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For a doubly stochastic matrix with a fractional entry, form the row-column bipartite graph of entries strictly between zero and one. Every incident vertex has degree at least two, so there is an even cycle. A matrix supported on that cycle with alternating entries $+1,-1$ has zero row and column sums. A sufficiently small positive multiple can be both added and subtracted while preserving entry bounds. The original matrix is then the midpoint of two distinct doubly stochastic matrices, so it is not an extreme point.

Every doubly stochastic matrix is a convex combination of permutation matrices. Equivalently, the extreme points of the doubly stochastic convex polytope are exactly the permutation matrices. The extreme-point assertion follows from an alternating-cycle perturbation of a doubly stochastic matrix; the convex-hull assertion follows because a bounded finite-dimensional polytope is the convex hull of its vertices. Permutation matrices are extreme since their zero entries force the same zeros in every convex decomposition.

For an integer $0\le k\le n$, the capped simplexis the convex hull of the zero-one vectors having exactly $k$ entries equal to one. It is a convex polytope whose extreme points are precisely these vectors.

If a feasible vector has a fractional coordinate, the integer coordinate sum forces at least two fractional coordinates. Perturbing one upward and the other downward by a sufficiently small amount gives two distinct feasible vectors whose midpoint is the original vector. It is therefore not an extreme point. Conversely, a zero-one vector cannot be a nontrivial convex combination of points of $[0,1{]}^n$, since each coordinate is already at a bound.

Every point of a compact metrizable convex subset of a Hausdorff locally convex space is the barycenter of a Borel probability measure concentrated on its extreme points. The theorem is an existence assertion; uniqueness requires additional hypotheses such as a simplex structure.

Every allocation $(x,y)$ in the convex feasible region for a resource budget is a convex combination of its three extreme points:The condition $x+2y\le 6$ is exactly ${\lambda }_1\ge 0$; nonnegative $x,y$ give the other two nonnegative coefficients. Thus the half-space description and the convex hull description encode the same triangle. At $(2,1)$ all three coefficients are $1/3$, and at $(3,9/8)$ they are $1/8$, $1/2$, $3/8$. The green and orange points in the budget diagram therefore have two complementary interpretations: direct resource constraints and mixtures of extreme choices.

The three extreme points alone are not a convex set, because their midpoints are absent. Their convex hull fills the whole triangle. More generally, Caratheodory theorem bounds how many generating points are needed for one convex combination in a finite-dimensional real vector space; its geometric content is that a complicated collection can describe each attainable average with a short representation.

Consider a stipulated continuous-allocation problem. Amounts $x,y$ of two activities use $x+2y$ units of a common resource, with budget six. The feasible regionis a triangle with extreme points $(0,0)$, $(6,0)$ and $(0,3)$. It is the intersection of three closed half-spaces, hence a closed convex set. The general intersection argument is short: a line segment between two points in every constraint set belongs to every constraint set, so it belongs to their intersection.^[ref]

For an original numerical check, take $A=(1,2)$ and $B=(5,1/4)$. Their resource totals are $5$ and $11/2$. Every convex combination $Z(t)=(1-t)A+tB$ is nonnegative and has resource totalIn particular the midpoint is $(3,9/8)$, using $21/4$ units. This verifies the full line segment, rather than just a plotted sample of it.

This is useful whenever divisible quantities share linear limits: the feasible choices remain compatible under averaging. A new linear equation also preserves this property, because averaging two identical constraint values gives the same value. The affine subspace and closed half-space viewpoints explain this without requiring a separate argument for every budget.

An exposed point is a point uniquely selected by maximizing a linear functional. In ${\mathbb{R}}^d$, a point $p$ of a convex set $C$ is exposed if some vector $c$ satisfiesGeometrically, a supporting hyperplane touches the convex set only at $p$. Every exposed point is an extreme point: a convex combination that attains the maximum can put positive weight only on maximizers, and there is only one.^[ref]

The unit disk calculation in extreme point exposes every point of the unit circle. For a corner $(a,b)$ of the square, $c=(a,b)$ exposes it: $ax+by$ attains two only at $(a,b)$. However, maximizing $x$ exposes no single point on the right edge, since all of them have the same objective value. A linear functional may have several optimal points, even when some of them are extreme.

