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Hilbert spaces and the spectral theorem

Inner products, orthonormal bases, Fourier series, and eigenvalues of operators.

An inner product gives angles and projections; an orthonormal basis expands any vector: Fourier series are exactly this in L². Self-adjoint operators have real spectrum and an eigen-decomposition, the infinite-dimensional version of diagonalising a symmetric matrix. Picture it: the symmetric matrix [[2,1],[1,2]] stretching along two perpendicular eigenvectors. Think it: quantum observables are self-adjoint operators; measurement outcomes are their spectrum.

Ýüklenen mysal · eigenvalues of [[2,1],[1,2]]

Linear-Algebra · Eigenvalues

Eigenvalues of [[2,1],[1,2]]

\left[\begin{matrix}2 & 1\\1 & 2\end{matrix}\right]

Adım adım

  1. \det(A - \lambda I) = 0

    Eigenvalues are the roots of the characteristic polynomial.

  2. \det\left[\begin{matrix}2 - \lambda & 1\\1 & 2 - \lambda\end{matrix}\right] = 0

    Subtract λ from the diagonal.

  3. \lambda^{2} - 4 \lambda + 3 = 0

    Expand the determinant.

  4. \left(\lambda - 3\right) \left(\lambda - 1\right) = 0

    Factor.

  5. \lambda = 3, \lambda = 1

    Eigenvalues (with multiplicity).

  6. \lambda = 1:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}-1\\1\end{matrix}\right]

    Solve (A − 1I)v = 0 for a basis eigenvector.

  7. \lambda = 3:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}1\\1\end{matrix}\right]

    Solve (A − 3I)v = 0 for a basis eigenvector.

Jawaby görkez
\lambda = 3,\; \lambda = 1

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Nädip: Hilbert spaces and the spectral theorem

  1. Eigenvalues are the roots of the characteristic polynomial.
  2. Subtract λ from the diagonal.
  3. Expand the determinant.
  4. Factor.
  5. Eigenvalues (with multiplicity).
  6. Solve (A − 1I)v = 0 for a basis eigenvector.
  7. Solve (A − 3I)v = 0 for a basis eigenvector.

Adamlar soraýan soraglar

What is a Hilbert space?

A vector space with an inner product (so lengths and angles make sense) that is complete (no missing limit points). Square-integrable functions form one; quantum states live in one.

_Ýaşa Functional Analysis

Normed and Banach spaces