research30 min studyLesson 84 in Mathematics

Spectra and resolvents

Replace a determinant test with invertibility, stability, and continuous-spectrum examples.

On this page

What you’ll understand

  • Define the resolvent using a bounded, everywhere-defined inverse.
  • Distinguish an eigenvalue from a continuous spectral point.
  • Explain why a large resolvent can signal amplification away from eigenvalues.

A determinant detects a singular square matrix, but no general determinant exists for an arbitrary infinite-dimensional operator. Spectrum replaces that test with a question about whether a shifted operator can be inverted stably for every allowed right-hand side. The inverse, when it exists, is the resolvent. Its norm measures response to forcing. This viewpoint reveals both spectral points with no eigenvectors and large responses far from any eigenvalue for nonnormal operators.

Define the inverse before naming its exceptional set

For a bounded operator A on a complex Hilbert space H, a complex number z belongs to the resolvent set when A−zI is bijective and its inverse is bounded on all of H. We use the convention R(z)=(A−zI)^{-1}; some texts reverse the sign. The spectrum is the complement of that set. For bounded A it is a nonempty compact subset of the complex plane. The bounded inverse theorem makes boundedness automatic once bijectivity between Banach spaces is established.

For a closed unbounded operator, the shifted map acts from its domain D(A) to H. A resolvent inverse must be defined on every vector of H, take values in D(A), and be bounded when regarded as a map H→H. The domain is part of the operator. Altering boundary conditions can change both the resolvent and spectrum even if the differential expression is unchanged. Later lessons will build that distinction rather than treating a derivative as a matrix with infinitely many rows.

R(z)=(A−zI)−1,ρ(A)={z:R(z) exists and is bounded on H},σ(A)=C∖ρ(A)R(z)=(A-zI)^{-1},\qquad \rho(A)=\{z:R(z)\text{ exists and is bounded on }H\},\qquad \sigma(A)=\mathbb C\setminus\rho(A)
For an unbounded closed operator, A−zI has domain D(A), and the inverse must solve the equation for every right-hand side in H.↔ Scroll wide equations horizontally.

A spectral point need not have an eigenvector

Consider multiplication by x on L2(0,1). For z outside [0,1], division by x−z is bounded, so the shifted equation is uniquely solvable. At a real λ inside [0,1], an eigenfunction would have to vanish outside the singleton x=λ, hence represent zero in L2. Nevertheless functions concentrated near λ have unit norm while their shifted outputs become arbitrarily small. This rules out a bounded inverse and places every λ in [0,1] in the spectrum.

For this self-adjoint multiplication operator, those points are continuous spectrum: the shifted map is injective, its range is dense, and it is not onto. Density follows because the orthogonal complement of the range is the adjoint kernel, which is zero. Non-surjectivity can be checked with a right-hand side whose division by x−λ is not square integrable. This precise terminology describes a failure of stable inversion; it does not mean that an arbitrary operator's spectrum must be one continuous interval.

(Af)(x)=xf(x),σ(A)=[0,1],ker⁡(A−λI)={0}(λ∈[0,1])(Af)(x)=xf(x),\qquad \sigma(A)=[0,1],\qquad \ker(A-\lambda I)=\{0\}\quad(\lambda\in[0,1])
The space uses Lebesgue measure, so a function supported on one point is zero as an L2 element. Concentrated approximate eigenvectors are different from genuine eigenvectors.↔ Scroll wide equations horizontally.

The resolvent measures more than eigenvalue distance

If |z| exceeds the norm of a bounded A, a Neumann series in A/z constructs the resolvent. This bounds the spectrum inside the norm disk. For a self-adjoint operator, the resolvent norm equals the reciprocal distance to the spectrum whenever z lies in the resolvent set. For a general nonnormal operator that exact equality fails: nearly dependent response directions can cause large amplification even when the spectral points are distant. This motivates pseudospectra in stability analysis.

A large resolvent does not by itself prove exponential growth of every time trajectory. It quantifies the sensitivity of a shifted solve; translating it into time-evolution statements needs additional assumptions and estimates. Similarly, numerical eigenvalues from a truncated matrix may miss resolvent amplification or approximate a continuous spectrum deceptively. Always state the operator, its domain, the chosen norm, and whether the claim concerns exact eigenvectors, approximate ones, or response to forcing.

R(z)=−1z∑n=0∞(Az)n,∣z∣>∥A∥R(z)=-\frac1z\sum_{n=0}^{\infty}\left(\frac Az\right)^n,\qquad |z|>\|A\|
This series converges in operator norm for bounded A. Its minus sign follows from the convention R(z)=(A−zI)^{-1}.↔ Scroll wide equations horizontally.

Large response with unchanged eigenvalues

Take A=(01000)A=\begin{pmatrix}0&10\\0&0\end{pmatrix}. Solve (A−I)x=b for b=(0,1), and compare the response norm with the distance from 1 to the spectrum.

  1. A is upper triangular with both eigenvalues zero, so the distance from z=1 to its spectrum is exactly one.

  2. The equations are −x₁+10x₂=0 and −x₂=1. Hence x₂=−1 and x₁=−10.

  3. The unit forcing has response norm √101, exceeding ten. Thus the resolvent norm at z=1 is at least √101, despite the spectral distance being one.

The result

Nonnormality allows strong forcing amplification that the eigenvalue locations alone do not reveal.

Explore the model

Interactive lab

Interactive lab · spectral

Finite modes are not the full spectrum

Pause at one time and increase mode count N while comparing coefficient decay p. The family Σ sin(nx−t)/nᵖ shows explicit finite basis components. Ask which features remain fixed when N changes. This lab neither computes a resolvent nor represents the continuous spectrum of multiplication by x.

Finite modes are not the full spectrumA finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Current readings: Sum at x = π/2: 0.867; Smallest wavelength: 2π / 5; Highest-mode coefficient: 0.2. All parameters are adjustable with the labeled controls below.01.573.144.716.28−2.56−1.2801.282.56x · radiansFinite sine sum
Finite sumHighest included mode
Sum at x = π/2
0.867
Smallest wavelength
2π / 5
Highest-mode coefficient
0.2
t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA plotted finite collection of modes is a chosen representation, not evidence that all spectral points are eigenvalues.

Check your understanding

Why does multiplication by x on L2(0,1) have no nonzero eigenfunction at λ=1/2?

Try it yourself

For D=diag(1,4), find the norm of (D−2I)^{-1} in the Euclidean norm. Explain why the nonnormal example does not contradict it.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Invert each diagonal entry and take the largest absolute multiplier.

Compare with the solution

D−2I=diag(−1,2), so its inverse is diag(−1,1/2) and has norm one. The distance from 2 to {1,4} is one, agreeing with the self-adjoint formula. The shear example is nonnormal, so that exact spectral-distance formula does not apply there.

Take it with you

  • The spectrum detects failure of a bounded inverse, not just failure of injectivity.
  • Continuous spectrum can arise without any normalizable eigenvectors.
  • Resolvent amplification can exceed reciprocal eigenvalue distance for nonnormal operators.

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