Finding the set of eigenvalues and the spectrum of an operator

Introducing the operator. Let's consider the following operator: $$T \in B(H), T: l^2 \to l^2,$$ $$T(x_1, x_2, x_3, \dots)= (x_2, x_3, \dots).$$

The problem. We are to find all the eigenvalues (the point spectrum $\sigma_{p}(T)$) and the spectrum
($\sigma(T)$).

All I know is that: $$\delta(T) = \{ \lambda \in \mathbb{C}: (\lambda I - T)^{-1} \in B(H) \}$$ Knowing the set above we can find $\sigma(T)$ using this dependence: $$\sigma(T) = \mathbb{C} \setminus \delta(T)$$

I would like to ask for some explanation what is the difference between them (the two spectrums) because I think I don't understand those objects properly.

I would like also to ask if there is a method to find $\sigma_{p}(T)$ and $\sigma(T)$.

• Here and here might be useful. – mattos Jun 17 '18 at 11:28

The point spectrum $\sigma_p(T)$ is a subset of the spectrum $\sigma(T)$. The spectrum consists of all $\lambda \in \mathbb{C}$ such that $\lambda I -T$ is not invertible in $B(H)$. Now this can mean that $\lambda I -T$ is not injective or that it is not surjective (Note that for finite-dimensional spaces these two are equivalent). In first case $\lambda$ is in the point spectrum $\sigma_p = \{\lambda \in \mathbb{C} : \lambda I-T \text{ is not injective}\}$.