# Spectral theory of compact operators

In functional analysis, compact operators are linear operators that map bounded sets to relatively compact sets. The set of compact operators acting on a Hilbert space H is the closure of the set 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 summarizes 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 matrix case.

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

## 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(λiA)m where Ker(λiA)m = Ker(λiA)m+1. Furthermore, the poles of the resolvent function ζ → (ζA)−1 coincide with the set of eigenvalues of A.

## Compact operators

### 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 λσ(C) is an eigenvalue of C.

ii) For all nonzero λσ(C), there exist m such that Ker(λiC)m = Ker(λiC)m+1, and this subspace is finite-dimensional.

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

iv) σ(C) is countable.

v) Every nonzero λσ(C) is a pole of the resolvent function ζ → (ζC)−1.

### Proof

#### Preliminary Lemmas

The theorem claims several properties of the operator λC where λ ≠ 0. Without loss of generality, it can be assumed that λ = 1. Therefore we consider IC, I being the identity operator. The proof will require two lemmas.

Lemma 1 (Riesz's lemma). Let X be a Banach space and YX, YX, be a closed subspace. For all ε > 0, there exists xX such that ||x|| = 1 and

$1-\varepsilon \leq d(x,Y)\leq 1$ 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.

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

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

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

#### Concluding the Proof

i) Without loss of generality, assume λ = 1. λσ(C) not being an eigenvalue means (IC) is injective but not surjective. By Lemma 2, Y1 = Ran(IC) is a closed proper subspace of X. Since (IC) is injective, Y2 = (IC)Y1 is again a closed proper subspace of Y1. Define Yn = Ran(IC)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 ynYn such that d(yn, Yn+1) > ½. Compactness of C means {C yn} must contain a norm convergent subsequence. But for n < m

$\left\|Cy_{n}-Cy_{m}\right\|=\left\|(C-I)y_{n}+y_{n}-(C-I)y_{m}-y_{m}\right\|$ 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(λiA)n} is an increasing sequence of closed subspaces. The theorem claims it stops. Suppose it does not stop, i.e. the inclusion Ker(λiA)nKer(λiA)n+1 is proper for all n. By lemma 1, there exists a sequence {yn}n ≥ 2 of unit vectors such that ynYn and d(yn, Yn − 1) > ½. As before, compactness of C means {C yn} must contain a norm convergent subsequence. But for n < m

$\|Cy_{n}-Cy_{m}\|=\|(C-I)y_{n}+y_{n}-(C-I)y_{m}-y_{m}\|$ 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(λiA)n} must terminate at some finite m.

Using the definition of the Kernel, we can show that the unit sphere of Ker(λiC) is compact, so that Ker(λiC) is finite-dimensional. Ker(λiC)n is finite-dimensional for the same reason.

iii) Suppose there exist infinite (at least countable) 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 ynYn and d(yn, Yn − 1) > ½. Then for n < m

$\left\|Cy_{n}-Cy_{m}\right\|=\left\|(C-\lambda _{n})y_{n}+\lambda _{n}y_{n}-(C-\lambda _{m})y_{m}-\lambda _{m}y_{m}\right\|.$ But

$(C-\lambda _{n})y_{n}+\lambda _{n}y_{n}-(C-\lambda _{m})y_{m}\in Y_{m-1},$ therefore ||Cyn − Cym|| > ε/2, a contradiction.

So we have that there are only finite distinct eigenvalues outside any ball centered at zero. This immediately gives us that zero is the only possible limit point of eigenvalues and there are at most countable distinct eigenvalues (see iv).

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

$\bigcup _{n}\left\{|\lambda |>{\tfrac {1}{n}}\right\}=\bigcup _{n}S_{n}.$ 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.

### 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(λC)ν = Ker(λC)ν + 1. By inspecting the Jordan form, we see that (λC)ν = 0 while (λC)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(λC)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.