LINEAR
ALGEBRA
AND
ITS
APPLICATIOXS
5. 65-108
On the Number of Solutions to the Complementarity
Spanning Properties of Complementary
Cones
KATTA
65
(1972)
Problem and
G. MURTY
The
University
Ann
Arbor,
of
Michigan
Michigan
Communicated
by
G. B. Da&zig
ABSTRACT
The relationship
between
the number of solutions to the complementarity
problem,
w = Mz + q,
ma
the right-hand
constant
proved
work
in this
The
number
vector
of solutions
condition
subdeterminants
of A4 are strictly
number
(odd
or even)
equal
M
positive.
If
either
the jth
negative
space.
them
-
vectors
vector
There
after
properties
replacing
obtained
M >
all principal
feasible
0, there
and
is at least
one
elements
of these solutions
is an odd
subdeterminants
of M
solutions
are nondegenerate.
is a constant
in Rn,
Also,
are
has the same parity
if the number
for each 4 E R”,
to know
-
I.?
for R”, an orthant
by
in this manner
., A.,},where
{A.,,.
of the unit
are thus
It is interesting
Orthants
for all 4 E Rn
The necessary
then
that
of com-
constant
is
is a P-matrix.
column
I.j.
When
of complementary
In the Cartesian system of coordinates
by a set of n-column
is finite
for each Q E R” is that all principal
and, in this case, the number
solutions
The main results
for each q E Rn if and only if all the diagonal
for all 4 E Rn which
and
problem
of M are nonzero.
to be unique
4 is nondegenerate.
feasible
to one
subdeterminants
are strictly
then the number
plementary
M are explored.
below.
to the complementarity
solution
positive;
whenever
nonzero,
M
feasible
WTZ = 0,
0,
4 and the matrix
for this solution
of
complementary
z>
are summarized
if and only if all the principal
sufficient
0.
matrix
2” orthants
of order
in Rn,
what
properties
some
given
is a convex
for eachj
n (denoted
and they
column
vector
-
1 to n, A.j
by I.j)
partition
these orthants
are called complementary
cone generated
=
is
or its
the whole
possess if we obtain
M.j
cones,
for j =
1 to n.
and their spanning
are studied.
Copyright
0
1972 by
American
Elsevier
Publishing
Company,
Inc.
66
KATTA
G. MURTY
1. INTRODUCTION
1.1.
The
of finding
complementary
column
vectors
quadratic
programming
problem
is that
w = (wJ E R” and z = (zi) E Rn satisfying
w=Mz+qn
w>o,
z>o,
wTz=O,
(1)
where M = (mij) is a given n x n square matrix,
column vector,
and wT denotes the transpose
real Euclidean
1.2.
Because
zv, z are nonnegative,
Thus, if one of the variables
tarity problem
Consider
the quadratic
the other should
known as the complemen-
programming
function
any optimal
solving
ze,-
Conversely,
in the quadratic
value in the quadratic
to finding
program
solution
also solves the
if the minimum
programming
to it also solves
(1) is equivalent
Hence the problem
Mz = q,
(Zen;z), then that
problem.
solution
problem
~~2%
to
(1) has any solution
programming
the objective
objective
i = 1,. . , n.
of order n.
subject
Thus,
for each
is sometimes
minimize
then
the constraint
zG”z = 0 will be referred to as the complemen-
tarity condition, and the problem
If Eq.
1
of W. Rn is the n-dimensional
in the pair wi, zi is positive,
be zero. Hence the constraint
quadratic
x
space.
n
~~2 = C, wlizi = 0 3 z8$zi= 0
i=l
1.3.
q = (qi) is a given n
problem
value for
is zero,
(1).
out whether
the minimum
above is zero or strictly
(1) is known as the complementary
positive.
quadratic program-
ming problem.
1.4.
Cottle
the problems
and Dantzig
[l]
and Lemke
in linear programming,
convex
also the problem of finding a Nash equilibrium
[8, 91 have shown that
quadratic
programming,
point of a bimatrix
all
and
game, can
ON
THE
COMPLERlENTARITY
be posed in the form of Eq.
[K].
Lemke
solving
and Howson
(1) which
Lemke
solution
Lemke
is based
on pivot
of solutions
positive
4.
of solutions
The motivation
of (1) see Scarf
a simple algorithm
have
shown
in the
conditions
for
for this problem
or if M is a
principal
Also see [16,
to (1) to the properties
(1) has a
diagonal.
on M and 4 under which
is to examine
11, 7, 31.
the relationship
of the
of the given matrices
M and
was provided
asked me to try and prove or construct
that
of M are positive
elements
to (1) is finite.
In this paper our main interest
number
[l]
determinants
with
[S] has also given sufficient
the number
applications
steps.
and Dantzig
if all the principal
matrix
F or other
(1).
[S, lo] have developed
[S] and Cottle
nonnegative
67
PROBLEM
by Gale
counterexamples
[4] when he
to the following
conjectures:
(a) M is a P-matrix
a unique
solution
if and only if the complementarity
(b) M > 0 is a Q-matrix
(c) If M is a Q-matrix,
of solutions
whenever
The result
2. NOTATION
2.1.
problem
if and only if mii > 0 for all i = 1,. . . , n.
the complementarity
4 is nondegenerate
of the investigation
AND
problem has an odd number
with respect
is the present
to M.
work.
PRELIMINARIES
If A is any matrix,
AT denotes
its transpose.
Ai. denotes
ith row vector of A and A.j denotes the jth column vector of A.
the unit
2.2.
A square
matrix
are strictly
M is called a P-matrix
positive.
degenerate if every matrix
The square
A obtained
by taking
for each j = 1,. . . , n is nonsingular.
M is a nondegenerate
are nonzero.
matrix
combination
2.4.
linear
definition
vectors
in R”.
M is
for all 4 E Rn.
The convex
by pos{A}.
Thus
as a nonnegative
linear
in A.
L(q) C Rzn is the linear
constraints
vectors
in A is denoted
if x can be expressed
of the column
Suppose
equality
vectors
M.j or
is that
subdeterminants
(1) has a solution
set of column
by the column
if and only
in Eq.
sub-
M is called non-
A.j to be either
An equivalent
if and only if all its principal
Let A be any finite
cone generated
x E pos(A}
if all its principal
matrix
M is said to be degenerate if it is not nondegenerate.
called a Q-matrix if the problem
2.3.
the
I denotes
matrix.
determinants
l.j
has
for each 9 E Rn.
manifold
determined
by the
68
KATTA
G. MURTY
ze-MMz=q,
without
any nonnegativity
constraints.
if and only if it satisfies
Eq.
(2).
The vector
For convenience
we write down the
vector
1
75
2.5.
solutions
The
convex
as
R2”
E
Z
[
polyhedron
(7~; z).
K(q) C L(q) is the
set
of all feasible
(w ; z) which satisfy
w-MMz=q,
z 3 0.
w 3 0,
2.6.
A basic feasible solution is a feasible
that the column
vectors
strictly
are linearly
positive
is an extreme
2.7.
in Eq.
2.8.
independent.
point of the convex
solution
is a solution
condition
to Eq.
wj and -
1,.
. , n.
each
M.j is associated
n } such
are
solution
K(q) and vice versa [2, 51.
complementary
wi, zi constitute
a complemen-
with zj.
I.j
is associated
Thus the pair (l.j,
vectors
of the
with the
-
M.j)
in (1).
set of column vectors is a set of column vectors
that
A.j
pair
The corresponding
tary set of variables.
W% = 0. A complementary
vector
pair of column
Thus any set of column
column vectors.
vectors.
basic feasible
in the pair is the complement
(1) the column
A comjdemedary
{A.,,j=l,...,
from
and zj which
(1) and vice versa.
For each i = 1,. . . , n the variables
are the jth complementary
2.9.
(ZPI;z) E K(q) such
uj
A complementary feasible solution is a feasible solution (w; z) E K(q)
In the system
variable
Every
polyhedron
tary pair, and each of the variables
other.
solution
(3) of the variables
which satisfies the complementarity
feasible
(3)
is either
vectors
of vectors
I.i
or -
containing
M.,
for each
exactly
one vector
is a complementary
set of variables
i =
set
of
is called a complemen-
Hence there are 2n complementary
sets of column
ON
THE
COMPLEMENTARITY
2.10.
Each
combination
solution
69
PROBLEM
to Eq.
(1) represents
of some complementary
set of column
if {A. j} is a complementary
Conversely,
q=
where
kPjA.,,
linear
vectors.
set of column
vectors
and if
= 1,. . .,n.},
qEpOs{A.j,j
i.e.,
4 as a nonnegative
pj > 0
for each j,
j=l
then a solution
the column
other
to (1) is obtained
A.? equal
variables
the variables
in (zu; z) not in this complementary
The pos cone generated
is known
by setting
associated
to fij for i = 1,. . . , n, respectively,
by any complementary
as a complementary
Thus
cone.
with
and all the
set equal
to zero.
set of column vectors
there
are 2n complementary
cones, and the union of all these cones is the set of all q for which (1) has
a solution.
All the results related
plementarity
problem
overlapping
properties
2.11.
Any
to the number
of distinct
can be interpreted
solutions
in terms
of the complementary
cones
either w, or Z, for each r, is known as a subcomjdementary
The column vectors
constitute
associated
a subcomplementary
pair of variables
(wi, zi) is the left-out
in the subcomplementary
2.12.
An almost
set {yi,.
complementary
where
yT is
set of variables.
with a subcomplementary
set of column vectors.
and
and vice versa.
{yr, . . . , Y~_~, Y~+~,. . , y,},
set of variables
of the com-
of the spanning
set of variables
The complementary
pair of variables
complementary
. , Y~_~, Y~+~,. . . , y,}.
feasible solution is a feasible
solution
(w ; z) E K(q) such that
w T z = wizi
i.e.,
2.13.
The
wjzj = 0
i is any integer
2.14.
for all
j fi,
i,
for some
set C,(q) is the almost complementary
C,(q) = {(w; ii): (w; z) E K(q),
where
for some
If x E R”,
pos{x}
from
wTz = w,zi,
set defined
wizj = 0
for
1 to n.
x # 0, then
= {y:
i.e.,
i.
the yay generated
y = iix
for some
by x is
L > O}.
by
i # i},
70
KATTA
If x1, x2 E Rn, x1 # 0, then
2.15.
