Compact operator on Hilbert space

In the mathematical discipline of functional analysis, the concept of a compact operator on Hilbert space is an extension of the concept of a matrix acting on a finite-dimensional vector space; in Hilbert space, compact operators are precisely the closure of finite-rank operators in the topology induced by the operator norm. As such, results from matrix theory can sometimes be extended to compact operators using similar arguments. By contrast, the study of general operators on infinite-dimensional spaces often requires a genuinely different approach.
For example, the spectral theory of compact operators on Banach spaces takes a form that is very similar to the Jordan canonical form of matrices. In the context of Hilbert spaces, a square matrix is unitarily diagonalizable if and only if it is normal. A corresponding result holds for normal compact operators on Hilbert spaces. More generally, the compactness assumption can be dropped. 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.
Some results for compact operators on Hilbert space will be discussed, starting with general properties before considering subclasses of compact operators.

Definition

Let be a Hilbert space and be the set of bounded operators on '. Then, an operator is said to be a compact operator''' if the image of each bounded set under is relatively compact.

Some general properties

We list in this section some general properties of compact operators.
If X and Y are separable Hilbert spaces, then T : X → Y is compact if and only if it is sequentially continuous when viewed as a map from X with the weak topology to Y. with the norm topology will be a Banach space, and the maps x** : Hom
The family of compact operators is a norm-closed, two-sided, *-ideal in L. Consequently, a compact operator T cannot have a bounded inverse if H is infinite-dimensional. If ST = TS = I, then the identity operator would be compact, a contradiction.
If sequences of bounded operators B_n → B, C_n → C in the strong operator topology and T is compact, then converges to in norm. For example, consider the Hilbert space with standard basis. Let P_m be the orthogonal projection on the linear span of. The sequence converges to the identity operator I strongly but not uniformly. Define T by T is compact, and, as claimed above, P_mT → IT = T in the uniform operator topology: for all x,
Notice each P_m is a finite-rank operator. Similar reasoning shows that if T is compact, then T is the uniform limit of some sequence of finite-rank operators.
By the norm-closedness of the ideal of compact operators, the converse is also true.
The quotient C*-algebra of L modulo the compact operators is called the Calkin algebra, in which one can consider properties of an operator up to compact perturbation.

Compact self-adjoint operator

A bounded operator T on a Hilbert space H is said to be self-adjoint if T = T*, or equivalently,
It follows that ⟨Tx, x⟩ is real for every x ∈ H, thus eigenvalues of T, when they exist, are real. 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 of the kernel. These basic facts play an important role in the proof of the spectral theorem below.
The classification result for Hermitian matrices is the spectral theorem: If M = M*, then M is unitarily diagonalizable, and the diagonalization of M has real entries. Let T be a compact self-adjoint operator on a Hilbert space H. We will prove the same statement for T: the operator T can be diagonalized by an orthonormal set of eigenvectors, each of which corresponds to a real eigenvalue.

Spectral theorem

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 of eigenvectors of T, with corresponding eigenvalues, such that.
In other words, a compact self-adjoint operator can be unitarily diagonalized. This is the spectral theorem.
When H is separable, one can mix the basis with a countable orthonormal basis for the kernel of T, and obtain an orthonormal basis for H, consisting of eigenvectors of T with real eigenvalues such that.
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 of H consisting of eigenvectors of T, with corresponding eigenvalues, such that.

The idea

Let us discuss first the finite-dimensional proof. 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 are invariant subspaces of T. The desired result is then obtained by induction for.
The existence of an eigenvector can be shown in two alternative ways:

One can argue algebraically: The characteristic polynomial of T has a complex root, therefore T has an eigenvalue with a corresponding eigenvector.
The eigenvalues can be characterized variationally: The largest eigenvalue is the maximum on the closed unit sphere of the function defined by.

