= Solution
<Self-adjointness> means $\langle Tv,w\rangle=\langle v,Tw\rangle$ for all $v,w\in V$, or $T=T^*$. Take the <inner product> linear in its first variable. If $Tv=\lambda v$ with $v\ne0$, then
$$
\lambda\|v\|^2=\langle Tv,v\rangle=\langle v,Tv\rangle=\overline\lambda\|v\|^2.
$$
Therefore \b[every <eigenvalue> is real].
If $U$ is invariant and $w\in U^\perp$, then for every $u\in U$, $Tu\in U$ and $\langle Tw,u\rangle=\langle w,Tu\rangle=0$. Hence $Tw\in U^\perp$, proving invariance of the <orthogonal complement>.
Induct on $d=\dim V$. The zero-dimensional case is empty. For $d>0$, choose an <eigenvector> and normalize it to a unit vector $e$. Its span is invariant, so its <orthogonal complement> is invariant as just proved. The restriction of $T$ to that complement is again <self-adjoint>. Its <dimension> is $d-1$, so <induction> supplies an <orthonormal basis> of <eigenvectors> there. Adjoining $e$ gives \b[an orthonormal <eigenbasis> of $V$]. This proves the <finite-dimensional spectral theorem> without assuming it in the argument.
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