Compact operator on Hilbert space

From Wikipedia, the free encyclopedia

In functional analysis, compact operators on Hilbert spaces are a direct extension of matrices: in the Hilbert spaces, they are the precisely the closure of finite rank operators in the uniform operator topology. As such, results from matrix theory can sometimes be extended to compact operators using similar arguments. In contrast, the study of general operators on infinite dimensional spaces often require 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 assumtion can be dropped. But, as stated above, the techniques used are less routine.)

This article will discuss a few results for compact operators on Hilbert space, starting with general properties before considering subclasses of compact operators.

Contents 1 Some general properties 2 Compact self adjoint operator 2.1 Spectral theorem 2.1.1 The idea 2.1.2 Details 2.2 Functional calculus 2.3 Simultaneous diagonalization 3 Compact normal operator 4 Unitary operator 5 See also 6 References

[edit] Some general properties

Definition Let H be a Hilbert space, L(H) be the bounded operators on H. T ∈ L(H) is a compact operator if the image of a bounded set under T is precompact.

Our first characterization result for compact operators is:

Theorem T is compact if and only if T maps weakly convergent sequences to norm convergence ones.

Proof: Suppose T is compact. Let {x_n} converges to x weakly. The claim is that Tx_n converges to Tx in norm. If this was not the case, there exists a subsequence

for some ε > 0. According to the uniform boundedness principle, {x_n} being weakly convergent implies that it is bounded. By the compactness assumption on T,

contains a norm-convergent subsequence, converging to some y. But norm-convergent implies weakly convergent, and, because the weak topology is Hausdorff, we can deduce that y = Tx. This is a contradiction.

Conversely, suppose T maps weakly convergent sequences to norm convergence ones. Let B ⊂ H be a bounded set. B is therefore weakly compact. So every sequence {Tb_n} where {b_n} ⊂ B contains a norm convergent subsequence. This proves the theorem.

The following states that the family of compact operators is a norm-closed, two-sided, *-ideal in L(H):

Theorem Let T be compact.

1. If S ∈ L(H), then both TS and ST are compact.
2. The adjoint of T, T*, is also compact.
3. If {T_k} is a sequence of compact operators and T_k → S in the uniform operator topology, then S is compact.

Proof: We will repeatedly apply the previous theorem, and the fact that a bounded operator is also continuous in the weak topology.

1. Let {x_n} converges to x weakly. Since S is continuous in the weak topology, Sx_n converges to Sx weakly. By compactness of T, TSx_n converges to TSx in norm. On the other hand, it is trivial that STx_n converges to STx in norm.
2. Let {x_n} converges to x weakly, then ||T*(x_n - x)||² = <T*(x_n - x), T*(x_n - x)> ≤ ||x_n - x|| · ||TT*(x_n - x)||. T is compact means TT* is also. Therefore the right hand side of this estimate tends to 0. This proves the claim.
3. This can be shown using a triangularization argument: again let {x_n} converges to x weakly, then

In the above estimate, the first, leftmost, term tends to 0 because {x_n} is bounded and T_k → S in operator norm. Second term tends to 0 because each T_k is compact. It is also clear that the third term vanishes in the limit. Thus the theorem is proven.

As a corollary, a compact operator T cannot have a bounded inverse. If ST = TS = I, then by the above theorem, the identity operator is compact, a contradiction. Alternatively, if {x_n} is an orthonormal sequence, then compactness means that Tx_n converges to 0 ∈ H. So T is not bounded below, therefore not invertible. In other words, the complex number 0 lies in the approximate point spectrum of a compact operator T.

Theorem Suppose {S_n} ⊂ L(H) and S_n → S in the strong operator topology. If T is compact, then S_nT converges to ST in norm.

Proof: Let B be the (closed) unit sphere and T(B) be its image under T. The theorem states that S_n converges to S uniformly on T(B). By assumption T(B) is precompact. We can apply the following standard argument that passes from pointwise convergence to uniform convergence on compact sets: for all ε > 0, there exists a finite cover of open balls B_ε(y_i) for the closure of T(B). For any y ∈ T(B), let ||y_i - y|| < ε, then

Choosing m such that for all m > n and for all i, ||S_ny_i - Sy_i|| < ε, then above becomes

But according to the uniform boundedness principle,

The proves the theorem.

