A matrix is totally uni-modular if the determinant of any (square) sub-matrix is {+1, 0, -1}. My question is, "Is there a way to transform(linear or non) a general matrix into a totally uni-modular matrix?" or, "Are there only certain matrices that can be transformed in such a way?" This is for an application in Linear (convex) optimization.

Thanks, Clark

  • $\begingroup$ Surely you want to impose some additional condition on the desired transformation, at least to exclude trivial transformations like the one sending all matrices to the zero matrix, and probably to exclude various other trivialities as well. $\endgroup$ – Andreas Blass Jun 19 '12 at 0:50
  • $\begingroup$ Yes, I thought I might need to clarify. My goal is to find a "smart" defensible method. A linear transformation is what I would like, but I am not very hopeful of that for a general case. I was thinking was more along the lines of some product/summation formula that takes the sign() of the final result. $\endgroup$ – Clark Jun 19 '12 at 0:55
  • $\begingroup$ This question is also posted at math.stackexchange.com/questions/160123/…. $\endgroup$ – Chris Godsil Jun 19 '12 at 1:22
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    $\begingroup$ You can place every matrix into Smith normal form, which, over a field, is totally unimodular, through multiplication by an invertible matrix on the left and an invertible matrix on the right. $\endgroup$ – Will Sawin Jun 19 '12 at 1:49
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    $\begingroup$ There are only finitely many totally unimodular matrices of a certain size. Therefore, assuming a partition into semialgebraic sets, some totally unimodular matrices must correspond to sets with interior. Which sets? What happens at the boundary of the sets? $\endgroup$ – Will Sawin Jun 19 '12 at 1:51

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