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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

Edit:

As a linear-based method, I was thinking more of multiplying the starting matrix by a Gaussian matrix (or some other distribution) and then using the sign() function to restrict the values to {-1,1} with small values probably set to 0. I mostly want to see it anyone knows any transformations for restricting values.

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What properties would you like the map to have? Zero map certainly satisfies your condition, and is even linear, but I doubt it's what you're looking for. Your question is too vague. –  tomasz Jun 19 '12 at 1:07
    
I agree, what properties is it supposed to satisfy? Given an integer Matrix $A$, it is transformable to a diagonal integer matrix and if we normalize each row, we get a totally unimodular matrix. But this transformation,for example, wouldn't preserve integrality of a polytope defined by A. What do you want to do with this? –  Tim Seguine Jan 20 '13 at 14:48
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1 Answer

Since every entry of a totally unimodular matrix is $\pm 1$ or $0$, there are only finitely many such matrices of any given size, and only countably many in all. So there is certainly no one-to-one transformation from general matrices to totally unimodular matrices.

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I believe the author meant to talk about integral matrices. –  tomasz Jun 19 '12 at 1:06
    
I was pretty sure there wasn't a one-to-one transformation, but I don't really care if the transformation retains uniqueness, so maybe something like a linear/simple transformation matched with the sign() function that will be TU (at least some of the time). Thanks Robert. –  Clark Jun 19 '12 at 1:13
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