AN ABSTRACT OF THE THESIS OF
BERYL MANSFIELD GREEN for the
(Name)
in
DOCTOR OF PHILOSOPHY
(Degree)
MATHEMATICS, ALGEBRA presented on
(Major)
(Date)
/6 /44ci
Title: CHARACTERIZATIONS OF MATRICES FOR WHICH CERTAIN
DETERMINAN3,.,'SL EQUALITIES HOLD
Abstract Approved:
Redacted for Privacy
Dr. David H. Carlson
An
nxn
matrix with complex elements for which all princi-
pal minors are non-negative and for which the inequality (1 t):
A(avp)A(ar,f3) < A(a)A(P)
for all a, RC {1,
,n} holds is defined
to be a generalized Hadamard or GH-matrix where A(a)
is the
principal minor of A with rows and columns indexed by a.
It
is known that all positive definite matrices, M-matrices and totally
non-negative matrices are GH-matrices. (1' ) generalizes (2'):
A(ct. .43) < A(a, )A( 13)
for all a, pc {1,
,
n}
for which a
13 = (i)
and (4')
det A < H
a...
i=1
Fan proved, using a lattice theorem, that (3'), Szasz's inequality, holds for GH-matrices with all principal minors positive. It is
shown in Chapter III that (3') holds for GH-matrices. Carlson proved
(5'), Mir sky's inequality, for the same class of matrices with all
principal minors positive.
Cases of strict equality in (1') through (5') are referred to as
(1) through (5) respectively.
In Chapter II it is shown for n x n matrices with complex
elements that (1) and (2) are equivalent and a characterization of matrices for which (1) and (2) hold is given.
Define
indexed by
{1,
,
A(a) to be the minor of A with rows and columns
a,p
respectively where
a, p
are ordered subsets of
n}.
Chapter III deals with the characterization of weakly sign-
symmetric matrices, i. e. , square matrices for which
A(: ij)A(: ii) > 0 for all a {1
j, TO C
with
i,jI a,
which
are GH-matrices and which are not GH-matrices.
In Chapter IV it is shown that equalities (1) through (5) are
> 0 for all
equivalent for GH-matrices with a..
1,1
i e {1,
n}.
In addition a determin.antal inequality proved by Marcus and Minc for
positive definite Hermitian matrices is proved for GH-matrices with
conditions for equality established.
In Chapter V some similar properties of totally non-negative
matrices and some work of Koteljanskii are investigated.
Characterizations of Matrices for Which Certain
Determinantal Equalities Hold
by
Beryl Mansfield Green
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Doctor of Philosophy
June 1969
APPROVED:
Redacted for Privacy
Associate Professor of Mathematics
in charge of major
Redacted for Privacy
Acting Chairman of Department of MatlAematics
Redacted for Privacy
Dean of Graduate School
Date thesis is presented
Typed by Clover Redfern for
April 10, 1969
Beryl Mansfield Green
ACKNOWLEDGMENT
The writer wishes to thank his committee members for their
kindness and assistance. In particular, appreciation is expressed to
Professor David Carlson who, through his valuable suggestions and
astute ability to pose the correct questions at the correct time has
constantly shown the direction for this research. Last, but not
least, appreciation is expressed to La Verne who with bachelors degree summa cum laude has beautified the hours of this research.
TABLE OF CONTENTS
Chapter
Page
I. OH-MATRICES
1. 1. Historical Survey
1. 2. Preliminaries
1
1
3
II. PROPERTIES OF SQUARE MATRICES
2. 1. The Support Matrix
2. 2. A Characterization
2. 3. Determinantal Equalities
III.
CHARACTERIZATION OF GH-MATRICES
3. 1. Characterization as WSS-Matrices
3. 2. Some Properties of GH-Matrices
3. 3. Some Properties of WSS-Matrices
3. 4. WSS-Matrices Which are Not GH-Matrices
IV. DETERMINANTAL EQUALITIES FOR OH-MATRICES
4. 1. Subadditive Functions of Lattices
4. 2. Determinantal Equalities
4. 3. Further Inequalities and Equalities
4. 4. A General Inequality
V. TNN-MATRICES
5. 1. A Characterization
5. 2. A Canonical Form
5. 3. Determinantal Equalities
BIBLIOGRAPHY
7
12
16
18
18
19
24
29
32
32
33
41
49
55
55
60
67
73
CHARACTERIZATIONS OF MATRICES FOR WHICH
CERTAIN DETERMINANTAL EQUALITIES HOLD
GH-MATRICES
I.
1.1. Historical Survey
Hadamard (1893) proved that if A = (a..)
13
is a positive semi-
definite Hermitian matrix, then
det A< H
i=1
a. i .
Since that time many proofs of this inequality and related inequalities
have been published. Szasz (1917) proved that if A is a positive
definite Hermitian matrix and if Pk
is the product of all k-rowed
principal minors of A then
1
1
1
(n-1)
1
Pl>P2
>p 3
(n2-1
>P
>
n-1)
(n2)
n-1
> pn.
Mirsky (1957) proved for the same class of matrices that if we define
PO =
1
1
1
-n
-n
n-1
1
Pn(n-k+1) Pk- 1(k-
)
Pnn-k Pk
n-1
(
k
)
1 < k < n-1.
2
Gantmacher and Krein (1960) and Koteljanskii (1950) studied the
class of sign-symmetric matrices which includes as subclasses the
totally non-negative, positive definite symmetric and M-matrices
(Ostrowski, 1937 ). Gantmacher and Krein (1960) referred to the in-
equality
det A <
II
i=1
a.
.
1,1
as Hadamard's inequality and proved this for positive definite sym-
metric matrices; they referred to
det A < A(1...p)A(p+1...n)
where
A(il. . . ik)
is the determinant of the principal subma.trix
with rows and columns indexed by {i1, ...,ik}, 1 < i
< ...<
as the generalized Hadamard inequality and established this for sign
symmetric matrices with all principal minors non-negative, totally
non-negative and positive definite symmetric matrices.
Gantmacher and Krein (1960) also essentially proved that
A(avf3)A(a(--43)< A(a)A(P)
for all a, 13C {1,
,
n}
for the same
classes of matrices. Fan (1960) proved that this latter inequality
holds for M-matrices.
Taussky (1958) proposed as a research problem the unification
of several aspects of the theories of positive matrices and positive
3
definite symmetric matrices. Carlson (1967) showed that for matri-
ces with all principal minors positive, the classes of sign-symmetric
matrices and matrices for which A(av43)A(ar13)< A(a)A(13) for all
a, P C
{1,
..., n}
are equivalent and mentioned that the latter class
includes the positive semidefinite symmetric, M-matrices and totally
non-negative matrices.
Fan (1967) proved that Szasz 's inequality holds for matrices in
the latter class which have all principal minors positive. Carlson
(1968a) proved Mirsky's inequality for the same class.
It seems appropriate to call square matrices with complex elements for which A(a) > 0 and A(avi3)A(anP)< A(a)A(P) for all
a, P C{i,. .
.
,
n},
generalized Hadamard or GH-matrices.
It has been the intention of the writer to study GH-matrices and
in particular, cases of strict equality in the inequalities mentioned
above. This has led to a study, under the same hypotheses, of some
properties of square matrices in general (Chapter II) and of totally
non-negative matrices (Chapter V).
1. 2.
Preliminaries
Before continuing, some definitions and basic facts must be
stated, however only those well-known definitions and theorems found
in a standard matrix theory textbook which are needed for a particular
reason will be included.
4
Our notation will follow closely that used by Marcus and Minc
(1965b), Carlson (1967) and Fan (1967).
