 Solution

For a Hermitian operator $A$, every eigenvalue is real. Indeed, if $Av=\lambda v$ with $v\ne 0$, self-adjointness gives $\lambda ⟨v,v⟩=⟨Av,v⟩=⟨v,Av⟩=\overline{\lambda }⟨v,v⟩$. 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 $w\perp v$, $⟨Aw,v⟩=⟨w,Av⟩=0$. 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 a Hermitian operator is diagonal in an orthonormal basis, with real diagonal entries.