I'm trying to prove that a linear problem has integer extreme points. Looking at the matrix structures, I guess the easiest way to prove this is by showing that this matrix is totally unimodular. My matrix (and its structutes) can be described as follows.
\begin{equation}B= \begin{pmatrix} I & A &0\\ I & 0 & A \end{pmatrix} \end{equation}
Such that matrix $A\in R^{m\times n}$ has the consecutive ones property and, thus, is totally unimodular. Can I show that matrix $B$ is also totally unimodular?
The matrix
\begin{equation} \begin{pmatrix} A &0\\ 0 & A \end{pmatrix} \end{equation}
also has the consecutive ones property and is TUM. It remains TUM if you append one large identity matrix, but you add two smaller identity matrices, which can ruin TUM. Just generating random matrices $A$ quickly results in counterexamples, for example $$A=\begin{pmatrix}0 & 0 & 1 & 1 & 1\\ 1 & 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 1 & 1\\ 1 & 0 & 0 & 0 & 0\\ 1 & 1 & 1 & 1 & 0\end{pmatrix}$$ The submatrix of rows 1,3,5,6,10 and columns 1,5,7,10,13 has determinant 2.