Note. In the finite-dimensional case, part of the first approach works in much greater generality; any square matrix, not necessarily Hermitian, has an eigenvector. This is simply not true for general operators on Hilbert spaces. In infinite dimensions, it is also not immediate how to generalize the concept of the characteristic polynomial.
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. We first sketch the argument for matrices.
Since the closed unit sphere S in R²ⁿ is compact, and f is continuous, f is compact on the real line, therefore f attains a maximum on S, at some unit vector y. By Lagrange's multiplier theorem, y satisfies
for some λ. By Hermiticity,.
Alternatively, let z ∈ Cⁿ be any vector. Notice that if a unit vector y maximizes ⟨Tx, x⟩ on the unit sphere, it also maximizes the Rayleigh quotient:
Consider the function:
By calculus,, i.e.,
Define:
After some algebra the above expression becomes
But z is arbitrary, therefore. This is the crux of proof for spectral theorem in the matricial case.
Note that while the Lagrange multipliers generalize to the infinite-dimensional case, the compactness of the unit sphere is lost. This is where the assumption that the operator T be compact is useful.

Details

Claim If T is a compact self-adjoint operator on a non-zero Hilbert space H and
then m or −m is an eigenvalue of T.
If, then T = 0 by the polarization identity, and this case is clear. Consider the function
Replacing T by −T if necessary, one may assume that the supremum of f on the closed unit ball B ⊂ H is equal to. If f attains its maximum m on B at some unit vector y, then, by the same argument used for matrices, y is an eigenvector of T, with corresponding eigenvalue =.
By the Banach–Alaoglu theorem and the reflexivity of H, the closed unit ball B is weakly compact. Also, the compactness of T means that T : X with the weak topology → X with the norm topology is continuous. 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. By maximality, which in turn implies that y also maximizes the Rayleigh quotient g. This shows that y is an eigenvector of T, and ends the proof of the claim.
Note. The compactness of T is crucial. In general, f need not be continuous for the weak topology on the unit ball B. For example, let T be the identity operator, which is not compact when H is infinite-dimensional. Take any orthonormal sequence. Then y_n converges to 0 weakly, but lim f = 1 ≠ 0 = f.
Let T be a compact operator on a Hilbert space H. A finite or countably infinite orthonormal sequence of eigenvectors of T, with corresponding non-zero eigenvalues, is constructed by induction as follows. Let H₀ = H and T₀ = T. If m = 0, then T = 0 and the construction stops without producing any eigenvector e_n. Suppose that orthonormal eigenvectors of T have been found. Then is invariant under T, and by self-adjointness, the orthogonal complement H_n of E_n is an invariant subspace of T. Let T_n denote the restriction of T to H_n. If m = 0, then T_n = 0, and the construction stops. Otherwise, by the claim applied to T_n, there is a norm one eigenvector e_n of T in H_n, with corresponding non-zero eigenvalue λ_n =.
Let F = ^⊥, where 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 e_m−1, then F= H_m and S = T_m = 0 by construction. In the infinite case, compactness of T and the weak-convergence of e_n to 0 imply that, therefore. Since F is contained in H_n for every n, it follows that m ≤ m = |λ_n| for every n, hence m = 0. This implies again that.
The fact that S = 0 means that F is contained in the kernel of T. Conversely, if x ∈ ker then by self-adjointness, x is orthogonal to every eigenvector with non-zero eigenvalue. It follows that, and that is an orthonormal basis for the orthogonal complement of the kernel of T. One can complete the diagonalization of T by selecting an orthonormal basis of the kernel. This proves the spectral theorem.
A shorter but more abstract proof goes as follows: by Zorn's lemma, select U to be a maximal subset of H with the following three properties: all elements of U are eigenvectors of T, they have norm one, and any two distinct elements of U are orthogonal. Let F be the orthogonal complement of the linear span of U. If F ≠, 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 ∪ contradicts the maximality of U. It follows that F =, hence span is dense in H. This shows that U is an orthonormal basis of H consisting of eigenvectors of T.