When is the adjacency matrix of a simple undirected graph totally unimodular?

Question

I would like to know the graphs for which the (node-node) adjacency matrix is totally unimodular. Is the following true?

The adjacency matrix of G is totally unimodular if and only if G is a tree.

Answer 1

Every tree has a totally unimodular adjacency matrix, but the converse is not true. The matrix $\begin{pmatrix}0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0\end{pmatrix}$ is totally unimodular, but the graph it represents is not a tree.

More generally, a graph has a totally unimodular adjacency matrix if and only if it can be written in the form $\begin{pmatrix}0 & B \\ B^T & 0\end{pmatrix}$ where $B$ is a totally unimodular matrix whose entries are zeros and/or ones.

Answer 2

EDIT: See Akbari and Kirkland, On unimodular graphs, Linear Algebra and its Applications, Volume 421, Issue 1, 1 February 2007, Pages 3-15. According to the abstract, "Graphs whose adjacency matrices are totally unimodular are also characterized."

That paper is available via https://www.sciencedirect.com/science/article/pii/S0024379506004708

Ignore what follows, left here for historical reasons:

Wikipedia says, "The unoriented incidence matrix of a bipartite graph, which is the coefficient matrix for bipartite matching, is totally unimodular (TU). (The unoriented incidence matrix of a non-bipartite graph is not TU.)"