The spectral theorem for self-adjoint operators
Understand spectral projections, functional calculus, and the limits of an eigenvector-only picture.
On this page
The question for today
What you’ll understand
- Interpret a projection-valued measure as an operator version of spectral coordinates.
- Evaluate a function of a finite self-adjoint matrix.
- Separate the compact eigenbasis theorem from the general spectral theorem.
A Hermitian matrix can be diagonalized in an orthonormal basis, turning matrix operations into scalar operations on eigenvalues. Infinite-dimensional self-adjoint operators retain that simplifying principle, but their spectral coordinates may be continuous. The general spectral theorem therefore uses projections associated with measurable sets rather than demanding an eigenvector at each spectral value. This lesson develops that picture through finite calculations and multiplication operators; it is a starting point for reading the theorem, not a proof of its full measure-theoretic machinery.
From orthogonal eigenspaces to spectral projections
For a finite self-adjoint matrix A, each distinct eigenvalue λ has an orthogonal projection P_λ onto its eigenspace. These projections add to the identity, are pairwise orthogonal, and reconstruct A as a sum of λP_λ. Degenerate eigenvalues require one projection onto the whole eigenspace, not an arbitrary preferred eigenvector. The operator is unchanged when an orthonormal basis within a degenerate eigenspace is rotated. Spectral projections capture the geometry without that coordinate ambiguity.
A projection-valued measure E assigns an orthogonal projection E(B) to each Borel subset B of the real line. Disjoint sets correspond to orthogonal projections, and countable additivity is interpreted in the strong operator sense. For a bounded self-adjoint operator, A is the spectral integral of λ against E. Multiplication by x on L2(0,1) has E(B)f equal to the indicator of B times f. Its projections are meaningful even though it has no nonzero eigenfunctions.
Functional calculus applies scalar rules safely
For a bounded Borel function g, the functional calculus defines g(A) by integrating g against the spectral measure. In finite dimensions this simply replaces each eigenvalue λ by g(λ) while preserving the spectral projections. One can define a positive square root, a resolvent, or a time-evolution operator without expanding an arbitrary matrix power by hand. If g is continuous on the compact spectrum of a bounded self-adjoint A, its operator norm is the maximum of |g| on that spectrum.
Scalar expressions still have domains. The inverse A^{-1} is a bounded everywhere-defined operator only when zero is outside the spectrum. A square root as a positive self-adjoint operator requires A to be nonnegative. For an unbounded self-adjoint A, an unbounded function g produces an operator defined only on vectors with a finite weighted spectral integral. Writing g(A) does not make that integrability condition disappear, and composing unbounded functions requires compatible domains.
An eigenbasis is a special spectral representation
For a compact self-adjoint operator, the Hilbert space is the orthogonal direct sum of its kernel and the closure of the span of its nonzero eigenvectors. Its nonzero eigenvalues are real and can only accumulate at zero. This is an eigenbasis representation after including a basis of the kernel. General bounded self-adjoint operators need not be compact, and their spectral measures may contain a continuous part. Importing a discrete sum into that situation can erase the main feature of the spectrum.
Spectral measures also organize observations. For a normalized x, μ_x is a probability measure over spectral values. In a finite state it assigns weights equal to squared projection lengths. This is a mathematical probability distribution associated with a particular vector and observable. Its use in quantum theory adds physical interpretation, but the spectral theorem itself does not simulate measurements or choose an outcome. The finite-mode lab below illustrates a representation choice and truncation, not a construction of arbitrary spectral measures.
Work it through
Take a square root spectrally
For , find its positive square root and verify it.
The orthonormal eigenvectors are e_+=(1,1)/√2 and e_-=(1,−1)/√2, with eigenvalues 9 and 1 respectively. Both are positive.
Replace these eigenvalues by 3 and 1. Thus A^{1/2}=3P_++P_-; assembling the projections gives the real matrix with diagonal entries 2 and off-diagonal entries 1.
Squaring that matrix gives diagonal entries 2²+1²=5 and off-diagonal entries 2·1+1·2=4, recovering A. Its eigenvalues 3 and 1 verify that the chosen square root is positive.
The positive square root is [[2,1],[1,2]]. Other matrix square roots can exist, but positivity selects this one uniquely.
The living notebook
Explore the model
Interactive lab · spectral
A finite spectral representation
Pause playback and compare N=2 with N=6 at the same coefficient decay p. Identify which shorter spatial scales appear. The lab adds the finite sum Σ sin(nx−t)/nᵖ with prescribed coefficients; it does not diagonalize A or generate a projection-valued measure.
- Sum at x = π/2
- 0.867
- Smallest wavelength
- 2π / 5
- Highest-mode coefficient
- 0.2
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 forDiscrete modes are one useful spectral representation, but the general theorem also accommodates continuous spectral measures.
Pause & reason
Check your understanding
Your turn
Try it yourself
For D=diag(1,4) and normalized x=(1,1)/√2, compute μ_x at the two eigenvalues and the expectation of D².
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Square the component magnitudes to obtain the spectral weights, then apply the function g(λ)=λ².
Compare with the solution
Each eigenvalue has weight 1/2. The expectation is (1/2)·1²+(1/2)·4²=17/2=8.5. Directly, D²=diag(1,16) gives the same inner product. The expectation of D is 5/2, whose square 25/4 is different from 17/2; functional calculus does not interchange an expectation with a nonlinear function.
Keep the idea
Take it with you
- Spectral projections generalize orthogonal eigenspace coordinates.
- Functional calculus applies scalar functions while preserving domain requirements.
- A continuous spectral part cannot be replaced by a fictitious eigenbasis.
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