Let $S=\{1,\ldots, m\}$ and $A_1, \ldots, A_n$ be subsets of S. Let $B=(b_{ij})_{m\times n} = \{0,1\}_{m\times n}$ be a matrix such that $$ b_{ij}= \begin{cases} 1 & i \in A_j \\ 0 & \mbox{otherwise} \end{cases}$$ Is matrix $B$ totally unimodular?
I do not know whether $B$ is totally unimodular. Can anyone have an answer? Thank you in advance!
In general $B$ is not TUM (totally unimodular). As an example you can take $m=3, n=4$ and $A_1=\{1\}$, $A_2=\{1,2\}$, $A_3=\{1,3\}$, $A_4=\{2,3\}$. Then $$B=\begin{pmatrix} 1&1&1&0\\ 0&1&0&1\\ 0&0&1&1 \end{pmatrix}$$ and $$det(\tilde{B})=\begin{pmatrix} 1&1&0\\ 1&0&1\\ 0&1&1 \end{pmatrix}=-2$$ On the other hand if $A_1=\{1\}$, $A_2=\{2\}$, $A_3=\{3\}$, $A_4=\{1,2,3\}$ then $$B=\begin{pmatrix} 1&0&0&1\\ 0&1&0&1\\ 0&0&1&1 \end{pmatrix}$$ is TUM, because it has the consecutive one’s property on the columns (see here for the definition). Therefore the TUM property of $B$ depends on the choose you make for the subsets of $S$.