Spectral theory of compact operators

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In functional analysis, compact operators are linear operators that map bounded sets to precompact ones. Compact operators acting on a Hilbert space H is the closure of finite rank operators in the uniform operator topology. In general, operators on infinite dimensional spaces feature properties that do not appear in the finite dimensional case, i.e. for matrices. The family of compact operators are notable in that they share as much similarity with matrices as one can expect from a general operator. In particular, the spectral properties of compact operators resemble those of square matrices.

This article first summarize the corresponding results from the matrix case before discussing the spectral properties of compact operators. The reader will see that most statements transfer verbatim from the matricial case.

The spectral theory of compact operators was first developed by F. Riesz.

Contents

[edit] Spectral theory of matrices

Further information: Jordan canonical form

The classification result for square matrices is the Jordan canonical form, which states the following:

Theorem Let A be an n × n complex matrix, i.e. A a linear operator acting on Cn. If λ1...λk are the distinct eigenvalues of A, then Cn can be decomposed into the invariant subspaces of A

\mathbb{C}^n = \oplus _{i = 1} ^k Y_i.

The subspace Yi = Ker(λi - A)m where Ker(λi - A)m = Ker(λi - A)m+1. Furthermore, the poles of the resolvent function ζ → (ζ - A)-1 coincide with the set of eigenvalues of A.

[edit] Compact operators

[edit] Statement

Let X be a Banach space, C be a compact operator acting on X, and σ(C) be the spectrum of C. The spectral properties of C are:

Theorem

i) Every nonzero λ ∈ σ(C) is an eigenvalue of C.

ii) For all nonzero λ ∈ σ(C), there exist m such that Ker(λi - A)m = Ker(λi - A)m+1.

iii) The eigenvalues can only accumulate at 0. If the dimension of X is not finite, then σ(C) must accumulate at 0.

iv) σ(C) is countable.

v) Every nonzero λ ∈ σ(C) is a pole of the resolvent function ζ → (ζ - C)-1.

[edit] Proof

The theorem claims several properties of the operator λ - C where λ ≠ 0. Without loss of generality, it can be assumed that λ = 1. Therefore we consider I - C, I being the identity operator. The proof will require two lemmas. The first is called Riesz's lemma:

Lemma 1 Let X be a Banach space and Y ⊂ X, Y ≠ X, be a closed subspace. For all ε > 0, there exists x ∈ X such that ||x|| = 1 and

1 - \epsilon \le d(x, Y) \le 1

where d(x, Y) is the distance from x to Y.

This fact will be used repeatedly in the argument leading to the theorem. Notice that when X is a Hilbert space, the lemma is trivial.

Another useful fact is:

Lemma 2 If C is compact, then Ran(I - C) is closed.

Proof: Let (I - C)xn → y in norm. If {xn} is bounded, then there exist a weakly convergent subsequence xn k. Compactness of C implies C xn k is norm convergent. So xn k = (I - C)xn k + C xn k is norm convergent, to some x. This gives (I - C)xn k → (I - C)x = y. The same argument goes through if the distances d(xn, Ker(I - C)) is bounded.

But d(xn, Ker(I - C)) must be bounded. Suppose this is not the case. Pass now to the quotient map of (I - C), still denoted by (I - C), on X/Ker(I - C). The quotient norm on X/Ker(I - C) is still denoted by ||·||, and {xn} are now viewed as representatives of their equivalence classes in the quotient space. Take a subsequence {xn k} such that ||xn k|| > k and define a sequence of unit vectors by zn k = xn k/||xn k||. Again we would have (I - C)zn k → (I - C)z for some z. Since ||(I - C)zn k|| = ||(I - C)xn k||/ ||xn k|| → 0, we have (I - C)z = 0 i.e. z ∈ Ker(I - C). Since we passed to the quotient map, z = 0. This is impossible because z is the norm limit of a sequence of unit vectors. Thus the lemma is proven.

