## QOTD: total unimodularity

### May 8, 2012

I started reading Matousek’s discrete geometry book. Specifically, chapter 12 on applications of high-dimensional polytopes. Mostly because he apparently draws a connection between graphs and the Brunn-Minkowski theory, and I’m interested to see exactly what that connection is and if it has any interesting implications.

I just finished the proof of the weak perfect graph conjecture (it’s been a while since I read a nontrivial pure math proof, so I haven’t actually *digested* it entirely). Now I’m on the exercises for that portion. So, here’s an interesting question involving the concept of total unimodularity, which is apparently one of those foundational concepts in an area called polyhedral combinatorics.

A matrix \(\mat{A}\) is called totally unimodular if every square submatrix of \(\mat{A}\) has determinant 0 or \(\pm 1.\) The question is to show that every nonsingular totally unimodular \(n \times n\) matrix maps the lattice \(\mathbb{Z}^n\) bijectively onto itself.

**Update (solution)**

It’s easy peasy. Clearly each entry of \(\mat{A}\) is in \(\{-1,0,1\}\), so \(\mat{A}\) maps \(\mathbb{Z}^n\) into itself. The fact that the mapping is bijective follows easily from Cramer’s rule, the fact that \(|\mathrm{det}(\mat{A})| = 1,\) and the fact that all the minors of \(\mat{A}\) are integral.