Spectrum of an element

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.

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.