Functional Analysis 33 | Spectrum of Compact Operators [dark version]

TL;DR

Compact operators are defined by mapping bounded sets to sets whose closure is compact, and it suffices to check the unit ball.

Briefing Cornell Notes

Briefing

Compact operators behave like “infinite-dimensional matrices”: they take bounded sets to sets whose closure is compact, and that finiteness-like property sharply constrains what can happen to their spectrum. For a bounded linear operator T on a Banach space X, compactness forces the spectrum to be countable at worst, and—crucially—zero must always lie in the spectrum. This already marks a major departure from the finite-dimensional case, where the spectrum is a finite, nonempty set and zero need not be special.

More precisely, for a compact operator T on an infinite-dimensional Banach space, the spectrum can be finite or countably infinite. If it is infinite, its nonzero spectral values must accumulate only at 0: there cannot be any other accumulation point in the complex plane. In practical terms, the spectrum away from 0 can be listed as eigenvalues {Λk} with Λk → 0. Every nonzero spectral value is an eigenvalue (belongs to the point spectrum), and its eigenspace is finite-dimensional—again echoing the matrix situation. Zero is the only possible “limit point,” but it may or may not be an eigenvalue; when it is an eigenvalue, its eigenspace could be finite-dimensional or infinite-dimensional.

The transcript then grounds these abstract facts with a concrete example on the Hilbert space L2. Define an operator T by (Tx)j = (1/j) xj, i.e., T scales the j-th coordinate by 1/j. This operator is bounded and compact because the image of the unit ball sits inside a “Hilbert cube”: vectors y in L2 whose j-th component satisfies |yj| ≤ 1/j. That set is compact in L2, so the closure of T(B1) remains compact, matching the definition of compactness.

Because T acts diagonally in the standard coordinate basis, its spectrum is easy to read off like a diagonal matrix. The eigenvalues are exactly the diagonal entries: 1, 1/2, 1/3, …, and each corresponds to the coordinate vector e_k, giving an eigenspace that is one-dimensional. The eigenvalues form a sequence that converges to 0, so 0 is included in the spectrum as the only accumulation point. In this example, 0 is not an eigenvalue; instead, it appears as part of the continuous spectrum. The overall picture is that compact operators can have infinitely many eigenvalues, but they must “pile up” at 0, and the nonzero part of the spectrum remains eigenvalue-driven with finite-dimensional eigenspaces.

The discussion closes by noting that diagonal structure makes the example straightforward, and that extending such diagonalization ideas to general compact operators is the next step in the series.

Cornell Notes

Compact operators on a Banach space send bounded sets to sets with compact closure, which forces strong restrictions on their spectrum. For a compact operator T, the spectrum is always nonempty and is either finite or countably infinite. In the infinite-dimensional setting, 0 always belongs to the spectrum, and any nonzero spectral values must be eigenvalues whose eigenspaces are finite-dimensional. If there are infinitely many nonzero eigenvalues {Λk}, they can accumulate only at 0, meaning Λk → 0 and no other accumulation point exists. Zero may or may not be an eigenvalue; when it is, its eigenspace can be finite- or infinite-dimensional.

Why does compactness force 0 to be in the spectrum for operators on infinite-dimensional spaces?

Compactness prevents the operator from behaving like an invertible map on “infinitely many directions” without losing control. The key spectral consequence stated is that for compact operators on infinite-dimensional Banach spaces, 0 is always an element of the spectrum. Intuitively, if nonzero spectral values existed without 0, the operator would resemble a finite-dimensional invertible situation, but compactness rules that out in infinite dimensions.

What can the spectrum of a compact operator look like away from 0?

Away from 0, the spectrum is countable and consists entirely of eigenvalues. Every nonzero Λ in the spectrum lies in the point spectrum, so there exists a nonzero eigenvector x with Tx = Λx. Moreover, the eigenspace ker(T − ΛI) is finite-dimensional for each nonzero Λ.

If a compact operator has infinitely many nonzero eigenvalues, where can they accumulate?

They can only accumulate at 0. Formally, the closure of the set of eigenvalues in C has no limit points except 0. Equivalently, the nonzero eigenvalues can be listed as a sequence Λk with Λk → 0, and there cannot be any other accumulation point in the complex plane.

How does the example T(x)j = (1/j)xj on L2 illustrate the general spectral rules?

The operator is diagonal in the coordinate basis, so its eigenvalues are exactly the diagonal entries: 1, 1/2, 1/3, …, with eigenvectors ek (the k-th coordinate vector). Each eigenspace is one-dimensional. The eigenvalues form a sequence converging to 0, so 0 lies in the spectrum. In this specific diagonal example, 0 is not an eigenvalue; it appears as continuous spectrum.

Can 0 be an eigenvalue for compact operators, and what determines the eigenspace size?

Yes, 0 may or may not be an eigenvalue for a compact operator. If 0 is an eigenvalue, the eigenspace ker(T) can be finite-dimensional or infinite-dimensional. The transcript emphasizes that only nonzero spectral points are guaranteed to have finite-dimensional eigenspaces; 0 is the exception where both possibilities occur.

Review Questions

For a compact operator T on an infinite-dimensional Banach space, what are the only possible accumulation points of its spectrum?
Why must every nonzero spectral value of a compact operator be an eigenvalue, and what can be said about the dimension of its eigenspace?
In the L2 example with (Tx)j = (1/j)xj, identify the eigenvalues and explain why 0 belongs to the spectrum even though it is not an eigenvalue.

Key Points

1
Compact operators are defined by mapping bounded sets to sets whose closure is compact, and it suffices to check the unit ball.
2
For compact operators on infinite-dimensional spaces, the spectrum is nonempty and 0 always belongs to it.
3
The spectrum of a compact operator is either finite or countably infinite; if infinite, nonzero spectral values form a sequence converging to 0.
4
Every nonzero spectral value of a compact operator is an eigenvalue, and its eigenspace is finite-dimensional.
5
Zero may or may not be an eigenvalue; if it is, its eigenspace can be finite- or infinite-dimensional.
6
In the L2 example T(x)j = (1/j)xj, the eigenvalues are 1/k with eigenvectors ek, and 0 is in the spectrum as continuous spectrum.

Highlights

Compactness turns spectral behavior into a “matrix-like” pattern: nonzero spectrum consists of eigenvalues with finite-dimensional eigenspaces.

For compact operators, any infinite list of eigenvalues must converge to 0—no other accumulation point is allowed.

A diagonal compact operator on L2 with entries 1/j has spectrum {1, 1/2, 1/3, …} ∪ {0}, with 0 appearing as continuous spectrum.

Even though 0 always lies in the spectrum, it need not be an eigenvalue; that distinction matters for the spectral type.

Topics

Compact Operators
Spectrum
Eigenvalues
Hilbert Cube
Continuous Spectrum