Is there anything known about which binary (square) matrices allow to be filled up with non-negative real numbers (in the non-zero positions) such that each row and each column sums up to one? E.g. for $$M=\begin{pmatrix}0&0&1\\0&0&1\\1&1&0\end{pmatrix}$$ this is not possible, even though there are no all-zero rows or columns.
Which binary matrices can be filled with numbers such that all rows and columns sum up to $1$?
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linear-algebra
matrices
discrete-mathematics
2 Answers
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If you want rows to sum to one, that is a Markov matrix (although wikipedia calls it a Stochastic matrix). If you want both rows and columns, it is called a doubly stochastic Markov matrix.
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It's exactly those matrices with non-zero permanent.
Proof: if $\text{perm}(M)>0$ then there exists $\sigma\in{\cal S}_n$ such that $\prod_i a_{i\sigma(i)}=1$ and putting $1$s at all positions $(i,\sigma(i))$ yields a (permutation) matrix where each row and column sums up to $1$.
On the other side, if there is some $M$ with $n_{ij}>0$ only if $m_{ij}=1$ where each row and column sums up to $1$, fix $i,j$ with $n_{1j}>0$. Then cut out the $i$th row and $j$ column, name the result $N$ and conclude by some induction argument that there exists $\tau\in{\cal S}_{n-1}$ such that $\prod_i n_{i\sigma'(i)}=1$. So $$\sigma'(i'):=\begin{cases}\tau(i')&i'