---
tag: A1G25R99PY
kind: statement
role: theorem
name: Spectral theorem for compact self-adjoint operators
msc: [47B07, 47B15, 47A75]
uses: [HM2HHE3J51, M2NEP1K5AE, 6HSJBJ8TTQ, 0BQAA2BK3Q, 0N6960W0BV, 0S23E6YE83, 0S9GGTR0EX, 06J4PXRG22, JEPDM5B5FC]
about: [M2NEP1K5AE]
sources:
  - url: https://en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space
---

## Statement

Let $H$ be a real or complex [[Hilbert space|HM2HHE3J51]].
Let $T$ be a [[compact operator|M2NEP1K5AE]] on $H$ with $T = T^*$, where $T^*$ is the [[adjoint|6HSJBJ8TTQ]] of $T$.
Then there are a finite or countable orthonormal family $(e_n)_{n \in N}$ in $H$ and nonzero real numbers $(\lambda_n)_{n \in N}$ such that:

1. $T e_n = \lambda_n e_n$ for every $n \in N$;
2. $\abs{\lambda_1} \ge \abs{\lambda_2} \ge \cdots$, and $\lambda_n \to 0$ if $N$ is infinite;
3. $T x = \sum_{n \in N} \lambda_n \ip{x}{e_n} e_n$ for every $x \in H$;
4. $H$ is the orthogonal direct sum of the [[kernel|0S23E6YE83]] $\ker T$ and the closed span of $(e_n)$.

## Remarks

The proof starts from the fact that $\norm{T} = \sup_{\norm{x} = 1} \abs{\ip{Tx}{x}}$ for self-adjoint $T$.
Compactness makes this supremum attained, at an [[eigenvector|0N6960W0BV]] with eigenvalue $\pm \norm{T}$.
The orthogonal complement of this eigenvector is invariant under $T$, and one repeats the argument there; the eigenvalues found tend to $0$ by the [[spectral theory of compact operators|JEPDM5B5FC]].
Adding an [[orthonormal basis|0S9GGTR0EX]] of $\ker T$ to $(e_n)$ gives an orthonormal basis of $H$ made of eigenvectors of $T$.
The nonzero [[eigenvalues|0BQAA2BK3Q]] of $T$ are exactly the numbers $\lambda_n$, and each appears as many times as the dimension of its eigenspace, which is finite.
Example: the operator $(Tf)(x) = \int_0^1 \min(x, y) f(y) \, dy$ on $L^2([0,1])$ is compact and self-adjoint, with eigenvalues $\big((n - \tfrac{1}{2}) \pi\big)^{-2}$ and eigenfunctions $\sqrt{2} \sin\big((n - \tfrac{1}{2}) \pi x\big)$, $n \ge 1$.
Without self-adjointness the theorem fails: the Volterra operator $(Vf)(x) = \int_0^x f(y) \, dy$ on $L^2([0,1])$ is compact but has no eigenvalue at all.
In finite dimension, this is the [[spectral theorem|06J4PXRG22]] for self-adjoint linear maps.

## Explanations

### L1 human

A symmetric matrix can be diagonalized in an orthonormal basis of eigenvectors.
This theorem says the same for compact self-adjoint operators on infinite-dimensional spaces.
The operator acts by stretching along infinitely many perpendicular axes, by factors that shrink to zero.

### L2 human

The theorem is the main tool for solving linear integral equations with symmetric kernels, and for boundary value problems such as the Sturm–Liouville problem through their Green's operators.
It shows that the eigenfunctions of such problems form an orthonormal basis, which justifies expansions in Fourier-type series.
Applied to $T^* T$ for an arbitrary compact operator $T$, it gives the singular values and the Schmidt decomposition of $T$.
For noncompact self-adjoint operators, eigenvectors may not exist, and one needs the spectral theorem in its measure form.