{y:
is the half-line
y = x2 + ilxl
through
The column vector
2.16.
the set
for some
x2 parallel
G. MURTY
13
0}
to the ray generated
by x1.
q is said to be nondegenerate with respect to
M if and only if, for all (Zen;.z) E L(q), at most n of the 2n variables
are zero.
Equivalently,
not lie in any subspace
(I : -
M).
Otherwise
which
are degenerate
q is nondegenerate
with respect
generated
1) or less column
belong
Two basic feasible
2.17.
to a finite
solutions
adjacent extreme points of K(q)
and
(w2; z2) has
a unique
extreme
points
of K(q).
extreme
points
of K(q)
vectors
of
the set of all q
of subspaces
of Rn.
(Al; zi) and (w2; z2) are said to be
if every
convex
combination
as a convex
The line segment
is called
Thus
number
representation
If K(q) is nonempty
2.18.
by (n -
q is said to be degenerate.
{zP,J~,
~$1
to M if it does
joining
of (WI; zl)
combination
of
any pair of adjacent
an edge of K(q).
and unbounded,
any basic feasible solution
of
w-M,z=O,
is known
as an extreme
homogeneous
solution
of Eq.
(3).
Any half-line
through a basic feasible solution in K(q) parallel through the ray generated
by an extreme
homogeneous
is called an unboztnded
on the half-line
combination
of basic feasible
of extreme
Consider
of (3) lies in K(q).
edge (or extreme half-line)
has a unique
combination
2.19.
solution
representation
solutions
homogeneous
the set of equality
constraint
in this system
of K(q) if every
point
as the sum of a convex
of K(q) and a nonnegative
solutions
constraints
w=Mz+q.
The ith
Such a half-line
linear
of (3).
Eq.
(2) again:
(2)
is
wi = Mi.z + qi.
Pi)
ON
THE
COMPLEMENTARITY
A principal
(i)
pivot in the position
Solve equation
.> z,
&+1,..
zi in terms
(ii)
other
of zi,.
equations
. , z~_~,wi, zifl,.
(2)
in
wi,
expressing
for zi in (i) in each
of the
(i, i) in (2) can only be performed
of this principal
one
complementary
principal
obtained
pivot in position
The result
(%I~,zi), and we get a transformed
in
. , z~_~,
in (2) by this equation,
in (2).
form as (a), but the left-hand
set
of the following steps:
zi in terms of zi, .
. ., 2,.
the expression
Thus a principal
rnii # 0.
(i, i) in (2) consists
(2i) for the variable
and replace the ith equation
Substitute
71
PROBLEM
pivot is to exchange
system of equations
set of variables
component
(the
pairs
of variables
If a series
of principal
if
the variables
which has the same
in it differ from the left-hand
ith).
However,
remains
unchanged
as a result
the
set
of
are performed
on the
the
of a
pivot.
2.20.
(2), then
pivots
it will be transformed
into
lvv + q,
21 =
where each pair
variables
(tii, vi) is a permutation
(wi, zi).
system
the system
A complementary
of the complementary
feasible
solution
pair of
to (4) is a solution
to the system
u>o,
2.21.
solutions
We notice
that
v>,o,
there
to (1) and solutions
is a one-to-one
to (5).
say.
correspondence
For example,
from (1) by making only one principal
(5)
UT-0= 0.
between
suppose (5) is obtained
pivot in which wi, zr are exchanged,
Then
6,;:
solves
(l)~zi=(z^,,~,,,...,~,,);
zl = (zi&, &, . . . 5,)
)
In general,
variables
since u, v in (5) are such that
(zeti,zi), we can construct
to each solution
solves
(tiui; vi) is a permutation
a solution
(G,; 2) to (1) by taking
(5).
of the
(a; 6) to (5) corresponding
the same permutation,
and vice
versa.
Thus the number
of solutions
to (1) is invariant
under principal
pivots.
72
KATTA
2.22.
Suppose
D is a nonsingular
09 ; t1 be the variables
columns
in problem
of the principal
the variables.
Then
submatrix
principal
submatrix
(1) corresponding
G. MURTY
of M.
D, and let w2; E2 denote
(1) can be written
Let
to the rows and
in the partitioned
the rest of
form:
w1 = DE1 + Ct2 + ql,
w2 = EP + F12 + q2,
[$o,
[$o,
A block principal pivot on the principal submatrix
the problem
into the equivalent
51 = D-i&
-
02 = ED-l&
[$o,
D consists in transforming
form:
DFCE2
+ (F -
+ (-
D-Iql),
ED-Y&J2
[;&o,
+ (q2 -
pivot can only be performed
result
principal
equivalent
problem
of variables
variables
pivot
exchanging
unchanged
(09; El) can be performed
to (7) and vice versa.
2.23.
striking
principal
principal
pivot.
in (ol;
the
pivots
El) one at a time.
to (1) is invariant
pivots.
submatrix
columns.
and 9 = (qi,, . . . , qiJT.
the i,, . . . , i,th
by
rows and all
,5 = (Zi,,. . . , ZiJT
Then
o=Nt+
9,
0 > 0,
E b 0,
as a principal
of M of order s, obtained
Let
cO=(zQ,,...,QT,
is known
sets
pairs of
to (l), then (El, G2; G1, {“) is a solution
Thus the number of solutions
Let N be a principal
. . , i,th
The
into an
pivot exchanging
by a series of principal
pair of variables
off from M all the rows excepting
but the i,,
of the block
the block principal
each complementary
block
the problem
The set of complementary
as a result
If (6?, G2; [i, [“) is a solution
under
if D is nonsingular.
is to transform
are exchanged.
M is nondegenerate,
sets of variables
(7)
of the same form in which the complementary
(09; li)
remains
When
ED-lql),
(P)Tc01+(U2)Tt2=0.
The block principal
of the block
(6)
(o~)~P+(~~)~~~=O.
OYE = 0,
subproblem of (1) in the variables
(8)
(w; E).
ON
THE
COMPLEMENTARITY
2.24.
Suppose
73
PROBLEM
(6; z^)is a complementary
feasible solution to (1) such
that
&= 0
for all
i # i,
or
i,, . . . ,
or
i,.
Let I% = (l;i!, . . , 6~~~)~and $ = (ii,, . . . , ii,).
Then, from the definition
of the principal
(Lj; l) solves
2.25.
subproblem
(8), we see that
If Y is any integer,
its parity
integer or even if r is an even integer.
it is said to be of constant parity
(8).
is said to be odd if I is an odd
When considering
if all the numbers
a set of integers,
in the set have the
same parity.
2.26.
A set of cones in R” whose union is R” is said to form a partition
of Rn if each cone in the set has a nonempty
of the interiors
3.
FINITENESS
3.1.
interior and the intersection
of any two cones in the set is empty.
OF THE NUMBER
Lemke
[8]
has
OF COMPLEMENTARY
shown
that
the
FEASIBLE
number
of
feasible solutions is finite whenever q is nondegenerate
Here we determine
the number
3.2.
the necessary
of solutions
THEOREM.
and sufficient
SOLUTIONS
complementary
with respect to M.
conditions
under which
to (1) is finite for each q E Rn.
The n$Lmber of complementary
feasible
solutions
is
finite for all q E Rn if and only if M is nondegenerate.
Proof.
Suppose
there exists a q E R” such that (1) has an infinite
number of distinct solutions.
negative linear combination
There are only 2” distinct
if (1) has an infinite
complementary
Each solution to (1) represents q as a nonof some complementary
complementary
number
set of column
yi 3 0
has an infinite
number
system
of n equations
number
of solutions,
of distinct
vectors
of distinct
solutions,
Thus,
there must exist
a
{A .j, j = 1,. . . , n} such that
for each
j = 1,. , , n,
solutions.
in n nonnegative
set of column vectors.
sets of column vectors.
Equation
variables.
then the set of column
vectors
(9)
(9) is a square
If (9) has an infinite
{A.j,
j = 1,. . , n}
74
KATTA
G. MURTY
must be linearly dependent.
Since {A .j} is a complementary
set of column
vectors, this implies by the definition
in 2.2 that M is degenerate.
To prove the converse, suppose M is degenerate.
We will show that
this implies the existence of a 4 # 0 for which (1) has an infinite number
of solutions.
Case 1. Suppose one of the column vectors of M, say M.,, is zero.
Then let q = (0, 1, 1,. . , 1)r. Then (ZPJ;z) = (0, 1, 1,. . . , 1; u, 0, 0,. . . , 0)
is a complementary
feasible solution for any cc > 0. Thus there are an
infinite number of distinct complementary
feasible solutions when q =
(0, 1, 1,. . . , 1)r in this case.
Case 2. Suppose M., # 0.
Since M is not nondegenerate,
there
exists a complementary
set of columns, say {A .j, j = 1,. . , n}, which
is linearly dependent.
So there exists CC= (cr,, . . , c(,)r # 0 such that
glA.&= 0.
Also A., is either 1.i or - M.,, and hence in this case A., # 0.
If foci
A.j = 0, let q = A., # 0. Then every (ze);z) obtained
by
setting the variable associated with A.1 equal to 1 + CC,the variable
associated with A.j equal to M for j # 1, and all other variables in (ZU;Z)
equal to zero, is a complementary
feasible solution for any M > 0. Hence
there are an infinite number of distinct complementary
feasible solutions
when q = A., # 0 in this case.
If cTs1 A.j # 0, let q = c,“=l
A.j.
Let
aj<O
= + 00
if there
does not exist
any
tcj < 0.
So 0 > 0. Every (ze);Z) obtained by setting the variable associated with
A.j equal to 1 + kj for j = 1,. . ., n and all the other variables in (XV;Z)
equal to zero is a complementary
feasible solution for any ;1 such that
0 < il 6 8. Hence there are an infinite number of distinct complementary
feasible solutions when q = c,“=l A.j # 0 in this case.
Hence if M is degenerate there exists a q # 0 for which (1) has an
infinite number of distinct solutions.
ON THE
COMPLEMENTARITY
3.3. COROLLARY.
there are an infinite
3.4.COROLLARY.
an infinite number
PROBLEM
75
If M is degenerate, there exists a q # 0 for which
number
of distinct complementary
feasible solutions.
If M is degenerate, the set of all q for which (1) has
of solutions is a subset of the union
of all complementary
comes which have empty interior.
4. UNIQUENESS
OF THE
COMPLEMENTARY
4.1. We now examine
solution
FEASIBLE
the question
of when
SOLUTION
Eq.
(1) has a unique
for each q E Rn, while M is fixed.
4.2. THEOREM.
The system (1) has a unique solution /or each q E Rn
if and only if M is a P-matrix.