All matrices are assumed to be square. All matrices are assumed to have complex elements unless specifically stated otherwise.
is the determinant of the subrnatrix of A
Definition 1. 2. 1. A(a)
with rows indexed by a and columns indexed by
are ordered subsets of
is the empty set
(1),
A(a) = A(a), A(a)
{1,
a
aj
i, j
a.
where
1
a, 13
if
and
A(ai.)
where
p
A[a]
A(au{i})
is the principal submatrix of A with rows
and columns indexed by a.
Definition 1. 2. 2.
sequences w
Remark 1. 2. 3.
For
1 < m _< n, Q T11,11.
(w "
1'
wm)
1 <w <
<w
m
< n.
Cases of strict equality in the following inequalities
will be referred to as (1),
(2),
... , (9), respectively.
1. 2. 3. 1.
A(a..)P)A(a.,-,13) < A(a)A(13)
1. 2. 3. 2.
A(ciLiP)< A(a)A(13)
with
is the set of all
a
P=
.
for all
for all
a, 13
a, 13 C 11,
{1,
,
,
1
n-1
1. 2. 3. 3. (Szasz's Inequality) Pk (k-1) < Pk+1
for
1 < k < n-1
P=
where
(n-1)
A(a)
II
aeOk, n
det A < a
1. 2. 3. 4.
2, 2... an, n
l, 1
-n
n-1 -1
1. 2.3. 5.
(Mirsky's Inequality) P n
for
1. 2. 3. 6.
in-11
(
k 1
--n+
Pk-1
k
< Pnn-k Pk'
1 < k < n-
P(n-k)(n-k+1)
n
Pk-1
n-1
k- 1)
max
1. 2. 3.7.
a E Qm,
< Pk
(n-1)
A(a)
for
for
1< k< n-1.
1<m_
< n.
n
1
max
1. 2. 3. 8.
A(a)
m+1
<
( max A(a )
aEQ
m+1, n
for
1 _< m < n-1.
1. 2. 3. 9.
'a.
1,
t=1
where
,
m,n
1 < t < n,
12
.1
j=1
are the eigenvalues of A.
6
Definition 1. 2. 3. A diagonal of an n x n matrix is a sequence
a
a-(1n,, cr(n)
where
is a permutation of
o-
{1,
,
}.
The diagonal corresponding to the identity permutation is called the
main or principal diagonal. A diagonal is called non-zero if all its
elements are non-zero.
Definition 1. 2. 4. Let A = (a.
.)
be a nxn Matrix then the sup-
port of A is the n x n matrix
if
s.
a.1, 3.
.
1, 3
SA = (s.1,3.)
where
is contained in a non-zero diagonal of A.
s.
.
1,3
= a.
.
1,3
Otherwise
= O.
Definition 1. 2. 5.
Definition 1. Z. 6.
dia A = dia (al, /,
SA
, an, n).
is at most dia A if
SA
is zero or
SA = dia A.
Definition 1. 2. 7,
Let n> 1.
An
nxn
matrix A is called a
cyclic permutation of a diagonal matrix if A can be obtained from
a diagonal matrix by a cyclic permutation of length
umns of A.
n
on the col-
7
II.
2. 1.
PROPERTIES OF SQUARE MATRICES
The Support Matrix
Theorem 2. 1. 1.
det SA = det A.
Proof: This follows directly from the definition of
Remark 2. 1. 2.
SAB
SA SB
as A = (
1
0
1
1),
B=(
1
1
01
SA.
1
1
AB = (1 2)
),
shows.
Theorem 2. 1. 3. det SA det SB = det SAB.
Proof:
det SA det SB = det A det B = det AB = det SAB°
Theorem 2. 1. 4. If
S
is at most dia. A then
A
det A = det SA = a 1,1 .. an,n .
Proof: By Theorem 2. 1. 1,
Lemma 2. 1. 5. Two distinct cycles
each of length
g
{1,
,
n}
determine a cyclic permutation
{1,
,
n}
of length
Proof: Since
qi(b)
g(b).
where
p
g
LI)
For this
b
such that qis(b) = g(b).
=b
g(b)
0< n-s < n-1.
implies
det A =1,1
a.an,n
X
on a subset of
b E {1,
n}
there exists a least positive
g(b)
s < n.
The mapping
on
1 < p.< n.
there exists
qi(b)
n
implies that
Thus
1<s<n
s>1
such that
5E {1,..., n}
and
so that
.
8
tlin- 1 (b)
4)s+1 03)
X: b
length
p = n-s+1
and
for all a C {1 , .
.
. ,
b
)
is an orbit of
1 < p = n-s+1 < n.
If (2) holds, then
Theorem 2. 1. 6.
4in
S
is at most
A[a]
dia A[a]
n}.
Proof: The proof is by induction on n.
If
dia A = A = SA so the result holds. If n = 2,
al, 1a2, 2 = A(1)A(2) = A(1, 2) = al, 1a2, 2 - al, 2a2, 1
n=
implies that
al, 2 = 0 or a2, 1 = 0. Thus the result holds.
Suppose the conclusion holds for k = n - 1
and let A be an
n x n matrix. If
, n}
is a permutation on
a-
of length t where
are disjoint and
=
1,
,
summand of
id.
LIJ
1<t<n
Then)
=0
with an orbit
where
Cr = LI)o
4,
and
is a cyclic permutation of length t on
det A[p]
a1,0_(1). .. an, cr (n)
then
{1,
is a 'non-principal"
ip( i.t)
and therefore is zero by hypothesis. Thus
unless
a-
is the identity of a cyclic permuta-
tion of length n.
By Lemma 2. 1. 5 and the induction hypothesis there exists at
most one cyclic permutation
.. an,o-(n) 1 0.
a
a-
of length
n
such that
It follows from this and
= A(1... n-1)A(n) = det A = a1,1... a
sgn
an,cr(n)
thata
a l,o-(1)
...
9
n,o-(n)
for all
=0
of length
n.
A and the theorem follows by
Thus the conclusion holds for
induction.
Lemma 2. 1. 7. Consider a binary relation
X = {1, 2,
on the set
f
in which for every i E X there is an ordered pair
,n}
(i,f(i)) E f with
i
F(i).
Then there is a subset of
f
which is a
permutation of a subset of X with an orbit of length greater than
Proof: Consider the mapping
where
F1(i)
Since
{1,F(1),F2(1),...,F11-1(1)}C
ments,
Fs(1)
F(F1-1(1))
=
(1)
for all integers k, 0< k < n.
Fk(1)
Ft(1) for some
=
F0
1=
1.
X and X
n
ele-
s, t where 0< t < s < n.
Let
contains
m be the smallest positive integer such that Fm(1) = F (1) for
t with
some
0< t < m-1.
Then the mapping
t
F(1)--.
Ft+1(1)
Fm-1(1).-Fm(1)
tation of length m - t
on
Ft+1(1)
Ft(1),
m> t + 1
{Ft(1),
Ft(1)
=
Fm-1(1)} C X
and since
so that m - t> 1.
Lemma 2. 1.8. If A(a) = 0 for all a C {1,
for all
is a cyclic permu-
.. ,n}
then
5=
A[a]
a C {1, ... ,n}.
Proof: A(a) = 0 for all a C {1,
particular that aii = 0 for all i e {1,
orem 2. 1. 6, SAta = 0 for all
a C {1,
,n}
,
n}.
implies (2); and in
Therefore by The-
10
Lemma 2. 1. 9. If det A 4 0 and
then
SA[ a] --
for all a C{i,..., n}
0
is a cyclic permutation of a diagonal matrix.
SA
Proof: Since det A 4 0, A has at least one non-zero diagonal.
No non-zero diagonal of A
corresponds to a permutation which is
the product of cycles of length less than
n
since by hypothesis all
principal submatrices of order less than
n
have zero support.
