TITLE: Definition of reducible matrix and relation with not strongly connected digraph
QUESTION [8 upvotes]: 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.   

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.

REPLY [10 votes]: 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 $A_{i_a\, j_b} \neq 0$.