Let $n\in \mathbb N_{>0}$ and $A\in M_{n\times n}(\{0,1\})$ a zero-one matrix. The sequences space \begin{equation} \Sigma_A:=\{s\in[n]^{\mathbb Z}\;:\; A_{s_{i}s_{i+1}}=1 \;\;\forall i\in\mathbb Z\} \end{equation} together with the shift operator $T:\Sigma_A\to\Sigma_A$ with $T(s)_i=s_{i+1}$ is apparently called a topological Markov shift or subshift of finite type.
The definition seems flawed to me. Suppose $A_{ij}=0$ except, say $A_{k_0,k_1}=1$. Then $\Sigma_A$ is empty, because if $s_i=k_0$, then $s_{i+1}=k_1$ and what can $s_{i+2}$ be. There is no transition rule for $k_1$ (i.e. $A_{k_1,j}=0$ for all $j$). In other words I can only understand the definition for what is called a transitive Markov shift, where $A$ is irreducible and the corresponding graph is strongly connected.
It is usually assumed that the defining graph is essential; that is, every vertex is both the source of at least one edge and the target of one edge - equivalently the corresponding matrix has no columns or rows of zeros. This is a perfectly fine assumption to make because if it were the case, then just remove the stranded vertex (resp. the corresponding row and column where one of them is all zeros).
Transitivity (equiv. irreducibility of $A$) is often assumed but not a required assumption. If $\Sigma_A$ is not transitive, then the corresponding graph can be decomposed into a finite number of communicating classes of states and the matrix $A$ can be rewritten (by conjugating with permutation matrices) into block triangular form such that the diagonal blocks each correspond to a communicating class. Each block corresponds to an irreducible component of the graph (equivalently a transitive subsystem of $\Sigma_A$).
This is all covered in Section 4.4 of Lind and Marcus, An Introduction to Symbolic Dynamics and Coding.