How to prove that this matrix is total unimodular This matrix is total unimodular (tested by a computer program).
1 1 1 1 -1 -1 -1 -1
0 1 1 1  0 -1 -1 -1
0 0 1 1  0  0 -1 -1
0 0 0 1  0  0  0 -1
1 0 0 0 -1  0  0  0
1 1 0 0 -1 -1  0  0
1 1 1 0 -1 -1 -1  0
1 1 1 1 -1 -1 -1 -1

Is there way to prove theoretical that this matrix is total unimodular?
I already tried some stuff from the wikipedia article, but the one criteria fails, because it has more than at most two non-zero entries per column.
 A: You may ignore the four columns on the right because they are negatives of the four columns on the left. You may also ignore the last row because it is identical to the first. Therefore, your matrix is totally unimodular if and only if the $7\times5$ matrix
$$
B=\pmatrix{1&1&1&1\\ 0&1&1&1\\ 0&0&1&1\\ 0&0&0&1\\ 1&0&0&0\\ 1&1&0&0\\ 1&1&1&0}
$$
is totally unimodular. However, $B$ is totally unimodular if and only if $C=DB$ is totally unimodular where $D=\operatorname{diag}(-1,-1,-1,-1,+1,+1,+1)$. Now, on each column of $C$, its entries are arranged in ascending order. By a result of Fujishige (1984; see the seventh result in the subsection "Common totally unimodular matrices" in Wikipedia), such a matrix $C$ is totally unimodular if and only if its every $2\times2$ submatrix has determinant 0, -1 or 1. You may continue from here.
A: As pointed out by user1551, you only need to prove that the $7\times 5$ matrix $B$ formed by the first 5 columns, where we swap the sign of the 5th column, is totally unimodular. This is immediate, because $B$ has the consecutive ones properties on the rows (that is, in each rows the 1 entries appear consecutively). 
See the  Wikipedia entry for "Unimodular matrix", at point 3 in the section "Common totally unimodular matrices". 
