Prove this matrix is or is not totally unimodular

Question

Is there an easy way to show or disprove the matrix

$\begin{bmatrix} 1&1& 0& 0&1 &0 &0 \\0&0&1&1&0&1&0 \\0&0&0&0&1&1&1 \\1& 0& 0&0&1& 0&0 \\0&1&0&0&1& 0&0 \\0&0&1&0&0&1&0 \\0&0&0&1&0&1&0 \\0&0&0&0&1&0&1 \\0&0&0&0&0&1&1 \end{bmatrix}$

is totally unimodular?

If you look at the matrix, almost all the rows have two non-zero entries and only $1/3$rd of the rows have three non-zero entries. So $2/3$rd of the matrix is directly totally unimodular. It is the rows with three non-zero entries which seem a little tricky.

The matrix isn't large. Just check all the square submatrices and see if any have determinant not $0,1,-1.$ — coffeemath, Apr 12 '23 at 23:12
@IonzaIeggiera That is not for Total unimodularity but only unimodularity. — Turbo, Apr 12 '23 at 23:54
@Turbo Ah, yes. I had conflated the latter concept with the former. — lonza leggiera, Apr 13 '23 at 01:11

score 1 · Accepted Answer · answered Apr 13 '23 at 03:38

1

Reorder the rows so that the columns satisfy the consecutive ones property:

0   1   0   0   1   0   0
1   1   0   0   1   0   0
1   0   0   0   1   0   0
0   0   0   0   1   0   1
0   0   0   0   1   1   1
0   0   0   0   0   1   1
0   0   1   0   0   1   0
0   0   1   1   0   1   0
0   0   0   1   0   1   0

answered Apr 13 '23 at 03:38

RobPratt

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I accepted but what is the consecutive ones property? – Turbo Apr 13 '23 at 16:02
1

It means that all the ones in each column appear consecutively, and it is a well-known sufficient condition for TU. – RobPratt Apr 13 '23 at 17:17
See a proof of the "consecutive ones" property here. – Jean Marie Apr 16 '23 at 08:03

Prove this matrix is or is not totally unimodular

1 Answers1