Unimodular matrix

1. 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.) More generally, in the appendix to a paper by Heller and Tompkins,[2] A.J. Hoffman and D. Gale prove the following. Let ${\displaystyle A}$ be an m by n matrix whose rows can be partitioned into two disjoint sets ${\displaystyle B}$ and ${\displaystyle C}$. Then the following four conditions together are sufficient for A to be totally unimodular:

• Every entry in ${\displaystyle A}$ is 0, +1, or −1;
• Every column of ${\displaystyle A}$ contains at most two non-zero (i.e., +1 or −1) entries;
• If two non-zero entries in a column of ${\displaystyle A}$ have the same sign, then the row of one is in ${\displaystyle B}$, and the other in ${\displaystyle C}$;
• If two non-zero entries in a column of ${\displaystyle A}$ have opposite signs, then the rows of both are in ${\displaystyle B}$, or both in ${\displaystyle C}$.

It was realized later that these conditions define an incidence matrix of a balanced signed graph; thus, this example says that the incidence matrix of a signed graph is totally unimodular if the signed graph is balanced. The converse is valid for signed graphs without half edges (this generalizes the property of the unoriented incidence matrix of a graph).[3]

2. The constraints of maximum flow and minimum cost flow problems yield a coefficient matrix with these properties (and with empty C). Thus, such network flow problems with bounded integer capacities have an integral optimal value. Note that this does not apply to multi-commodity flow problems, in which it is possible to have fractional optimal value even with bounded integer capacities.

3. The consecutive-ones property: if A is (or can be permuted into) a 0-1 matrix in which for every row, the 1s appear consecutively, then A is TU. (The same holds for columns since the transpose of a TU matrix is also TU.) [4]

4. Every network matrix is TU. The rows of a network matrix correspond to a tree T = (V, R), each of whose arcs has an arbitrary orientation (it is not necessary that there exist a root vertex r such that the tree is "rooted into r" or "out of r").The columns correspond to another set C of arcs on the same vertex set V. To compute the entry at row R and column C = st, look at the s-to-t path P in T; then the entry is:

• +1 if arc R appears forward in P,
• −1 if arc R appears backwards in P,
• 0 if arc R does not appear in P.

See more in Schrijver (2003).

5. Ghouila-Houri showed that a matrix is TU iff for every subset R of rows, there is an assignment ${\displaystyle s:R\to \pm 1}$ of signs to rows so that the signed sum ${\displaystyle \sum _{r\in R}s(r)r}$ (which is a row vector of the same width as the matrix) has all its entries in ${\displaystyle \{0,\pm 1\}}$ (i.e. the row-submatrix has discrepancy at most one). This and several other if-and-only-if characterizations are proven in Schrijver (1998).

6. Hoffman and Kruskal[5] proved the following theorem. Suppose ${\displaystyle G}$ is a directed graph without 2-dicycles, ${\displaystyle P}$ is the set of all dipaths in ${\displaystyle G}$, and ${\displaystyle A}$ is the 0-1 incidence matrix of ${\displaystyle V(G)}$ versus ${\displaystyle P}$. Then ${\displaystyle A}$ is totally unimodular if and only if every simple arbitrarily-oriented cycle in ${\displaystyle G}$ consists of alternating forwards and backwards arcs.

7. Suppose a matrix has 0-(${\displaystyle \pm }$1) entries and in each column, the entries are non-decreasing from top to bottom (so all −1s are on top, then 0s, then 1s are on the bottom). Fujishige showed[6] that the matrix is TU iff every 2-by-2 submatrix has determinant in ${\displaystyle 0,\pm 1}$.

8. Seymour (1980)[7] proved a full characterization of all TU matrices, which we describe here only informally. Seymour's theorem is that a matrix is TU if and only if it is a certain natural combination of some network matrices and some copies of a particular 5-by-5 TU matrix.

1. The following matrix is totally unimodular:

${\displaystyle A=\left[{\begin{array}{rrrrrr}-1&-1&0&0&0&+1\\+1&0&-1&-1&0&0\\0&+1&+1&0&-1&0\\0&0&0&+1&+1&-1\end{array}}\right].}$

This matrix arises as the coefficient matrix of the constraints in the linear programming formulation of the maximum flow problem on the following network:

2. Any matrix of the form

${\displaystyle A=\left[{\begin{array}{ccccc}&\vdots &&\vdots \\\dotsb &+1&\dotsb &+1&\dotsb \\&\vdots &&\vdots \\\dotsb &+1&\dotsb &-1&\dotsb \\&\vdots &&\vdots \end{array}}\right].}$

is not totally unimodular, since it has a square submatrix of determinant −2.