That A has at most one non-zero diagonal corresponding to a
cyclic permutation of length n
follows by hypothesis and Lemma
2. 1.5.
Thus
is a cyclic permutation of a diagonal matrix.
SA
Theorem 2. 1. 10. If A(a) = 0 for all
a
{1,
,
and
n}
det A4 0
then A is a cyclic permutation of a diagonal matrix.
Proof:
is a cyclic permutation of a diagonal matrix by
SA
detA = al,cr(1 )a2,o-(2) ' a n,cr(n) where
Lemmata 2. 1. 8, 2. 1. 9. Let
is a cycle of length n, and suppose for some
CT
aj.,k 4 0
s(j) = k.
plie s
where
k
k 4 o.(j).
n-s < n-1
implies that
r(j)
0-n(j) = j
There exists
so that
k
and since
1 < s.
s < n.
0< n.-s,
j, k E
se {1,
n}
This and
Therefore
1 < p=n-s+1 < n.
such that
a.. = 0 imJ,J
-s < -1
implies
The mapping
f: j_crso).(rs+10)_...
tyn-1(j)
on
contrary to the hypothesis that A[a]
a=
crs(j),
,Crn-1(j)}
is a cycle of length p=n-s+1
11
has zero support.
Thus A =
is a cyclic permutation of a diago-
SA
nal matrix.
Theorem 2. 1. 11.
if and only if
Let A be an
nxn
for all a
dia A[ a]
SA[
(4):
A=
matri.X. Then
a.
11
Odet
0
{1,
,
n}
SA
dia A
and
.
i=1
Proof: Suppose
(4) holds.
5A[]
a
= dia. A [a]
for a C {1,..., n} and that
If A has a non-zero, non-principal diagonal it must be
cyclic in length n. It follows from Lemma 2. 1. 5 that A has at most
one such diagonal. It then follows from the hypothesis that
det A =
ri
a,
is 1,
1
that A has none.
Thus
.
1
SA = dia A.
Conversely
det A = det SA = det dia A =
11
a.
.
i=1 1,1
implies (4).
SA = dia A
implies, by definition,
det A
0.
It follows
trivially that SA[ a] = dia A[a] for all 1 x 1 principal submatrices. Suppose that for some a with cardinality k, 1 < k < n
12
that
SAL r
where
a.
a) / dia A[c].
a = {iv ...
and
ik}
are subsequences of the sequence
a, p
r-, p = 4),
diagonal
Let
A[a]
a k._.) p =
..
a.
1,(r(i1).
{1,
... , n}
n}
with
has a non-zero, non-principal
and therefore A has a non-zero,
ik,cr(ik)
non-principal diagonal a.
i
an,in
which
.. a.
11,o-ak,crak)aik+
is a cOntradiction.
1
k+1
A Characterization
2. 2.
Theorem 2. 2.1. If for all a C {1, ... ,n}, n> 2
dia Ara),
and
P = {itc+r
T
S
A[a]
is at most
then A = PTP-1 where P is a permutation matrix
is a lower triangular matrix.
Proof, The proof is by induction on n.
sult is trivially true. For n = 2,
For n = 1
the hypothesis and Theorem
2.1.4 give that a1,1a2,2 = det A = a1,1a2,2 - a1,2a2,1.
Therefore
P = I and the
a1,2 = 0 we may take
P= 01
=
0
we
may
take
a2,1
(1 0) and the result
a12 = 0 or a21 =
result holds. If
the re-
0.
If
holds.
k = n-i
Suppose that the conclusion holds for
an
urnn
n x n matrix.
3o
of
and let
A be
By Lemma 2. 1. 7 and hypothesis there is a col-
A which has at most one non-zero element and this
element must lie on the diagonal. Thus a.1,3 = 0 for this j0
.
fixed and all
i
such that i
j0,
1
n.
Let
P1
be the
13
permutation matrix obtained from the identity matrix by interchanging
the
joth
columns, then PiAPi =
and nth
1
has the form
0
kk
X
13-11AP1
a,
j0
.
0
where Akk is a kxk matrix. By the induction hypothesis there
exists a permutation matrix P2 and a lower triangular matrix T,
such that
or
Akk = P2T 1P-21
P=
T1 = P-1Akk P2 .
Let
0
P2
P1
then
P-1AP
=
(P1
0)
2
-1) A (P (P2 O)
1
1
0
-1
(P2
0
Akk
o)
X
1
T1
\x 2
0
1
1
1
(Pa 0)
01
a.
o'
=T
a.
30'30 I
.
is lower triangular thus A = PTP-1.
P -1
AkkP2
2
X2
a.
.
30'3 0
14
is an n x n triangular matrix then
Lemma 2. 2. 2. If A
A(ak.43)A(ani3)
for all a, PC {1,
A(a)A(13)
Proof: Since A is triangular,
A(y) =
a
Therefore if
n}.
NC {1,
ai
a.1 1,11
i
=1,
is triangular for
A[ y]
},
,
,
k
p=
Then
,
,i
,1
= (a. .... a.1
... a.
.
1
1
,
ikDA({i
... a.
.
1
,
1kk pp
A(avP)A(ar-NP) = A({i
= (a.
let
{1,
a.,For
2' 2
,1
PP
.
.
m,1m
1
.
... a.
... a.1
)(a.
.
,1
m/
)(a.
l
mm pp
1
,
91
.
,1
1
mm
,1
mm
kk
1,ip,lp
= A(a)A(13).
Lemma 2. 2. 3. Let
matrices,
B
and
,
PAP-1
P
performing the permutation
trix.
A, B, P are n x n
where
is the permutation matrix obtained by
on the columns of the identity ma-
Q
Then for all a C {1,
,
n}, B(a) = A(Q(a)).
Proof; P a permutation matrix implies P-1
A = (a. .), B = (b. .)
1,3
1,3
b.
.
1,3
and
P = (P. .)
1,3
t=3.
Pt = (P.3,1.)
then
t t,k )P.
P. a
=
k
=
.
Pt.
and
Let
15
For each
P.
3'
there is one and only one index Q(j) such that
j
P.
and
1
for
=0
1,
Li)
k
Therefore
Q(j).
Pi,tat,Q(j)
t-=1
For each
there is one and only one index Q(i)
i
P.=
1
Q(i)
b.
1,t
Let
= aQ(i),Q0).
i
1
1
m
1
i
awii),Q0.1
T
then
Q(i1),Qam)
= A(Q(a)).
aQ(im),Q(.
Q(im),Q(im)
rn rn
Theorem 2. 2.4. If A = PTP1
trix and
Therefore
so
1,im
11
b.
Q(i).
m) C {1,
,Q(im)}
Q(a) = {Q(i ),
B(a) =
=0 for t
P.
and
such that
where
P
is a permutation ma-
is a triangular matrix then (2) holds.
Proof: Let Q be the permutation of the columns of the identity matrix determined by P.
Then by Lemmata 2. 2. 2 and 2. 2. 3,
A(a)A(3) = T(Q(a))T(Q(P)) = T(Q(a)vQ(13))for all a,PC
..,n},
P = (13-
Q (a) v Q(P) = {Q(i)lie a} L.) {Q(i)liE (3} = {Q(i0 I ie a v 131= Q(aL)P).
Therefore A( )A(P) = T(Q(ct)vQ(P)) = T(C2(a k_ip) =
Theorem 2. 2. 5. The following are equivalent for an n x n matrix:
(i) (2) holds.
16
is at most dia A[al for all a. c
SA[a.
A = PTP- 1
where
n}.
,
is permutation matrix and
P
T
is a
triangular matrix.
