Spectral theory of compact operators

Spectral theory of matrices

The classical 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

:\mathbf{C}^n = \bigoplus _{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.

Compact operators

Statement

Every nonzero λ ∈ σ(C) is an eigenvalue of C. For all nonzero λ ∈ σ(C), there exists m such that Ker((λ − C)m) = Ker((λ − C)m+1), and this subspace is finite-dimensional. It is called the generalized eigenspace for λ. The eigenvalues can only accumulate at 0. σ(C) is at most countably infinite. If the dimension of X is not finite, then σ(C) must contain 0. Every nonzero λ ∈ σ(C) is a pole of the resolvent function ζ → (ζ − C)−1.

Proof

;Preliminary Lemmas

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.

For all ε 0, there exists x ∈ X such that |x| = 1 and

:1 - \varepsilon \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.

Let (I-C) x_n \to y in norm.

If d\left(x_n, \operatorname{Ker}(I-C)\right) is bounded, then there exists a sequence m_n \in \operatorname{Ker}(I-C) such that (x_n - m_n)_n is bounded, and we still have (I-C)(x_n - m_n) \to y.

So WLOG, (x_n)n is bounded. Then compactness of C implies that there exists a subsequence x{n_k} such that C x_{n_k} is norm convergent. So x_{n_k}=(I-C) x_{n_k}+C x_{n_k} is norm convergent, to some x. Thus y = (I-C) x \in \operatorname{Ran}(I-C).

Now we show that d\left(x_n, \operatorname{Ker}(I-C)\right) is bounded.

If not, then select a divergent subsequence x_n, and define vectors z_n = x_n / |x_n + \operatorname{Ker}(I-C)|_{X/\operatorname{Ker}(I-C)}.

Since |x_n + \operatorname{Ker}(I-C)|_{X/\operatorname{Ker}(I-C)} = d\left(x_n, \operatorname{Ker}(I-C)\right) is unbounded, we have (I-C)z_n \to 0. Further, we also have that z_n + M is a sequence of unit vectors in X/\operatorname{Ker}(I-C).

So by the previous half of the proof, there exists a convergent subsequence z_{n_k} \to z, such that (I-C)z = 0, so z \in \operatorname{Ker}(I-C), so z + \operatorname{Ker}(I-C) is a zero vector, contradiction.

;Concluding the Proof i) Without loss of generality, assume λ = 1.

Assume 1 is not an eigenvalue, then I - C is injective. Since it is bounded, but has no bounded inverse, it is not surjective, by the bounded inverse theorem.

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, since I-C is injective. By Riesz's lemma, we can choose unit vectors yn ∈ Yn such that d(yn, Y**n+1) ½. Compactness of C means {C yn} has a convergent subsequence. But for n

: \left | C y_n - C y_m \right | = \left | (C-I) y_n + y_n - (C-I) y_m - y_m \right |

and notice that (C-I) y_n \in Y_{n+1}, ; (C-I) y_m \in Y_{m+1} \subset Y_{n+2}, ; y_m \in Y_{m} \subset Y_{n+1}, thus

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

which implies |Cyn − Cym| ½, so it cannot have a convergent subsequence. Contradiction. ▮

ii) Consider the sequence { Yn = Ker(* λ* − C)n} of closed subspaces. It satisfies Ker(\lambda - C)^1 \subset Ker(\lambda - C)^2 \subset \cdots. If we ever have some Ker(\lambda - C)^n = Ker(\lambda - C)^{n+1}, then the sequence stops increasing from there on.

The theorem claims it stops increasing after finitely many steps. Suppose it does not stop, i.e. the inclusion Ker(* λ* − C)n ⊂ Ker(* λ* − C)n+1 is proper for all n. By Riesz's lemma, there exists a sequence {yn}n ≥ 2 of unit vectors such that yn ∈ Y**n and d(yn, Y**n − 1) ½. As before, compactness of C means {C yn} must contain a norm convergent subsequence. But for n

:| 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(* λ* − C)n} must terminate at some finite m.

Ker(* λ* − C)n is compact by induction on n. By Riesz's lemma, this means it is finite-dimensional.

For n = 1, given any sequence x_n \in Ker(\lambda - C) with norm bounded by 1, by compactness of C, there exists a subsequence such that x_{n_k} = Cx_{n_k} / \lambda \to y. Thus, the closed unit ball in Ker(\lambda - C) is compact.

Induct. Let x_n \in Ker(\lambda - C)^{n+1} be a sequence in the unit ball with norm bounded by 1. Now, (\lambda - C)x_{n_k} is contained within a ball in Ker(\lambda - C)^{n}, which is compact by induction. We also use the compactness of the operator. So we take subsequences twice, to obtain some x_{n_k}, such that Cx_{n_k} and (\lambda - C)x_{n_k} are both convergent, so x_{n_k} is also convergent. ▮

iii) Suppose there exist infinite (at least countable) distinct {λn} in the spectrum, such that |λn| ε for all n. We derive a contradiction, thus concluding that there are no nonzero accumulation points.

By part i, they are eigenvalues. Pick corresponding eigenvectors {xn}.

Define Yn = span{x1...xn}. The sequence {Yn} is a strictly increasing sequence. Choose unit vectors such that yn ∈ Y**n and d(yn, Y**n − 1) ½. Then for n

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

Since (C- \lambda_n) y_n \in Y_{n-1} \subset Y_{m-1}, ; \lambda_n y_n \in Y_n \subset Y_{m-1}, ; (C- \lambda_m) y_m \subset Y_{m-1}, we have

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

therefore |Cyn − Cym| ε/2, a contradiction. ▮

iv) By iii) and the Cantor–Bendixson theorem. ▮

v) If \sigma(C) does not contain zero, then C has a bounded inverse, so I = C^{-1}C is compact, so X is finite-dimensional. ▮

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

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 ≠ μ.

• 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}.

Operators with compact power

If B is an operator on a Banach space X such that Bn is compact for some n, then the theorem proven above also holds for B.

References

• John B. Conway, A course in functional analysis, Graduate Texts in Mathematics 96, Springer 1990.