$A \in M_n$ is reducible if and only if there is a permutation $i_1, ... , i_n$ of $1,... , n$

Question

Let $A \in M_n$. $A$ is called reducible if $A$ is permutation similar to a matrix of the form

$$\begin{pmatrix} B & 0\\ C &D \end{pmatrix} $$ where $B$ and $D$ are square matrices.

Reading a book I got stuck in the following lines:

Obviously, $A \in M_n$ is reducible if and only if there is a permutation $i_1, ... , i_n$ of $1,... , n$ and an integer $s$ with $1 \leq s \leq n-1$ such that $A [i_1,. . . , i_s | i_{s+1} , . .. , i_n] = 0$.

I am confused with... $A [i_1,. . . , i_s | i_{s+1} , . .. , i_n] = 0$...what does this means?

Answer 1

It means that there are $s$ rows with indexes $i_1$,\dots, $i_s$ and $n-s$ columns with indexes $i_{s+1}$,\dots, $i_n$ such that the submatrix of these rows and columns is a zero matrix.

Answer 2

It means that the submatrix of $A$ given by $$ \begin{bmatrix} a_{i_1,i_{s+1}} & \cdots & a_{i_1,i_n}\\ \vdots & & \vdots\\ a_{i_s,i_{s+1}} & \cdots & a_{i_s,i_n} \end{bmatrix} $$ is equal to the zero matrix.