Proof:
(i) implies (ii) by Theorem 2. 1. 6.
) implies (iii) by Theorem 2. 2. 1.
(iii) implies (i) by Theorem 2. 2. 4.
Corollary 2. 2.6. If a.
>0
for all i e [1,... , n},
SA
is at most dia A,
and (2) then
det A> 0.
Proof: (2) implies
by Theorem
Thus by Theorem 2. 1. 4, detA = a1,1... an,n > 0.
2. 1. 6.
2. 3. Determinantal Equalities
Theorem 2. 3. 1. Equalities (1) and (2) are equivalent.
Proof: Since
a, RC {1,
If
a
(--N p
A(41) = 1,
and
}
si
(1) implies (2).
a
r-N
13 = cl) ,
then
a.
and
Conversely, if
then (2) implies (1) in this case.
13
can be indexed so that
i
where 0< s < r.
, ir}, p = : {ir-s
{il'
t}
(2) holds, Theorems Z. 2. 5, 2. 1. 1 and Z. 1.4 imply that
a.=
A(y) = det SA[yi = det (dia A[y]) for all
y c {1,..., n}.
Since
Thus
17
A (aLip)A
= (a.1,1
= (a.
rmi3)
... a.
..
= A(a)A(13).
.
1,ir 1
,1
)(a.
t
1
r-s ,i r-s
a.Ca.
.i
1
1
,i
... a.
.
r 0. r
... a.
Remark 2. 3. 2. By Theorem 2. 1. 4.
(1), (2),
)
)(a.
,it
r-s
-s
... a.
)]
r r
(1) or (2) implies (4). However
,
(3) or (5) are not necessary for (4) as consideration of
A=
(1
1
1
1
1
0
1
0-1
shows.
Remark 2. 3. 3. Equalities (3) and (5) are not necessary for (1) or
(2).
()
For consider
100
A= 01 0
0
.
0 -1
Remark 2. 3. 4. (1), (2) and (5) are not necessary for (3). For con-
sider
A=
(11
0
2i
0
i
1
-2 -i
Remark 2. 3. 5. (1) and (2) are not necessary for (5), for consider
18
III.
3. 1.
CHARACTERIZATION OF GH-MATRICES
Characterization as WSS-Matrices
Definition 3.1.1. An n x n matrix is called a weakly sign
symmetric matrix or WSS-matrix if A( a i,) A( a j) > 0 for all
aj
ai
, n}
where i, j a. These are equivalent to the signfib {J}C {1,
symmetric matrices referred to previously.
Definition 3. 1. 2. An
n x n matrix is called a generalized Hada-
mard matrix or a GH-matrix if A(a) > 0 and
A(aA.../ 13)A(ar--)13) < A(a)A(13)
for all a,
13 C
,
n}.
We shall need Sylvester's identity (Gantmacher, 1960) restated
as:
Theorem 3. 1. 3. (Sylvester's Identity):
with complex elements. For
where
of
a
d
be
= A( a i.)
p
for
a
{1,
i, j e {1,
and let q = n-p.
A
is an n x n matrix
... , n}
define
D :---
(d.
.),
... , n} - a. Let the cardinality
Then
det D = (A(a))1cl
nxn
matrix with complex ele-
det A.
Carlson (1967) proved:
Theorem 3. 1.4.
Let A be an
ments and with A(a) > 0 for all
matrix if and only if
Theorem 3. 1. 5.
A
a C {1,
,
n}.
A
is a GH-
is a WSS-matrix.
Let A be an
nxn
matrix with complex
19
elements and with A(a) > 0 for all a C {1,
u}.
A is a GH-
matrix if and only if (1) A is a WSS-matrix and (2) A(a) = 0 and
implies A(13) = 0 for all a, p c ti,
aC
Proof: Suppose
a, PC {1,
a.
j
,
}.
A(a,L)13)A(a(-\13) < A(a)A(13)
Let i, j e {1,
Consider Ala, i, jj.
,
and choose
n}
Define
for all
dPq
= A( a P)
aq
so that
a
for p,q
fi, j
By Theorem 3.1. 3 and hypothesis,
- d..d.j i =
d..d.j
j
Thus
A(ai;)A(a
aj a
j}) = A(a)A(aij)< A(ai)A(aj)= d..d...
11 33
d..d. > 0.
If
1J 31
A(a) =0 and a c p then
< A(P) = A(ak...)03-allA(arm(P-a))< A(a)A(13-0,
Therefore
A(13) = 0.
Conversely, let
a, 13 C {1,
0 = A(aL)13)A(arf3)< A(a)A(13).
all principal minors of
orem 3.1. 4
3. 2.
If
., n}.
A(aA..)13)
If
A(av13) = 0 then
then by hypothesis
0
are positive. Therefore by The-
A[a,L13]
A(a,LiP)A(ar--)13)< A(a)A(13).
Some Properties of GH-Matrices
Definition 3. 2.1. A functional
case
=
f
f(a v (3) + f(a A13) < f (a) + f(f3)
is subadditive on a lattice
for all
Fan (1967) proved the following:
a, f3
E
L.
L
in
20
Theorem 3. 2. Z.
ment
(1),
Let
and let
be a distributive lattice with the least ele-
L
be a subadditive functional defined on
f
L
such that f(0) > 0.
Let x1' x2' . , x be p(> 2) elements of
such that x.1 A x. = CI) for 1 < i < j < p. If
L
.3
(a)
1
1<k<p
(b)
)
12
,<i
i<
for
, x.
f(
and
T=
for
1<k<p
for
1<k<
,
then
2T >
+
k- 1
Tk+1
-1
and
S
k
p- 1
(
> Sk+1
-1
k-1 )
(
k
for
1 < k < p- 1.
)
Remark 3. 2. 3. As an application of Theorem 3. 2. 2, Fan (1967) de-
fines a functional f on the lattice of subsets of
GH-matrix with A(a) > 0 for all
f(a) = log A(a).
Then
f
a C {I,
,
n}
{1,
,
for a
such that
is sulDadditive and the theorem gives (d)
from which Szasz's inequality follows.
Remark 3. 2. 4. If A is a OH-matrix then A + E II need not be
21
generalized Hadamard for a real number
For consider the GH-
E.
matrix
A=
1
1
1
(1
1
0)
1
.
1
A + eI = (1+E
1
1
1
1
1+E
0
1+E
1
fails to be weakly sign-symmetric for all E > 0 since
2
1
3
A(1 3)A( 1 2) = -1)(e) < 0.
1
Theorem 3. 2. 5. If A is a GH-matrix then Szasz's inequality holds
for
A.
Proof: Let
Pk+1 =
0
k
be an arbitrary integer,
1 < k < n-1.
If
then since A is generalized Hadamard,
1
1
n-1
(k-1) > P (n-1) = 0.
Pk
k+1
If
Pk+1 > 0 then A generalized Hadamard implies that all
principal minors of A with order equal to or less than k + 1 are
greater than zero. Let a E Qk+1,n and let
22
R
ak
=
A().
H
pEQk,k4.1
PCa
Since
Ala]
is generalized Hadamard and A(a) > 0,
equality holds for A[a]
Szasz's in-.
by Remark 3.2.3, Therefore
1
1
R
(k-1) = R r(> A(a).
ak
a.k
Since each principal minor of A with order
duct
R
ak
minor A(a)
k
as a factor exactly once for each
appears in the proordered principal
k+1
which contains it, and since there are
n-k of these,
multiplying these together gives
i\
n-k
(k-1)
ak
a.E Qk+1,n
>
rt
A(a)
P
iaEQk+1,n
Thus
(n-1 )
n-k
n-1
k
(k-1 )
Pk
= Pk
>
.
