Vertex of a linear polyhedron ($\mathrm{rank}{A}_{I(p)}=n$)

For a linear polyhedron $P=\{x\in {\mathbb{R}}^n:Ax\le b\}$, a vertex of a linear polyhedron is a feasible point $p$ at which the normals of the active constraints span ${\mathbb{R}}^n$. Equivalently, $n$ linearly independent active constraints uniquely determine $p$. A linear polyhedron may be unbounded or lower-dimensional; the definition applies in those cases as well as to a convex polytope.

Such a point is an extreme point. Indeed, if $p=\lambda u+(1-\lambda )v$ with $u,v\in P$ and $0<\lambda <1$, every active constraint at $p$ must also be an equality at $u$ and $v$. Selecting $n$ independent active rows forces $u=v=p$. Conversely, if the active normals do not span ${\mathbb{R}}^n$, take a nonzero vector $h$ annihilated by all active rows. The finitely many inactive inequalities have positive slack, so both $p+\epsilon h$ and $p-\epsilon h$ remain feasible for sufficiently small $\epsilon >0$. Their midpoint is $p$, so $p$ is not an extreme point. Thus these notions coincide for a linear polyhedron.