spectral theory of compact operators

{{further|Jordan canonical form}}

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...λ_k are the distinct eigenvalues of A, then Cⁿ can be decomposed into the invariant subspaces of A

:$\backslash mathbf\{C\}^n\; =\; \backslash bigoplus\; \_\{i\; =\; 1\}^k\; Y\_i.$

The subspace Y_i = Ker(λ_i − A)^m where Ker(λ_i − A)^m = Ker(λ_i − A)^m+1. Furthermore, the poles of the resolvent function ζ → (ζ − A)⁻¹ coincide with the set of eigenvalues of A.

{{Math theorem|name=Theorem|math_statement=Let X be a Banach space, C be a compact operator acting on X, and σ(C) be the spectrum of C.

1. Every nonzero λ ∈ σ(C) is an eigenvalue of C.
2. 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.
3. The eigenvalues can only accumulate at 0.
4. σ(C) is at most countably infinite.
5. If the dimension of X is not finite, then σ(C) must contain 0.
6. Every nonzero λ ∈ σ(C) is a pole of the resolvent function ζ → (ζ − C)⁻¹.

}}

;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.

{{Math theorem|name=Lemma 1|note=Riesz's lemma|math_statement=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 $\backslash |$x$\backslash |$ = 1 and

:$1\; -\; \backslash varepsilon\; \backslash le\; d(x,\; Y)\; \backslash le\; 1$

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.

{{Math theorem|name=Lemma 2|math_statement=If C is compact, then Ran(I − C) is closed.}}

{{Math proof|drop=hidden|proof=

Let $(I-C)\; x\_n\; \backslash to\; y$ in norm.

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

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 $y\; =\; (I-C)\; x\; \backslash in\; \backslash operatorname\{Ran\}(I-C)$.

Now we show that $d\backslash left(x\_n,\; \backslash operatorname\{Ker\}(I-C)\backslash right)$ is bounded.

If not, then select a divergent subsequence $x\_n$, and define vectors $z\_n\; =\; x\_n\; /\; \backslash |x\_n\; +\; \backslash operatorname\{Ker\}(I-C)\backslash |\_\{X/\backslash operatorname\{Ker\}(I-C)\}$.

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

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

}}

;Concluding the Proof

{{Math proof|drop=hidden|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 Y_n = Ran(I − C)ⁿ. Consider the decreasing sequence of subspaces

:$Y\_1\; \backslash supset\; \backslash cdots\; \backslash supset\; Y\_n\; \backslash cdots\; \backslash supset\; Y\_m\; \backslash cdots$

where all inclusions are proper, since $I-C$ is injective. By Riesz's lemma, we can choose unit vectors y_n ∈ Y_n such that d(y_n, Y_n+1) > ½.

Compactness of C means {C y_n} has a convergent subsequence. But for n < m

:$\backslash left\; \backslash |\; C\; y\_n\; -\; C\; y\_m\; \backslash right\; \backslash |\; =\; \backslash left\; \backslash |\; (C-I)\; y\_n\; +\; y\_n\; -\; (C-I)\; y\_m\; -\; y\_m\; \backslash right\; \backslash |$

and notice that $(C-I)\; y\_n\; \backslash in\; Y\_\{n+1\},\; \backslash ;\; (C-I)\; y\_m\; \backslash in\; Y\_\{m+1\}\; \backslash subset\; Y\_\{n+2\},\; \backslash ;\; y\_m\; \backslash in\; Y\_\{m\}\; \backslash subset\; Y\_\{n+1\}$, thus

:$(C-I)\; y\_n\; -\; (C-I)\; y\_m\; -\; y\_m\; \backslash in\; Y\_\{n+1\},$

which implies $\backslash |$Cy_n − Cy_m$\backslash |$ > ½, so it cannot have a convergent subsequence. Contradiction. ▮

ii) Consider the sequence { Y_n = Ker( λ − A)ⁿ} of closed subspaces. It satisfies $Ker(\backslash lambda\; -\; A)^1\; \backslash subset\; Ker(\backslash lambda\; -\; A)^2\; \backslash subset\; \backslash cdots$. If we ever have some $Ker(\backslash lambda\; -\; A)^n\; =\; Ker(\backslash lambda\; -\; A)^\{n+1\}$, 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 − A)ⁿ ⊂ Ker( λ − A)ⁿ⁺¹ is proper for all n.