An extreme point of a convex set is a point that cannot be made by genuinely mixing two different points of the set. For a convex set $C$ in a real vector space, the condition on $p\in C$ isThus $p$ lies on no open line segment between two distinct points of $C$. It is enough to exclude midpoint representations: if $p$ lies inside a line segment, two nearby points on opposite sides have midpoint $p$.^[ref]

The square $[-1,1{]}^2$ has just four extreme points, at $(\pm 1,\pm 1)$. In a convex combination giving a corner, each coordinate is already at its largest or smallest allowed value, forcing both input points to agree with that corner. In contrast, the boundary point $(1,0)$ is not extreme:It is the midpoint of two other points on the same edge. An interior point also admits a small line segment through it.

The unit disk $D=\{z\in {\mathbb{R}}^2:\Vert z{\Vert }_2\le 1\}$ has a different shape: every point on its unit circle is extreme. For a unit vector $p$, the Cauchy-Schwarz inequality givesand equality is possible only at $z=p$. If a convex combination equals $p$, applying this linear functional forces both ingredients to attain that equality. Both must therefore be $p$. Extremality need not look like a sharp corner.

These points help describe all allowable mixtures. Every $(x,y)\in [-1,1{]}^2$ is a convex combination of its four corners, with the weight on $(a,b)\in \{-1,1{\}}^2$ equal to $(1+ax)(1+by)/4$. For the unit disk, write $z=rp$, with $0\le r\le 1$ and $p$ a unit vector; at zero choose any $p$. ThenTwo extreme points suffice for that mixture. The Krein-Milman theorem gives a much broader reconstruction: a nonempty compact convex set in a locally convex space whose topology is Hausdorff is the closed convex hull of its extreme points.

Mark Krein and David Milman published their reconstruction theorem in 1940. Milman's colleagues describe that result and its subsequent use as a method of representing convex sets through their extreme points.^[ref]

For the real l-infinity sequence space, every coordinate of an extreme point must be an endpoint of $[-1,1]$; any slack coordinate permits opposite perturbations. For the real absolutely summable sequence space, a unit vector with two nonzero coordinates permits a transfer of mass between them, so only signed coordinate vectors are extreme. For the real l2 sequence space, the parallelogram law makes every unit vector extreme. In all three cases interior points of the unit ball are nonextreme.

The density operators on a finite-dimensional Hilbert space form a convex set. Its extreme points are exactly the rank-one projectors, hence exactly the pure states. To prove this, suppose $|\psi ⟩⟨\psi |={\sum }_i{a}_i{\rho }_i$ with $a_i>0$ and ${\rho }_i$ positive semidefinite. For $w\perp \psi$, $0={\sum }_i{a}_i⟨w|{\rho }_i|w⟩$ is a sum of nonnegative numbers, so every term vanishes. Diagonalizing ${\rho }_i$ shows that $⟨w|{\rho }_i|w⟩=0$ implies ${\rho }_{i}w=0$. Thus each ${\rho }_i$ is supported on the line spanned by $\psi$, and its unit trace forces ${\rho }_i=|\psi ⟩⟨\psi |$.

Conversely, a density operator of rank at least two has a spectral decomposition with at least two positive eigenvalues. If $\lambda$ is one of them, $0<\lambda <1$ and $\rho =\lambda |e⟩⟨e|+(1-\lambda )\sigma$, where $\sigma$ is the normalized remainder. These are distinct density operators, proving that $\rho$ is not an extreme point.

For complex $C(K)$ on a compact Hausdorff space, the extreme points of its dual unit ball are precisely unimodular multiples of Dirac measures. A measure whose variation measure has mass on two disjoint sets splits as a nontrivial convex combination of normalized restrictions and is not extreme. This description is the key to the Banach–Stone theorem.

The distinction appears at a smooth join between a straight segment and a curved boundary. Consider the convex hullThe upper tangent from $P$ meets the unit circle at $T=(1/2,\sqrt{3}/2)$. The supporting line isEvery point of the unit disk has $T\cdot z\le 1$, with equality only at $T$. Hence the convex face of $C$ where equality holds is exactly the line segment $TP$.

The point $T$ is extreme. If a nontrivial convex combination of two points of $C$ equals $T$, both must belong to the convex face $TP$. Since $T$ is an endpoint of that line segment, both must equal $T$.

