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Homeproof of Birkhoff-von Neumann theorem

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# proof of Birkhoff-von Neumann theorem

First, we prove the following lemma:

Lemma:

A convex combination of doubly stochastic matrices is doubly stochastic.
Proof:

Let $\{A_{i}\}_{{i=1}}^{m}$ be a collection of $n\times n$ doubly-stochastic matrices, and suppose $\{\lambda_{i}\}_{{i=1}}^{m}$ is a collection of scalars satisfying
$\sum_{{i=1}}^{m}\lambda_{i}=1$ and $\lambda_{i}\geq 0$ for each $i=1,\ldots,m$.
We claim that $A=\sum_{{i=1}}^{m}\lambda_{i}A_{i}$ is doubly stochastic.

Take any $i\in\{1,\ldots,m\}$. Since $A_{i}$ is doubly stochastic, each of its rows and columns sum to 1. Thus each of the rows and columns of $\lambda_{i}A_{i}$ sum to $\lambda_{i}$.

By the definition of elementwise summation, given matrices $N=M_{1}+M_{2}$, the sum of the entries in the $i$th column of $N$ is clearly the sum of the sums of entries of the $i$th columns of $M_{1}$ and $M_{2}$ respectively. A similar result holds for the $j$th row.

Hence the sum of the entries in the $i$th column of $A$ is the sum of the sums of entries of the $i$th columns of $\lambda_{k}A_{k}$ for each $i$, that is, $\sum_{{k=1}}^{m}\lambda_{k}=1$. The sum of entries of the $j$th row of $A$ is the same. Hence $A$ is doubly stochastic.

Observe that since a permutation matrix has a single nonzero entry, equal to 1, in each row and column, so the sum of entries in any row or column must be 1. So a permutation matrix is doubly stochastic, and on applying the lemma, we see that a convex combination of permutation matrices is doubly stochastic.
This provides one direction of our proof; we now prove the more difficult direction: suppose $B$ is doubly-stochastic. Define a weighted graph $G=(V,E)$ with vertex set $V=\{r_{1},\ldots,r_{n},c_{1},\ldots,c_{n}\}$, edge set $E$, where $e_{{ij}}=(r_{i},c_{i})\in E$ if $B_{{ij}}\neq 0$, and edge weight $\omega$, where $\omega(e_{{ij}})=B_{{ij}}$.
Clearly $G$ is a bipartite graph, with partitions $R=\{r_{1},\ldots,r_{n}\}$ and $C=\{c_{1},\ldots,c_{n}\}$, since the only edges in $E$ are between $r_{i}$ and $c_{j}$ for some $i,j\in\{1,\ldots,n\}$. Furthermore, since $B_{{ij}}\geq 0$, then $\omega(e)>0$ for every $e\in E$.
For any $A\subset V$ define $N(A)$, the neighbourhood of $A$, to be the set of vertices $u\in V$ such that there is some $v\in A$ such that $(u,v)\in E$.
We claim that, for any $v\in V$, $\sum_{{u\in N(\{v\})}}\omega(u,v)=1$. Take any $v\in V$; either $v\in R$ or $v\in C$. Since $G$ is bipartite, $v\in R$ implies $N(\{v\})\subset C$, and $v\in C$ implies $N(\{v\})\subset R$. Now,

$\begin{array}[]{c}{\displaystyle v=r_{i}\implies\sum_{{u\in N(r_{i})}}\omega(r% _{i},u)=\sum_{{\stackrel{j=1,}{e_{{ij}}\in E}}}^{n}\omega(r_{i},c_{j})=\sum_{{% \stackrel{j=1,}{B_{{ij}}\neq 0}}}^{n}B_{{ij}}=\sum_{{j=1}}^{n}B_{{ij}}=1}\\ {\displaystyle v=c_{j}\implies\sum_{{u\in N(c_{j})}}\omega(u,c_{j})=\sum_{{% \stackrel{i=1,}{e_{{ij}}\in E}}}^{n}\omega(r_{i},c_{j})=\sum_{{\stackrel{i=1,}% {B_{{ij}}\neq 0}}}^{n}B_{{ij}}=\sum_{{i=1}}^{n}B_{{ij}}=1}\end{array}$ |

since $B$ is doubly stochastic. Now, take any $A\subset R$. We have

$\sum_{{\stackrel{v\in A}{w\in N(A)}}}\omega(v,w)=\sum_{{v\in A}}\sum_{{w\in N(% \{v\})}}\omega(v,w)=\sum_{{v\in A}}1=|A|$ |

