Given the polyhedron $P = \{v \in \mathbb R^2 \mid Av \le b\}$ with $A = \begin{bmatrix} -1 & -1 \\ 2 & -1 \\ -1 & 2 \\ 1 & 2 \end{bmatrix}$ and $b = \begin{bmatrix} 0 \\ 1 \\ 1 \\ 2 \end{bmatrix}$, I want to find the edges of P.
I know that is $F$ is a non-empty face then $F=P_I$ for some $I \subset \{1,2,3,4\}$, where $P_I = \{x \in P \mid A_I x = b_I\}$ ($A_I$ and $b_I$ are obtained by deleting rows not in $I$). Also, if there exist $z \in P_I$ such that $I(z) =\{i \in I \mid a_i z = b_i \} = I$ then $\text {dim} P_I = d - \text {rk} A_I$. (in this case $d=2$).
How can I find the edges of $P$ using this knowledge or any other method ?
I see that all pair of rows of $A$ are linearly independent.
You could just draw the inequalities and intersect them, and then find the edges of P as lines intersection points.