Resolvent and Spectrum of an Operator — Mathematics Knowledge System

Overview

The spectrum of a bounded linear operator T on a Banach space generalises the set of eigenvalues from linear algebra. The spectrum sigma(T) consists of all scalars lambda for which T - lambda I is not invertible. The resolvent set rho(T) is its complement, and on rho(T) the resolvent operator R(lambda, T) = (T - lambda I)^{-1} is a bounded operator-valued analytic function. Spectral theory underpins the spectral theorem for self-adjoint operators, perturbation theory, and the study of differential equations.

Intuition

For a finite matrix, the spectrum is exactly the set of eigenvalues. In infinite dimensions, there are more ways to fail invertibility: the operator could fail to be injective (eigenvalue), or injective but not have dense range (residual spectrum), or have dense but non-closed range (continuous spectrum). The resolvent R(lambda, T) = (T - lambda I)^{-1} encodes all spectral information: it is analytic outside sigma(T) and has singularities on sigma(T). Spectral theory is the study of these singularities.

Formal Definition

Definition

Let T in B(X) where X is a complex Banach space. The resolvent set rho(T) = {lambda in C : T - lambda I is bijective with bounded inverse}. The spectrum sigma(T) = C \ rho(T). The resolvent operator is R(lambda, T) = (T - lambda I)^{-1} in B(X) for lambda in rho(T). The spectrum decomposes: the point spectrum sigma_p(T) consists of eigenvalues (T - lambda I not injective), the continuous spectrum sigma_c(T) where T - lambda I is injective with dense but not closed range, and the residual spectrum sigma_r(T) where T - lambda I is injective with non-dense range.

\rho(T) = \{\lambda \in \mathbb{C} : (T - \lambda I)^{-1} \in B(X)\}

Resolvent set

R(\lambda, T) = (T - \lambda I)^{-1},\quad \lambda \in \rho(T)

Resolvent operator

R(\lambda, T) - R(\mu, T) = (\mu - \lambda)\, R(\lambda, T)\, R(\mu, T)

Resolvent identity

\sigma(T) \subseteq \overline{B}(0, \|T\|),\quad r(T) = \sup_{\lambda \in \sigma(T)} |\lambda| = \lim_{n\to\infty}\|T^n\|^{1/n}

Spectral radius

Notation

Notation	Meaning
$\sigma(T)$	Spectrum of operator T
$\rho(T)$	Resolvent set of T
$R(\lambda, T)$	Resolvent operator at lambda
$r(T)$	Spectral radius of T
$\sigma_p(T)$	Point spectrum (eigenvalues)

Theorems

Theorem 4.1: Theorem 4.1

For T in B(X), sigma(T) is a non-empty compact subset of C contained in the disk |lambda| <= ||T||. The resolvent R(lambda, T) is an analytic B(X)-valued function on rho(T).

Theorem 4.2: Theorem 4.2 (Resolvent Identity)

For lambda, mu in rho(T): R(lambda,T) - R(mu,T) = (mu - lambda) R(lambda,T) R(mu,T). In particular, R(lambda,T) and R(mu,T) commute.

Theorem 4.3: Theorem 4.3 (Spectral Mapping Theorem)

For p a polynomial, sigma(p(T)) = p(sigma(T)) = {p(lambda) : lambda in sigma(T)}. More generally, for f analytic on a neighbourhood of sigma(T), sigma(f(T)) = f(sigma(T)).

Worked Examples

1
Point spectrum: If Rx = lambda x then (0,x_1,x_2,...) = (lambda x_1, lambda x_2,...). The first component gives 0 = lambda x_1. If lambda != 0 then x_1 = 0, then x_2 = 0, etc., forcing x = 0. If lambda = 0, Rx = 0 requires x = 0. So sigma_p(R) = {}.
2
For |lambda| > 1: ||(R - lambda I)|| >= |lambda| - ||R|| = |lambda| - 1 > 0 and (R - lambda I) is invertible by a Neumann argument (or directly: (R - lambda I)^{-1} exists and is bounded). So |lambda| > 1 implies lambda in rho(R).
3
For |lambda| < 1: The adjoint R* is the left shift: R*(x_1,x_2,...) = (x_2,x_3,...). We check R*e_1 = 0 where e_1 = (1,0,0,...), so 0 is an eigenvalue of R*, meaning 0 in sigma_p(R*). For any lambda with |lambda| < 1, one shows lambda is an eigenvalue of R*, hence lambda in sigma_r(R) subset sigma(R).
4
For |lambda| = 1: A limiting argument shows lambda in sigma(R). Combining, sigma(R) = {lambda : |lambda| <= 1}.
$\sigma(R) = \overline{\mathbb{D}} = \{\lambda \in \mathbb{C} : |\lambda| \le 1\}$

✓ Answer

The spectrum of the right shift on l^2 is the closed unit disk, with empty point spectrum. The spectral radius is 1 = ||R||.

Practice Problems

Mediumproof writing

Show that the spectrum sigma(T) of T in B(X) is a closed subset of C.

MediumMultiple choice

For T in B(X), which is always true about sigma(T)?

sigma(T) is emptysigma(T) contains only eigenvaluessigma(T) is a non-empty compact subset of Csigma(T) = {lambda : |lambda| = ||T||}

Quiz

Summary

The spectrum sigma(T) = C \ rho(T) generalises eigenvalues to infinite-dimensional spaces.
sigma(T) is always a non-empty compact subset of the disk |lambda| <= ||T||.
The resolvent R(lambda,T) = (T - lambda I)^{-1} is analytic in lambda on rho(T).
The resolvent identity: R(lambda,T) - R(mu,T) = (mu - lambda)R(lambda,T)R(mu,T).
The spectrum decomposes into point, continuous, and residual parts.

References

BookRudin, W. — Functional Analysis (2nd ed.), Chapter 10
BookKato, T. — Perturbation Theory for Linear Operators, Chapter III
Websiteen.wikipedia.org