Proof.
A proof of this theorem
given in [12].
the sign reversal
not a P-matrix.
based on mathematical
induction
is
Here we give a much simpler proof due to Gale based on
property
of matrices
Then by Theorem
discussed
in [6].
Suppose
M is
2 of [6] there exists an x E R”, x # 0,
such that y = Mx and, for each i = 1 to n, xi and yi have opposite
signs
(i.e., Xiyi < 0).
Let
if
yi > 0,
if
yi<O;
if
yi >, 0,
=-yyi
if
yi<O;
x.+
z zzzxi
if
xi > 0,
if
xi<O;
if
xi >, 0,
if
xi (
Yi+ = Yi
=o
yi- = 0
=o
xi- = 0
= -
xi
0.
Then
yi=yg+-yyi-,
yi+&yi->O;
yi+yi-=
0
for
i = 1 to n;
(10)
76
xi
KATTA
zzz
xi+ -
x.t
xi+&xi->
J
xi+xi-
0;
= 0
Since xiyi < 0 for each i = 1 to n, we verify
for each i = 1 to n.
that
i = 1 to n.
(11)
xi+yi+ = xiyi-
= 0
So
(y+)TX+ = (y-)Tx-
Since y = Mx,
for
G. XIUKTY
= 0.
(12)
we have
(13)
y+-Mx+=y--Mx-zq".
x+ #
x # 0,
From (lo)-(14)
x-.
(14)
we conclude that, when 4 = 4, (1) has two distinct solutions,
namely,
(ze);z) = (y’;
x’)
(20; 2) = (y-;
x-).
and
Thus, if M is not a P-matrix,
distinct
there exists a 4 E Rn for which (1) has two
solutions.
Now suppose M is a P-matrix.
has at least one solution
for which (1) has two distinct
(z&and, since these two solutions
plementarity
Then,
by Theorem
for each q E R”.
solutions,
Suppose
namely,
6,) = M(z -
2)
are distinct,
Z-
6 of [I], Eq.
there exists
(a; 2) and (G,; 2).
z^# 0.
From
the com-
condition
(15) and (16), we verify
to n and, since Z not a P-matrix,
unique
z^f
that
(zB~-
0, this implies
which is a contradiction.
Gi)(Zi -
by Theorem
(16)
ii) < 0 for all i = 1
2 of [B] that
So the solution
M is
to (1) must be
for each q E R”.
4.3. COROLLARY.
Keeping
M fixed, if (1) has at most one solution for
each q E Rn, then M is a P-matrix
qER”.
Then
(15)
(&?)T.z= (G)‘Y = 0,
using
(1)
a q E Rn
and (1) has exactly one solution for any
ON THE
COMPLEMENTARITY
77
We have shown in the first part of the proof of Theorem
Proof.
that,
PROBLEM
if M is not a P-matrix,
then there exists
q = 4, (1) has at least two distinct
M has to be a P-matrix
If
4.4. COROLLARY.
a 4 E R” such that,
solutions.
4.2
when
So, under the hypothesis,
and the corollary
follows
from Theorem
4.2.
(1) has a unique solution for each q E R”, then any
subproblem of (1) has a unique solution for each of its right-hand
pGxipa1
constant vectors.
This
Proof.
principal
follows
submatrix
4.5.
The
COROLLARY.
Paragraphs
from
Theorem
of a P-matrix
4.2
and
the
fact
that
every
is also a P-matrix.
set of complementary
2.9 and 2.10 forms a partition
cones
obtained
as in
of R” if and only if M is a
P-matrix.
If M is a P-matrix,
Proof.
(1) has a unique
solution.
in the whole space Rn.
hence
every
Since M is a P-matrix,
complementary
by Paragraph
then by Theorem
interiors
2.10 the system
(1) has a unique
complementary
Conversely,
complementary
linearly
R”,
this
Also,
every
cone does not
Hence
and hence,
solution.
independent
implies
a partition
Each
(1) has a unique
After Theorem
Thrall,
complementary
2.10,
solution
cones which is equivalent
cone by
vectors
is
cones partition
corresponding
cone.
of a
for all such q, (1)
set of column
to each
So (1) has a unique
4.2, M must be a P-matrix.
4.2 was conjectured
and Wesler
of Rn, every
and hence M must
lies in the interior
by Paragraph
solution for each q E Rn and hence, by Theorem
4.6. Note.
interior
and, since the set of complementary
that
the
of R”.
q E R” which
complementary
of the
Therefore
cones form a partition
q E R” lying on the face of any complementary
by Samelson,
the intersection
cones is empty.
cone does not lie in any other
property
and
Also, since
for each q E Rn
cone must have a nonempty
be nondegenerate.
has a unique
forms
interior.
solution
cone.
if the complementary
complementary
the partition
cones
cones
of a complementary
of any pair of complementary
set of complementary
for each q E Rn,
it is nondegenerate
cone has a nonempty
any q E R” which lies in the interior
lie in any other
4.2,
So the union of all the complementary
and proved, a theorem
[14] on the partition
of Rn by convex
to it and to Corollarv 4.5 has come to our notice.
KATTA
78
G. MURTY
Their proof was based on geometric considerations,
while the proof given
here is based on the properties of P-matrices.
However, in Theorems
7.2 and 7.10, we adopt
geometric
procedures
much
like theirs.
4.7. As a generalization
of Corollary 4.4, it is interesting
to check
whether every principal submatrix of a Q-matrix is also a Q-matrix.
This
Let
is not necessarily
true, as the following example illustrates.
-1
lV=
i
2
2
-I’
I
(17)
In Fig. 1 each complementary
cone is indicated by a dotted line segment
In this case, we verify that the union of
running across its generators.
the complementary
cones is the whole space, R2.
So, by Paragraph 2.10, the matrix M in (17) is a Q-matrix.
Now we
examine the principal submatrix
obtained by striking off the first row
and column.
N = (-
1).
ON
THE
This
COMPLEMENTARITY
is not
a Q-matrix,
because
the associated
has no solution if q2 < 0. However,
every principal
under
submatrix
Corollary
principal
if M is a Q-matrix
of M is also a Q-matrix.
of the solution
of “uniqueness”
of the solution.
4.8. THEOREM.
This is discussed
Su$pose
solutiolz for each q E R”.
and M > 0, then
This will be proved
M has the property
Keep q2,.
to (1) also affects
the
below.
that (1) has a unique
, qn fixed but let q1 vary.
be the value of z1 in the solution to (1) as a function of ql.
monotonic
subproblem
5.3.
The property
nature
79
PROBLEM
decreasing in ql, and it is strictly monotonic
Let zl(ql)
Then z,(q,)
is
decreasing in the
region in which it is fiositive.
Proof.
fixed.
solution
Proof by contradiction.
Let 2 = (pe,.
, qJT, which is held
Pick any value for q1 and let /3 > 0 be arbitrary.
Let (~2; 2) be the
to (1) when
(G; Z) be the solution
to (1) when
91$-p
q=
[
9
1.
2, and zl(ql + 8) = 3,.
Let
6 = (z&, . . .) z&J’,
i = (&, . . , z”,)‘,
6 = (Gz,.
l=
.) zc,,)‘,
(&I,. . ) Z”,)T,
and
m., = (wLz1,.. .,m,JT.
If 2, > 0 we wish to show that 2, < 2,.
2, > 0.
By complementarity,
4=
Suppose
Gr = z&r= 0.
41 +
[
Then,
I
P + m11z^,
3 +
m.,z^, ’
if
not;
then 2, >
80
KATTA
(1) has two solutions,
G. MURTY
namely,
and
contradicting
the uniqueness
> 0, then
if z,(qi)
It remains
all ,B > 0.
Then,
of the solution to (1) for each CJE R”.
z,(q, + P) < z,(q,)
to be shown
Suppose
not;
that
if z,(g,)
then z,(qi)
= 0 then
z,(qi + p) = 0 for
= f, = 0 and zi(g, + 8) = 2, > 0.
if
41+/3
q=
[
(1) has two solutions,
which
is again
4.9.
solution
-
Hence,
for any /3 > 0.
M.,},
4.10.
1’
namely,
a contradicti’on.
We now show that,
if M is a Q-matrix
when q is any element
then
9
of the set {I.i,
and (1) has a unique
I.,, . . . , I.,;
-
M.,,
. .,
M is a P-matrix.
Let
THEOREM.
[zj] = union of all com$demerztary cones which codain
-
M.j as a generator,
[wj] = union of all co~mplementary cones which contain I.j as a generator.
If I.j 6 [zj] and -
M., $ [wj] for each i = 1 to n and M is a Q-matrix,
then
M is a P-matrix.
Proof.
4.11.
obtained
that
Let
N be the
by striking
principal
submatrix
of M
off the first row and column
N is also a Q-matrix.
of order
of M.
(n -- l),
We now show
ON
THE
CO1\IPLERIENTARITY
Suppose
not.
the principal
81
PROBLEM
Let u = (‘wz,. . . , ZLZJ,)~
and 6 = (z2,. . . , z,)~.
subproblem
in (w; t),
OJ b
If N is not a Q-matrix,
which
CUT&$
= 0.
,530,
0,
there exists
Consider
is to solve
a i E P-l
(18)
such that,
when 9 = i,
(18) has no solution.
Let
If (1) has a solution
(W; 2) when q = q, then Zi > 0 in it, as otherwise
(6; E) would be a solution
to (18).
{$7:
S=
.’ ,
only to complementary
Since
are only a finite
there
there
must
feasible
number
exist
every
point
on the line
cc real}
( 9 1
corresponds
one is convex,
Thus
solutions
in which
of complementary
an tcO such that
zi > 0.
cones and each
the half-line
(B:,=(;)+B1.,, 820)
lies entirely
in a complementary
complementary
feasible
solution
line we must have zi > 0.
plementary
pos(1.i)
cone
cone.
By the argument
corresponding
-
M.,
is a generator.
also lies in the same complementary
we conclude
to any point on this half-
This implies that this half-line
for which
is a contradiction.
all principal
submatrices
lies in a com-
This
implies
that
cone, i.e., I., E [zi], which
Hence N must be a Q-matrix.
that
above, in every
By a similar argument
of M of order (PZ-
1) must
be Q-matrices.
4.12.
Let .M be the principal submatrix
by striking
that
JV
of n/rof order (PZ-
off the first two rows and columns from M.
must
Suppose
be a Q-matrix
not.
its right-hand
also.
