TITLE: Spectrum vs eigenvalues
QUESTION [8 upvotes]: Easy question about linear operators - in physics (often) the terms spectrum and (set of) eigenvalues of an operator are used interchangeably.
I'd like a simple compare and contrast to know the difference according to mathematicians.
Many thanks!

REPLY [4 votes]: I'd like to add something further in the infinite-dimensional case than the other responses, which I think is important if we're going to talk about eigenvalues vs. spectrum. I'm going to specify how they're related.
As mentioned by @RobertFurber, we can have elements of the spectrum which are not eigenvalues. Let $T:X\rightarrow X$ be a bounded operator on a Banach space. The spectrum $\sigma(T)$ can be decomposed into three pieces:

The point spectrum: this consists of $\lambda\in\sigma(T)$ so that $T-\lambda I$ is not injective. In this case, $\lambda$ is called an eigenvalue of $T$.

The continuous spectrum: this consists of $\lambda\in \sigma(T)$ such that $T-\lambda I$ is one-to-one, and the range of $T-\lambda I$ is dense in $X$, but it is not equal to $X$.

The residual spectrum: this consists of $\lambda\in \sigma(T)$ such that $T-\lambda I$ is one-to-one, and such that $T-\lambda I$ is not dense in $X$.


While the spectrum is always non-empty, we can have elements with empty point spectrum. As an example, if $X=L^2([0,1])$ and $M:X\rightarrow X$ is the multiplication operator defined by sending $f\in X$ to the function $Mf(x)=xf(x)$ with $x\in[0,1]$,  then $\sigma(M)=[0,1],$ and each $\lambda\in [0,1]$ is part of the continuous spectrum (this is an easy exercise, and you can also find it on this site).
An example where every element of the spectrum (except potentially zero) is an eigenvalue is the case of a compact, self-adjoint operator on a Hilbert space.