I want to find the corners, faces and the dimension of the faces of a polyhedron. I would very much appreciate someone looking over my work, especially the part concerning the definitions of the faces.
The polyhedron $\mathcal{P}\subseteq\mathbb{R}^2$ is defined by the inequalities: $\quad x_1+x_2\geq 1,\quad x_1-x_2\geq 3$
This gives us matrix $A$ and vector $b$ with:
$A=\begin{pmatrix} -1\quad -1\\-1\qquad1\\\end{pmatrix},\qquad b=\begin{pmatrix} -1 \\-3\end{pmatrix}\quad$ and $\quad Ax\leq b$
$\mathcal{P}$ has one corner and to find it, we must see where the two inequalities intersect:
$1-x_1=-3+x_1\iff x_1=2 \implies x_2=-1$
Now to prove that $z:=(2,-1)^\text{T}$ really is a corner:
$\begin{pmatrix} -1\quad -1\\-1\qquad1\\\end{pmatrix}\begin{pmatrix} 2 \\-1\end{pmatrix}=\begin{pmatrix} -1 \\-3\end{pmatrix}$
$\implies$ eq$(\{z\})=(1,2)\implies \text{rank}(A_\text{eq(z)})=\text{rank}\begin{pmatrix} -1\quad -1\\-1\qquad1\\\end{pmatrix}=2$
Since $\mathcal{P}\subseteq\mathbb{R}^n$ where n=2=rank, $z$ is a corner of $\mathcal{P}$.
The polyhedron has two faces $F_1$ and $F_2$, which can be confirmed by sketching the two inequalities.
Intuitively, I would think that we could write $F_1$ and $F_2$ the following way:
$F_1=\mathcal{P}\cap\{(x_1,x_2):x_1+x_2\geq 1 \land x_1\geq 2\},\qquad$similarly:
$F_2=\mathcal{P}\cap\{(x_1,x_2):x_1-x_2\geq 3 \land x_1\geq 2\}$
The dimension of $F_1$ and $F_2$ is equal to $1$, as dim($F)=$dim$(\mathcal{P})-1=2-1=1$