I have the following matrix
$$A=\pmatrix{1&0&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\\ 1&1&0&0&0&0\\0&0&1&1&0&0\\0&0&0&0&1&1\\ -1&0&-1&0&-1&0\\ 0&-1&0&-1&0&-1}$$
By using Matlab, I believe it is a totally unimodular matrix. However, I am not able to prove this property. Does anyone give me some hints on this? Much appreciated.
Suppose that the matrix has a submatrix with determinant $\notin\{-1,0,1\}$ whose top row is a subrow of row $1$. Since there is only one $1$ in row $1$, it must be the case that the matrix formed by deleting row $1$ also fails to be totally unimodular. Similarly, we can discard the next $5$ rows of $A$ and we have only to prove that the resulting matrix (which I will also call $A$ for conformity with the Wikipedia article) is totally unimodular. The wiki says,
Let $A$ be an $m\times n$ matrix whose rows can be partitioned into two disjoint sets$B$ and $C.$ Then the following four conditions together are sufficient for $A$ to be totally unimodular:
Every entry in $A$ is $0, +1, \text{ or }−1$;
Every column of $A$ contains at most two non-zero (i.e., $+1$ or $-1$) entries;
If two non-zero entries in a column of $A$ have the same sign, then the row of one is in $B$, and the other in $C$;
If two non-zero entries in a column of $A$ have opposite signs, $B,$ or both in $C.$
In our (new) matrix $A$ we can let $B$ comprise the first $3$ rows and $C$ the last $2$ rows, and it is clear the the conditions are satisfied.