Then there exists
in (~a,. . . , w,; z3,.
a Q E Rnh2 such that the subproblem
. , z,) has no complementary
constant
vector
2) obtained
We will now show
is Q.
feasible
solution
when
KATT.4
82
G. MtJRTY
Let
Then, for any il, p real, (1) has no solutions in which both zr and ze are
equal to zero, when q = 4 by paragraph
2.24.
4.13.
the line
Fix il = Ai, ,u = pi, where ill > 0 and p1 > 0, and consider
&:
Points
zi and
cones
entire
q=
[:::I,
u real].
on this line have complementary
feasible solutions in which both
zs cannot be zero together.
Since the number of complementary
is finite and each is convex, there must exist an CI,,such that the
half-line
(20)
is in a complementary
cone. Suppose this half-line
cone pos{I.,, - M.,, A.,,. . ., A.,}.
Then
3LJ.i + ,/_Q.,~pos{I.~,
- M.,,A.,,.
is in the complementary
. ., A.,}.
Suppose
&I.,
+ ,uJ.s
= cri1.i + ~e(-
M.,) + i ajA+
j=3
where c(i, ua,. . . , CI, > 0.
If tci > 1i, then, if we put (A, p) = (0, pi), q of (19) will have a complementary
feasible solution in which both zi = ze = 0, which is a contradiction to paragraph
4.12.
of
(A, - cc,)I., + pi1.e lies in the intersection
If c(i < 11, then
pas{- M.,, A.,,. . ., A.,} with pos{I.i, I.,}.
We note that pos{I.i, 1,s)
cannot entirely lie in pas{ - M.z, A .3,. . . , A .n} because then 1.i E [zi]
and I., E [za], contradicting
the hypothesis.
So pos{I.,, I.,} and
ON THE
COMPLEMENThRITY
pas{-
M.,,
unless
1, ,U are such
A.,,.
intersect
. .) A.,}
83
PROBLEhI
in a half-line
and, when
(A, ,u) >, 0,
that
we have
AI., + ,uI.,$pos(I.,,
Similarly
we see that,
a complementary
must
cone
a finite
when
M.,
tors.
But
pos{I.,,
in pos(1.n
this
This implies
1,s)
is a closed
excepting
all of pos{I.r,
cone which has both
So N
every
From
submatrix
diagonal
both -
M.,
-
-
M., and
in pos{I.i,
I.,>
belong to the union
M., as genera-
and -
cone and, if it contains
all points
of
number
of half-lines,
that
lies in some com-
M., and -
I.,
M., as generators,
the hypothesis
By Tucker’s
By a similar
of M must
then
which
of the theorem
[17]
we can show that
Hence
be strictly
all the elements
positive.
we see that
every
matrix
A?!,
a series of principal pivots (as in Paragraph
that all its diagonal
theorem
argument
of M is a Q-matrix.
from M by performing
2.20) has the property
that
of half-lines
in (20) can
to the hypothesis.
in the principal
obtained
I.,}.
must be a Q-matrix.
principal
all the points
Th’is implies
it contains
in
3L1.i + ,~1.,
lies in the union of
the half-line
those lying on a finite
plementary
is a contradiction
+ ,uil.,)
1.2}r
that
cones containing
union
is contained
then
cone for which both
those lying on a finite number
of all complementary
. , B.,},
I.,}.
(&I.,
in a complementary
are generators.
excepting
in pos{I.,,
of half-lines
only be contained
-
M.,, I.,, B.,,
(Ai, ,~r) 3 0, unless
number
A.,,. . .,A.,}.
for (A, ,u) 3 0, if AI., + $.a
pas{ -
lie on some half-line
Hence,
-M.,,
elements
(see also Lemma
are strictly
6.1 in [13])
positive.
this implies
M is a P-matrix.
4.14.
efficient
It may be possible
Note.
algorithm
a P-matrix
for testing
5.1.
4.10 to develop
a given real square
matrix
an
M is
or not.
5. ON THE Q-NATURE
Suppose
each i and j.
to use Theorem
whether
OF NONNEGATIVE
the square
matrix
MATRICES
M is nonnegative,
This case is of particular
interest
because
i.e., rnij > 0 for
the problem
of
84
KATTA
finding a Nash equilibrium
as a problem
point of a bimatrix
game can be formulated
of the form (1) in which M 3 0.
to know when such a matrix is a Q-matrix.
G. MURTY
See [l].
It is of interest
The following theorem
discusses
this question.
Let M > 0.
5.2. THEOREM.
for each i = 1,.
Proof
[l]
Proof
if and only if m,, > 0
., n.
This
of sufficiency.
in a corollary
has been proved
under their
= 0.
Theorem
Suppose
of necessity.
of M, say, m,,
M is a Q-matrix
Then
one of the principal
we will construct
4 = 4, (1) has no solution.
by Cottle
and Dantzig
5.
diagonal
elements
a 4 E R” such that
when
Let
1, 1, 1,. . ., 1)T.
4^= (-
(21)
If 4 = 4, (1) becomes
w=Mz+q”>
w 3 0,
In any feasible
solution,
WTZ= 0.
z b 0,
(22)
wi = Mi.z + ii and so wi > 0 since Mi. 3 0,
z > 0, and gi = 1 > 0, for each i = 2 to n. Hence in any complementary
feasible
solution
However,
of (22) we must
wl
implies
Thus
that
have zi = 0 for each i = 2 to n.
if za = * * * = z, = 0,
wi = -
=
M,.z + g1
1 < 0 since m,,
(22) has no complementary
= 0 and ii = feasible
solution
1.
and hence,
when
4 = 4, (1) has no solution.
The
example
in (21) is due to Gale.
5.3. COROLLARY.
If M 3 0 and M is a Q-matrix,
then all principal
submatrices of M are also Q-matrices.
5.4. COROLLARY.
If M > 0, the union of all the complementary
is the whole space Rn if and only if m,, > 0 for all i = 1 to n.
cones
ON THE
COMPLEMENTARITY
85
PROBLEM
6. ON THE CONSTANTPARITY OF THE NUMBER OF COMPLEMENTARYFEASIBLE
SOLUTIONS
6.1.
We now examine
q varies
over
how the number
of solutions
to (1) varies
as
R” while M is fixed.
If M is nondegenerate, then the number of complementary
6.2. THEOREM.
feasible solutions has the same parity for all q E Rn which are nondegenerate
with respect to M.
In the proof of this theorem
Proof.
by Lemke
in
6.3.
Resdts
(i)
(ii)
Then
For
each
i = 1,.
. , n, the
that,
on every
solution
set
C,(Q)
or unbounded)
of
in C,(q).
solution
and y”E R”,
edge of C,(q).
Suppose
then
F is an
Let
13>0},
and (vJ~; 2”) is an extreme
homo-
of K(g”).
Along this unbounded
edge F, wi = wil + Bwi2, and it remains bounded
for all 0 > 0 only if wi2 = 0.
(1) has an infinite
ze,=wl-
number
“ilI.i
which is a contradiction
6.5.
if M is nondegenerate
(zu,;z)=(w~+~w~;z~+~z~),
where (~IzJ~;
zl) is a basic feasible
Thus,
complementary
edges in C,(q) differs from the number
unbounded
edge of K(Q) contained
F={(w;z):
Theorem
almost
to M:
when q = 9.
to (1) by an even number.
We now prove
unbounded
with respect
of solutions
or is the union of some edges (bounded
wwiis unbounded
geneous
number
The number of unbounded
of solutions
6.4.
If g is nondegenerate
(Lemke).
(1) has a finite
either is empty
K(q).
(iii)
we use some of the results proved
[S].
If wi2 = 0, then, if we put q = g of solutions,
z = z1 + 8z2
+ tw;
w~~I.~,
namely,
to the hypothesis
that
for all
0 > 0,
M is nondegenerate
by
3.2.
on every
unbounded
edge of C,(p),
We now use the result obtained
4”E Rn is nondegenerate
with respect
wi is unbounded.
in Paragraph
6.4 to show that, if
to M and ct is any real number
such
86
KATTA
that
i = g + cd.,
number
is also nondegenerate
of unbounded
Pick
any unbounded
Y = max
it is easily
verified
F1=((w;z):
I
0,
on this edge.
Let
ze$ + a
-
z4Yi21.
0>v}
that
there
exists
to each unbounded
an unbounded
edge F in C,(q).
~r)1.~ we can establish
between
the unbounded
edge F1 in
Conversely,
a correspondence
edges in C,(g) and those of C,(q).
correspondence
6.6.
Now, to continue
vectors
by
between
This establishes
a l-to-l
edges in C,(q) and those in C,(g).
Hence both C,(q) and C,(d) must have the same number of unbounded
two-column
the
that
shown
4”= d + (-
unbounded
then
edge in C,(i).
we have
C,(d) corresponding
treating
to M,
are the same.
(w;z)=(zd+0w2+cd.i;z1+~z2),
is an unbounded
Thus
respect
edge
By 6.4, wi2 > 0 and ‘wi is unbounded
Then
with
edges in C,(q”) and C,(g)
G. AXURTY
the proof of Theorem
edges.
6.2, let 4 and q be any
in Rn both of which are nondegenerate
with respect
to M.
By Paragraph
of the number
6.5 and (i) and (iii) of 6.3, we conclude
of solutions
4 one component
to M both
to (1) does not change
at a time so that it remains
before
and after
the
nondegenerate
it becomes
Hence
of being nondegenerate
equal
whenever
6.7.
Thus
of solutions
the number
q is nondegenerate
Note.
with respect
at a time, retaining
to M throughout,
until
to S.
the number
4 = @ or g.
with respect
alteration.
It is always possible to alter y”by one component
the property
that the parity
if we alter the vector
The
with respect
assumption
dropped
from the hypothesis
example
below.
to (1) has the same parity
of solutions
that
M
of Theorem
Let
M=
to M.
is nondegenerate
cannot
be
6.2, as can be seen from the
01
1
[ 12’
whether
to (1) has the same parity
ON
THE
Then
COMPLEMENTARITY
(1) is to solve
Wl
w2
1
0
0
1
Only complementary
22
21
0
=4
-1
-1
Wl> WZ,Zl, 22 >
Fig.
87
PROBLEM
lqr
-2
q2
WlZl +
0,
w2z2 =
cones with nonempty
0.
interiors
are indicated
in
2.
[I
1
4=
leads to one solution
to (1) and
2=
leads to two solutions
with
respect
Here
The
i-7
-2
to (1) even though
both these
are nondegenerate
to M here.
