A unimodular matrix $M$ is a square integer matrix having determinant $+1$ or $−1$. A totally unimodular matrix (TU matrix) is a matrix for which every square non-singular submatrix is unimodular. A totally unimodular matrix need not be square itself. Obviously, any totally unimodular matrix has only $0$, $+1$ or $−1$ entries.

Now suppose a $n\times n$ non-singular matrix $A$ is totally unimodular. Can we prove that $A^{-1}$ is also totally unimodular? Or if it is not correct, can we have a counterexample?

  • $\begingroup$ it seems that invertible TU matrices also form a group like unimodular matrices...but I don't have a proof yet. $\endgroup$ – Suvrit Apr 19 '13 at 19:13
  • $\begingroup$ Doesn't this follow from looking at the adjoint or cofactor marix? Gerhard "Ask Me About Binary Matrices" Paseman, 2013.04.19 $\endgroup$ – Gerhard Paseman Apr 19 '13 at 22:01
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    $\begingroup$ S. Sra: Good luck with that proof. You might consider 1 1;0 1 meanwhile. Gerhard "Not Grouplike Under Matrix Multiplication" Paseman, 2013.04.19 $\endgroup$ – Gerhard Paseman Apr 19 '13 at 22:11
  • $\begingroup$ @Gerhard: The inverse of the matrix that you've mentioned is also TU...so I don't understand your comment??? $\endgroup$ – Suvrit Apr 19 '13 at 22:25
  • $\begingroup$ S. Sra, I am suggesting that I don't know what group structure you are placing on the set of matrices. For the two notions of multiplication I considered, the matrix I gave does not help form a group. Gerhard "How Do You Multiply Them?" Paseman, 2013.04.19 $\endgroup$ – Gerhard Paseman Apr 19 '13 at 22:52

The answer is yes, because if $B=A^{-1}$, then we have an equality between minors: $$B(I,J)=\pm\frac{A(J^c,I^c)}{\det A},$$ for every subsets $I,J\subset[[1,n]]$ of same cardinals. This formula generalizes that giving the entries of $A^{-1}$ in terms of minors of $A$. The $\pm$ sign is not essential to prove the stability of the TU class under inversion.

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    $\begingroup$ ah, good old Cramer's rule (essentially!) $\endgroup$ – Suvrit Apr 19 '13 at 23:03
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    $\begingroup$ Thank you very much. But I think the formula should be $B(I,J) = \pm \frac{A(J^c, I^c)}{\det A}$. $\endgroup$ – qianchi Apr 20 '13 at 2:50
  • $\begingroup$ @qianchi. Of course you're right. I'll edit. $\endgroup$ – Denis Serre Apr 20 '13 at 3:39

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