Let $n\in\Bbb{N}$ and $\mathcal{P}=\mathcal{P}(A,b)\subseteq\Bbb{R}^n$ be a polyhedron with $\dim(\mathcal{P})=n$. Let $Z$ be the row index vector of $A$ and $i$ the index of a row. Let $\mathcal{P}_i=\mathcal{P}(A_{Z\backslash \{i\}},b_{Z\backslash \{i\}})$. Show that:$$\mathcal{P}\neq \mathcal{P}_i\iff \text{there exists a non-trivial face }\mathcal{F}\text{ of }\mathcal{P}, \text{with }\;eq(\mathcal{F})=\{i\} $$Here $eq(\mathcal{F})$ is the row index of the matrix equation $Ax=b$, that describes $\mathcal{F}$.
$"\impliedby"$ Assume that there exists a non-trivial face $\mathcal{F}$ of $\mathcal{P}$.
Then there exists an $i\in Z$ sucht that for a row vector $a_i$ of $A$, $a_i^Tx=b_i$ describes span$(\mathcal{F})$. This is true only for one row: the $i$-th row of $Ax=b$. The intersections with the other equations describes the boundaries of $\mathcal{F}$ but without the equation in row $i$ there is no supporting hyperplane for $\mathcal{F}$ and therefore $\mathcal{F}$ is not a face of the projection $\mathcal{P}_i=\mathcal{P}(A_{Z\backslash \{i\}},b_{Z\backslash \{i\}})$, which means $\mathcal{P}\neq\mathcal{P}_i$
$"\implies"$ Now assume $\mathcal{P}\neq\mathcal{P}_i$
Because of the definition: $\mathcal{P}_i=\mathcal{P}(A_{Z\backslash \{i\}},b_{Z\backslash \{i\}})$, we know that $P_i$ is a projection of $\mathcal{P}$ (and therefore also a polyhedron). By our assumption $\mathcal{P}\neq\mathcal{P}_i$ , therefore the difference of $\mathcal{P}$ and $\mathcal{P}_i$ are some non-trivial points of $\mathcal{P}$ that aren't projected onto $\mathcal{P}_i$. Consequently, and because $\mathcal{P}$ is a polyhedron, row $i$ of matrix equation $Ax=b$ must be non-trivial and because it is non-trivial it must describe a (non-trivial) face $\mathcal{F}$ of $\mathcal{P}$ (with eq$(\mathcal{F})=\{i\})$.
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