The following theorem (Aitken 1956) will be needed.
Theorem 3. 2. 6.
(Jacobi's Theorem): If
A
is an n x n matrix
23
then for arbitrary
B = adj A,
and
i
1<
<... < i
< n,
k <k <... < k
1
i
1
kv
v
p
1
v=1
v=1
i
1, 2, ...
p = 1,
kt
ii
ii
n-p
(det A)P- 1
A
...kp
where
1k'
n- 11
1
1
,n
i
0. , . 0n-p form a complete set of indices
and
p
1
as do kk and
1
define
-1
(det
k' ...k'n-p .
I
1
If
det A
0
and
1.
Theorem 3. 2. 7, If A is a GH-matrix then
B = adj A is gener-
alized Hadamard.
Proof: We will show that B has all principal minors non-
negative, is weakly sign-symmetric and if B(a) = 0 and
aCPC {1, .
,
n}
then
B((3)
O.
That the principal minors of
B
are non-negative follows from
their relationship given in Theorem 3. 2. 6Ao the non-negative principal minors of A.
To show
ality let a = {1,
B weakly sign-symmetric, without loss of generp-2}, i = p-1, j = p.
By Theorem 3. 2. 6,
24
B(
a3
If
B
B(
p-1, p+1,
P+1'
A(
p-1,p+1,...,n
a 1 - A P'
p, p+1,
n ,n)(det A)2(13-2)> 0.
det A =0 then by Theorem 3. 2.6 all principal minors of
with order greater than one are zero. If det A> 0 all principal
minors of A are greater than zero since A is generalized Hadamard, thus by Theorem 3. 2. 6 all principal minors of
B
are greater
than zero.
Therefore if B( a) = 0 for some a C {1,
aC RC {1,
1,
then
,
n}
and
B(13) = 0.
Remark 3. 2, 8. Consider the GH-matrix
A=
Let
(c. .)
and
c
4,1
be the second compound of
= A(
3
1
4
2) = -1
so that (c)
1,3
1
A( 2) = 1
c14
, =
34
is not a GH-matrix. There-
Then
fore, in general, the compounds of a GH-matrix are not generalized
Hadamard,
3. 3.
Some Properties of WSS-Matrices
Theorem 3. 3. 1.
Let A be an n x n WSS -matrixwith all prin-
cipal minors non-negative. If for some
while
A(1... p+1) = 0,
1 < p < n-2, A(1...p) > 0
p,
then A(1... p+2)
0
and at least one of
25
A(
1... p p+2) ,
1... p p+1
A(
Proof: Let
p p+1
1,
is zero.
1.. p p+2
d.
1,3
1
=
,
p
1
p+1, p+2
1,j
1.p
.. 3
Then by
.
Sylvester's identity,
d
det D = d
= -d
p+1, p+2dp+2, p+1
d
p+1
d
p+1
= A(1 ...
p+1, p+2
p+2, p+2
p)A(1... p+2) > 0 .
Since by hypothesis
d
d
p+1, p+2p+2, p+1
=A
1... p p+1)A(1... p
1... p p+2
it follows that equality holds so that
d
p+2, p+1
d
p+2
1... p p+1 ) > 0
p+1, p+2
= 0
or
=0.
Also since A(1... p) > 0
then
A(1... p)A(1... p+2) = 0
implies that A(1... p+2) = 0.
Remark 3. 3. 2. Given a set of
vectors
V,
W,
n
vectors
V,
any set of
n+1
each of which is a linear combination of the vectors
is linearly dependent.
Lemma 3. 3. 3.
Let A be an
n x n WSS-matrix with all principal
minors non-negative. If A(a) > 0 for all a E Qn2,n
for all
a E Qn_i,
then the rank of A is n-2.
and
A(a) = 0
26
Proof: Since A(1.
A(1. .. n) = 0, Theorem
1... n- 2 n
A(
1... n-2 n-1) = 0 or
and
n- 2) > 0
3. 3. 1 gives det A = 0 and either
1... n 2 n- ) 0. Suppose A( I.. . n-2 n-1) = 0.
A(
1... n-2 n
1... n- 2 n
Let X be the,
1-
rows of
submatrix of A consisting of the first n-1
since A(1... n
2n- 1
.
and
=0
1-1)
A( 1..
n-2
m of X
0, any colun
n)
is a linear combination of the first n-2 columns of
(n-1) x (n-1) minors of
Remark 3. 3. 2, all
Let
X
Thus by
X.
are zero,
be the submatrix of A consisting of the last n-1
Y
columns of A.
column of
Then
A.
Yt
1
n-1
A(2.. .n) = 0 and A( 2...n ) 0, any
is a linear combination of the middle n-2 columns.
Since
Thus by Remark 3. 3. 2 all
(n-1) x (n-I) minors of
Y1
are zero.
be the submatrix of A consisting of rows 1,3,...,n.
1 3... n
A(1 3... n) = 0 and A( 2 3... n) = 0 since it is a minor of Y1.
1
Thus by Remark 3. 3. 2, all (n-1) x (n-1) minors of Z are zero.
Let
Z
Let Yn be the submatrix of A consisting of the first n-1
1 3. ..
lies in Z it is zero, and
columns of A. Since A( 2... n-1)
n1
Thus, by Remark 3.3. 2, all
A(1 2...n-1) = 0.
ors of
Yn
(n-1) x (n-1) min-
are zero.
Finally let
by omitting the
Yk
kth
column of A for
A(1... k- 1 k+1... n) = 0 and
Remark 3. 3. 2 all
Since any
columns of A obtained
consist of the n-1
A(
.
1 < k < n.
n-1
1..k-1
.
k+1..
(n-1) x (n-1) minors of
Yk
Then
) = 0.
Thus by
are zero.
(n-1) x (n-1) minor of A lies in Yk for some
27
k,
1 < k< n,
(n-1) x (n-1) minors of A
all
are zero. Thus A
has rank n-2.
Let A
Theorem 3. 3. 4.
be an
If for some
cipal minors non-negative.
for all a
of A
E
Q
is
for all a
A(a) = 0
p = n-1
E
Q
A(a) > 0
then the rank
p+1,n ,
the result is obvious. Therefore assume
Let
B
a+1
=A11
Pl.
where
1< p < n,
p,
p.
Proof: If
p < n-1.
and
p,n
WSS-matrix with all prin-
nxn
1 < a <... < a p+1
< n,
trary (p+1) x (p+1) minor of
1 < (3
A.
Pp+1
<.... < (3 p+1 < n be an arbi-
By Lemma 3. 3. 3 the principal
submatrices
=
for
p+1}
i E {1
rank p.
with
AB.
pi
a1...ap+1 pi
a1.. a p+1 p.
aj
for all
j
E
{1,
Therefore all (p+1) x (p+1) minors of
Each column of
B
is a column of a (p+1) x (p+1)
,
p+1}
B.
have
are zero.
minor of some
being a column of A(a.1... ap+1) in case Pi = a3. for some
Thus in either case the columns of B are linear combinations of
B.,
the columns of A(a1...ap+1)
so that by Remark 3. 3. 2,
B = 0.
28
Thus all
rank
(p+1) x (p+1)
minors of A are zero so that A has
p.
Lemma 3. 3. 5. Suppose A
is a 4 x 4 WSS-matrix with A(a) > 0
for all a C {1, 2, 3, 4} and a.
2x2
some
.
for all i E {1, 2, 3, 4}.
>0
principal minor is zero, then det A = 0.
Proof: Without loss of generality suppose A(1 2) = 0,
by Theorem 3. 3. 1,
det A = 0
or all
A(1 2 3) = 0
2 x 2 principal minors of
A(2 3) = 0
so again by Theorem 3. 3. 1,
Thus either det A = 0 or all
A(2 3 4) = 0.
