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Lecture Notes on Total Unimodularity
Fabio D’Andreagiovanni
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
n
2. The vertices of the polyhedron P = {xR : Ax=b, x0n} are
m
integral for every bZ
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)
If A is unimodular then the vertices of the polyhedron
n
m
P = {xR : Ax=b, x0n} are integral for every bZ
o
n
x vertex of P = {xR : Ax=b, x0n}



Proof:
o
x basic feasible solution
there exists (m x m) square submatrix B of A with det(b)=0
m

such that x =
xBR
n-m
xNR
We have Ax=b  BxB+NxN=b
and A = (B N)
and
o
x=
o
m
-1
o
n-m
x BR
x NR
=
B b
0n-m
 0n
-1
B = adj(B) / det(B) where adj(B) is the adjunct matrix of B:
B-1 integer matrix
A unimodular matrix  |det(B)|=1 
 x0Zn for every bZm
A integer matrix
-1
B bZ
m
for every bZ
m
On unimodularity and integral vertices (2 3)
n
If the vertices of the polyhedron P = {xR : Ax=b, x0n} 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
Proof:
Let B be an (m x m) square submatrix of A with det(b)=0 and denote
by B
-1
i
-1
the i-th column of the corresponding inverse matrix B .
We prove that the generic i-th column B
-1
i
has integer components:
• Let tZm be an integer vector such that t+B-1i  0m
• Let b(t)=Bt+ei with ei being the i-th unit vector, then
-1
-1
B b(t)
=
0n-m



B (Bt+ei)
0n-m
-1
=
t + B ei
=
0n-m
t+B
-1
i
0n-m
 0n
basic feasible solution of the system Ax=b(t), x  0n
n
Vertex of P = {xR : Ax=b(t), x  0n}
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
-1
has an integer inverse matrix B , then A is unimodular
Proof:
If B is an (m x m) square submatrix of A with det(b)=0, then B
-1
is an
integer matrix

-1
|det(B)| and |det(B )| are integers
-1
-1
Since |det(B)| |det(B )| = |det(BB )| = 1

|det(B)| =|det(B )| = 1

A is unimodular
-1
Quod erat demonstrandum
Vertices and standard form
Theorem THF2: Let A be an (m x n) matrix and b an m-dimensional
0
n
vector. x is a vertex of the polyhedron P = {xR : Ax=b, x0n} if and
only if the point
0
n
0
m
x R
=
y R
STD
is a vertex of the polyhedron P
x
0
b-Ax
0
n+m
= {(x,y)R
: Ax+Is=b, x0n, y0m}