We next give an example demonstrating the theorem. Consider the Hilbert space l²(N), with standard basis {e_n}. Let P_m be the orthogonal projection on the linear span of {e₁...e_m}. Clearly the sequence {P_m} converges to the identity operator I strongly but not uniformly. Define T by Te_n = (1/n)² · e_n. T is compact, and, as claimed by the theorem, P_mT → I T = T in the uniform operator topology: for all x,

Notice each P_m is a finite rank operator. Similar reasoning shows that:

Theorem 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.

An algebraic property of the family of compact operators, L_c(H), is

Theorem The compact operators is the minimal norm-closed, two-sided, *-ideal of L(H).

Proof: Let Î be a norm-closed, two-sided, *-ideal of L(H). We will show that Î contains all rank-1 operators, therefore finite rank operators. The norm-closedness of Î then proves the claim. Since Î is non-trivial, there exists a non-zero operator T ∈ Î such that Tf = g where f and g are nonzero. For any h, k ∈ H, define

then, for all x,

In other words, BTA ∈ Î is the rank one operator that maps h to k. So Î indeed contains the compact operators.

The above argument also shows that, without the norm-closedness assumption, the finite rank operators are a minimal two-sided *-ideal in L(H).

The quotient algebra L(H)/L_c(H) is called the Calkin algebra, which allows one to study properties of an operator up to compact perturbation.

[edit] Compact self adjoint operator

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. Let T be a compact self adjoint operator on a Hilbert space H. We will prove the same statement for T: T can be diagonalized by an orthonormal set of eigenvectors, each of which correspond to a real eigenvalue.

[edit] Spectral theorem

[edit] The idea

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 compliment of x are invariant subspaces of T. The desired result is then obtained by iteration. The existence of an eigenvector can be shown in at least two ways:

1. One can argue algebraically: The characteric polynomial of T has a complex root, therefore T has an eigenvalue with a corresponding eigenvector. Or,
2. The eigenvalues can be characterized variationally: The largest eigenvalue is the maximum of the following function on the closed unit ball: f: R²ⁿ → R defined by f(x) = x*Tx = <Tx, x>.

Note In the finite dimensional case, the existence of eigenvectors can be taken for granted. Any square matrix, not necessarily Hermitian, has an eigenvector. This is not true for operators on general Hilbert spaces.

The spectral theorem for the compact self adjoint case can be obtained analogously: one finds an eigenvector by extending either finite-dimensional argument above, then apply induction. This section takes the variational approach. We first sketch the argument for matrices.

Since the closed unit ball B in R²ⁿ is compact, and f is continuous, f(B) is compact on the real line, therefore f attains a maximum on B, at some unit vector y. By Lagrange's mutipliers theorem, y satisfies

for some λ. By Hermiticity, Ty = λy.

However, the Lagrange multipliers do not generalize easily to the infinite dimensional case. Alternatively, let z ∈ Cⁿ be any vector. Notice that if y maximizes <Tx, x>, it also maximizes

Consider the function h: R → R, h(t) = g(y + t z). By calculus, h' (0) = 0, i.e.,

Let m = <Ty, y>/<y, y>. After some algebra the above expression becomes (Re denotes the real part of a complex number)

But z is arbitrary, therefore (T - m)(y) = 0. This is the crux of proof for spectral theorem in the matricial case.

[edit] Details

Let T be a compact self adjoint operator on a Hilbert space H. As in the case of matrices, we consider the function f: H → R defined by f(x) = <Tx, x>. Restrict f to the closed unit ball B ⊂ H. If we can show f attains a maximum at some unit vector y, then, by the same argument used for matrices, y is an eigenvector with corresponding eigenvalue <Ty, y>.

By Banach-Alaoglu theorem and reflexivity of H, B is weakly compact. Also, the compactness of T means the image of a weakly convergent sequences under T is norm convergent. This in turn implies f is continuous in the weak topology on H. The image of B under f is therefore compact in the real line and f attains a maximum m at some y ∈ B.

By maximality, ||y|| = 1, which in turn implies y also maximizes g(x) (same g from above). This shows that y is an eigenvector with corresponding eigenvalue <Ty, y>.

