Mathematics.

functional analysis

Spectral Theory of Operators

Real Analysis120 minDifficulty9 out of 10

You should know: hilbert spaces, eigenvalues and eigenvectors

Overview

Spectral theory extends the notion of eigenvalues and eigenvectors from finite-dimensional linear algebra to operators on infinite-dimensional Banach and Hilbert spaces. The spectrum of an operator \(T\) replaces the finite set of eigenvalues and decomposes into the point spectrum (eigenvalues), continuous spectrum, and residual spectrum. For bounded self-adjoint operators on Hilbert spaces, the spectral theorem provides a complete decomposition analogous to diagonalisation: every such operator can be realised as a multiplication operator on an \(L^2\) space, encoded by a projection-valued measure.

Intuition

In finite dimensions, every symmetric matrix can be diagonalised. Spectral theory asks: can we diagonalise an infinite-dimensional operator? For compact self-adjoint operators the answer is an exact analogue of the finite case—eigenvalues accumulate only at 0, and eigenvectors form a complete orthonormal basis. For general bounded self-adjoint operators, diagonalisation requires a continuum of generalised eigenvectors encoded by a spectral measure. The spectrum is the set of values at which the operator almost fails to be invertible.

Formal Definition

Definition

Let \(T \in \mathcal{B}(X)\) be a bounded linear operator on a complex Banach space \(X\). The resolvent set is \(\rho(T) = \{\lambda \in \mathbb{C} : (T - \lambda I)^{-1} \in \mathcal{B}(X)\}\). The spectrum is \(\sigma(T) = \mathbb{C} \setminus \rho(T)\). The spectrum decomposes as: point spectrum \(\sigma_p(T)\) where \(T - \lambda I\) is not injective; continuous spectrum \(\sigma_c(T)\) where \(T - \lambda I\) is injective with dense but not closed range; residual spectrum \(\sigma_r(T)\) where \(T - \lambda I\) is injective but the range is not dense.

σ(T)={λC:(TλI) is not boundedly invertible}\sigma(T) = \{\lambda \in \mathbb{C} : (T - \lambda I) \text{ is not boundedly invertible}\}
Spectrum
σ(T)=σp(T)σc(T)σr(T)(disjoint)\sigma(T) = \sigma_p(T) \cup \sigma_c(T) \cup \sigma_r(T) \quad (\text{disjoint})
Spectral decomposition
r(T)=supλσ(T)λ=limnTn1/nr(T) = \sup_{\lambda \in \sigma(T)} |\lambda| = \lim_{n\to\infty} \|T^n\|^{1/n}
Spectral radius formula

Properties

Spectrum of self-adjoint operator is real

IfTB(H)isselfadjoint(T=T),thenσ(T)R.If T \in \mathcal{B}(H) is self-adjoint (T^* = T), then \sigma(T) \subset \mathbb{R}.

Spectrum of unitary operator lies on unit circle

IfUisunitary(UU=UU=I),thenσ(U){λ:λ=1}.If U is unitary (U^*U = UU^* = I), then \sigma(U) \subset \{\lambda : |\lambda| = 1\}.

Eigenvectors for distinct eigenvalues of self-adjoint operators are orthogonal

IfTe1=λ1e1andTe2=λ2e2withλ1λ2andTselfadjoint,thene1,e2=0.If Te_1 = \lambda_1 e_1 and Te_2 = \lambda_2 e_2 with \lambda_1 \neq \lambda_2 and T self-adjoint, then \langle e_1, e_2 \rangle = 0.

Theorems

Theorem 5.1: Spectral Radius Formula
ForTB(X),thespectralradiussatisfiesr(T)=limnTn1/nT.For T \in \mathcal{B}(X), the spectral radius satisfies r(T) = \lim_{n\to\infty} \|T^n\|^{1/n} \le \|T\|.
Theorem 5.2: Spectrum is compact and nonempty
ForanyTB(X)onacomplexBanachspace,σ(T)isanonemptycompactsubsetof{λ:λT}.For any T \in \mathcal{B}(X) on a complex Banach space, \sigma(T) is a nonempty compact subset of \{\lambda : |\lambda| \le \|T\|\}.
Theorem 5.3: Spectral Theorem for Bounded Self-Adjoint Operators
IfTisaboundedselfadjointoperatoronaHilbertspaceH,thenthereexistsauniqueprojectionvaluedmeasureEonσ(T)RsuchthatT=σ(T)λdE(λ).If T is a bounded self-adjoint operator on a Hilbert space H, then there exists a unique projection-valued measure E on \sigma(T) \subset \mathbb{R} such that T = \int_{\sigma(T)} \lambda\,dE(\lambda).
Theorem 5.4: Spectral Theorem for Compact Self-Adjoint Operators
IfTisacompactselfadjointoperatoronaseparableHilbertspaceH,thenHhasanorthonormalbasis{en}ofeigenvectorsofTwithcorrespondingrealeigenvaluesλn0.If T is a compact self-adjoint operator on a separable Hilbert space H, then H has an orthonormal basis \{e_n\} of eigenvectors of T with corresponding real eigenvalues \lambda_n \to 0.

