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Let $p$ be a vertex of a linear polyhedron and choose $n$ independent active constraints, written $Bp=d$, with $B$ invertible. Suppose $p=\lambda u+(1-\lambda )v$, where $u,v\in P$ and $0<\lambda <1$. Each selected inequality has $B_{i}u\le d_i$ and $B_{i}v\le d_i$, whileSince both weights are positive, equality in this weighted upper bound forces $B_{i}u=B_{i}v=d_i$ for every selected row. Hence $Bu=Bv=d$, and invertibility gives $u=v=p$. Therefore every vertex of a linear polyhedron is an extreme point.

An extreme point $p\in P$ cannot be written as a nontrivial convex combination of two distinct points of $P$: if $p=\lambda u+(1-\lambda )v$ with $u,v\in P$ and $0<\lambda <1$, then $u=v=p$.

A vertex of a linear polyhedron is a feasible point at which the active constraints include $n$ linearly independent row normals. Equivalently, some $n$ independent bounding hyperplanes intersect uniquely at that point. This definition does not require $P$ to be bounded or full-dimensional. For a linear polyhedron, it is also equivalent to the point uniquely maximizing a suitable linear function: sum the selected active inequalities to construct one such linear function.

No. For the linear polyhedron $P=[0,1{]}^2$, take $c=(1,0{)}^T$ and $x^{\ast }=(1,1/2{)}^T$. The maximum of $c^{T}x$ is one, attained at $x^{\ast }$. Yetso $x^{\ast }$ is not an extreme point, hence not a vertex of a linear polyhedron. An optimal face can contain many points; an arbitrary maximizer need not be one of its vertices of a linear polyhedron.

The point $(1,4)$ satisfies dual feasibility and supplies an upper bound of $13$ by weak duality. Directly, take one copy of the first constraint and four copies of the second:Since $w\ge 0$, the original objective function is at most $13-2w\le 13$. The feasible point $(2,3,0,0)$ attains $13$. HenceThere are also nonvertex optima. In terms of the slack variables, the exact objective function identity isEquality requires $s_1=s_2=w=0$, leaving the entire optimal faceNegative $x$ or $y$ on this ray are allowed. This illustrates why the single vertex of a linear polyhedron does not imply a unique optimum.

Only $z,w$ have printed nonnegativity restrictions; $x,y$ are unrestricted. In four-dimensional space, a vertex of a linear polyhedron needs four independent active constraints. There are exactly four available constraints, so all must be active. Solving them givesand therefore the only possible vertex of a linear polyhedron is $(x,y,z,w)=(2,3,0,0)$. It is feasible, and its four active normals are independent, so it is indeed a vertex of a linear polyhedron.

One can also see uniqueness geometrically from the slack variables $s_1=1-2x+y-5w$ and $s_2=3-y-z-w$. The feasible region is in affine bijection with the nonnegative orthant in $(s_1,s_2,z,w)$:That orthant has only the origin as an extreme point, giving the same unique vertex of a linear polyhedron.

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.