A(1 2 3)
2x2
A(1 24)
A(1 3) = 0
2x2
principal
Suppose they are
principal minors of A are zero so that A
has rank one by Theorem 3. 3. 4. Thus
Theorem 3. 3. 6. If A
and
A(1 3 4) = 0 and
minors of A(1 3 4) and A(2 3 4) are zero.
zero. Then all
then
Thus either
A(1 2 4) = 0.
and
are zero. Suppose they are zero. Then in particular
and
If
det A = 0.
is an n x n, 1 < n < 4,
all principal minors non-negative and
WSS-matrix with
a..
> 0 for all
1,1
i E {1
n},
then A is a GH-matrix.
Proof: The result is obvious for n = 1, n = 2 by Theorem
3. 1. 5.
Let
n = 3.
Theorem 3. 3. 1,
If some
2x2
principal minor is zero then by
det A = 0.
Thus by Theorem 3. 1. 5,
A
is a GH-matrix. If no 2 x 2
29
principal minors of A are zero then by Theorem 3. 1. 5
A
is a
GH-matrix.
Let n = 4.
If
A has a zero 2 x 2 principal minor, then
det A
by Lemma 3. 3. 5,
0 and since all 3 x 3 principal sub-
matrices are generalized Hadamard,
orem 3. 1. 5. Suppose A(a) > 0
zero
A
is a GH-matrix by The-
for all a e Q2,4.
If
3 x 3 principal minor then by Theorem 3. 3. 1,
A has a
det A = 0.
Therefore whether A has zero 3 x 3 principal minors or not
A
is a GH-matrix by Theorem 3. 1. 5.
3. 4.
WSS-Matrices Which are Not GH-Matrices
Remark 3. 4. 1.
/1
1
1
1
1
1
1 21 1
1
1
2
2
1
\1
1
2
2
2
1
2
2
\2
(1
1
1
1
0001
-1
0
0
and
010
001
100
0
are examples of such matrices. Therefore it follows from this and
Theorem 3. 3. 6 that for all such matrices with
i E {1
n},
1,1
for all
n > 5.
Theorem 3. 4. 2.
A(a) > 0
a. . > 0
Let A by an
for all a C {1,
.
,
n x n matrix such that
det A> 0,
n}, A is weakly sign-symmetric and
30
and all principal submatrices of A with order less than
n
are
generalized Hadamard. Then the following are equivalent:
is not generalized Hadamard.
A
There exists
{1,
a, P C
.
.
. ,
where
n}
a
c
p,
A(a) = 0,
A(13) > 0.
B = adj A has a principal submatrix
B
[qJ]
which is a
cyclic permutation of a diagonal matrix.
Proof: (a) and (b) are equivalent by the contrapositive to Theorem 3. 1. 5.
By the hypothesis that all principal submatrices
Suppose (b).
are generalized Hadamard
a
c
p,
A(a) = 0
show that all
n-11.
0,
is less than n.
and the order of A[a]
S = tY1 Y
C
{1,
...
y
By hypothesis that all (n-1)x (n-1) principal submatrices
there is some
Theorem 3. 1. 5 implies
such that A(yo) = 0.
yo E S
Then it is clear that
Y C Yo.
yo
(Th y
Since
contains
Let
y
yo
be an
yo, and jo=
arbitrary member of S, io = {1, ... ,n}
Yo
We will
,n} where the cardinality of
are generalized Hadamard and A( a) = 0,
and
i.e. , A(13) = det A > 0.
, n1,
(n.. 1)(n-1) and (n-Z)x(n-Z) principal minors of A
are zero. Let
is
p
n-2 elements of
y.
{1,
det A> 0 all (n-2) x (n-2) principal
minors of A[y0] are zero by Theorem 3. 3. 1. Therefore
A(yorThy) = 0
and since
Yo (Thv C y
and
A[y]
is generalized
,
n}
31
Hadarnard,
A(y) = 0.
Thus, all (n-1)x (n-1) principal minors of
A are zero. It follows, from this and the fact that det A > 0 that
all (n-2)x (n-2) principal minors are zero, by Theorem 3. 3. 1.
Thus, by Theorem 3. 2. 6, all diagonal elements and all
principal minors of
Let
k
Adj A
B
are zero.
be the minimum value,
such that there
1 < k < n,
of order
exists some principal submatrix
B[p]
of
such that all principal minors of
B[ii]
of order less than
zero, and
B(Lp) > 0.
2x2
By Lemma 2. 1. 10,
Adj A
B
Bkid
k
are
k
is a cyclic permu-
tation of a diagonal matrix.
Conversely, suppose (c). Since
B[Li]
and therefore
Adj B
has a zero diagonal element, Theorem 3. 2. 6 implies that A has an
(n-1) x (n-1) zero principal minor. Since det A> 0,
Since by Theorem 3. 1. 5,
A
(b) holds.
is generalized Hadamard if and
only if all principal submatrices of A are generalized Hadamard,
the following result holds.
Theorem 3. 4. 3.
a C {1,
n}.
Let
A
A
be a WSS-matrix with A(a) > 0 for all
is not a GH-matrix if and only if A has a non--
singular principal submatrix satisfying (c).
32
IV. DETERMINANTAL EQUALITIES FOR GH-MATRICES
4.1. Suba.dditive Functions of Lattices
Definition 4. 1.1. Define a functional
in case
f
f(avP) + f(cLA P) = f(a) + f(P)
to be additive on a lattice
for all
a, 13 E L.
With this definition, Fan's (1967) proof of Theorem 3. 2. 2 also
proves the following:
Theorem 4.1. 2. If
is additive on L
f
f(c) = 0
and
then under
the remaining hypothesis of Theorem 3. 2. 2 strict equality holds in
(c) and (d) in Theorem 3. 2. 2.
As shown by Carlson (1968a), Mirsky's inequality follows by
Theorem 3. 2. 2 for GH-matrices with all principal minors positive.
The same argument employing Theorem 4.1. 2 gives:
Theorem 4. 1. 3. If A is a GH-matrix for which all principal
minors are positive, then (1) implies (5).
Theorem 4. 1.4. Suppose
element
cl)
L
is a distributive lattice with a least
in which every interval is a complemented sublattice.
Let f be a functional on L with f(c) = 0.
f
is additive on
L
if and only if f(xvy) = f(x) + f(y) for all x, y E L with x Ay = 11).
Proof: Suppose
f
is additive, then since
f(xv y) + f(xAy) = f(xVy) = f(x) + f(y)
f(40) = 0,
for all x, y E L with x A y =
33
Conversely, let x, yE L and first suppose x, y are com-
then xvy=y
parable. Without loss of generality suppose x < y,
and
Therefore
x A y = x.
Suppose
x A y = 4)
f(x) + f(y) = f(xAy) + f(xvy).
then since f(4)) = 0, f(xvy) + f(xAy) = f(xvy)
= f(x) + f(y).
Suppose
ZEL
are incomparable and
x, y
such that
(xAy) v z = x
and
xAy
There is a
ci).
since by hypotli-
(xAy) Az = cl),
sis the interval
[11,x]
is a complemented sublattice. Since
then
z<x
so that x Az = z.
(xAy) v z = x
Z A Y = (xA z)Ay = (xAy)A z = (I).
then
y v z < yv x.
Since
Thus
y < y Vx
Since
and z<x< y vx
(yVz) A x = (yA x) V (z Ax)
(yAx) v z = x
thenyv z>x and since yvz>y then y vz>yv x. Thus
yvz=xvy.