M is degenerate
argument
1
in Paragraph
because
the matrix
6.4 fails.
(I.,,
-
M.,)
is singular.
88
KATTA
6.8.
Note.
The assumption
M cannot
be dropped
seen from
Fig. 3.
that q is nondegenerate
from the hypothesis
of Theorem
G.MURTY
with respect to
6.2, as can be
Let
M=
1
1
2
i2
1’
..
..
..
.
l.
..a**.
‘.
)I,
- M.]
FIG. 3
When q is nondegenerate
with respect to M the number
(1) is an odd number but, when q = -
M.,,
of solutions
(1) has exactly
to
two distinct
solutions.
6.9. COROLLARY.
number
of solutions
with respect
6.10.
If M is nodegenerate
to (1) is an even number
and not a Q-matrix,
then the
for all q which are nondegenerate
to M.
Proof.
By Paragraph
2.10, the set of all q for which (1) has
a solution is the union of the Zn complementary
cones.
Each complemen-
tary cone is a closed set in Rn, and hence their union (being a union of
a finite number of closed sets) is itself closed.
(1) has no solution is the complement
set.
Because M is not a Q-matrix,
The set of all q for which
of this union, and thus is an open
this open set is nonempty.
the set of all q for which (1) has no solution
is a nonempty
Therefore
open cone.
ON
THE
COMPLEMENTARITY
6.11.
By Paragraph
89
PROBLEM
2.16, the set of all q which
are degenerate
is
the union of a finite number of subspaces of Rn, each of which has dimension < (n -
1). Hence the set of all q which are degenerate with respect
to M has no interior.
6.12.
By 6.10 and 6.11, we conclude
degenerate
with respect
solutions.
that there must exist a 9 non-
to M, for which
Now, by applying
Theorem
(1) has no solution,
6.2, we conclude
i.e., zero
that the number
of solutions to (1) has the same parity as zero, i.e., even parity, whenever
q is nondegenerate
6.13.
Corollary
Note.
as seen from
6.14.
with respect
Example
6.9 is not necessarily
of Corollary
unless M 3 0.
This is discussed
I.
WITH
ON
PROBLEMS
true if M is degenerate,
6.7.
The converse
Note.
to M.
A CONSTANT
6.9 is not necessarily
true
in Note 8.17.
NUMBER
OF
COMPLEMENTARY
FEASIBLE
SOLUTIONS
7.1.
solutions
Here we show that, if the number
is a constant
of complementary
for all nonzero 2 E Rn, then that constant
feasible
is equal
to one and M is a P-matrix.
We also show that, if the number of com-
plementary
is a constant
feasible
solutions
with respect to M, then that constant
the complementary
for all q E R” nondegenerate
is equal to one and in this case all
cones which have nonempty
interiors form a partition
of R”.
7.2.
THEOREX
If the number of complementary
feasible solutions is
a constant for all q E R”, q # 0, then M is a P-matrix
and that constant
is equal to one.
Proof.
7.3.
Whatever
M may be, (1) always has at least one solution
every q > 0 (the solution is w = q; z = 0).
If M is not a Q-matrix,
exists a 4 # 0 for which (1) has no solution
at all.
Q-matrix,
the number
of solutions
q # 0. Thus, under the hypothesis
7.4.
to (1) cannot
Hence, if M is not a
be a constant
for all
of Theorem 7.2, M must be a Q-matrix.
The number of complementary
ever q is nondegenerate
for
there
feasible solutions
with respect to M.
is finite when-
By Corollary 3.3, if M is not
I<ATTr\
90
nondegenerate,
of solutions.
G. BIUKTY
there exists a q # 0 for which (1) has an infinite
number
Since the number of solutions to (1) is a constant for any q # 0, M
must therefore be a nondegenerate
matrix.
Hence, all the principal
submatrices
of M are also nondegenerate.
Also, every subcomplementary
set of column vectors is linearly independent
and every complementary
cone has a nonempty
interior.
set of column
7.5. Let {A.,,. . ., A.,_l} be any subcomplementary
vectors.
We now show that the hyperplane
generated
by this subcomplementary
set of column vectors strictly separates the points representing
the left-out complementary
pair of column vectors I., and
-M.,.
Then the interiors
of the complementary
cones
Suppose not.
pos{Aq,. . .) A .n_l, I.,} and pos{A .i, . . , A .n_l, - M.,} have a nonempty
intersection.
Consider two hyperplanes
(see Fig. 4) each of which is
/
generated by
(I, - M). If
is a subspace
independent,
a linearly independent
subset of (VZ- 1) column vectors of
these two hyperplanes
are distinct, then their intersection
of dimension (n - 2). The set (A.,, . . , A.,_,} is linearly
and there are only a finite number of subspaces generated
ON
THE
COMPLEMENTARITY
by subsets
pick
of (n -
91
PROBLEM
1) or less column
vectors
of (I, -
M).
So we can
a point
n-1
4" =
I,A.i,
C
j=l
where ill,.
by (PZsection
. , ii,..., are all > 0, such that q* is not in any subspace generated
2) or less column
of pos{A.r,
the hyperplane
(n -
through
1) linearly
vectors
. . , A.,_l}
of (I, -
column
vectors
So, if q* is also on the hyperplane
tary set of column
hyperplane
vectors,
through
through
in the hyperplane
vectors,
then
through
that
set
through
from
by a subset
of (I, -
of
M).
some other subcomplemen-
then that hyperplane
A.,,
on the hyperplane
which is distinct
and is generated
A.,, . . . , A.,_,
independent
M) and also not in the inter-
with a hyperplane
must coincide
with the
Since both I., and - M., are not
, , A.,_,.
A.,,. . ., A.,_,, we conclude that, if q* lies
some other
must
subcomplementary
be of the
form
{B.,,.
set of column
. ., B.n_l},
where
LV.~} for i = 1 to n - 1.
The intersection of the cones pos(A.r, . . , A.n_l, I.n> and pos{A.r, . . . ,
A.,_,, - M.,} has a nonempty interior.
Hence, by the choice of q*, we
B.j E {I.j,
-
can find an CC> 0 sufficiently
small
that,
if we pick
4^= q” + cd.,,
then
4 is nondegenerate
of both
the cones.
degenerate
with respect
to M and it is in the intersection
By the nondegeneracy
for all but a finite number
an cc0 > 0 sufficiently
small
of (i, 4 + &I.,
is also non-
So we could pick
of values of A,.
that
n-1
q = 2 &A.
j +
PI.,
j=l
is nondegenerate
with respect
to M for all fi satisfying
/I # 0, -
q, <
P < x0. Hence, if we had chosen our original M.so small that 0 < cc < cc,,,
and 4 is in the interior of both pas{ A.,,
. , -4 .+r, I.n} and pas{ A. 1,. . ,
A.
n-17 -
M.n},
then
S-1
y”=
~iljA.j-d.,
j=l
is outside both these complementary
4 by the hyperplane
which is degenerate
through
cones and is strictly
A.,, . ., A.,_,.
on the closed line segment
separated
from
Also q* is the only point
joining
the points 4 and q.
KATTA
92
G. MUIiTY
Thus, if g lies in any complementary
cone in which 4 does not lie,
then q* must lie on a face of that complementary
cone. Hence by the
choice of q* we conclude that, if Q lies in any complementary
cone in
which 4 does not lie, then that cone must be of the form pos{Ls.,, . . . , B.,},
where B.j is contained in {I.j, - M.j> for j = 1 to n, and the hyperplane
through
B.,, . . , B.,_,
must coincide with the hyperplane
through
A.,,. . ., A.,_,.
But the hyperplane
through A.,,.
., A.,_, separates g
from both I., and - M.,.
So p cannot lie in any complementary
cone
in which fi does not lie.
Hence the number of solutions to (1) when q = y” is strictly less (at
least by two) than the number when q = 4, leading to a contradiction.
By a similar argument,
we verify that the hyperplane
through any
subcomplementary
set of column vectors strictly separates the points
representing
the left-out complementary
pair of column vectors.
7.6. We now show that the principal subproblem of (1) in the variables
property.
The
(q,.
., w,-1; 21,. . . >z,-1 ) satisfies a similar separation
column vector corresponding
to zwj in this subproblem
is the jth column
vector of the unit matrix of order (E - l), which we denote by 9.j, and
the column vector corresponding to zj in this subproblem is - (mrj, msI, . . ,
mn_l,j)T
which we denote by - m.j. We note that the column vectors in
the subproblem
are obtained by deleting the last component
from the
column vectors in the original problem.
set of
Let {a.r,. . ., u.,_~, a.,+l,. ., u.,_1 > be any subcomplementary
column vectors in the subproblem.
We want to show that the hyperplane
in IV-l through these column vectors strictly separates Y.i and - m.,
Let A., be the column vector corresponding
to n.,, r = 1,. . , i - 1,
n - 1, in the original problem.
Then {A.,, . , A.i_l, A.i+l,
i+ l,...,
1s a subcomplementary
set in the original problem.
By
. ., A.,_l,I.,}
Paragraph 7.5 the hyperplane in Rn through these column vectors strictly
separates I.i and - M.i.
Suppose this hyperplane
is
DX=O,
where
D = (d,,
. , , d,) and X E R”.
DA.,
= 0,
DA.,_1 = 0,
Then
ON
THE
COMPLEMENTARITY
>0
DI.i
93
PROBLEM
and
D(-
AI.<) (0,
DA.<,_, = 0,
Now
DI.,
= 0,
D1.n
=
equations
imply
d = (d,,.
da.,
= 0,
da.,_l
= 0,
d9.i
> 0
da.,+l
= 0,
da.,_,
= 0.
and
Hence
and it strictly
the
subproblem
By a similar
argument
(1) of order
(n -
7.7.
strictly
pair of column
7.8.
and
For
through
the
any
hr = (wznll)f
is easily
0.
property.
subproblem
of
problem
of
N), if N is nondegenerate
subcomplementary
in the problem,
hypothesis
separation
property.
the points representing
vectors
of the
m.,.
complementarity
(I, -
in R”-l
. , a.,_,}
that every principal
vectors
every
-
a similar
the separation
hypothesis.
The induction
By nondegeneracy,
9.i
also satisfies
1, with column
separates
dx = 0 is the hyperplane
separates
we can verify
and if the hyperplane
vectors
d, = 0,
WZ.J < 0,
set {a.,, . . . , a.,_l, a.,+l,.
