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  "body": "\n## Statement\n\nLet $n \\ge 1$, and let $P \\subseteq \\R^n$ be nonempty.\nThe following are equivalent.\n\n1. There are a real $m \\times n$ matrix $A$ and $b \\in \\R^m$, for some $m \\ge 0$, with $P = \\set{x \\in \\R^n : A x \\le b}$, where $\\le$ is taken coordinatewise.\n2. There are finite sets $V, W \\subseteq \\R^n$ with $V \\ne \\emptyset$ and $P = \\mathrm{conv}(V) + \\mathrm{cone}(W)$.\n\nHere $\\mathrm{conv}(V)$ is the [[convex hull|0FQSDC2EQ2]] of $V$, and $\\mathrm{cone}(W)$ is the [[convex cone|6VE4KA12VD]] of all nonnegative linear combinations of $W$, with $\\mathrm{cone}(\\emptyset) = \\set{0}$.\nIn particular, $P$ is a [[convex polytope|300K6CB1KH]] if and only if it is a bounded set of the form in 1.\n\n## Remarks\n\nFor 2 implies 1, write $P$ as the projection to $\\R^n$ of the set of all $(x, \\lambda, \\mu)$ with $x = \\sum_v \\lambda_v v + \\sum_w \\mu_w w$, $\\lambda, \\mu \\ge 0$ and $\\sum_v \\lambda_v = 1$.\nFourier–Motzkin elimination shows that the projection of a set given by finitely many linear inequalities is again such a set.\nFor 1 implies 2, one can apply the first direction to a suitable polar cone.\nThe polytope case follows, because a bounded set $\\mathrm{conv}(V) + \\mathrm{cone}(W)$ must have $\\mathrm{cone}(W) = \\set{0}$.\nExample: the square $[0, 1]^2$ is $\\set{x : 0 \\le x_1 \\le 1, \\ 0 \\le x_2 \\le 1}$, given by four inequalities, and also $\\mathrm{conv}\\set{(0, 0), (1, 0), (0, 1), (1, 1)}$.\nThe quadrant $\\set{x : x_1 \\ge 0, x_2 \\ge 0}$ is $\\mathrm{conv}\\set{0} + \\mathrm{cone}\\set{(1, 0), (0, 1)}$.\nFiniteness is needed: the closed unit disc is an intersection of infinitely many half-planes and the convex hull of infinitely many points, but of neither form above.\n\n## Explanations\n\n### L1 human\n\nThere are two ways to describe a convex polytope: by its corners, or by the flat walls that bound it.\nThe theorem says that both ways describe the same shapes.\nAny finite list of points spans a shape that can be cut out by finitely many walls.\nAny bounded region cut out by finitely many walls is spanned by finitely many corners.\n\n### L2 human\n\nThe two descriptions are the [[V-representation and H-representation|W8P2G5R02N]] of a polytope.\nLinear programming uses both: the feasible region is given by inequalities, while the simplex method moves between vertices.\nChanging from one description to the other is the vertex and facet enumeration problem.\nThe sizes can differ a lot: the cube $[0, 1]^n$ needs $2n$ inequalities but has $2^n$ vertices.\nThe theorem is a finite-dimensional cousin of Farkas' lemma, and the two can be derived from each other.\n",
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