By hypothesis and the fact that
f(y) + f(z) = f(yAz) = f(xvy).
zAy
Also since
ci),
it follows that
(x Ay) A z = cl)
,
it follows
that f(xAy) + f(z) = f((xAy)vz) = f(x).
Combining the last two results gives
i. e. ,
4. 2.
f
f(x) + f(y)=f(xAy)+f(xvy),
is additive.
Determinantal Equalities
The result stated in Theorem 4. 2. 3 follows from Theorem 2.3.1.
However the different proof below in the lattice setting is given for
completeness.
34
Lemma 4. 2.1. Let A be a GH-matrix.
y=
If
> 0, 1 < i <
a.
I
1,1
a
then
y
and
a, 13 C {1,
n}.
Suppose
0 = A(aL)P)A(orR) = Mo)A(R) = 0.
if and only if
B(a)B().
B(ak..)13)B(anP)
Proof: Let
a, 13
p-
subscript i E a
or
Suppose
B = A[y].
a C,13 < y . A(ak..)13)A(ar13) = A(a)A(P)
{i}a
Let
y.
and
If
n}.
{1,
Then
a.
.
12 1
p(ty there exists a
a L.)
= 0;
a L)p
i
implies that
Therefore by Theorem 3.1. 5 either A( a) = 0
{i} C P
or A() = 0,
C
Thus
A(aL.)13) = 0.
0 = A(avr3)A(anP)
= A(e.)A(13) = 0.
If
a
PC y
B(t.i) = A(1P)
for all
then a C y, PC y,
LP
a em
C y, A(av13)A(ar43)
RCN.
Since
B(ak-JP)B(ar43) = B(a13(3)
= A(a)A(13).
Remark 4. 2. 2.
Thus in order to characterize OH-matrices for
which A(av13)A(arq3) = Ma)A(P)
for all a,
restrict attention to matrices for which
Theorem 4. 2. 3. If A
a1 .
pc {1,,. n}
.>
0
for all
we may
n}.
iE
is a GH-matrix then (1) and (2) are equiva-
lent.
Proof: Since A(0) = 1,
(1) implies (2).
Conversely accord-
ing to Lemma 4. 2. 1 it is sufficient to assume that a
i E {1, ...,n}.
>0
Therefore by Theorem 2. 2. 5 and hypothesis,
for all
35
S
A
= dia A so by Theorem 2. 1. 4,
by Theorem 3. 1. 5,
det A =
a1,1"
for all a C {1,
A(a) > 0
the functional defined on the lattice of subsets of
f(a) = log A(a)
for
an,n > 0. Thus
Let f be
, n}.
{1,
,
n}
by
aC{1,.,n}.
Then by Theorem 4. 1. 4,
is additive, i. e.
f
f(a,../P) + f(an) = f(a) + f(p) for all
a, 13 C {1,
,
n}.
Therefore
log A(a) + log A(arP) = log A(a) + log A() implies
A(ak.43)A(arm3) = A(a)A(P)
Theorem 4. 2. 4. If A
for all a, p c
,
n}.
is a GH-matrix then (3) and (4) are equiva-
lent.
Proof: If
a.
1,1
=0
for some
then by Theorem 3. 1. 5 there
i
is a principal minor of every order k for
1 <k < n
a..
> 0 for all i E {l,
1,1
(3) and (4) hold. Thus suppose
If (3), i. e.
n-1
n-1
)
k- 1 )
Pk
for
Pk+1
1 < k < n-
then in particular,
n-1
0 )
a1,1... an,n =P1
so that (4) holds.
(n-1n_i)
= Pn
= det A
so that both
n}
.
36
det A = a1,1... an,n > 0 and Theorem 3. 1. 5 im-
Conversely,
plies that A(a,) > 0 for all
on the lattice of subsets of
Then by hypothesis
a C {1,
{1,
by the rule f(a) = log A(a).
n}
,
Define the functional
n}.
is subadditive so by Theorem 3. 3. 2,
f
n-1
k-1
,
n- 1
)
)
>
for
Pk+1
1 < k < n-
;
(4) and transitivity imply (3).
If A
Theorem 4. 2. 5.
is a GH-matrix then (1) or (2) implies (3)
and (4).
Proof: As already shown (1) and (2) are equivalent and (3) and
(4) are equivalent. If a.1,1 = 0 for some
.
E
{1,
n}
then by
A has a zero principal minor of every order
Theorem 3. 1. 5,
1 < p < n.
i
Therefore (3) and (4) hold.
Suppose
Theorem 2. 2. 5,
a.
.
1,1
> 0 for all
i
E
{1,
n}.
Then by (2) and
so that det A = a1,1... an,n,
SA = dia A
or (4)
holds and therefore (3) holds.
(1
Remark 4. 2. 6.
1
1
00
0)
0
0
shows that (3) or (4) or (6) does not
imply (1) or (2).
Lemma 4. 2. 7. Let A be a GH-matrix and suppose that for some
f
37
fixed
1 < k < n-1,
k,
1
1
n-1
n-1
(
(k-1)
= Pk+1
Pk
)
k
Then for all a e Qk+1,n'
1
k
k-1
A(a).
A(p)
=
a
PE Qk,k+1
Ca
Proof: Szasz s inequality holds for A[a]
GH-matrix for all a E Qk+1,n.
since
A[ a]
is a
Thus,
1
(k=1
R
< A(a)
ak
implies
1
(
(Pk
k
n-k
1
(k-1)
-1
aEQk+1,n
<
ak
A(a) =
aEQk+1,n
implies
n-1
n-1
(k-1)
Pk
H
(
<Pk+1
)
Pk+1
38
Since equality holds in the last statement, equality must hold in
1
(k-1) < A( )
Ra
k
for all
a E Qk+1, n.
Theorem 4. 2. 8. For an arbitrary positive integer n,
ai > 0
is an n x n GH-matrix with
suppose A
for all i E {1, ... , n}. Then
.
(3) implies (2). Therefore (1), (2), (3), and (4) are equivalent.
Proof: We will show that
a C {1,
,
Since all
2x2
n}.
S
for all
= dia A[a ]
r
A[a]
is one, the result is trivial.
If the cardinality of a
principal submatrices are generalized Hadamard,
Szasz's inequality holds.
Let
a=
then A(a)< a.
i2}
.
a.
11,11
.
.
12,12
(3) implies
a1,1*.. an,n =P 2
Thus inequality cannot occur in A(a) < a.
.
a.
so that either
.
11,11 12,12
a. .
11,12
0
or a.
.
12,11
=0
and
principal submatrices A[a]
Suppose
for all
2x2
.
SA[a] = dia A[a]
trices A[a] where
SA[a] = dia A[a]
1<k<
no
for all k x k principal subma<n
and consider any (no+1)x (n+1)
39
Let Rm be the product of all m x m,
principal submatrix A[p].
1 < m< n +1,
principal submatrices of A[p],
then since
S=
dia A[a] ,
A[ a]
By (3) and Lemma 4. Z. 7,
1
1
no
no
(n -1)
A(P) = Rn
no-1)
°
no-1
=
II
a.
iE p
so that (4) holds for A[p].
SA[] = dia A[P].
Thus
.
1,1
Therefore by Theorem 2. 1. 11
SA[a]
= dia Ala]
for all a C {1,
,
n}
so by Theorem 2. 2. 5, (2) holds.
That (1), (2), (3) and (4) are equivalent is the combination of the
results of Theorems 4. 2. 3, 4. 2. 4, 4. 2. 5 and the result above.
Remark 4. 2. 9. If det A> 0
Theorem 4. 2. 10. If A
then (5) and (6) are equivalent.
is a GH-matrix and det A = 0 then (3),
(4) and (6) are equivalent.
Proof: (3) and (4) are equivalent as shown in Theorem 4. 2. 4.