Yet $T$ is not exposed. A nonzero linear functional maximizing at $T$ on $C$ must also maximize there on the included unit disk. Its vector must be a positive multiple of $T$, by the equality condition in the Cauchy-Schwarz inequality. It then gives $P$ the same value as $T$. No linear functional can select $T$ alone. Being an extreme point is a property of possible mixtures; being an exposed point adds the ability to distinguish it by one linear objective.

A convex set $F\subseteq C$ is a face of $C$ if $ty+(1-t)z\in F$, with $y,z\in C$ and $0<t<1$, implies $y,z\in F$. A singleton face is an extreme point. A face of a face is a face of the original convex set.

For an integer $0\le k\le n$, the fantope is the convex hull of rank-$k$ orthogonal projection matrices in ${\mathbb{R}}^n$. Equivalently, it consists of real symmetric matrices with eigenvalues in $[0,1]$ and matrix trace $k$. To prove the equivalence, apply the spectral theorem for real symmetric matrices and express the eigenvalue vector as a convex combination of the zero-one extreme points of the capped simplex. The sum of the largest eigenvalues is its support function.

For the degree-constrained linear programming relaxation of a matching in a graph, retain precisely the edges with fractional coordinates. In a bipartite graph, an even graph cycle in this subgraph permits alternating positive and negative perturbations preserving all vertex sums. If the subgraph is a forest, use a graph path between two tree leaves: its endpoints have positive slack and its internal sums remain unchanged. Either perturbation expresses the original feasible point as the midpoint of two distinct feasible points. Hence a fractional feasible point cannot be an extreme point, proving integrality of all extreme points of the bipartite degree relaxation.

At an extreme point of $\{x:Ax\ge b\}$, the active rows span the ambient dimension: otherwise a nonzero direction orthogonal to all active rows permits small feasible moves of both signs. Select $n$ independent active rows to form an integer matrix $B$. Its nonzero determinant has absolute value at least one. Cramer's rule and the determinant expansion bound each coordinate by $n!U^n$, where $U$ bounds the entries of $A$ and $b$. The vector $b$ may be real; only the denominator matrix needs to be integer.

If a compact convex set is the closed convex hull of a subset $S$, every extreme point lies in the closure of $S$. A neighborhood avoiding $S$ around a putative missing extreme point gives finitely many closed convex caps covering $S$ and excluding that point. The convex hull of a finite union of compact convex sets is compact: group the terms by their set and use the simplex parametrization. The point must lie in that hull, contrary to extremality. The closures are in the given Hausdorff locally convex topology.

A nonempty compact face of a convex set in a Hausdorff locally convex space contains an extreme point. Chains of compact faces have nonempty compact intersection. A minimal face supplied by the Zorn lemma must be a singleton, since a separating linear functional would otherwise expose a proper maximizer face. This is the existence step in the Krein-Milman theorem.

Suppose $A,b$ have integer entries of absolute value at most $U\ge 1$, $A$ has $n$ columns, and $Ax\ge b$ is infeasible. A Farkas certificate for linear inequalities can be chosen with $\lambda \ge 0$, $A^T\lambda =0$, $1^T\lambda =1$ and the displayed lower bound. Normalize an infeasibility certificate to obtain the last equality, then maximize $b^T\lambda$ over the resulting compact linear polyhedron. A maximizing extreme point has at most $n+1$ positive entries: otherwise the augmented columns $(A_i^T,1)$ on its support would be dependent, allowing feasible perturbations in both directions. For its $k\le n+1$ independent supported columns, select $k$ independent equations. Cramer's rule expresses the supported entries with a common nonzero integer determinant denominator. That determinant has absolute value at most $k!U^k\le [(n+1)U{]}^{n+1}$. The positive value $b^T\lambda$ therefore has a positive integer numerator over a denominator of at most this size, proving the bound.

The set of minimizers of a linear objective on a linear polyhedron. It is a face: if a strict convex combination of two feasible points minimizes the objective, both points must also minimize it. Hence an extreme point of an optimal face is an extreme point of the original polyhedron. Successively minimizing further linear objectives restricts to nested faces, whose vertices remain original vertices. This preserves uniform rational denominator bounds in exact feasibility-oracle optimization.