Total Unimodularity and partition of the rows

Question

I have a question regarding a theorem about total unimodularity and partition of the rows of a matrix.

The theorem says that given a matrix $A \in \mathbb{Z}^{m\times n}$ , A is totally unimodular iff for any $I \subset [m]$ there exist a partition of the set I, $(I_1,I_2)$, such that $\sum_{i\in I_1}a_{i}-\sum_{i\in I_2}a_{i} \in \lbrace 0,\pm1 \rbrace$

However I look at the matrix $A= \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix}$, and I know it's unimodular, however if I choose $I = \lbrace R_2,R_3\rbrace$ and $I_1 = R_2, I_2 = R_3$ then I get that the sum is -2 thus indicating that the matrix is not totally unimodular.

How am I doing it wrong?

Answer 1

The theorem as stated is false, which can be shown by the matrix you included. I think you are trying to use the theorem relating equitable bi-colorings of TU matrices:

Theorem: A matrix $A$ is TU $\Leftrightarrow$ for any submatrix of $A$, there exists an equitable bi-coloring.

To use this theorem, consider any sub-matrix of a TU matrix $A$. Then you can designate red and blue rows such that in each column, the sum of the cells in the red rows minus the sums of the cells in the blue rows is $\pm1$, or 0. Furthermore, there is a coloring satisfying this, such that the number of red rows and the number of blue rows differ by at most one.

In your example, you are coloring the rows and summing along the rows, while you need to color the rows and sum along the columns.