Spectral theorem for compact self-adjoint operators

Remarks

The proof starts from the fact that $\norm{T} = \sup_{\norm{x} = 1} \abs{\ip{Tx}{x}}$ for self-adjoint $T$. Compactness makes this supremum attained, at an eigenvector with eigenvalue $\pm \norm{T}$. The orthogonal complement of this eigenvector is invariant under $T$, and one repeats the argument there; the eigenvalues found tend to $0$ by the spectral theory of compact operators. Adding an orthonormal basis of $\ker T$ to $(e_n)$ gives an orthonormal basis of $H$ made of eigenvectors of $T$. The nonzero eigenvalues of $T$ are exactly the numbers $\lambda_n$, and each appears as many times as the dimension of its eigenspace, which is finite. Example: the operator $(Tf)(x) = \int_0^1 \min(x, y) f(y) \, dy$ on $L^2([0,1])$ is compact and self-adjoint, with eigenvalues $\big((n - \tfrac{1}{2}) \pi\big)^{-2}$ and eigenfunctions $\sqrt{2} \sin\big((n - \tfrac{1}{2}) \pi x\big)$, $n \ge 1$. Without self-adjointness the theorem fails: the Volterra operator $(Vf)(x) = \int_0^x f(y) \, dy$ on $L^2([0,1])$ is compact but has no eigenvalue at all. In finite dimension, this is the spectral theorem for self-adjoint linear maps.

Going deeper

For a student

The theorem is the main tool for solving linear integral equations with symmetric kernels, and for boundary value problems such as the Sturm–Liouville problem through their Green's operators. It shows that the eigenfunctions of such problems form an orthonormal basis, which justifies expansions in Fourier-type series. Applied to $T^* T$ for an arbitrary compact operator $T$, it gives the singular values and the Schmidt decomposition of $T$. For noncompact self-adjoint operators, eigenvectors may not exist, and one needs the spectral theorem in its measure form.