By Riesz's lemma, there exists a sequence {y_n}_{n ≥ 2} of unit vectors such that y_n ∈ Y_n and d(y_n, Y_{n − 1}) > ½.

As before, compactness of C means {C y_n} must contain a norm convergent subsequence. But for n < m

:$\backslash |\; C\; y\_n\; -\; C\; y\_m\; \backslash |\; =\; \backslash |\; (C-I)\; y\_n\; +\; y\_n\; -\; (C-I)\; y\_m\; -\; y\_m\; \backslash |$

and notice that

:$(C-I)\; y\_n\; +\; y\_n\; -\; (C-I)\; y\_m\; \backslash in\; Y\_\{m-1\},$

which implies $\backslash |$Cy_n − Cy_m$\backslash |$ > ½.

This is a contradiction, and so the sequence { Y_n = Ker( λ − A)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 $x\_n\; \backslash in\; Ker(\backslash lambda\; -\; C)$ with norm bounded by 1, by compactness of $C$, there exists a subsequence such that $x\_\{n\_k\}\; =\; Cx\_\{n\_k\}\; /\; \backslash lambda\; \backslash to\; y$. Thus, the closed unit ball in $Ker(\backslash lambda\; -\; C)$ is compact.

Induct. Let $x\_n\; \backslash in\; Ker(\backslash lambda\; -\; C)^\{n+1\}$ be a sequence in the unit ball with norm bounded by 1. Now, $(\backslash lambda\; -\; C)x\_\{n\_k\}$ is contained within a ball in $Ker(\backslash lambda\; -\; C)^\{n\}$, 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 $(\backslash lambda\; -\; C)x\_\{n\_k\}$ are both convergent, so $x\_\{n\_k\}$ is also convergent.

▮

iii) Suppose there exist infinite (at least countable) distinct {λ_n} in the spectrum, such that $|$λ_n$|$ > 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 {x_n}.

Define Y_n = span{x₁...x_n}.

The sequence {Y_n} is a strictly increasing sequence.

Choose unit vectors such that y_n ∈ Yⁿ and d(y_n, Y^{n − 1}) > ½. Then for n < m

:$\backslash left\; \backslash |\; C\; y\_n\; -\; C\; y\_m\; \backslash right\; \backslash |\; =\; \backslash left\; \backslash |\; (C-\; \backslash lambda\_n)\; y\_n\; +\; \backslash lambda\_n\; y\_n\; -\; (C-\; \backslash lambda\_m)\; y\_m\; -\; \backslash lambda\_m\; y\_m\; \backslash right\; \backslash |.$

Since $(C-\; \backslash lambda\_n)\; y\_n\; \backslash in\; Y\_\{n-1\}\; \backslash subset\; Y\_\{m-1\},\; \backslash ;\; \backslash lambda\_n\; y\_n\; \backslash in\; Y\_n\; \backslash subset\; Y\_\{m-1\},\; \backslash ;\; (C-\; \backslash lambda\_m)\; y\_m\; \backslash subset\; Y\_\{m-1\}$, we have

:$(C-\; \backslash lambda\_n)\; y\_n\; +\; \backslash lambda\_n\; y\_n\; -\; (C-\; \backslash lambda\_m)\; y\_m\; \backslash in\; Y\_\{m-1\},$

therefore $\backslash |$Cy_n − Cy_m$\backslash |$ > ε/2, a contradiction. ▮

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

v) If $\backslash sigma(C)$ 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. ▮

}}

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

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

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

:$E(\backslash lambda)\; (\backslash lambda\; -\; C)^\{\backslash nu\}\; =\; (\backslash lambda\; -\; C)^\{\backslash nu\}E(\backslash lambda)\; =\; 0.$

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}.

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.

• Spectral theorem
• Spectral theory of normal C*-algebras

{{Reflist}}

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

{{Functional analysis}}

{{Spectral theory}}

Category:Functional analysis

Category:Spectral theory

Category:Linear operators