5

Is there any (theoretic) way I can prove the matrix is totally unimodular? I have tested it by Matlab and know it is TU, however I cannot prove it.

-1 -1 -1 -1  0  0  0  0  0  0  0  0
 0  0  0  0 -1 -1 -1 -1  0  0  0  0
 0  0  0  0  0  0  0  0 -1 -1 -1 -1
 1  1  1  1  0  0  0  0  0  0  0  0
 0  0  0  0  1  1  1  1  0  0  0  0
 0  0  0  0  0  0  0  0  1  1  1  1
 1  0  0  0  1  0  0  0  1  0  0  0
 0  1  0  0  0  1  0  0  0  1  0  0
 0  0  1  0  0  0  1  0  0  0  1  0
 0  0  0  1  0  0  0  1  0  0  0  1
Rodrigo de Azevedo
  • 23,788
remo
  • 441
  • Yes, But I must prove that any square submatrix of it has a $det\in{-1,0,1}$. You can see the wiki for the definition. – remo Jun 23 '13 at 17:54
  • Ah sorry, I didn't know that "totally unimodular" allowed for $det = 0 $. – Calvin Lin Jun 23 '13 at 18:00

2 Answers2

8

First, observe that since $R_1= - R_4$, $R_2= - R_5$, $R_3 = - R_6 $, then if any sub matrix uses any of these pairs of rows, then the determinant must be 0. Hence, (WLOG) we may assume that rows $1, 2, 3$ are removed (and replaced with rows 4, 5, 6 respectively in the sub matrix determinant calculation). Now call this matrix $A$.

If you look up the Wikipedia article, you will see a sufficient condition for Totally Unimodular matrices:

  1. Every column of contains at most two non-zero entries.
  2. Every entry in is 0, +1, or −1
  3. If two non-zero entries in a column of $A$ have the same sign, then the row of one is in $B$, and the other in $C$.
  4. If two non-zero entries in a column of $A$ have opposite signs, then the rows of both are in $B$, or both in $C$.

Observe that if $B$ is the set of rows 4, 5, 6 and $C$ is the set of rows 7, 8, 9, 10, then this will satisfy the conditions. Hence we are done.

Calvin Lin
  • 77,906
  • @remo I came up with it when I started reading about totally unimodular matrices. Off the top of my head, I don't know a proof of the quoted result, though since it's in the appendix of a paper, it should be easy. – Calvin Lin Jun 23 '13 at 18:19
0

Here is another, perhaps easier way, of seing this. Your matrix is:

\begin{equation} \begin{pmatrix}\begin{pmatrix}-1 & -1 & -1 & -1 \\ 1 & 1 & 1 & 1 \end{pmatrix} \otimes Id \\ Id \otimes \begin{pmatrix}1 & 1 & 1 \end{pmatrix} \end{pmatrix} \end{equation}

Geoffrey Negiar
  • 103
  • 2
    The stacking of two totally unimodular matrices on top of each other is not totally unimodular in general. More explanation is required. – daw Nov 21 '19 at 07:58