Fabio D’Andreagiovanni Lecture Notes on Total Unimodularity

Total Unimodularity Definition 1: an (m x n) matrix A is unimodular iff for every (m x m) square submatrix B of A it holds det (B) {-1, 0, 1} Definition 2: an (m x n) matrix A is totally unimodular iff for every (p x p) square submatrix B of A with p>0 it holds: det (B) {-1, 0, 1} Unimodular but not totally unimodular Not Unimodular Unimodular and totally unimodular

On unimodularity and integral vertices Theorem THF1: let A be an (m x n) integer matrix A such that rank(A) = m. The following statements are equivalent: 1.A is unimodular 2.The vertices of the polyhedron P = {xR n : Ax=b, x0 n } are integral for every bZ m 3.Every (m x m) square submatrix B of A that is non-singular has an integer inverse matrix B -1 Proof: we prove the equivalence showing that  (12)  (23)  (31)

On unimodularity and integral vertices (1 2) there exists (m x m) square submatrix B of A with det(b)=0 Proof: x o vertex of P = {xR n : Ax=b, x0 n }     If A is unimodular then the vertices of the polyhedron P = {xR n : Ax=b, x0 n } are integral for every bZ m x o basic feasible solution such that x = and A = (B N) x B R m x N R n-m We have Ax=b Bx B +Nx N =b and x o = = 0 n  x o B R m x o N R n-m B -1 b 0 n-m B -1 = adj(B) / det(B) where adj(B) is the adjunct matrix of B: A integer matrix   B -1 integer matrix A unimodular matrix |det(B)|=1  B -1 bZ m for every bZ m  x 0 Z n for every bZ m 

On unimodularity and integral vertices (2 3) Proof: If the vertices of the polyhedron P = {xR n : Ax=b, x0 n } are integral for every bZ m, then every (m x m) square submatrix B of A that is non-singular has an integer inverse matrix B -1 basic feasible solution of the system Ax=b(t), x 0 n   Let B be an (m x m) square submatrix of A with det(b)=0 and denote by B -1 i the i-th column of the corresponding inverse matrix B -1. We prove that the generic i-th column B -1 i has integer components: Let tZ m be an integer vector such that t+B -1 i 0 m Let b(t)=Bt+e i with e i being the i-th unit vector, then B -1 b(t) 0 n-m = B -1 (Bt+e i ) 0 n-m = t + B -1 e i 0 n-m = t + B -1 i 0 n-m 0 n Vertex of P = {xR n : Ax=b(t), x 0 n }  t+B -1 i is an integer vector  B -1 i is an integer vector 

On unimodularity and integral vertices (3 1)  If every (m x m) square submatrix B of A that is non-singular has an integer inverse matrix B -1, then A is unimodular If B is an (m x m) square submatrix of A with det(b)=0, then B -1 is an integer matrix Proof: |det(B)| and |det(B -1 )| are integers  Since |det(B)| |det(B -1 )| = |det(BB -1 )| = 1 |det(B)| =|det(B -1 )| = 1  A is unimodular  Quod erat demonstrandum

Vertices and standard form Theorem THF2: Let A be an (m x n) matrix and b an m-dimensional vector. x 0 is a vertex of the polyhedron P = {xR n : Ax=b, x0 n } if and only if the point is a vertex of the polyhedron P STD = {(x,y)R n+m : Ax+Is=b, x0 n, y0 m } x0Rny0Rmx0Rny0Rm x 0 b-Ax 0 =