1) satisfies
Induction
order Y < n -
d(-
Thus
the subcomplementary
subproblem
Because
., d,_,).
that
x = (x1,. . . , x,_~) E R”-l.
through
tary
0.
Let
= d, = 0.
preceding
Let
DA.+,
set of column
the left-out
then
verified
complemen-
N is a P-matrix.
for the case Y = 1.
Since Y = 1, the subcomplementary
set is the null set and hence the hyperplane
through the subcomplementary
KATTA
94
set is the singleton
consisting
of the origin itself.
the points on R1 representing
1 and -
So N = (?nii) > 0 and hence
7.9.
Hence,
determinants
Since
of M of order
and 7.6,
1 are strictly
sub-
So M-l
as
= Q,
74 3 0,
2 b 0,
principal
positive.
of M # 0.
(1) can be written
M-lw
every
Thus all principal
by 7.4, determinant
the constraints
z -
hypothesis
< n -
~zii < 0.
in this case.
1 is a P-matrix.
M is nondegenerate
Thus
exists.
by the induction
this separates
nz,,, we should have -
is a P-matrix
of M or order n -
submatrix
Since
G. MUKTY
zT7P.J
= 0,
(23)
where
Q=
If (1) has a constant
then
(23) has a constant
or order (n -
of the principal
of solutions
for every
number
of solutions
for each Q E h’“, Q # 0.
Hence
> 0.
4 E R”, q # 0,
all principalsubdeterminants
positive.
obtained
Let tc be the value
by striking
off the first
Then
%l
determinant
=
(24)
-~~~;~.~~
of M
Thus, by (24), the determinant
So all principal
positive.
of M-l
from M-l.
‘J.
But tc > 0, mi,
used previously,
1) or less are strictly
subdeterminant
row and column
Mplq.
number
Hence, by the arguments
of M-l
-
subdeterminants
of M is also strictly
of M are strictly
positive.
M is a P-matrix.
So the induction
verified
hypothesis
for Y = 1 in 7.8.
Thus,
by 7.5, Theorem
7.10.
THEOREM.
7.7 holds when Y = n also.
Hence
by induction
It has been
it holds for all n.
7.2 is true for all n.
If the number of complementary
feasible solutions
is a constant for all q E R”, which are nondegenerate with respect to M, then
that constant is equal to one and, in this case, the set of conzplementary cones
with nonemptv
interiors forms a partition
for R”.
Proof.
7.11.
Whatever
M may be, (1) has a solution
always find a q nondegenerate
with respect
if q > 0 and we can
to M which is nonnegative.
ON
THE
COMPLEMENTARITY
If M is not a Q-matrix
by 6.12, we can find a q nondegenerate
with respect
Hence, if the number
to M, for which (1) has no solution.
(I) is a constant
95
PROBLEM
for all q nondegenerate
with respect
of solutions
to
to M, M must be
a Q-matrix.
7.12.
. ., A.,_l, LI.~+~,. ., A.n} b e any subcomplementary
Let {A.,,.
set of column
If the two complementary
vectors.
LI.,_~, I.i, YI.~+~,. . ., A.,}
and
have nonempty
then the intersection
empty.
interiors,
pos(A.i,.
For, if their interiors
. ., A.i_l,
cones
pos{A.i,.
M.i, A.i+l,.
-
of their interiors
have a nonempty
intersection,
,
. ., A.,)
must be
by an argu-
ment similar to that in 7.5 we can find two points 4, y”both nondegenerate
with respect
to M, such that (1) has two more solutions
its number
of solutions
when
q = p, leading
when q = @ than
to a contradiction
of the
hypothesis.
Thus,
exists
through
complementary
that
every subcomplementary
a hyperplane
separating
the points
pair of column
the number
vectors.
of complementary
every q nondegenerate with respect
this separation
se$aration
7.13.
7.14.
problem
Consequently
there
since we only know
solutions
is a constant
to M, it is not possible
As in 7.6, it can be shown that
a similar weak separation
in the subproblems
Suppose
has been obtained
principal
However,
feasible
vectors,
to the left-out
for
to claim that
we refer to this as the zyeak
$ro$erty.
(1) satisfies
be strict
will be strict.
set of column
corresponding
subproblem
The separation
of
may not
also.
the problem
by performing
pivot on the system
(25) also satisfies
all the principal
every principal
property.
a series of principal
(1).
Then,
the weak separation
subproblems
pivots,
or a block
by 2.21 and 2.22, we see that
property
and consequently
of (25) also satisfy a similar weak separation
property.
7.15.
r < n -
Induction hypothesis.
1 satisfying
every subcomplementary
the weak
In any complementarity
separation
property
set of column vectors
problem of order
(i.e.,
there exists
that
through
a hyperplane
96
KATTA
which separates
the interiors
section,
the left-out
complementary
pair of column
of any pair of complementary
then
both
complementary
G. MUKTY
vectors),
cones have a nonempty
cones
are equal
if
inter-
to the nonnegative
orthant
7.16.
The induction
hypothesis
In this case, M = (mii)
is easily verified
and the weak separation
only if m,, 3 0, and hence
the weak
tary
cones have
7.17.
We
a nonempty
now show
cones in the original
property
separation
implies the stronger result that the interiors
for the case Y = 1.
holds, if and
property
in this
of no two distinct
case
complemen-
intersection.
that,
problem,
if the
interiors
of two
complementary
both of which have the rays generated
by
two or more column vectors of the unit matrix in common, have a nonempty
intersection,
of R”.
then
Suppose
the two cones
Let Y.a,
. ., B.,).
I.%, B.,,.
are equal
the two cones are pos{I.,,
to the nonnegative
1.s, A .s,
a.,, b., for Y = 3 to n represent
vectors I.s, A.,, B., with their first components
interiors
of pos{Y.s,
a.s,. . . , a.,}
of Rn-l, and complementary
and pos{Y.,,
wn, 22,. . .) z,) have a nonempty
hypothesis
b.,, .
b.,,
intersection.
the column
Then the relative
both
. , b.,}
subproblem
subsets
in (w,,
. .,
By 7.13 and the induction
7.15, this implies that both po~{Y.~, a.s, . . , a.,} and pos{S.s,
, b.,}
second
removed.
cones of the principal
orthant
. , A .%} and pos{I.,,
.
are equal
to the nth
and (1.r
rows of both
B.,
I.,
to the nonnegative
the matrices
are positive
B.,)
...
orthant
of R”-l.
(I.,
multiples
j I.,
Hence
A.,
the
. . . A.,)
of row vectors
of the
unit matrix.
Now, considering
similarly
(2% W3,. . .> 7fJ,; Zl,Z3>.
these matrices (1.i 1.a
also positive
multiples
.> zn),
A.,
the principal
we
*. . A.,)
permutation
and pos{I.,,
7.18.
I.,,
Now suppose
A.,,.
. ., A.,}
that
intersection.
and pos{A.i,
Make a block principal
with
the column
unit vectors
first
and (I.,
I.,
B.,
vectors
rows of both
...
of the unit matrix.
are positive multiples
the interiors
are
Hence the
of the column
. , A .,)
orthant
of R”.
of two complementary
cones,
by at least two column vectors in common,
Suppose
A.,, B.,,.
these
cones
are pos{A.i,
A.z,
. ., B.,}.
pivot so that each of the variables
A.,, A.s,. . . , A.,
in the transformed
B.,)
I.,, A .3,.
., B.,} are equal to the nonnegative
which have the rays generated
a nonempty
the
matrices and both the cones pos{I.,,
B.,,.
have
in the variables
that
of the row vectors
column vectors of both these matrices
vectorsof
subproblem
conclude
problem.
become
associated
The transformed
associated
with
problem
the
also
ON THE
COMPLEMENTXRITY
satisfies
the weak
conclude
(A.,
that
A., : ...
matrices
separation
under
this
in common,
the interiors
principal
is a column
pivot
which
..,E.,}
of Rn.
cones
which contain
cones be pos{T.r,
into
vectors
has the same property
vector
A.,, .
generated
by com-
only one column vector
of which have a nonempty
vector
of column
under a block
we
matrices
is transformed
multiples
two complementary
sets of column vectors
column
of the
..,A.,}andpos{A.i,A.,,B.a,.
orthant
plementary
the common
each
.. * j B.,)
B.,
themselves
must be equal to the nonnegative
the complementary
and, by using 7.17,
pivot
This is possible
matrices
Now consider
by 7.14
are positive
impliesthatbothpo~{A.r,A.,,A.~,.
7.19.
97
principal
(A., : A.,
vectors
matrix.
only if each of these
property
block
and
: A.,)
whose column
of a permutation
PROBLEM
intersection.
Suppose
of the unit matrix.
, A .n} and pos{I.i,
Let
B.,, . . ,
B.,} where, for each i = 2 to n, (A.j, B.j) is a permutation of (l.j, - M.J.
through (I.r, B.,,. . ., B.,} separates the points
If this separation is strict, then, since the
represented by A., and B.,.
interiors of pos{I.,, A.,, . . . , A.,} and pos(l.r, B.,, . . . , B.,) have a nonempty intersection,
the interiors of pos{1.,, A.,, . , A.,} and pos{I.r, A.,,
B.,, . . ., B.,} must have a nonempty intersection
too.
By 7.18, this
., I?.,}, which
implies that pos{I.,, A.,, A.,, . . ., A.,} = pos{I.,, A.,, B.,,.
contradicts the hypothesis that the interiors of pos(1.r, A.,, . . . , A.,} and
for the interiors of
pos{I.,,
B.,,. . ., B.,} have a nonempty intersection,
pos{I.r, A.,, B.,,.
., B.,} and pos{I.,, B.,, B.,,. . ., B.,} are disjoint, by
the separation property.
Thus A., must be on the hyperplane through
{I+ B.,,.
., B.,}.
By a similar argument, this implies that A., lies on
the hyperplane through {I.i, B.,, . . . , B.,_l, B.,+l,. . . , B.,} and B., lies on
the hyperplane through {I.i, A.,, . . ., A.,_,, A.r,_l,.
., A.,} for Y = 2 to n.
This implies that the two complementary
cones pos{I.r, A.,, . , A.,}
and pos(I.,,
B.,,
. . , B.,} are equal. Also, since the pair (A.j, B.J is a
By 7.12 the hyperplane
permutation
of (I.j,
rays generated
both
of these
cones
In general,
nonempty
vectors
-
M.j)
by I.,, .
for each i = 2 to n, we conclude
, I.,
are equal
are all extreme
to the nonnegative
if the two complementary
intersection
containing
are generated
one column
nonnegative
orthant
in common,
based
on block
that these two complementary
of Rn.
orthant
that
the
Hence
of Rn.
cones whose interiors
by complementary
vector
the I.j, we could use an argument
in 7.18 and conclude
rays of this cone.
have
a
sets of column
which is not one of
principal
pivoting
as
cones must equal the
KATTA
98
7.20.
