Suppose $A=(a_{ij})$ is a $n\times m$ matrix with nonnegative entries such that \begin{eqnarray} \label{fone} \sum_{j=1}^m a_{ij} &=& 1, \quad i=1,\ldots, n, \\ \label{ftwo} \sum_{i=1}^n a_{ij} &=& 1, \quad j=1,\ldots, m. \end{eqnarray}Then $n=m$ .

This is seen by summing equation () over $i=1,\ldots, n$ and equation () over $j=1,\ldots, m$ . Then \begin{eqnarray*} \sum_{i=1}^n \sum_{j=1}^m a_{ij} &=& n, \\ \sum_{i=1}^n \sum_{j=1}^m a_{ij} &=& m, \end{eqnarray*}and since the right hand sides coincide, it follows that $n=m$ .