Given an $n$ by $n$ matrix $A$ whose rows and columns sum to $m \in \mathbb N$ and entries are nonnegative integers, does there exist a permutation matrix $P$ such that $A - P$ has only nonnegative entries?

If this is true, then we can write $A$ as the sum of $m$ permutation matrices, and I'll have a good Thanksgiving.


If we can find a permutation $\sigma$ such that $A_{i,\sigma(i)} \neq 0$, for every $1 \leq i \leq n$, then we are done, because the associated permutation matrix will satisfy what you want.

So we consider the following bipartite graph $G$ on $X \sqcup Y$, $X = Y = \{1,\ldots,n\}$, and we set $i \sim j$ if and only if $A_{i,j} \neq 0$. Any perfect matching on $G$ will give us the desired permutation, so lets check that we can satisfy Hall's condition in this graph.

Let $I \subseteq X$, and we have that $N(I) = \{j : \exists i \in I, i \sim j \}$. First notice that for any $i$ by definition of $A$ and construction of $G$ we have that $$\sum_{i \sim j}A_{i,j} = m$$ So we have the following $$\begin{align} |I| &= \frac{1}{m} \sum_{i \in I} \sum_{j \in N(I)} A_{i,j}\\ &= \frac{1}{m} \sum_{j \in N(I)} \sum_{i \in I} A_{i,j} \quad \text{(We switched sumation order)} \\ &\leq \frac{1}{m} \sum_{j \in N(I)} \sum_{i \in X} A_{i,j} \quad \text{(We sum over all $X$, which is $m$)}\\ &= \frac{1}{m}m|N(I)| = |N(I)| \end{align}$$ Which is exactly Hall's condition, so $G$ has a perfect matching which gives us the permutation as we wanted.


By Birkhoff–von Neumann theorem, every doubly stoahastic matrix is a convex combination of permutation matrices. Therefore $\frac1mA=\sum_{k=1}^r c_kP_k$ for some permutation matrices $P_1,P_2,\ldots,P_r$ and some $c_1,c_2,\ldots,c_n>0$. Hence $A\ge mc_1P_1$ entrywise. As $A$ is an integer matrix, it follows that $A\ge P_1$ entrywise, i.e. $A-P_1$ is entrywise nonnegative.

  • $\begingroup$ @bof Some authors do prove Birkohff's theorem using essentially the above problem statement. More specifically, one can prove Birkohff theorem by first showing that a nonzero doubly stochastic matrix must contain a positive diagonal. However, there are other ways to prove the theorem, e.g. to show that the set of doubly stochastic matrices is convex and its extreme points are precisely the permutation matrices. $\endgroup$ – user1551 Nov 29 '13 at 1:34
  • $\begingroup$ In this case, if a doubly stochastic matrix $A$ is not a scalar multiple of permutation matrix, one can show that there is a closed walk between some $m\le n$ nodes and hence $A\pm\epsilon B$ is doubly stochastic for some nonzero matrix $B$ and for all small $\epsilon$. Such an argument does not rely on the existence of a positive diagonal. $\endgroup$ – user1551 Nov 29 '13 at 1:36
  • $\begingroup$ How do we know that $mc_j$ is an integer? Isn't it possible that $\{mc_1,\ldots, mc_r\} \cap \mathbb Z = \emptyset$? Does this matter? I think it does because if $0<|mc_j|<1$ then doesn't that mean that $A \geq P_j$ entrywise is false? $\endgroup$ – user28877 Nov 29 '13 at 19:36
  • $\begingroup$ @user710587 No one says that $mc_1$ is an integer. It is merely positive. However, since $A\ge mc_1P_1$ entrywise, if the $(i,j)$-th entry of $P_1$ is $1$, the $(i,j)$-th entry of $A$ must be positive. Since $A$ is an integer matrix, this means the entry must be at leasst $1$. Therefore, for all $1$s in $P_1$, the counterpart entries in $A$ must be at least $1$, i.e. $A\ge P_1$ entrywise. $\endgroup$ – user1551 Nov 29 '13 at 23:31

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