The only remaining
like pos{A
.r,
case is that
. . , A .n} and pos{B.,,
.
the pair (A.,, B.j)
is a permutation
together
contain
interiors
of these two complementary
Consider
one of these cones.
separates
of two complementary
. . , B.,}
where, for each i = 1 to n,
M.J.
-
{A.,, . . . , A.n},
If the separation
property
is strict,
. ., A.,}
of the two cones pas{ B.,, . . . , B.,}
= pos{B.,,
B.,}
intersection.
A.,,.
and p~s{A.~,.
is a face of
7.12, this hyperplane
then, since the interiors
intersection,
and pas{ B.,, A.,, . . . , A.,}
By 7.18, this implies that pos(B.,,
and hence the interiors of pas{ B.,,
. ., A.,),
are disjoint
. ., A.,}
the
intersection.
which
have a nonempty
. ., B.,}
of po~(A.~,.
the interiors
also have a nonempty
Suppose
cones have a nonempty
through
and pos{B.,,.
Thus these two cones
in the problem.
By the weak separation
A., and B.,.
cones
of (l.j,
all the 2n column vectors
the hyperplane
G. MURTY
by the separation
property,
. . .,
, B.,)
which is
a contradiction.
Therefore
similar
{A.,,.
{B.,,.
lies on the hyperplane
B.,
argument,
we conclude
of (l.j,
-
., B.,}
Then,
7.21.
also holds for
orthant
n.
By 7.15,
holds for all Y < n -
we conclude
cones
that,
property
have
cones must be equal to the nonnegative
lie in the nonnegative
of Rn.
orthant
solution.
7.22.
is a constant
7.11, we therefore
By
So, whenever
interiors
of Theorem
Thus
Example.
(=
this constant
with respect
the set of complementary
form a partition
is a constant
7.10 the number
that
both
for some 4, that q must
for all q nondegenerate
conclude
q is nondegenerate
intersection,
of R”. Thus, if the problem
feasible solutions
But under the hypothesis
then, if the interiors
a nonempty
orthant
1, it
if a complementarity
7.12;
has two or more complementary
to M.
by I.j is an extreme
Thus both of these cones are equal
hypothesis
the weak separation
solutions
. .,
of R”.
of any two complementary
tary feasible
through
Hence po~{A.~,.
and, since the pair (A.?, B.j) is a permutation
if the induction
7 =
satisfies
a
through
lies on the hyperplane
M.j) for each j = 1 to n, the ray generated
to the nonnegative
By
{A.,, . ., A.,}.
for each Y = 1 to n.
ray of this cone for each j = 1 to n.
problem
through
B., lies on the hyperplane
and A.,
. ., A.,_,, A.T+l>. . .> A.,}
. ., B.,_,, B.,+l,. . ., B.,}
= pos{B.,,.
A.,}
that
to M,
of complemenwith respect
must equal
1.
(1) has a unique
cones which have
nonempty
of R”.
A problem
where
the number
1) for all 4 which are nondegenerate
but not for all q # 0, is obtained
by taking
of solutions
to (1)
with respect
to M,
n = 2 and
ON THE
COMPLEMENT.L\RITY
99
PROBLEM
1
M=l
1
The
complementary
cones
1
1.
[
corresponding
to this
problem
which
have
.
:
.*
:
:
:
--M
1,
-
M.2
FIG. 5
nonempty
interiors
are depicted
in Fig. 5.
Here (1) has a unique solution
whenever
for any M > 0.
When
q=
(1) has an infinite
w = 0,
We also notice
hyperplane
that
through
number
[11,1
u > 0,
of solutions,
z = (2x, (1 -
I+),
namely,
for any
only the weak separation
-
M., separates
0 <A
property
I., and -
< 1.
holds,
since the
M.,, but not strictly.
100
KATTA
7.23.
A problem where the weak separation
Example.
is satisfied,
but in which the number
is not a constant
by taking
of complementary
for all q nondegenerate
0
complementary
nonempty
property
feasible
with respect
7.12
solutions
to M, is obtained
n = 2 and
M=
The
G. MURTY
interiors
cones
[
-1
-1
0
corresponding
are shown in Fig.
1
.
to this
problem
which
have
1.s)
and
6.
FIG. 6
The
pos(-
interiors
M.,,
-
AI.,)
of the
two
have
equal to the nonnegative
complementary
a nonempty
orthant
of R”.
cones
intersection,
pos{I.,,
and they
are both
M is not even a Q-matrix
in this
case.
7.24.
solutions
Example.
We noticed
that,
for any nonzero right-hand
subproblem
of (1) satisfies
given that (1) has a constant
a similar
if (1) has a constant
constant
vector,
property.
number of solutions
number
of
then every principal
However,
if we are only
for any 4 nondegenerate
ON THE
COMPLEMENTARITY
with respect
similar
to M, then principal
property.
subproblems
As an example,
In this case (1) has a unique
to M.
However,
solution
WI3 0,
has no solution
7.25.
[ -1
a
1
1
1’
whenever
its principal
0% -k
Wl =
0
of (1) may not satisfy
let n = 2,
M=
respect
101
PROBLEM
q is nondegenerate
subproblem
with
in (w,; z,),
41,
21 >
0,
Wl.zl
=
0,
when q1 < 0 and has one solution
if q1 > 0.
If (1) has a constant number of solutions for all q
COROLLARY.
which are nondegenerate with respect to M, then every principal
subproblem
of (1) has at most one solution for any of its right-hand constant vectors
which is nondegenerate
This
Proof.
subproblem
hence
and not nonnegative.
follows from the fact
of (1) also satisfies
the induction
hypothesis
So, if the right-hand
subproblem
that
in this case every
the weak separation
principal
property
7.12
and
7.15 holds for it.
constant
vector
is nondegenerate,
can have two or more solutions
the principal
only if the constant
vector
is nonnegative.
8. THE ODD
8.1.
number
NUMBER
THEOREM
Here we show that,
of complementary
q is nondegenerate
FOR NONNEGATIVE
Q-MATRICES
if M is a nonnegative
Q-matrix,
feasible
with respect
solutions
to M.
is an odd number
then the
whenever
This result may not hold if M is
not nonnegative.
8.2.
If
THEOREM.
complementary
M > 0 and is a Q-matrix,
then the number
of
feasible solutions is an odd number for any q nondegenerate
with respect to M.
Proof.
8.3.
Proof
is by induction
on n.
If n = 1, then M = (mll)
ml1 > 0 by Theorem
5.2.
Here
and M is a Q-matrix
q = (ql) and for each
if and only if
q E R1 there
is
KATTA
102
exactly
one complementary
feasible
solution.
G. MURTY
Hence Theorem
8.2 is true
when n = 1.
8.4.
Induction
Suppose
hy$othesis.
Theorem
8.2
complementarity
problems
of order (PZ-
this implies
Theorem
8.2 also holds for problems
8.5.
that
By
matrices.
Corollary
Consider
with the right-hand
5.4,
1) or less.
all principal
the principal
constants
= 9.
by the induction
all
of order n.
of M
If Z? is nondegenerate
hypothesis
for
are also Q-
in (ZJJ~,. . . , w,; z2,. . . , z,)
lem, then it has an odd number of complementary
9 = i,
true
We now show that
submatrices
subproblem
is
in the subprob-
feasible solutions
when
8.4.
Let
where
q1 > 0, be nondegenerate
with respect to M.
q1 > 0 and M 3 0, the variable zi must be equal to zero in
any solution to (1) when q = Q. Thus, if (G; 2) is a solution to (1) when
q = q, z”, = 0, and hence (Ga,. . ., G’,; T,, . ., 2,) is a complementary
Since
feasible
solution
Also,
solution
if
to the subproblem
(ws*,
when 9 = i.
is any
. , w,* ; z2*, . . . , z,*)
to the subproblem
when 9 = 9,
complementary
feasible
define
n
wi* =
21 +
2 WZljZj* > 0;
j=z
and then
solution
(z~i*, ~a*,.
. ., VJ,*; 0, zg*,. . ., z,*) is a complementary
problem when q = q.
Thus
every
complementary
leads to a complementary
versa.
feasible
to the original
feasible
feasible
solution
solution
of the original
problem
to the subproblem
and vice
Hence both problems must have the same number of complementary
feasible solutions.
By a similar
odd number
degenerate
Hence, when q = 4, (1) has an odd number of solutions.
argument
we conclude
of complementary
with respect
that
feasible
the original
solutions
problem
whenever
to M and at least one component
has an
q is non-
in the vector
q is positive.
It only remains
9 < 0.
to be shown that
the same result
holds even when
ON
THE
COMPLEMENTARITY
8.6.
We now show that,
the almost
to +
complementary
on every
2.5,
K(q)
unbounded
set C,(q),
co, while za,. . , z, remain
From
103
PROBLEM
both
is the set of all (ul; z) satisfying
i=l,...,n,
ze,> 0.
z b 0,
M is a Q-matrix,
Consider
wi and zi tend
finite.
ZUi= Mi.Z + qi,
Since
edge of K(q) lying in
the variables
(26)
m,, > 0 for all i = 1,. . . , n by Theorem
any unbounded
edge of K(q).
If all the
variables
remain
finite on this edge, then by (26) all the variables
remain
finite,
and hence
the edge cannot
wi,. . . , w, also
be an unbounded
edge.
on every unbounded
edge in K(q) at least one of the variables
must tend to +
If zi tends to +
the facts
that
tend to +
00.
all remain
Then,
complementary
bounded
on that
edge and consequently
Thus
on every
8.7.
Suppose
to +
M on that
with
the
respect
exists an c(~ > 0 such that for all tc > c~athe point q with respect
to M.
Hence
(4:
lies in the interior
of complementary
of unbounded
F={(w;z):
00 on that
edge.
variable
ze~i+ +
to M.
