If $S:(x_0,x_1,x_2,\ldots) \to (0,x_0,x_1,x_2,\ldots)$ Why does $0 \in \sigma (S)$? and what is $\dim(R(S))$ in $\ell^2$? Consider the unilateral shift $S$ on $\ell^2$ defined by $$(x_0,x_1,x_2,\ldots) \to (0,x_0,x_1,x_2,\ldots)$$  Why dose $0$ is eigenvalue of $S$? and what is $\dim(R(S))$ in $\ell^2$?( Range of $S$)

The reason that $0$ is in the spectrum of $S$ is that $S$ is not invertible.  Indeed, this is clear by seeing that $S$ is not bijective.  The element $(1,0,0,…)$ is not in the range of $S$, for example.
However the range of $S$ has (Hilbert space dimension) equal to $\ell^2$.  If $e_0,e_1,…$ is the canonical orthonormal basis for $\ell^2$ (where $e_i$ is $1$ in the $i$-th position and $0$ elsewhere), then the range of $S$ has orthonormal basis $e_1,e_2,…$ (just $e_0$ removed).
It should be mentioned that $0$ is NOT an eigenvalue of $S$, but is just in the spectrum.  For if there were $(x_0,x_1,x_2,…) \in \ell^2$ with $S(x_0,x_1,x_2,…)=0$, then each $x_i=0$.  Another way to see this is that $S$ is an isometry.
