The spectral theorems form a cornerstone of functional analysis. They are a vast generalization to infinite-dimensional Hilbert spaces of a basic result in linear algebra: an Hermitian matrix can be diagonalized or conjugated to a diagonal matrix with real entries along the diagonal.
There is a caveat, though: if we consider a separable Hilbert space then we can choose a countable orthonormal (Hilbert) basis of , a linear operator then has a matrix representation in this basis just as in finite dimensional linear algebra. The spectral theorem does not say that for every selfadjoint there is a basis so that has a diagonal matrix with respect to it. The situation that comes closest to the analogy of finite dimensions is the spectral theorem for compact operators.
There are several versions of the spectral theorem, or several spectral theorems, differing in the kind of operator considered (bounded or unbounded, selfadjoint or normal) and the phrasing of the statement (via spectral measures, multiplication operator norm), which is why this page does not consist of one statement only. Plus there are several different strategies to prove e.g. the spectral theorem for bounded selfadjoint operators, so there is no canonical way to prove it (none is particularly simple).
This is a short description of a way to prove the spectral theorem for bounded, selfadjoint operators in the spectral measure form, it may serve as a memory hook:
Let be a Hilbert space and A be an selfadjoint operator on . Then we can form the smallest von Neumann algebra generated by both A and , the identity operator. This is an abelian algebra, that is, via the Gelfand representation?, the abelian version of the Gelfand-Naimark theorem, isometric isomorph to the algebra of continuous complex valued functions of a compact Hausdorff space. The spectral integral of A now becomes the Lebesgue (or Riemann) integral of a continuous function.
theorem: there is a one to one correspondence of (bounded or unbounded) selfadjoint operators and spectral measures E such that
A = \integral \lambda E(d\lambda)
A is bounded iff E is bounded.
As stated on the spectral measure page, given a resolution of the identity one can make sense of for any function that is Borel measurable. Since we have for a selfadjoint operator for , one can define a bounded operator for every bounded Borel measurable function .
The special case of compact operators is mentioned here because it comes closest to the finite dimensional situation of diagonalizing Hermitian matrixes.
theorem: let be a compact selfadjoint operator, then the spectrum of consists of eigenvalues only, and all eigenvalues have finite multiplicity. Therefore, the spectral theorem says in this case, that
A = \sum \lambda_k |x\rangle \langle x|
where is an eigenvalue (each eigenvalue appears n times if n is its multiplicity) and an eigenvector with eigenvalue .
The converse statement is of course true, too, given a sequence in convergent to , the sum above (with arbitrary ) defines a normal compact operator, and a selfadjoint compact operator iff all are real. It is possible to generalize the statement to compact operators on (not necessarily complete) normed spaces.
(…) normal operator (…)
For the case of symmetric real matrices there is a simple proof using Lagrange multiplier techniques.