X=(xij)
\sumixij=\sumjxij=1,
Thus, a doubly stochastic matrix is both left stochastic and right stochastic.[1]
Indeed, any matrix that is both left and right stochastic must be square: if every row sums to 1 then the sum of all entries in the matrix must be equal to the number of rows, and since the same holds for columns, the number of rows and columns must be equal.
See main article: Birkhoff polytope. The class of
n x n
Bn
(n-1)2
n2
2n-1
2n-1
2n
The Birkhoff–von Neumann theorem (often known simply as Birkhoff's theorem[2]) states that the polytope
Bn
n x n
Bn
X
\theta1,\ldots,\thetak\ge0,
| k | |
| \sum | |
| i=1 |
\thetai=1
P1,\ldots,Pk
X=\theta1P1+ … +\thetakPk.
This representation is known as the Birkhoff–von Neumann decomposition, and may not be unique. It is often described as a real-valued generalization of Kőnig's theorem, where the correspondence is established through adjacency matrices of graphs.
n=2
n
n!/nn
1/n
Let X be a doubly stochastic matrix. Then we will show that there exists a permutation matrix P such that xij ≠ 0 whenever pij ≠ 0. Thus if we let λ be the smallest xij corresponding to a non-zero pij, the difference X – λP will be a scalar multiple of a doubly stochastic matrix and will have at least one more zero cell than X. Accordingly we may successively reduce the number of non-zero cells in X by removing scalar multiples of permutation matrices until we arrive at the zero matrix, at which point we will have constructed a convex combination of permutation matrices equal to the original X.[8]
For instance if
| X= | 1 |
| 12 |
\begin{pmatrix}7&0&5\ 2&6&4\ 3&6&3\end{pmatrix}
P=\begin{pmatrix}0&0&1\ 1&0&0\ 0&1&0\end{pmatrix}
λ=
| 2 | |
| 12 |
X-λP=
| 1 | |
| 12 |
\begin{pmatrix}7&0&3\ 0&6&4\ 3&4&3\end{pmatrix}
Proof: Construct a bipartite graph in which the rows of X are listed in one part and the columns in the other, and in which row i is connected to column j iff xij ≠ 0. Let A be any set of rows, and define A as the set of columns joined to rows in A in the graph. We want to express the sizes |A| and |A| of the two sets in terms of the xij.
For every i in A, the sum over j in A of xij is 1, since all columns j for which xij ≠ 0 are included in A, and X is doubly stochastic; hence |A| is the sum over all i ∈ A, j ∈ A of xij.
Meanwhile |A| is the sum over all i (whether or not in A) and all j in A of xij ; and this is ≥ the corresponding sum in which the i are limited to rows in A. Hence |A| ≥ |A|.
It follows that the conditions of Hall's marriage theorem are satisfied, and that we can therefore find a set of edges in the graph which join each row in X to exactly one (distinct) column. These edges define a permutation matrix whose non-zero cells correspond to non-zero cells in X.
There is a simple generalisation to matrices with more columns and rows such that the i th row sum is equal to ri (a positive integer), the column sums are equal to 1, and all cells are non-negative (the sum of the row sums being equal to the number of columns). Any matrix in this form can be expressed as a convex combination of matrices in the same form made up of 0s and 1s. The proof is to replace the i th row of the original matrix by ri separate rows, each equal to the original row divided by ri ; to apply Birkhoff's theorem to the resulting square matrix; and at the end to additively recombine the ri rows into a single i th row.
In the same way it is possible to replicate columns as well as rows, but the result of recombination is not necessarily limited to 0s and 1s. A different generalisation (with a significantly harder proof) has been put forward by R. M. Caron et al.[9]