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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.

An extreme point of a convex set cannot be written as a nontrivial convex combination of two distinct members. If a permutation matrix $P=\theta B+(1-\theta )C$ with $0<\theta <1$ and $B,C$ doubly stochastic, each zero entry of $P$ forces the corresponding entries of $B,C$ to be zero by nonnegativity. The unique possible nonzero entry in each row must then be one by the row sum. Thus $B=C=P$, proving that every permutation matrix is extreme.

Conversely, let $A$ be a doubly stochastic matrix that is not a permutation matrix. At least one entry lies strictly between zero and one. Form a bipartite graph whose two vertex classes are rows and columns, with edges precisely at these fractional entries. Every incident row has at least two fractional entries: a single fractional entry, with all other entries zero or one, could not give row sum one. The same holds for columns.

A finite nonempty graph whose incident vertices all have degree at least two contains a cycle. In a bipartite graph that cycle has even length. Assign alternating signs $+1,-1$ to its entries, giving a nonzero matrix $D$ with every row and column sum zero. Choose $\epsilon >0$ smaller than the distance of each cycle entry from both zero and one. Then $A+\epsilon D$ and $A-\epsilon D$ remain doubly stochastic matrices, are distinct, and satisfySo $A$ is not extreme. ThereforeThis alternating-cycle perturbation of a doubly stochastic matrix proves the extreme-point part of the Birkhoff-von Neumann theorem directly.