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6.4 Spectral theorem for compact self-adjoint operators

6.4 Spectral theorem for compact self-adjoint operators

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
🧐Functional Analysis
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The Spectral Theorem for Compact Self-Adjoint Operators is a powerful tool in functional analysis. It shows that these operators have an orthonormal basis of eigenvectors, with real eigenvalues converging to zero if infinite.

This theorem allows us to represent compact self-adjoint operators in a canonical form, diagonalize them, and solve eigenvalue problems. It's crucial for understanding operator behavior and properties in Hilbert spaces.

Spectral Theorem for Compact Self-Adjoint Operators

Spectral theorem for compact operators

  • States that if TT is a compact self-adjoint operator on a Hilbert space HH, then HH has an orthonormal basis consisting of eigenvectors of TT
  • Eigenvalues of TT are real and converge to 0 if there are infinitely many (λ1,λ2,...\lambda_1, \lambda_2, ...)
  • TT can be represented as Tx=n=1λnx,enenTx = \sum_{n=1}^{\infty} \lambda_n \langle x, e_n \rangle e_n, where λn\lambda_n are the eigenvalues and ene_n are the corresponding eigenvectors
  • Provides a canonical form for compact self-adjoint operators (T=n=1λn,enenT = \sum_{n=1}^{\infty} \lambda_n \langle \cdot, e_n \rangle e_n)
  • Allows for the diagonalization of compact self-adjoint operators (UTU=DU^{*}TU = D, where DD is diagonal)
  • Enables the solution of eigenvalue problems involving compact self-adjoint operators (Tx=λxTx = \lambda x)
  • Establishes a connection between the operator's properties and its eigenvalues and eigenvectors (λn0\lambda_n \to 0, {en}\{e_n\} orthonormal basis)

Orthonormal basis of eigenvectors

  • Let TT be a compact self-adjoint operator on a Hilbert space HH
  • Consider the eigenvalue problem Tx=λxTx = \lambda x
  • Eigenvalues are real since TT is self-adjoint (Tx,y=x,Ty\langle Tx, y \rangle = \langle x, Ty \rangle)
  • Eigenvectors corresponding to distinct eigenvalues are orthogonal (ei,ej=0\langle e_i, e_j \rangle = 0 for iji \neq j)
  • If λ0\lambda \neq 0 is an eigenvalue, then the eigenspace Eλ={xH:Tx=λx}E_{\lambda} = \{x \in H : Tx = \lambda x\} is finite-dimensional because TλIT - \lambda I is compact and has a finite-dimensional null space
  • Choose an orthonormal basis for each eigenspace EλE_{\lambda} ({e1,...,ek}\{e_1, ..., e_k\} for EλE_{\lambda})
  • The union of these bases forms an orthonormal set in HH ({e1,e2,...}\{e_1, e_2, ...\})
  • If HH is not spanned by the eigenvectors, consider the orthogonal complement of the span of eigenvectors (Hspan{e1,e2,...}H \ominus \text{span}\{e_1, e_2, ...\})
    • TT restricted to this complement is compact and self-adjoint, so 0 is the only possible eigenvalue
  • Conclude that the union of the orthonormal bases for the eigenspaces forms an orthonormal basis for HH

Spectral decomposition of operators

  • Let TT be a compact self-adjoint operator on a Hilbert space HH
  • Find the eigenvalues λn\lambda_n and corresponding orthonormal eigenvectors ene_n of TT
  • Express TT as Tx=n=1λnx,enenTx = \sum_{n=1}^{\infty} \lambda_n \langle x, e_n \rangle e_n, which is the spectral decomposition of TT
  • The spectral decomposition can be truncated if there are finitely many eigenvalues (Tx=n=1Nλnx,enenTx = \sum_{n=1}^{N} \lambda_n \langle x, e_n \rangle e_n)
  • Provides a way to represent the operator in terms of its eigenvalues and eigenvectors (T=n=1λn,enenT = \sum_{n=1}^{\infty} \lambda_n \langle \cdot, e_n \rangle e_n)
  • Useful for understanding the behavior and properties of the operator (Tx=n=1λnx,enenTx = \sum_{n=1}^{\infty} \lambda_n \langle x, e_n \rangle e_n)

Applications of spectral theorem

  • Given a compact self-adjoint operator TT, use the spectral theorem to find the eigenvalues and eigenvectors of TT and express TT in terms of its spectral decomposition
  • Diagonalization:
    1. Let {λn}\{\lambda_n\} be the eigenvalues and {en}\{e_n\} be the corresponding orthonormal eigenvectors of TT
    2. Define U:HHU: H \to H by Uen=enUe_n = e_n (identity on the basis)
    3. Then UTU=DU^{*}TU = D, where DD is a diagonal operator with λn\lambda_n on the diagonal
  • Solving eigenvalue problems:
    1. Use the spectral decomposition to express the eigenvalue problem Tx=λxTx = \lambda x as n=1λnx,enen=λx\sum_{n=1}^{\infty} \lambda_n \langle x, e_n \rangle e_n = \lambda x
    2. Solve for the eigenvalues and eigenvectors using the properties of the spectral decomposition (λ=λn\lambda = \lambda_n, x=enx = e_n)
  • Approximation of compact self-adjoint operators (Tn=1Nλn,enenT \approx \sum_{n=1}^{N} \lambda_n \langle \cdot, e_n \rangle e_n)
  • Analysis of integral equations with compact self-adjoint integral operators (Tx(s)=abK(s,t)x(t)dtTx(s) = \int_a^b K(s,t)x(t)dt)
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