---
tag: 4KN39NQZ02
kind: definition
name: Spectrum of an element
msc: [46H05, 47A10]
defines: EH0CZ19RKP
uses: [0XVEZDE9J8, 7SB03M850M, 0XPX5NC67E]
sources:
  - url: https://en.wikipedia.org/wiki/Spectrum_(functional_analysis)
---

## Definition

Let $\mathbb{K}$ be a field, let $A$ be an associative algebra over $\mathbb{K}$ with unit $1$, and let $a \in A$.
The *spectrum* of $a$ is the set
$$\sigma(a) = \set{\lambda \in \mathbb{K} : a - \lambda 1 \text{ is not invertible in } A}.$$
Here invertible means that $a - \lambda 1$ is a [[unit|0XVEZDE9J8]] of $A$.
If $A$ has no unit, the spectrum of $a$ is its spectrum in the [[unitization|7SB03M850M]] $A^+$.

## Remarks

The spectrum depends on the algebra: if $a$ lies in a subalgebra $B$ of $A$ that contains $1$, then $\sigma_A(a) \subseteq \sigma_B(a)$, and the inclusion can be strict.
In a non-unital algebra, $0$ always lies in the spectrum, because no $(0, a)$ is invertible in $A^+$.

Example: for $A = M_n(\C)$, $a - \lambda 1$ is invertible exactly when $\det(a - \lambda 1) \ne 0$.
So $\sigma(a)$ is the set of eigenvalues of $a$, the [[spectrum of the matrix|0XPX5NC67E]].

Example: for $A = C([0,1])$ with complex values and $f \in A$, $\sigma(f) = f([0,1])$.
If $\lambda \notin f([0,1])$, then $1/(f - \lambda)$ is continuous and inverts $f - \lambda$; if $\lambda = f(t_0)$, then $f - \lambda$ vanishes at $t_0$ and has no inverse.

For bounded operators on $\ell^2$, the right shift $S$ has no eigenvalues, but $0 \in \sigma(S)$ because $S$ is not surjective.
So in infinite dimension the spectrum can contain points that are not eigenvalues.