&\implies \left\{ \begin{array}{l}
        \forall i \in \{1,...,n_l\},\forall j \in \{n_l + 1, ..., N\}, a_{ij}' = 0, \\\forall i \in \{n_l+1,...,N\},\forall j \in \{1, ..., n_l\}, a_{ij}' = 0.
    \end{array} \right.\\
    &\implies \exists (B,C) \in \mathcal{M}_{n_l}(\mathbb{R})\times\mathcal{M}_{n_h}(\mathbb{R}),  A = \left (\begin{array}{c|c}
B & 0\\
\hline
0 & C 
\end{array}\right).\end{aligned}$$

So $1 \nrightarrow N$ et $N \nrightarrow 1$ with the matrix $A'$. Let $i = \sigma^{-1}(1)$ and $j = \sigma^{-1}(N)$, we have $i\nrightarrow j$ and $j\nrightarrow i$ with the matrix $A$.

#### Second case : $n_l + n_h < N$

Let $G_1 = \{1,...,n_l\}$, $G_2 = \{n_l + 1,...,N - n_l\}$, and $G_3 = \{N - n_h + 1,...,N\}$.

$$\begin{aligned}
    A X = X & \iff A'X' = X'.\\
    &\implies 
        \forall i \in \{1,...,N\},\sum_{j=1}^N a_{ij}' X_j' = X_i'.\\
    &\implies \left\{ \begin{array}{l}
        \forall i \in \{1,...,n_l\},\forall j \in \{n_l + 1, ..., N\}, a_{ij}' = 0 ,\\\forall i \in \{N-n_h+1,...,N\},\forall j \in \{1, ..., n_l\}, a_{ij}' = 0.
    \end{array} \right.\\
    &\implies \exists (B,C,D,E,F),  A = \left (\begin{array}{c|c|c}
B & 0 & 0\\
\hline
C & D & E \\
\hline
0 & 0 & F 
\end{array}\right).\end{aligned}$$

=\[ draw = black, thick, fill = white, minimum size = 4mm \] (2) [$G_2$]{}; (1) \[below 