fabio d’andreagiovanni lecture notes on total unimodularity
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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
On unimodularity and integral vertices (2 3)
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
On unimodularity and integral vertices (3 1)
If every (m x m) square submatrix B of A that is non-singularhas 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)| = 1A 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. 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=
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