---
tag: NXV8S3145K
kind: statement
role: theorem
name: Real spectrum of self-adjoint operators
msc: [47B15, 47A10]
uses: [HM2HHE3J51, 6HSJBJ8TTQ, BMKATJ802A, XWXZ0DGW3Z]
about: [BMKATJ802A]
sources:
  - url: https://en.wikipedia.org/wiki/Self-adjoint_operator
---

## Statement

Let $H$ be a complex [[Hilbert space|HM2HHE3J51]] and let $T \colon H \to H$ be a bounded linear operator with $T^* = T$, where $T^*$ is the [[Hilbert space adjoint|6HSJBJ8TTQ]].
Then the [[spectrum|BMKATJ802A]] $\sigma(T)$ is a subset of $\R$.
More precisely, with $m = \inf_{\norm{x} = 1} \ip{Tx}{x}$ and $M = \sup_{\norm{x} = 1} \ip{Tx}{x}$, one has $\sigma(T) \subseteq [m, M]$, and both $m$ and $M$ belong to $\sigma(T)$.

## Remarks

Proof idea: let $\lambda = a + ib$ with $a, b$ real and $b \ne 0$.
Because $T - a I$ is self-adjoint, $\norm{(T - \lambda I)x}^2 = \norm{(T - aI)x}^2 + b^2 \norm{x}^2 \ge b^2 \norm{x}^2$.
So $T - \lambda I$ is injective with closed range, and the orthogonal complement of its range is $\ker(T - \bar\lambda I) = \set{0}$ by the same estimate; hence $T - \lambda I$ is invertible.

Example: the multiplication operator $(Tf)(x) = x f(x)$ on $L^2[0,1]$ is self-adjoint, and its spectrum is $[0, 1]$, a real interval, although $T$ has no eigenvalues.

Self-adjointness is needed: the unilateral shift on $\ell^2$ is not self-adjoint, and its spectrum is the closed unit disc.
The finite-dimensional case is the fact that [[eigenvalues of a self-adjoint operator are real|0C4NT4GT3X]].
The same conclusion holds for unbounded self-adjoint operators, and for self-adjoint elements of a C*-algebra.

## Explanations

### L1 human

A self-adjoint operator is the infinite-dimensional analogue of a Hermitian matrix.
Its spectrum, the set of numbers $\lambda$ for which $T - \lambda I$ cannot be inverted, lies on the real line.
This is why self-adjoint operators model measurable quantities in quantum mechanics.

### L2 human

In infinite dimensions the spectrum can contain points that are not eigenvalues, so the finite-dimensional argument with an eigenvector does not suffice; the proof uses a lower bound instead.
The result is the first step towards the spectral theorem for self-adjoint operators, which represents $T$ as an integral of $\lambda$ over $\sigma(T) \subseteq \R$.
The numbers $m$ and $M$ are the ends of the closure of the [[numerical range|XWXZ0DGW3Z]], and $\norm{T} = \max(\abs{m}, \abs{M})$.