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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)

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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.

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Hi Eric: if you right click and "show as TeX" you can see how I do the following matrix: $A=\begin{pmatrix} 0&1&0\\ 1 &0&0\\0&0&-1 \end{pmatrix}$ –  rschwieb May 26 '12 at 0:39
    
Wow, that's much easier than anything else I've seen… do you know what package has the pmatrix environment? –  Eric Stucky May 31 '12 at 5:44
    
I'm afraid I don't know the specifics... it's just always worked when I tried it! –  rschwieb May 31 '12 at 11:00
    
@EricStucky It's in amsmath, I think. (I always include amsmath, so there's a small chance it's part of LaTeX itself, but I don't think so.) –  hoyland Aug 8 '12 at 19:13
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