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Suppose matrix $A \in \mathbb{R}^{m×n}$ has the consecutive ones property and, thus, is totally unimodular. Is the following block matrix also totally unimodular (TU)?

$$B = \begin{pmatrix} A & 0 & \dots & 0\\ 0 & A & \dots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \dots & A\\ I & I & \dots & I\end{pmatrix}$$

This type of matrix appears when dealing with the Coloring Problem over Interval Graphs, which can be solved in polynomial time, but I don't know if there is any result on the total unimodularity of such matrix.

A similar question was asked here (Is this block matrix also totally unimodular?), where $A$ was known to be TU. In this case the block matrix can be non-TU, however the counter-example given uses a matrix A that does not have the consecutive one's property.

Rodrigo de Azevedo
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Rafael Colares
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1 Answers1

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Great question. Thanks Rafael for pointing out in the comments below a counterexample which, unfortunately, invalidates any affirmative proof, including what I had mistakenly posted. The submatrix obtained in his counterexample is $$\begin{bmatrix} 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 1 \\ 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \end{bmatrix}.$$

Craveable Banana
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    Actually I have found a counter-example. Let A be the following matrix: A = \begin{bmatrix} I & I & I & I \ M & {\bf 0} & {\bf 0} & {\bf 0}\ {\bf 0} & M & {\bf 0} & {\bf 0}\ {\bf 0} & {\bf 0} & M & {\bf 0}\ {\bf 0} & {\bf 0} & {\bf 0} & M \end{bmatrix} where M = \begin{bmatrix} 1 & 1 & 1 & 0\ 0 & 1 & 1 & 1\ 0 & 0 & 1 & 1 \end{bmatrix}

    Then consider the submatrix composed of rows rows 2,4,5,7,9 and columns 2,3,4,6,8. Such submatrix has determinant -2...

     – Rafael Colares Mar 29 '19 at 08:54
  • Yeah I think you are right. I misunderstood the definition of submatrices. Will try to update answer accordingly. – Craveable Banana Apr 01 '19 at 04:41