Then
cones.
there
aI., is nondegenerate
(27)
We now show that
Let
(w;z)=(wl+ew2;zl+ez2),
(wil + 1920,~)(z~i+ 0Zi2) = 0,
e>oo)
Then
for all
i # 1,
for all
0 > 0,
Hence
(w,; z) = (0, wsi + &eQ,. . .) z4?Jnl
+ ow,2; Zll + ez,2,. . .) z,l + 82,2)
is a complementary
feasible
co.
cones in which this half-line lies is precisely
edges in C,(g).
edge in C,(q).
and zvi2 > 0 by 8.6.
, z, should
cr.>%)
of a set of complementary
the number
also
half-line
q=Q--cxI.l,
the number
be an unbounded
the entire
za,.
zi must tend to +
edge in C,(q)
q is nondegenerate
wi must
edge of K(q) lies
the variables
Hence
UJ~ also tends
unbounded
and fixed,
if any unbounded
set C,(g),
edge.
Thus
zi,. . . , z,
co on this edge, then from (26) and
M > 0, nzii > 0, and qi is finite
00 along this edge.
in the almost
5.2.
zi,. . . , z,
solution
for
104
KATTA
4 = q-
(zeil + 0z8i2)I.,
and, since wi2 > 0, as 19 varies
(4: 4 = q is eventually
from 0 to
the same half-line
has pos{I.,}
complementary
8 > O}
as in (27).
bI.,
cannot
as a generator.
lie in any complementary
For,
set of columns
{I?.,,.
y”-
= liI.1
El.,
13> 0
cx)
(Wil + l%Q)I.,,
Also, for any E > ~a, 4 which
for all
G. MURTY
if it does, there
cone
a sub-
such that
. ., B.,)
+ i
exists
&B.,
i=2
for some ii,.
. . , A, > 0. Then 4”-
the subcomplementary
that
4”Hence,
pos{A.i,
is nondegenerate
al.,
if the half-line
A.2,.
and A., =
with respect
A.,,
. . , A.,
Now we can express
M.i.
for some /P, B2 3 0.
lies in the subspace through
contradicting
the assumption
to M for all K > CQ,.
in (27) lies in some complementary
. , A .n}, then
-
(E + iii)I.,
set {I?.s, . . . , B.,},
must
be linearly
this half-line
cone, say
independent
as
Thus
q = cd.1 +
,$BilL4.i +
5 Pi2A.i
(@.- MO)
i=l
P3)
for any tl. > c(~.
Suppose
variable
(w; zT) is obtained
associated
and all the other variables
implies
wil = CQ, ml2 = 1, and the
A.i equal to Dir, i = 1,. . . , PZ
in (w; z) equal to zero, for r = 1, 2.
Then (27)
that
is an unbounded
Thus
every
edge in C,(q).
unbounded
cone in the interior
Hence
by setting
with the column vector
the number
of complementary
large number.
edge in C,(g) gives rise to a complementary
of which
the half-line
of unbounded
feasible solutions
in (27) lies and vice versa:
edges in C,(f)
for 4”-
is equal to the number
crl.,, where t( is a sufficiently
ON THE
COMPLEMENTARITY
8.8.
105
PROBLEM
Thus, for any pi such that i = (qi, q2,. . . , g”JT is nondegenerate
with respect to M, the number of unbounded
edges in C,(d) is a constant.
This number is equal to the number of complementary
half-line
(27) eventually
8.9.
cones in which the
lies as tc,, is made large.
By the nondegeneracy
of 4”we know that there exists a &, such
that, for all P > PO, (P, 4”s,. . , 4”J T is nondegenerate with respect to M.
Hence we can always pick a qi* > 0 such that q* = (qi*, &,, . . , Q,JT is
nondegenerate
plementary
with respect
Since qi* > 0, the number
to M.
feasible solutions
when q = q* is an odd number.
by 6.3(iii) the number of unbounded
edges in C,(q*) is an odd number, and
hence by 8.8 the number of unbounded
By 6.3(iii)
the number
is therefore
Hence,
original
of comTherefore
edges in C,(q) is an odd number.
of complementary
feasible
solutions
when 4 = q
an odd number.
under
problem
the induction
of order n.
hypothesis,
Theorem
8.2 holds
By 8.3 and by induction,
for the
Theorem
8.2 is
true for all 92.
8.10.
If M is a Q-matrix and if there exists a complemen-
COROLLARY.
tary set of column vectors (A.,,
that each of the remaining
of (I, -
. . , A.,}
A.,,.
., -A.,)
such
among the column vectors
Proof.
Transform
of the unit matrix
a block
principal
and Paragraphs
pivot.
i=l,...,
n,
with respect to M.
the column
vectors
for all
feasible solutions is an odd number for
then the number of comjdementary
all q which are nondegenerate
vectors
by making
A.,,.
into the column
. ., A.,
the necessary
Then Corollary
8.10 follows
principal
In the special case when n = 2, the restriction
THEOREM.
complementary
pivots
from Theorem
or
8.2
2.21 and 2.22.
be removed from the hypothesis
8.12.
. ., B.,
M) satisfies
B.j~pos{-
8.11.
which is linearly independent,
vectors B.,,
feasible
of Theorem
8.2.
This is discussed below.
If n = 2 and M is a Q-matrix
solutions
degenerate with respect to M.
that M > 0 can
is an odd number
then the number of
whenever
q is non-
106
KATTA G. MURTY
Proof.
8.13.
orthant
Case 1.
If pas{-
of Rz, Theorem
8.12
M.,,
-
M.,)
follows
is a subset
from Theorem
of the nonpositive
8.2.
8.14.
Case 2. If pos{ - I.,, - I,} C pos{ - M.i, - M.s}, the hy~50thesis that M is a Q-matrix implies that - M., and - M., are contained one
each in pos{l.i,
-
1,s)
and pos(-
1.i, I.,},
respectively.
We verify that in this case the number of complementary
is 1 or 3 for every
8.15.
that
Case 3.
exactly
q nondegenerate
with respect
Since M is a Q-matrix,
one of -
M.,
or -
M.,
(See
Fig.
7.)
feasible solutions
to M.
the only other
is contained
possibility
in the
interior
is
of
ON THE
pos(-
COMPLEMENTARITY
I.,,
(See Fig. 8.) Suppose it is -
I.,}.
-
107
PROBLEM
that M is a Q-matrix implies that either -
I.,).
M.i . Then the hypothesis
M., E pos{I.,,
pos{I.,,
feasible solutions is either 1 or 3 for all q nondegenerate
8.16.
M.,}
or -
M.2 E
In either case we verify that the number of complementary
n = 2, there exists
If
COROLLARY.
with respect to M.
a q nondegenerate
with
respect to M for which the number of cona@?ementary feasible solutions is at
most one.
8.17.
Note.
When
M > 0, and Corollary
As an example,
n >, 3, Theorem
verify
by verifying
that 4 = (1, 1, l)T is nondegenerate
-
M.,,
I.,},
with respect
complementary
q lies in each of the complementary
po~{I.~,
true if
that the union of all
cones is R3. Also, M is a nondegenerate
q = 4” there are four distinct
1.2, I.,},
necessarily
consider
It can be shown that this is a Q-matrix
the 8 complementary
8.2 is not
8.16 may not be true.
feasible
cones pos{I.,,
and pos(T.r, I.,,
-
matrix.
to M.
solutions,
When
because
I.,, 1.a}, pas{M.3}
We
M.,,
and in none
of
the others.
By Theorem
even number
a Q-matrix,
6.2, the number of complementary
for all q nondegenerate
this number
feasible solutions is an
with respect to M and, since M is
must be 3 2.
This shows that the converse
of Corollary
6.9 is not necessarily
true
unless M 2 0.
8.18.
Note.
When
n > 3, the number
of complementary
feasible
solutions can be > 2 for all q E Rn. The example in 8.17 shows this. Thus,
when n > 3, the complementary
than twice
cones can span the whole space more
around.
ACKNOWLEDGMENTS
Part of this work
is contained
in the dissertation
[12]
submitted
in partial
satisfaction of the requirements for the Ph. D. degree of the University of California,
Berkeley. Research performed while at the University of California has been partially
108
KATTA
supported
by the Office of Naval Research
National Science Foundation
Nonr-222(83).
and the
under Grant GP-8695 with the University of California.
I wish to express my gratitude
me to the problem
under Contract
G. MURTY
to Professor
and for giving so generously
David
Gale for having introduced
of time and valuable
suggestions.
REFERENCES
1 R. W. Cottle and G. B. Dantzig,
2 G. B. Dantzig, Linear
Princeton,
New Jersey,
3 G. B. Dantzig
Amsterdam,
4 D. Gale, Private
1(1968),
and A#.
and Extensions,
Princeton
103-125.
University
Press,
1963.
and R. W. Cottle,
North-Holland,
Linear Algebra
Programming
in Nonlinear
Programming
(J. Abadie,
cd.).
McGraw-Hill,
New York, 1960.
1967, pp. 55-73.
communication.
5 D. Gale, The Theory
of Lineav
Economic
Models,
6 D. Gale and H. Nikaido, Math. Ann. 159(1965), 81-93.
7 H. W. Kuhn, Proc. Nat. Acad. Sci., U.S.A.,
47(1961), 1657-1662.
8 C. E. Lemke, Management
Sci. 11(1965), 681-689.
9 C. E. Lemke, RPI Math. Report No. 76, Rensselaer Polytechnic
New York, 1967.
10 C. E. Lemke and J. T. Howson,
Jr., SIAM
J. Ap@Z. Math.
Institute,
(2) 12(1964), 413-423.
11 0. L. Mangasarian, SIAM
J. A$@. Math. 12(1964), 77%780.
12 K. G. Murty, “On the Number of Solutions to the Complementary
Programming
ORC
Problem,”
65-17, Operations
Research
Center,
of California, Berkeley, California, June 1968.
13 T. D. Parsons, “A Combinatorial Approach to Convex Quadratic
Doctoral
Dissertation,
Department
of Mathematics,
Troy,
Princeton
Quadratic
University
Programming,”
University,
May
1966.
14 H. Samelson, R. M. Thrall, and 0. Wesler, Proc. Amer. Math. Sot. 9(1958), 805-807.
15 H. E. Scarf, “Equilibrium
Points of Two Person Games and the Brouwer Fixed
of Fixed Points of a
Point Theorem”
(CF-51124), and “The Approximation
Continuous Map” (CF 216, November 1966), Cowles Foundation
Discussion
Papers.
Also SIAM
J. A#.
Math. 15(1967), 1328-1343.
16 A. W. Singleton, Proc. London Math. Sot. 16(1966), 519-536.
17 A. W. Tucker,
Received
March.
SIAM
1969
Rev.
5(1963),
305.