Spectrum vs eigenvalues

The other answers adequately address the finite-dimensional case. If we consider a Hilbert space $\newcommand{\Hil}{\mathcal{H}}\Hil$ of countable dimension with orthonormal basis $(e_i)_{i \in \mathbb{N}}$, we can define an operator $T : \Hil \rightarrow \Hil$ by $$ T(e_n) = \frac{1}{n}e_n $$ It is clear from this definition that each $e_n$ is an eigenvector with eigenvalue $\frac{1}{n}$, and as $(e_i)_{i \in \mathbb{N}}$ is an orthonormal basis, this is a complete set of eigenvectors. So the set of eigenvalues is $\{ \frac{1}{n} \mid n \in \mathbb{N} \}$.

However, $T$ is not invertible (one way to see this is that if we try to define an inverse we get an unbounded operator, another is that the vector $\sum_{n=1}^\infty \frac{1}{n}e_n$ is in $\Hil$ but not in the image of $T$), so $0$ is an element of its spectrum that is not an eigenvalue. In fact $\{ 0 \} \cup \{ \frac{1}{n} \mid n \in \mathbb{N} \}$ is exactly the spectrum of $T$.

The spectrum is always a non-empty closed subset of $\mathbb{C}$, while the set of eigenvalues does not need to be. Also, the set of eigenvalues can be empty (for example, this is the case for the operator $S(e_n) = e_{n+1}$).

The spectrum of a linear operator $T$ on a finite dimensional space is exactly the set of its eigenvalues.

It's defined as $\{\lambda:T-\lambda I$ is not invertible$\}$, and $T-\lambda I$ is not invertible in a finite dimensional space iff its kernel is nonzero, meaning exactly that $\lambda$ is an eigenvalue of $T$.

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:

1. 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 $T$.

2. 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 $X$.

3. 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, 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 zero) is an eigenvalue is the case of a compact, self-adjoint operator on a Hilbert space.