Definition of reducible matrix and relation with not strongly connected digraph I connot quite understand the definition of reducible matrix here.  
We know $A_{n\times n}$ is reducible, when there exists a permutation matrix $\textbf{P}$ such that:
$$P^TAP=\begin{bmatrix}X  & Y\\0  & Z\end{bmatrix},$$
where $X$ and $Z$ are both square.   


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*I cannot understand: $a_{i_{\alpha}j_{\beta}}=0, \ \ \forall \alpha = 1, \ldots ,\mu,\ \ \text{and} \ \  \beta = 1,\ldots, \nu$.  Could anyone provide a specific example?  

*How can we say if it is the case, then the corresponding digraph is not strongly connected.   
Here is one answer about this. If strongly connected digraph holds, there exists a path $i_1i_2,\ldots,i_n$. How to say this condition will violate $a_{i_{\alpha}j_{\beta}}=0$?  
Ex: Consider the strongly connected digraph: $1 \rightarrow 2 \rightarrow 3 \rightarrow 1$. $A$ could be chosen as $$A=\begin{bmatrix}0  & 2 & 0\\0  & 0 & 3\\ 4 & 0 & 0\end{bmatrix}$$   
I cannot grasp the structure of matrix corresponding to the digraph. 
 A: Consider your $3 \times 3$ matrix $A$, as a matrix with nodes  $\{1,2,3\}$. More specifically, you can consider your matrix $A$ as an adjacency matrix of a graph. Assume that $a_{12}, a_{23}, a_{31}$ are strictly positive elements. Then it holds:


Since we can reach any node of the graph starting from any node, the matrix $A$ is irreducible and the respective graph - let's say $G$ - is a strongly connected graph.


Consider the following case:

After some reordering (more strictly, you apply the transformation $P^TBP$), we take the matrix form you described, i.e. $$B = \begin{bmatrix} \color{purple} X & Y \\ \color{blue}{\mathbf0} & \color{red}Z \end{bmatrix}.$$ 
Also, consider the $2$ disjoint sets $V_1=\{1,3,4\}$ and $V_2 = \{2,5\}$. Using the notation of the link you provided, consider any $i_a \in V_2$ and any $j_\beta \in V_1$. Then, we have that $$B_{i_a\, j_\beta} = 0.$$
Thus, matrix $B$ is reducible and the corresponding graph is not a strongly connected graph. As you may have observed, in this case, the above condition for the strongly connected graphs does not hold. Indeed, if we start e.g. from node $2$ (or $5$), we can never reach state $3$ (or $1$ or $4$).

In the first case of the irreducible matrix $A$ get any partition of $\{1,2,3\}$  consisting of two disjoint sets $V_1, V_2$, e.g. $V_1 = \{1,3\}$ and $V_2 = \{2\}$. You can confirm  that there will always be at least one $i_a \in V_1$ and one $j_b \in V_2$ such that $A_{i_a\, j_b} \neq 0$. 