Let $B=N(A)$. But then clearly $A\subset N(B)$, by definition of neighbourhood. So

$|N(A)|=|B|=\sum_{{\stackrel{v\in B}{w\in N(B)}}}\omega(v,w)\geq\sum_{{% \stackrel{v\in B}{w\in A}}}\omega(v,w)=\sum_{{\stackrel{w\in A}{v\in N(A)}}}% \omega(v,w)=|A|$ |

So $|N(A)|\geq|A|$. We may therefore apply the graph-theoretic version of Hall’s marriage theorem to $G$ to conclude that $G$ has a perfect matching. So let $M\subset E$ be a perfect matching for $G$. Define an $n\times n$ matrix $P$ by

$P_{{ij}}=\left\{\begin{array}[]{ll}1&\text{if }e_{{ij}}\in M\\ 0&\text{otherwise}\end{array}\right.$ |

Note that $B_{{ij}}=0$ implies $P_{{ij}}=0$: if $B_{{ij}}=0$, then
$(r_{i},c_{j})\not\in E$, so $(r_{i},c_{j})\not\in M$, which implies $P_{{ij}}=0$.
Further, we claim that $P$ is a permutation matrix:

Let $i$ be any row of $P$. Since $M$ is a perfect matching of $G$, there exists $e_{0}\in M$ such that $r_{i}$ is an end of $e_{0}$. Let the other end be $c_{j}$ for some $i$; then $P_{{ij}}=1$.

Suppose $i_{i},i_{2}\in\{1,\ldots,n\}$ with $i_{1}\neq i_{2}$ and $P_{{i_{1},j}}=P_{{i_{2},j}}=1$ for some $j$. This implies $(r_{{i_{1}}},c_{j}),(r_{{i_{2}}},c_{j})\in M$, but this implies the vertex $i$ is the end of two distinct edges, which contradicts the fact that $M$ is a matching.

Hence, for each row and column of $P$, there is exactly one nonzero entry, whose value is 1. So $P$ is a permutation matrix.
Define $\displaystyle{\lambda=\min_{{i,j\in\{1,\ldots,n\}}}\{B_{{ij}}\mid P_{{ij}}\neq
0\}}$. We see that $\lambda>0$ since $B_{{ij}}\geq 0$, and $P_{{ij}}\neq 0\implies B_{{ij}}\neq 0$. Further, $\lambda=B_{{pq}}$ for some $p,q$.
Let $D=B-\lambda P$. If $D=0$, then $\lambda=1$ and $B$ is a permutation matrix, so we are done. Otherwise, note that $D$ is nonnegative; this is clear since $\lambda P_{{ij}}\leq\lambda\leq B_{{ij}}$ for any $B_{{ij}}\neq 0$.
Notice that $D_{{pq}}=B_{{pq}}-\lambda P_{{pq}}=\lambda-\lambda\cdot 1=0$.
Note that since every row and column of $B$ and $P$ sums to 1, that every row and column of $D=B-\lambda P$ sums to $1-\lambda$. Define $B^{{\prime}}=\frac{1}{1-\lambda}D$. Then every row and column of $B^{{\prime}}$ sums to 1, so $B^{{\prime}}$ is doubly stochastic.
Rearranging, we have $B=\lambda P+(1-\lambda)B^{{\prime}}$. Clearly $B_{{ij}}=0$ implies that $P_{{ij}}=0$ which implies that $B^{{\prime}}_{{ij}}=0$, so the zero entries of $B^{{\prime}}$ are a superset of those of $B^{{\prime}}$. But notice that $B^{{\prime}}_{{pq}}=\frac{1}{1-\lambda}D_{{pq}}=0$, so the zero entries of $B^{{\prime}}$ are a strict superset of those of $B$.
We have decomposed $B$ into a convex combination of a permutation matrix and another doubly stochastic matrix with strictly more zero entries than $B$. Thus we may apply this procedure repeatedly on the doubly stochastic matrix obtained from the previous step, and the number of zero entries will increase with each step. Since $B$ has at most $n^{2}$ nonzero entries, we will obtain a convex combination of permutation matrices in at most $n^{2}$ steps.
Thus $B$ is indeed expressible as a convex combination of permutation matrices.

# References

G. Birkhoff, “Tres observaciones sobre el algebra lineal,” Univ. Nac. Tucumán Rev, Ser. A, no. 5, pp147–151. (1946)

## Mathematics Subject Classification

15A51*no label found*

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