We are now ready to prove the theorem.

i) Without loss of generality, assume λ = 1. λ ∈ σ(C) not being an eigenvalue means (I - C) is injective but not surjective. By Lemma 2, Y1 = Ran(I - C) is a closed proper subspace of X. Since (I - C) is injective, Y2 = (I - C)Y1 is again a closed proper subspace of Y1. Define Yn = Ran(I - C)n. Consider the decreasing sequence of subspaces

Y_1 \supset \cdots \supset Y_n \cdots \supset Y_m \cdots

where all inclusions are proper. By lemma 1, we can choose unit vectors yn ∈ Yn such that d(yn, Yn+1) > ½. Compactness of C means {C yn} must contain a norm convergent subsequence. But for n < m

\| C y_n - C y_m \| = \| (C-I) y_n + y_n - (C-I) y_m - y_m \|

and notice that

(C-I) y_n - (C-I) y_m - y_m \in Y_{n+1},

which implies ||Cyn - Cym|| > ½. This is a contradiction, and so λ must be an eigenvalue.

ii) The sequence { Yn = Ker(λi - A)n} is an increasing sequence of closed subspaces. The theorem claims it stops. Suppose it does not stop, i.e. the inclusion Ker(λi - A)n ⊂ Ker(λi - A)n+1 is proper for all n. By lemma 1, there exists a sequence {yn}n ≥ 2 of unit vectors such that yn ∈ Yn and d(yn, Yn - 1) > ½. As before, compactness of C means {C yn} must contain a norm convergent subsequence. But for n < m

\| C y_n - C y_m \| = \| (C-I) y_n + y_n - (C-I) y_m - y_m \|

and notice that

(C-I) y_n + y_n - (C-I) y_m \in Y_{m-1},

which implies ||Cyn - Cym|| > ½. This is a contradiction, and so the sequence { Yn = Ker(λi - A)n} must terminate at some finite m.

iii) Suppose eigenvalues of C do not accumulate at 0. We can therefore assume that there exist a sequence of distinct eigenvalues {λn}, with corresponding eigenvectors {xn}, such that |λn| > ε for all n. Define Yn = span{x1...xn}. The sequence {Yn} is a strictly increasing sequence. Choose unit vectors such that yn ∈ Yn and d(yn, Yn - 1) = 1. Then for n < m

\| C y_n - C y_m \| = \| (C- \lambda_n) y_n + \lambda_n y_n - (C- \lambda_m) y_m - \lambda_m y_m \|.

But

(C- \lambda_n) y_n + \lambda_n y_n - (C- \lambda_m) y_m \in Y_{m-1},

therefore ||Cyn - Cym|| > ε, a contradiction.

iv) This is an immediate consequence of iii). The set of eigenvalues {λ} is the union

\cup_n \{ |\lambda| > \frac{1}{n} \} = \cup_n S_n .

Because σ(C) is a bounded set and the eigenvalues can only accumulate at 0, each Sn is finite, which gives the desired result.

v) As in the matrix case, this is a direct application of the holomorphic functional calculus.

[edit] Invariant subspaces

As in the matrix case, the above spectral properties lead to a decomposition of X into invariant subspaces of a compact operator C. Let λ ≠ 0 be an eigenvalue of C; so λ is an isolated point of σ(C). Using the holomorphic functional calculus, define the Riesz projection E(λ) by

\, E(\lambda) = {1\over 2\pi i}\int _{\gamma} (\xi - C)^{-1} d \xi

where γ is a Jordan contour that encloses only λ from σ(C). Let Y be the subspace Y = E(λ)X. C restricted to Y is a compact invertible operator with spectrum {λ}, therefore Y is finite dimensional. Let ν be such that Ker(λ - C)ν = Ker(λ - C)ν + 1. By inspecting the Jordan form, we see that (λ - C)ν = 0 while (λ - C)ν - 1 ≠ 0. The Laurent series of the resolvent mapping centered at λ shows that

\, E(\lambda) (\lambda - C)^{\nu} = (\lambda - C)^{\nu}E(\lambda) = 0.

So Y = Ker(λ - C)ν.

The E(λ) satisfy E(λ)2 = E(λ), so that they are indeed projection operators or spectral projections. By definition they commute with C. Moreover E(λ)E(μ) = 0 if λ and μ are distinct.

  • Let X(λ) = E(λ)X if λ is a non-zero eigenvalue. Thus X(λ) is a finite-dimensional invariant subspace, the generalised eigenspace of λ.
  • Let X(0) be the intersection of the kernels of the E(λ). Thus X(0) is a closed subspace invariant under C and the restriction of C to X(0) is a compact operator with spectrum {0}.

[edit] Operator with compact power

Let B be an operator on X such that Bn is compact for some n. The theorem proven above holds for B.

[edit] References

  • John B. Conway, A course in functional analysis, Graduate Texts in Mathematics 96, Springer 1990. ISBN 0-387-97245-5