Worked Examples

  1. Point spectrum: \(\lambda \in \sigma_p(M_\phi)\) iff \(\phi(x) = \lambda\) on a set of positive measure. Since \(\{x = \lambda\}\) has measure zero for every \(\lambda\), \(\sigma_p(M_\phi) = \emptyset\).

  2. Resolvent: \((M_\phi - \lambda I)^{-1}\) acts by multiplication by \(1/(x-\lambda)\). This is bounded iff \(1/(x-\lambda) \in L^\infty([0,1])\), which fails when \(\lambda \in [0,1]\).

    (MϕλI)1=ess-supx[0,1]1xλ\|(M_\phi - \lambda I)^{-1}\| = \text{ess-sup}_{x \in [0,1]} \frac{1}{|x - \lambda|}
  3. For \(\lambda \notin [0,1]\), \(1/(x-\lambda)\) is bounded on \([0,1]\), so \(\lambda \in \rho(M_\phi)\). Hence \(\sigma(M_\phi) = [0,1]\).

  4. Since \(M_\phi - \lambda I\) is injective for all \(\lambda \in [0,1]\) (multiplication by \(x-\lambda\) is injective on \(L^2\)) but not boundedly invertible, the entire interval \([0,1]\) is in \(\sigma_c(M_\phi)\).

Answer: \(\sigma(M_x) = [0,1]\) with purely continuous spectrum; no eigenvalues.

Practice Problems

Difficulty 8/10

Prove that the spectrum of a bounded operator on a complex Banach space is nonempty.

Difficulty 9/10

Compute \(\sigma(S)\) and \(\sigma(S^*)\) where \(S\) is the unilateral right shift on \(\ell^2\): \(S(x_1, x_2, \ldots) = (0, x_1, x_2, \ldots)\).

Difficulty 9/10

Prove that eigenvectors of a self-adjoint operator corresponding to distinct eigenvalues are orthogonal.

Common Mistakes

Common Mistake

The spectrum consists only of eigenvalues

In infinite dimensions the spectrum has three disjoint parts: point spectrum (eigenvalues), continuous spectrum, and residual spectrum. The multiplication operator \(M_x\) on \(L^2([0,1])\) has spectrum \([0,1]\) but no eigenvalues.

Common Mistake

Spectral radius always equals the operator norm

The spectral radius \(r(T) = \lim \|T^n\|^{1/n} \le \|T\|\); equality holds for normal operators (\(T^*T = TT^*\)) but fails for non-normal ones. The nilpotent matrix \(\begin{pmatrix}0&1\\0&0\end{pmatrix}\) has norm 1 but spectral radius 0.

Quiz

For a bounded self-adjoint operator on a Hilbert space, the spectrum is always a subset of:
The spectral radius \(r(T)\) of a bounded operator equals:
For the right shift operator on \(\ell^2\), the open unit disk \(\{|\lambda| < 1\}\) belongs to which part of the spectrum?

Summary

  • The spectrum \(\sigma(T)\) is the set of \(\lambda\) for which \(T - \lambda I\) fails to be boundedly invertible; it is always compact and nonempty for bounded operators on complex Banach spaces.
  • The spectrum decomposes into point spectrum (eigenvalues), continuous spectrum, and residual spectrum.
  • Spectral radius formula: \(r(T) = \lim_n \|T^n\|^{1/n}\); for self-adjoint operators \(r(T) = \|T\|\).
  • Self-adjoint operators on Hilbert spaces have real spectrum and admit the spectral theorem: \(T = \int \lambda\,dE(\lambda)\).
  • Compact self-adjoint operators have a countable spectrum accumulating only at 0, with eigenvectors forming a complete orthonormal basis.

References

  1. BookRudin, W. Functional Analysis. 2nd ed., McGraw-Hill, 1991. Chapters 10–12.
  2. BookReed, M. and Simon, B. Methods of Modern Mathematical Physics, Vol. 1. Academic Press, 1980.