= Solution
For a <Hermitian operator> $A$, every <eigenvalue> is real. Indeed, if $Av=\lambda v$ with $v\ne0$, self-adjointness gives $\lambda\langle v,v\rangle=\langle Av,v\rangle=\langle v,Av\rangle=\overline\lambda\langle v,v\rangle$. To prove the finite-dimensional <spectral theorem>, proceed by induction on the dimension. The <fundamental theorem of algebra> supplies a root of the characteristic polynomial, hence a unit eigenvector $v$. Its <orthogonal complement> is invariant: for $w\perp v$, $\langle Aw,v\rangle=\langle w,Av\rangle=0$. The restriction of $A$ to that complement is again Hermitian. Induction supplies an <orthonormal basis> of eigenvectors there; adjoining $v$ gives an orthonormal eigenbasis of the whole space. The zero-dimensional case is immediate. Consequently \b[a Hermitian operator is diagonal in an orthonormal basis, with real diagonal entries].
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