
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