Compact operator on Hilbert space
By contrast, the study of general operators on infinite-dimensional spaces often requires a genuinely different approach.As stated above, the techniques used to prove results, e.g., the spectral theorem, in the non-compact case are typically different, involving operator-valued measures on the spectrum.Similar reasoning shows that if T is compact, then T is the uniform limit of some sequence of finite-rank operators.When a closed linear subspace L of H is invariant under T, then the restriction of T to L is a self-adjoint operator on L, and furthermore, the orthogonal complement L⊥ of L is also invariant under T. For example, the space H can be decomposed as the orthogonal direct sum of two T–invariant closed linear subspaces: the kernel of T, and the orthogonal complement (ker T)⊥ of the kernel (which is equal to the closure of the range of T, for any bounded self-adjoint operator).These basic facts play an important role in the proof of the spectral theorem below.The classification result for Hermitian n × n matrices is the spectral theorem: If M = M*, then M is unitarily diagonalizable, and the diagonalization of M has real entries.Theorem For every compact self-adjoint operator T on a real or complex Hilbert space H, there exists an orthonormal basis of H consisting of eigenvectors of T. More specifically, the orthogonal complement of the kernel of T admits either a finite orthonormal basis of eigenvectors of T, or a countably infinite orthonormal basis {en} of eigenvectors of T, with corresponding eigenvalues {λn} ⊂ R, such that λn → 0.Corollary For every compact self-adjoint operator T on a real or complex separable infinite-dimensional Hilbert space H, there exists a countably infinite orthonormal basis {fn} of H consisting of eigenvectors of T, with corresponding eigenvalues {μn} ⊂ R, such that μn → 0.Proving the spectral theorem for a Hermitian n × n matrix T hinges on showing the existence of one eigenvector x.Once this is done, Hermiticity implies that both the linear span and orthogonal complement of x (of dimension n − 1) are invariant subspaces of T. The desired result is then obtained by induction forIn the finite-dimensional case, part of the first approach works in much greater generality; any square matrix, not necessarily Hermitian, has an eigenvector.The spectral theorem for the compact self-adjoint case can be obtained analogously: one finds an eigenvector by extending the second finite-dimensional argument above, then apply induction.Note that while the Lagrange multipliers generalize to the infinite-dimensional case, the compactness of the unit sphere is lost.If f attains its maximum m(T) on B at some unit vector y, then, by the same argument used for matrices, y is an eigenvector of T, with corresponding eigenvalue λ = ⟨λy, y⟩ = ⟨Ty, y⟩ = f(y) = m(T).These two facts imply that f is continuous on B equipped with the weak topology, and f attains therefore its maximum m on B at some y ∈ B.Let T be a compact operator on a Hilbert space H. A finite (possibly empty) or countably infinite orthonormal sequence {en} of eigenvectors of T, with corresponding non-zero eigenvalues, is constructed by induction as follows.Let F = (span{en})⊥, where {en} is the finite or infinite sequence constructed by the inductive process; by self-adjointness, F is invariant under T. Let S denote the restriction of T to F. If the process was stopped after finitely many steps, with the last vector em−1, then F= Hm and S = Tm = 0 by construction.In the infinite case, compactness of T and the weak-convergence of en to 0 imply that Ten = λnen → 0, therefore λn → 0.The fact that S = 0 means that F is contained in the kernel of T. Conversely, if x ∈ ker(T) then by self-adjointness, x is orthogonal to every eigenvector {en} with non-zero eigenvalue.If F ≠ {0}, it is a non-trivial invariant subspace of T, and by the initial claim, there must exist a norm one eigenvector y of T in F. But then U ∪ {y} contradicts the maximality of U.It follows that F = {0}, hence span(U) is dense in H. This shows that U is an orthonormal basis of H consisting of eigenvectors of T. If T is compact on an infinite-dimensional Hilbert space H, then T is not invertible, hence σ(T), the spectrum of T, always contains 0.The functional calculus map Φ is defined in a natural way: let {en} be an orthonormal basis of eigenvectors for H, with corresponding eigenvalues {λn}; for f ∈ C(σ(T)), the operator Φ(f), diagonal with respect to the orthonormal basis {en}, is defined by settingConversely, any homomorphism Ψ satisfying the requirements of the theorem must coincide with Φ when f is a polynomial.By the Weierstrass approximation theorem, polynomial functions are dense in C(σ(T)), and it follows that Ψ = Φ.The more general continuous functional calculus can be defined for any self-adjoint (or even normal, in the complex case) bounded linear operator on a Hilbert space.-invariant non-trivial closed subspace; and thus by the lemma above, therein would lie a common eigenvector for the operators (necessarily orthogonal to Q).Since the operators restricted to the eigenspaces (which are finite-dimensional) are automatically all compact, we can apply Theorem 1 to each of these, and find orthonormal bases Qσ for theThen if G were compact then there is a unique decomposition of H into a countable direct sum of finite-dimensional, irreducible, invariant subspaces (this is essentially diagonalisation of the family of operatorsIf G were not compact, but were abelian, then diagonalisation is not achieved, but we get a unique continuous decomposition of H into 1-dimensional invariant subspaces.However, if U is identity plus a compact perturbation, U has only a countable spectrum, containing 1 and possibly, a finite set or a sequence tending to 1 on the unit circle.