Compact self-adjoint operators are the closest infinite-dimensional analogue of real symmetric matrices. Compactness produces eigenvectors, and self-adjointness makes the eigenspaces orthogonal.
Quadratic form of a self-adjoint operator
For self-adjoint operators, the quadratic form \(\langle Tx,x\rangle\) contains spectral information. Maximizing it is the infinite-dimensional analogue of Rayleigh quotient arguments.
Lemma. If \(T=T^*\in\mathcal L(H)\), then \(\|T\|=\sup_{\|x\|=1}|\langle Tx,x\rangle|\).
Proof. The inequality \(\sup|\langle Tx,x\rangle|\le\|T\|\) is Cauchy-Schwarz. For the reverse inequality, use the polarization identity to express \(\langle Tx,y\rangle\) in terms of the quadratic form \(q(z)=\langle Tz,z\rangle\). Bounds on \(q\) over the unit sphere imply bounds on all matrix coefficients \(\langle Tx,y\rangle\), and taking the supremum over unit \(x,y\) gives \(\|T\|\le\sup_{\|z\|=1}|q(z)|\).
Existence of an eigenvalue
Compactness turns a maximizing sequence into a convergent subsequence. That compactness is what produces an eigenvector.
Theorem. If \(T\ne0\) is compact and self-adjoint on a Hilbert space \(H\), then either \(\|T\|\) or \(-\|T\|\) is an eigenvalue.
Proof. Choose unit vectors \(x_n\) such that \(|\langle Tx_n,x_n\rangle|\to\|T\|\). Passing to \(-T\) if necessary, assume \(\langle Tx_n,x_n\rangle\to\|T\|\). Compactness gives a subsequence, still denoted \(x_n\), such that \(Tx_n\to y\).
Compute \(\|(T-\|T\|I)x_n\|^2=\|Tx_n\|^2-2\|T\|\langle Tx_n,x_n\rangle+\|T\|^2\). Since \(\|Tx_n\|\le\|T\|\) and the middle term tends to \(2\|T\|^2\), this norm tends to \(0\). Thus \(\|T x_n-\|T\|x_n\|\to0\). Because \(Tx_n\to y\) and \(\|T\|>0\), \(x_n\to y/\|T\|\). Letting \(n\to\infty\) gives \(T(y/\|T\|)=\|T\|(y/\|T\|)\), and the limit vector has norm \(1\).
Orthogonal eigenspaces
Self-adjointness makes eigenspaces for different eigenvalues orthogonal. This gives the decomposition its Hilbert-space geometry.
Proposition. If \(T=T^*\) and \(Tx=\lambda x\), \(Ty=\mu y\) with \(\lambda\ne\mu\), then \(x\perp y\). Nonzero eigenspaces of a compact operator are finite-dimensional.
Proof. \(\lambda\langle x,y\rangle=\langle Tx,y\rangle=\langle x,Ty\rangle=\overline{\mu}\langle x,y\rangle\), and the eigenvalues of a self-adjoint operator are real, so \((\lambda-\mu)\langle x,y\rangle=0\). If a nonzero eigenspace for \(\lambda\ne0\) were infinite-dimensional, its unit ball would contain a sequence without convergent subsequence; applying \(T=\lambda I\) on that eigenspace would contradict compactness.
Compact self-adjoint spectral theorem
The spectral theorem for compact self-adjoint operators is the precise diagonalization result that survives from finite-dimensional symmetric matrices.
Theorem. Let \(T=T^*\) be compact on \(H\). There is an orthonormal set of eigenvectors \((e_j)\) for the nonzero eigenvalues \((\lambda_j)\) such that \(\lambda_j\to0\) if infinitely many occur, and \(H=\ker T\oplus\overline{\operatorname{span}}\{e_j\}\). Moreover \(Tx=\sum_j\lambda_j\langle x,e_j\rangle e_j\).
Proof. Use the eigenvalue-existence theorem on \(T\) to obtain an eigenvector for an eigenvalue of maximal absolute value. The orthogonal complement of its eigenspace is invariant under \(T\), because self-adjointness gives \(\langle Tx,e\rangle=\langle x,Te\rangle=0\). Repeat on the invariant complement. Nonzero eigenspaces are finite-dimensional, so choose orthonormal bases inside each.
If the process leaves a reducing subspace \(K\) on which \(T\) has no nonzero eigenvalue, then the restriction of \(T\) to \(K\) must be zero, otherwise the existence theorem would produce one. Nonzero eigenvalues can accumulate only at \(0\); if \(|\lambda_j|\ge\varepsilon>0\) for infinitely many orthonormal eigenvectors, compactness of \(T\) would make \((\lambda_je_j)\) have a convergent subsequence, impossible because these vectors stay separated. The expansion follows from the orthogonal decomposition and continuity of \(T\).
Positive compact operators
Positivity orders the spectrum and makes quadratic-form estimates especially transparent.
Corollary. If \(T\) is compact, self-adjoint, and positive, then all nonzero eigenvalues are positive and \(T\) has a positive square root \(T^{1/2}\) given by \(T^{1/2}x=\sum_j\lambda_j^{1/2}\langle x,e_j\rangle e_j\).
Proof. For an eigenvector \(e\), \(\lambda\|e\|^2=\langle Te,e\rangle\ge0\), so \(\lambda\ge0\). The displayed series defines a bounded operator because \(\sup_j\lambda_j^{1/2}<\infty\). Squaring it on the orthogonal decomposition gives \(T\), and positivity is immediate from the coefficients.
Mercer-type intuition
Kernel operators provide a concrete image: compact self-adjoint operators behave like infinite matrices with eigenfunction expansions.
Example. Symmetric compact integral operators are infinite-dimensional analogues of symmetric matrices. The eigenfunction expansion diagonalizes the kernel in \(L^2\), and truncating the expansion gives finite-rank approximations.