I know that if a matrix $A$ is Totally Unimodular (TU), then the matrix $(A\; I)$ is unimodular. Can I then say that the matrix $(A\; -I)$ is also TU? ($I$ is the identity block matrix)
I'm not sure how one does matrices in LaTeX, so I will use Matlab notation. $A-I$ need not be TU: consider the almost-permutation matrix $A =[ [0,1,0] ~~ [1,0,0] ~~ [0,0,-1] ]$. This is clearly unimodular, and it's not too difficult to check and see it is also TU. However, $A-I = [ [-1, 1,0] ~~ [1,-1,0] ~~ [0,0,-2] ]$. If we consider the bottom-right 2x2 corner, we have $[[-1,0] ~~[0,-2]]$, which has determinant 2; thus $A-I$ is not TU.