Editing Unimodular matrix (section)

===Common totally unimodular matrices===
1. The unoriented incidence matrix of a [[bipartite graph]], which is the coefficient matrix for bipartite [[matching (graph theory)|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,<ref>{{Citation|last=Heller|first=I.|title=Linear Inequalities and Related Systems|volume=38|pages=247–254|year=1956|editor-last=[[Harold W. Kuhn|Kuhn]]|editor-first=H.W.|series=Annals of Mathematics Studies|contribution=An Extension of a Theorem of Dantzig's|location=Princeton (NJ)|publisher=Princeton University Press|last2=Tompkins|first2=C.B.|editor2-last=[[Albert W. Tucker|Tucker]]|editor2-first=A.W.}}</ref> A.J. Hoffman and D. Gale prove the following. Let <math>A</math> be an ''m'' by ''n'' matrix whose rows can be partitioned into two [[disjoint sets]] <math>B</math> and  <math>C</math>. Then the following four conditions together are [[Necessary and sufficient conditions|sufficient]] for ''A'' to be totally unimodular:
* Every entry in <math>A</math> is 0, +1, or −1;
* Every column of <math>A</math> contains at most two non-zero (i.e., +1 or −1) entries;
* If two non-zero entries in a column of <math>A</math> have the same sign, then the row of one is in <math>B</math>, and the other in <math>C</math>;
* If two non-zero entries in a column of <math>A</math> have opposite signs, then the rows of both are in <math>B</math>, or both in <math>C</math>.
It was realized later that these conditions define an incidence matrix of a balanced [[Signed graph#Incidence matrix|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).<ref>T. Zaslavsky (1982), "Signed graphs," ''Discrete Applied Mathematics'' 4, pp. 401&ndash;406.</ref>

2. The [[Constraint (mathematics)|constraint]]s 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 problem]]s, 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.)
<ref>{{cite journal |last1=Fulkerson |first1=D. R. |last2=Gross |first2=O. A. |title=Incidence matrices and interval graphs. |journal=Pacific Journal of Mathematics |date=1965 |volume=15 |issue=3 |pages=835–855 |url=https://projecteuclid.org/euclid.pjm/1102995572 |language=en |issn=0030-8730}}</ref>

4. Every '''network matrix''' is TU. The rows of a network matrix correspond to a tree {{nowrap|1=''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 {{nowrap|1=''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 <math>s:R \to \pm1</math> of signs to rows so that the signed sum <math>\sum_{r \in R} s(r)r</math> (which is a row vector of the same width as the matrix) has all its entries in <math>\{0,\pm1\}</math> (i.e. the row-submatrix has [[Discrepancy of hypergraphs|discrepancy]] at most one). This and several other if-and-only-if characterizations are proven in Schrijver (1998).

6.
Hoffman and [[Joseph Kruskal|Kruskal]]<ref>{{Citation
  | last = Hoffman
  | first = A.J.
  | last2 = [[Joseph Kruskal|Kruskal]]
  | first2 = J.B.
  | contribution = Integral Boundary Points of Convex Polyhedra
  | editor-last = [[Harold W. Kuhn|Kuhn]]
  | editor-first = H.W.
  | editor2-last = [[Albert W. Tucker|Tucker]]
  | editor2-first = A.W.
  | title = Linear Inequalities and Related Systems
  | series = Annals of Mathematics Studies
  | publisher = Princeton University Press
  | location = Princeton (NJ)
  | pages = 223–246
  | volume = 38
  | year = 1956 }}</ref>
proved the following theorem. Suppose <math>G</math> is a [[directed graph]] without 2-dicycles, <math>P</math> is the set of all [[dipath]]s in <math>G</math>, and <math>A</math> is the 0-1 incidence matrix of <math>V(G)</math> versus <math>P</math>. Then <math>A</math> is totally unimodular if and only if every simple arbitrarily-oriented cycle in <math>G</math> consists of alternating forwards and backwards arcs.

7. Suppose a matrix has 0-(<math>\pm</math>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<ref>{{Citation
  | last = Fujishige
  | first = Satoru
  | title = A System of Linear inequalities with a Submodular Function on (0, ±1) Vectors
  | journal = Linear Algebra and Its Applications
  | pages = 253–266
  | volume = 63
  | year = 1984
  | doi=10.1016/0024-3795(84)90147-2| doi-access = free
  }}</ref>
that the matrix is TU iff every 2-by-2 submatrix has determinant in <math>0, \pm1</math>.

8. [[Paul Seymour (mathematician)|Seymour]] (1980)<ref>{{Citation
  | last = [[Paul Seymour (mathematician)|Seymour]]
  | first = P. D.
  | title = Decomposition of regular matroids
  | journal = [[Journal of Combinatorial Theory]] | series=Series B
  | pages = 305–359
  | volume = 28
  | issue = 3
  | year = 1980
  | doi = 10.1016/0095-8956(80)90075-1 | doi-access = }}</ref> 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.