fabio d’andreagiovanni lecture notes on total unimodularity

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 butnot totally unimodular

Not Unimodular Unimodular andtotally 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 = {xRn: Ax=b, x0n} are

integral for every bZm

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

(1 2) (2 3) (3 1)

On unimodularity and integral vertices (1 2)

there exists (m x m) square submatrix B of A with det(b)=0

Proof:

xo vertex of P = {xRn: Ax=b, x0n}

If A is unimodular then the vertices of the polyhedron P = {xRn: Ax=b, x0

n} are integral for every bZm

xo basic feasible solution

such that x = and A = (B N)

xBRm

xNRn-m

We have Ax=b BxB+Nx

N=b and xo= = 0

n

xoBRm

xoNRn-m

B-1b0

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-1bZm for every bZm

x0Zn for every bZm


Proof:

If the vertices of the polyhedron P = {xRn: Ax=b, x0n} are integral

for every bZm , 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 0n

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-1i has integer components:

• Let tZm be an integer vector such that t+B-1i 0

m

• Let b(t)=Bt+ei with e

i being the i-th unit vector, then

B-1b(t)0

n-m

=B-1(Bt+e

i)

0n-m

=t + B-1e

i

0n-m

=t + B-1

i

0n-m

0n

Vertex of P = {xRn: Ax=b(t), x 0n}

t+B-1i is an integer vector B-1

i is an integer vector

Vertices and standard form

Theorem THF2: Let A be an (m x n) matrix and b an m-dimensional

vector. x0 is a vertex of the polyhedron P = {xRn: Ax=b, x0n} if and

only if the point

is a vertex of the polyhedron PSTD = {(x,y)Rn+m: Ax+Is=b, x0n, y0

m}

x0Rn

y0Rm

x0

b-Ax0=

fabio d’andreagiovanni lecture notes on total unimodularity

Documents