Note The compactness of T is crucial here. In general, f need not be continuous in the weak topology. For example, let T be the identity operator, which is not compact. Take any orthnormal sequence {y_n}. Then y_n converges to 0 weakly but lim f(y_n) = 1 ≠ 0 = f(0).

By self adjointness, the orthogonal compliment of y, {y}^⊥, and span{y} are invariant subspaces of T. By induction, we obtain an orthonormal family {e_n} consisting of all eigenvectors of T. The set {e_n} form a Hilbert space basis: if (span{e_n})^⊥ is a nontrivial subspace, then applying the same procedure gives an eigenvector y not in {e_n}, contradicting the assumption that {e_n} contains all eigenvectors.

Let {λ_n} be the eigenvalues corresponding to {e_n}. Compactness of T and e_n → 0 means T e_n = λ_n e_n → 0. Therefore λ_n → 0.

The above can be summarized by:

Theorem For all compact self adjoint operator T on a Hilbert space H, there exists an orthonormal basis {e_n} consisting of eigenvectors of T, with corresponding eigenvalues {λ_n} ⊂ R, such that λ_n → 0.

In other words, a compact self adjoint operator can be unitarily diagonalized. This is the spectral theorem.

[edit] Functional calculus

The spectral theorem shows σ(T) = {λ_n} ⊂ R, the spectrum of T, consists of only eigenvalues of T, and 0 if 0 is not already an eigenvalue. The set σ(T) inherits a subspace topology from the real line.

Any spectral theorem can be reformulated in terms of a functional calculus. In the present context we have:

Theorem Let C(σ(T)) denote the C*-algebra of continuous functions on σ(T). There exists an isometric homomorphism Φ:C(σ(T)) → L(H) such that Φ(1) = I and, if f(λ) = λ, then Φ(f) = T. Moreoever, σ(f(T)) = f(σ(T)).

The functional calculus map Φ is defined in a natural way: for f ∈ C(σ(T)),

The isometry property comes from diagonalization: By the spectral theorem, the operator norm

Extending to polynomials then invoking the Stone-Weierstrass theorem shows Φ is an isometry.

The other properties of Φ can be readily verified.

[edit] Simultaneous diagonalization

It is known from linear algebra that any commuting family {T_α} of Hermitian matrices can be simultaneously (unitarily) diagonalized.

In the present context, the following is true:

Theorem Let {T_α} be a commuting family of self adjoint operators. Suppose one of them, say T_β is compact. Then there exists an orthonormal basis {e_n} consisting of common eigenvectors of {T_α} such that

The proof extends from the matrix case. Let S₁ be the eigenspace of T_β corresponding to λ₁. Since λ_n → 0, any eigenspace of T_β is finite dimensional. By the assumption that any pair of operators from the family commute, S₁ is an invariant subspace of every T_α. So, according to linear algebra, T_α can be simultaneously diagonalized when restricted to S₁. Induction finishes the argument.

[edit] Compact normal operator

The family of Hermitian matrices is a proper subset of matrices that are unitarily diagonalizable. A matrix M is diagonalizable if and only if it is normal, i.e. M*M = MM*. Similar statements hold for compact normal operators.

Let T be compact and T*T = TT*. Apply the Cartesian decomposition to T: define

The self adjoint compact operators R and J' = (1/i)J are called the real and imaginary parts of T respectively. T is compact means T*, consequently R and J' , are compact. Furthermore, the normality of T implies R and J' commute. Therefore they can be simultaneously diagonalized, from which follows the claim.

[edit] Unitary operator

The spectrum of an unitary operator U lies on the unit circle in the complex plane; it could be the entire unit circle. However, if U is identity plus a compact perturbation, U has only discrete spectrum. More precisely, suppose U = I + C where C is compact. UU* = U*U shows C is normal. Since UC = CU, they can be simultaneously diagonalized, therefore both have only discrete spectrum.

[edit] See also

• Singular value decomposition#Bounded operators on Hilbert spaces. The notion of singular values can be extended from matrices to compact operators.

• Decomposition of spectrum (functional analysis). If the compactness assumption is removed, operators need not have countable spectrum in general.

• Calkin algebra

[edit] References

• J. Blank, P. Exner, and M. Havlicek, Hilbert Space Operators in Quantum Physics, American Institute of Physics, 1994.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics I: Functional Analysis, Academic Press, 1972.