Is this block matrix also totally unimodular?

Question

Suppose matrix $A\in \mathbb R^{m\times n}$ is totally unimodular (TUM). Is the following matrix also TUM?

$$ \begin{pmatrix} A & 0 & 0 \\ 0 & A & 0 \\ 0 & 0 & A\\ I & I & I\\ \end{pmatrix} $$

Thanks.

Answer 1

Let $A \in R^{m \times n}$ be a TUM matrix.

A matrix $A$ is TUM if and only if every sub matrix of $A$ permits an equitable bi-coloring. That is, we can designate columns to be either red or blue and for each row, the sum of the red cells minus the sum of the blue cells is $\pm 1$, or 0. Furthermore this coloring is equitable, that is the number of red columns and the number of blue columns differs by at most 1.

Thus, we need to construct an equitable bi-coloring for an arbitrary sub-matrix of the block matrix, to show that it is TUM. Consider the matrix: $$ \begin{pmatrix} A & 0 & 0 \\ 0 & A & 0 \\ 0 & 0 & A \\ I & I & I \end{pmatrix} $$ Fix an arbitrary sub-matrix $B$ from the block matrix above.

Then if $B$ contains contents of no $A$ block. Then each row has at most 1 non-zero entry, so any coloring such that the number of red columns and the number of blue columns differs by at most 1 is a valid equitable bi-coloring.

If $B$ contains a portion of 1 $A$ block, then that sub-block of $A$ has an equitable bi-coloring. Color the corresponding columns in $B$ accordingly. Any remaining columns can be colored so that the total number of red columns and the total number of blue columns in $B$ differ by at most 1.

If $B$ contains a portion of 2 $A$ blocks, then each of these sub-blocks have an equitable bi-coloring. If both of these portions have odd width, pick the coloring of the first $A$ sub-block to have 1 more red column, and the coloring of the second $A$ sub-block to have 1 more blue column. This ensures that there are the same number of red and blue columns, in these sections of $B$. Alternatively, if at most one of the of these sub-blocks has odd width, then pick any coloring honoring the equitable bi-colorings given by the sub-blocks. Then, number of red columns and the number of blue columns differs by at most 1. Any remaining columns can be colored so that the total number of red columns and the total number of blue columns in $B$ differ by at most 1.

Finally, If $B$ contains a portion of 3 $A$ blocks then each of these sub-blocks have an equitable bi-coloring. If all of them have an even number of red and blue columns, then pick any coloring honoring the equitable bi-colorings given by the sub-blocks. If one of them is even, pick either coloring honoring the equitable bi-coloring given by the sub-block; pick the coloring of the first remaining $A$ sub-block to have 1 more red column, and the coloring of the second remaining $A$ sub-block to have 1 more blue column. If two of them are even, pick any coloring honoring the equitable bi-colorings given by the sub-blocks. If all of them are odd, pick the coloring of the first $A$ sub-block to have 1 more red column, the coloring of the second $A$ sub-block to have 1 more blue column, and the third coloring honoring either equitable bi-coloring given by the sub-block. In each of these cases, the number of red and blue columns can differ by at most one. Any remaining columns can then be colored so that the total number of red columns and the total number of blue columns in $B$ differ by at most 1.

Therefore there exists a equitable bi-coloring for any arbitrary sub-matrix of the block matrix. It is TU.

Answer 2

The counterexample is given by the following:

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

Consider the subset of columns { 1, 2, 3, 4, 6, 7, 8, 11, 12}.

To ensure the difference of the sums for each row is less than 2, it requires

C1 red ---> C4 blue---> C6 red

C1 red ---> C7 blue---> C8 red--->C11 blue--->C12 red

C6 and C12 are red and the sum for the last row of them is +2.