Difference between an eigenvalue and a spectral value What is the difference in the definition of a spectral value and an eigenvalue. My notes from functional analysis says

$\lambda$ is an eigenvalue of an operator $A$ if $\,\exists \, x \in \mathbb{C^n}$ such that $$Ax = \lambda x$$ This implies $(A - \lambda I)x = 0  \Rightarrow \ker(A - \lambda I) \neq {0}$. This is equivalent
  of saying that $A$ is not injective.

On the other hand, the definition of a spectral value is

$\lambda$ is called a spectral value of $A$ if $A - \lambda I$ is not
  invertible.

What is the difference here? How is it that some operators can have spectral values and not eigenvalues (eigenvalues $\subset$ spectrum(A)) and lastly, how do they conincide when the space is finite dimensional ?
 A: Consider the unilateral shift on $\ell^2(\mathbb N)$, i.e.
$$
S(a_1,a_2,\ldots)=(0,a_1,a_2,\ldots).
$$
It is easy to verify that $S$ is injective, so $0$ is not an eigenvalue of $S$. But $S$ is not invertible (because it is not surjective). So we have $0$ as an element of the spectrum of $S$, but not an eigenvalue. 
In finite dimension, two things happen: since the domain and codomain are of equal finite-dimension, an operator is injective if and only if it is surjective. And all linear operators are continuous. So if $T-\lambda I$ is not invertible, this means that it is not bijective; then it is not injective, so it has a nonzero kernel, i.e. there exists nonzero $x$ with $Tx=\lambda x$. Thus, in finite dimension the spectrum consists only of eigenvalues. 
A: As you acquire examples, please be sure to include this multiplication operator:
$$
                           X = L^{2}_{\mu}(-1,1),\\
                            (Af)(x) = xf(x).
$$
Here, $\mu$ is a finite Borel or Lesbesgue measure on $[-1,1]$. The reason to include this one in your collection of examples is that bounded selfadjoint operators with $\sigma(A) \subseteq [-1,1]$ can be essentially viewed as copies of this operator. If $\mu$ has an atom at $\lambda \in [-1,1]$ (i.e., $\mu\{\lambda\} \ne 0$,) then $\chi_{\{\lambda\}}$ is an eigenfunction of $A$ with
$$
           A\chi_{\{\lambda\}}=\lambda \chi_{\{\lambda\}}
$$
If $\lambda$ is not an atom of $\mu$, but $\lambda$ is in the support of $\mu$, then
$$
\begin{align}
               \|(A-\lambda I)\chi_{[\lambda-\epsilon,\lambda+\epsilon]}\|^{2} & = \int_{\lambda-\epsilon}^{\lambda+\epsilon}|\lambda-x|^{2}\,d\mu(x) \\
   & \le \epsilon^{2}\|\chi_{[\lambda-\epsilon,\lambda+\epsilon]}\|^{2}
\end{align}
$$
In other words,
$$
       u_{\epsilon}= \frac{1}{\|\chi_{[\lambda-\epsilon,\lambda+\epsilon]}\|}\chi_{[\lambda-\epsilon,\lambda+\epsilon]}
$$
is a unit vector for which $\|(A-\lambda I)u_{\epsilon}\| \le \epsilon$. So $u_{\epsilon}$ is an approximate eigenvector. Even though there is no eigenvector with eigenvalue $\lambda$, there is a unit vector which is nearly an eigenvector, to any desired degree of approximation. In this case $\lambda \in \sigma(A)$; the inverse of $A-\lambda I$ is unbounded and only densely-defined.
