III.86 The Spectrum

G. R. Allan

In the theory of LINEAR MAPS [I.3 §4.2], or operators, on a VECTOR SPACE [I.3 §2.3], the notions of EIGENVALUE AND EIGENVECROR [I.3 §4.3] play an important role. Recall that if V is a vector space (over or ) and if T : V → V is a linear mapping, then an eigenvector of T is a nonzero vector e in V such that T(e) = λe for some scalar λ then λ is the eigenvalue corresponding to the eigenvector e. If V is finite dimensional, then the eigenvalues are also the roots of the characteristic polynomial χ(t) = det(tI – T) of T. Because every nonconstant complex polynomial has a root (the so-called FUNDAMENTAL THEOREM OF ALGEBRA [V.13]), it follows that every linear operator on a finite-dimensional, complex vector space has at least one eigenvalue. If the scalar field is , then not all operators have eigenvectors (e.g., consider a rotation about the origin in ²).

The linear operators that arise in analysis usually act on infinite-dimensional spaces (see [III.50]). We consider continuous linear operators acting on a complex BANACH SPACE [III.62]; these will be referred to simply as operators (even though not all linear operators on an infinite-dimensional Banach space are continuous). We shall now see that, for X infinite dimensional, not every such operator has an eigenvalue.

Example 1. Let X be the Banach space C[0, 1], consisting of all continuous, complex-valued functions on the closed interval [0, 1] of the real line. The vector-space structure is the “natural” one (e.g., for f, g ∈ X the sum f + g is defined by setting (f + g) (t) = f(t) + g(t) for each t and the norm is the supremum norm, that is, the largest value of any |f(t)|).

Now let u be a continuous complex-valued function on [0, 1]. We can associate with it a multiplication operator M_u on C[0, 1] as follows. Given a function f, let M_u(f) be the function that takes t to u(t)f(t). It is clear that M_u is linear and continuous. We shall see that whether M_u has an eigenvalue depends on the choice of u. We consider two simple cases.

(i) Let u be the constant function u(t) ≡ k. Then evidently M_u has the single eigenvalue k and every (nonzero) function f in X is an eigenvector.

(ii) Let u(t) = t for all t. Suppose that the complex number λ is an eigenvalue of M_u. Then there is some f ∈ C[0, 1], not identically zero, such that u(t)f(t) = λf(t) and so (t – λ) f(t) = 0 for all t. But then f(t) = 0 for all t ≠ λ, so that, since f is continuous, f(t) = 0, contrary to hypothesis. So, for this choice of u, the operator M_u has no eigenvalue.

Let X be a complex Banach space and let T be an operator on X. Then T is said to be invertible if and only if there is some operator S on X for which ST = TS = I (here, ST is the composition of S and T, and I is the identity operator on X). It can be shown that T is invertible if and only if T is both injective (i.e., T(x) = 0 only for x = 0) and surjective (i.e., T(X) = X). The part here that is not just simple algebra is to show that if T is both injective and surjective, then the linear inverse T^–1 is a continuous operator. A complex number λ is an eigenvalue of T precisely if T – λI is not injective.

If V is a finite-dimensional space, then an injective operator T : V → V is necessarily also surjective, and hence invertible. For X infinite dimensional this implication is no longer valid.

Example 2. Let H be the HILBERT SPACE [III.3 7] ² that consists of all sequences of complex numbers such that . Let S be the “right-shift” operator defined by S(ξ₁, ξ₂, ξ₃, . . .) = (0, ξ¹, ξ₂, . . .). Then S is injective but not surjective. The “reverse shift” S*, defined by S*(ξ₁ ξ₂, ξ₃, . . .) = (ξ₂, ξ₃,. . .), is surjective but not injective.

With this example in mind, we make the following definition.

Definition 3. Let X be a complex Banach space and let T be an operator on X. The spectrum of T, denoted by Sp T (or σ(T)), is the set of all complex numbers λ such that T – λI is not invertible.

The following remarks should be clear.

(i) If X is finite dimensional, then Sp T is just the set of eigenvalues of T.

(ii) For general X, Sp T includes the set of eigenvalues of T, but may be larger (e.g., in example 2, 0 is not an eigenvalue of S, but 0 does belong to the spectrum of S).

It is easy to show that the spectrum is always a bounded and closed (i.e., COMPACT [III.9]) subset of . A rather deeper fact is that it is never empty: that is, there will always be some λ for which T – λI is not invertible. That is proved by applying LIOUVILLE’s THEOREM [I.3 §5.6] to the analytic operator-valued function λ (λI – T)^-1, defined for λ not in the spectrum of T.

Example 1 continued. We have already seen that not all multiplication operators have eigenvalues. However, they do have an easily described spectrum. Let M_u be such an operator and let S be the set of all values u(t) taken by the function u. Let μ = u(t₀) be one of these values and consider the operator M_u - μI. Given any function f in C[0, 1], the value of (M_u - μI)f at t₀ is u(t₀)f(t₀) - μf(t₀) = 0. It follows that M_u - μI is not surjective (for instance, the range of M_u - μI does not contain any nonzero constant function) and therefore μ belongs to the spectrum of M_u. Thus, S is contained in the spectrum of M_u; it is not hard to show that the two are in fact equal.

We may easily generalize this example to show that if K is any nonempty compact subset of , then there is a linear operator T with K as its spectrum. Let X be the space of continuous complex-valued functions defined on K, for each z ∈ K, let u(z) = z, and let T be the multiplication operator M_u, defined as it was when K was the set [0, 1].

The spectrum is central to most aspects of operator theory. We shall briefly mention a result about Hilbert space operators, known as the spectral theorem (there are a number of variations).

Let H be a Hilbert space with inner product 〈x,y〉. A continuous linear operator T on H is called Hermitkin if 〈Tx,y〉 = 〈x, Ty〉 for every x and y in H.

Examples 4.

(i) If H is finite dimensional, then a linear operator S on H is Hermitian if and only if, with respect to some (and hence every) ORTHONORMHL BASIS [III.37], S is represented by a Hermitian matrix (i.e., a matrix A with A = ^T).

(ii) On the Hilbert space L₂[0, 1], let M_u be the operator of multiplication by a continuous function u (just as in example 1, but now we apply M_u to functions in L₂[0,1] rather than just C[0, 1]). Then M_u is Hermitian if and only if u is real-valued.

If H is finite dimensional and T is a Hermitian operator on H, then H has an orthonormal basis consisting of eigenvectors of T (a “diagonal basis”). Equivalently, T = λ_jP_j, where {λ₁, . . . , λ_k} are the distinct eigenvalues of T and P_j is the orthogonal projection of H onto the eigenspace E_j ≡ {x ∈ H : Tx = λ_jx}.

If H is infinite dimensional and T is a Hermitian operator on H, then it is not generally true that H has a basis of eigenvectors. But, very importantly, the representation T = Σλ_jP_j does generalize to a representation T = ∫ λdP, a kind of integral with respect to a “projection-valued measure” on the spectrum of T.

There is an intermediate case, for so-called compact Hermitian operators, “compactness” being a kind of strong continuity, of great importance in applications. The technicalities are much simpler than in the general case, involving an infinite sum, rather than an integral. A very readable introduction may be found in Young (1988).

Further Reading

Young, N. 1988. An Introduction to Hilbert Space. Cambridge: Cambridge University Press.