If (3) holds then A has a zero principal minor of every order
40
so that (6) holds.
1_
<k<n
Conversely, if (6) then P
0 = (det A)(n-k)(n-k+1)
for
1 < k < n-1
det A = 0
and
n-1
n-1
k- )
Pk-11
k
= Pk
)
imply that A has a zero principal minor of every
order. Therefore (3) and (4) hold.
Remark 4. 2. 11. (6) does not imply (1) or (2) as A =
However if
a.
.
1,1
>0
110 )
1
shows.
10
000
for all i c {1, ... , n} then the following result
holds:
Theorem 4. 2. 12.
iE
n}
If
is a GH-matrix with 1,i
a.> 0 for all
A
then (1), (2), (3), (4), (5), and (6) are equivalent.
Proof: By Theorem 4. 2. 8, (1),
(2),
(3) and (4) are equivalent.
If (2) then Theorems 2. 2. 5 and 2. 1. 4 imply that
det A = det SA = det (dia A) = a1,1... an,n > 0.
orem 3. 1. 5,
A(a) > 0
for all a C
by Lemma 4. 1. 3. Since
{1,
,
n}
Therefore by Theso that (5) follows
(5) implies (6).
det A > 0,
Conversely suppose (6). Then in particular
n-1
(det A
(n- 1 )n
(
0
0
n-1
)
=P
1
1
1
)
= (a11.. . an,n)
-1
>0
41
implies
4. 3.
det A = a1,1* ..
a> 0.
Therefore (5) and (4) hold.
Further Inequalities and Equalities
Marcus and Minc (1965b) proved the following:
Theorem 4. 3. 1.
If
is a positive semi-definite Hermitian ma-
A
trix then 1. 2. 3. 8 holds with equality for a given m if and only if
either A has rank less than m or
identity, where
A(con)
max
=
A[
(3]
is a multiple of the
A(w).
e Qm+1,n
Lemma 4. 3. 2. If A
is a GH-matrix then 1. 2. 3. 7 holds:
max A(a) > (det A)n
1 _<m <_ n.
ae Qm,n
Equality holds if and only if either all principal minors of
with order equal to or greater than m
are zero or
SA
A
is a
multiple of the identity matrix.
Proof: If det A = 0
det A> 0.
then 1. 2. 3. 7 is obvious. Suppose
Szasz's inequality holds for A by Theorem 3. 2. 5 so by
transitivity, for an arbitrary m,
1 < m < n,
1
1
n-1
n-1
P (m-1)
(
=
IIA(
)
clE Qm, n
m-1 )
> det A.
42
max A(a) > A(a) so that
Also for all a E Qm,n,
aEQ m,n
>11
max
(aE Q
A(o )
.
aE Qm,n
Thus
(
max A(a) m
aE Q
max A(a)
=
n-1
n-1
m-1
EQ
cie Cm,n
m,n
> det A
)m1)
A(
ra,
so that 1. 2. 3.7 holds.
Suppose for some m, 1 < m < n,
(7).
If not all A(a) are equal for
that equality holds, i. e.
a E Qm,n
vnin)
n-1
(det A)(rn
then
= \(det A)n
( max A(a)
aE Q
m,n
n-1
(
>
A(a) > (det A)
H
EQ
which is a contradiction. Therefore
m-1)
m,n
A(a) = (det A)
for all
a E Qm,n
If the common valuc is A(a) = 0 for all a 0m,n then
since A is generalized Hadamard, all principal minors of A
with order equal to or greater than m are zero.
Suppose the common value is positive, then
det A > 0.
43
order to show that SA is a multiple of the identity matrix, we will
first show (7) for all k, m < k < n. Let y E Q
where
m+1,n
max
A(y) =
A(P)
.
PE Qm+1,n
Let
and aCy.
a E Qm, n
Then by 1. 2. 3. 7 and the hypothesis
that A is generalized Hadamard,
A(a) =
max A(a) =
aE Qm,n
max
A(a) > (A(y))m+1
ac Qm,rn+1
m+1
max
=
m
A(P)+1
>
m+1
((det A)
=
(detA) n= max A(a)
acQna,n
PE Qm+1,n
Therefore
m+1
max
pE Q
A(P) = (det A
m+1,ti
m+1
so that A(13) = (det A) n
for all
Q
m+1,n
Thus (7) holds for all k, m< k< n.
Now,
13
'
A(a) = (det A) n
for all
Qm,n implies
44
(
n
n-1
)
(
Pm =
m = (det A) m--
Ma) = (det A)n
H
CLE Q
1)
m,n
Or
n-1
,
m-1 ) = det A.
Thus by Szasz's inequality and transitivity
1
n-1
k, m <- k < nPk(k-1) = det A for all
Therefore
n-1
det (adj A
det A)n-1
so that (4) holds for
adj A.
= Pn- 1
=
Since
adj A
is generalized Hadamard
with all principal minors positive, (1), (2), (3), and (5) hold by Theorem 4. 2.12. Thus by Theorem 2. 2. 5 adj A = PTP-1
where
T
is a triangular and P is a permutation matrix. Since
adj A and
dia (adj A) = dia T,
the eigenvalues of
of
n-1,
adj A
T
T
have the same eigenvalues, and
lie on its principal diagonal, the eigenvalues
lie on its principal diagonal also. Since (7) holds for
45
n-1
A(a) = (det A) n
for all a E Qn-1,n'
i. e., all principal diagonal elements of adj A
are equal. Therefore all eigenvalues of adj A are equal and nonzero.
A has no zero eigenvalues.
det A> 0,
Since
eigenvalue of adj A then there exists x
(adj A)x =
X1 x.
0
characteristic equation of
A
is
(X-
det A
)n = 0
1
.
Therefore,
det A =
n- 1
.
1
By Theorem 3. 2. 6 for an arbitrary i,
.
(det A
Thus
1 < i < n,
adj A(1,..., i-1,i+1,...,n)
n ) -(n-2)
nA`l
implies
X1
1
1,1
X1x
Thus all eigenvalues of A are equal. The
is an eigenvalue of A.
)n
is an
det A
implies
X1
A
term ,iet A - ( det
X
X1
such that
Therefore x - detAA (adj A)x - detA A
Ax - det A x
a.
If
1
n_ 1
n_1
1
=>1
with constant
46
1 )n
(
det A = X1n-1 = X n-
1
=
i=1
so that (4) holds for A.
and (5) hold for A
a..
1,i
Thus by Theorem 4. 2. 12, (1), (2), (3),
so by Theorem 2. 2. 5
SA =
dia A - det A
Xi
Theorem 4. 3. 3. Let A be a GH-matrix with det A> O.
max A(a) = (det A
aeQm,n
(7)
for some
m,
Then
n
1 < m < n-1, if and only if (1), (2), (3), (4), and (5)
hold and all principal diagonal elements of A are equal.
Proof: The sufficiency follows from Lemma 4. 3. 2. The necessity follows from (4), i. e.
H
a.
i=1
.
1,1
= ( max
1 < i< n
Al
= det A
so that (7) holds for m =
Theorem 4. 3. 4. If A is a GH-matrix then
47
1
max A(a))
>( max
EQ
E Qm,n
for all m, 1 <_ m <_ n-1.
A( a)
m+1,n
Equality holds for a given m if and only
if either all principal minors of A with order equal to or greater
than m are zero or
wcQ
m+1, n
is a multiple of the identity, where
SA[w°]
satisfies
max
A(w°
A(w)
w E Qm+1
Proof: Let
let
wo E 0m+1,n
.
,n
be an arbitrary integer,
1 <m <_ n-1 and
such that
A(w0)
=
max
A(w).
weQm+1,n
For all a such that
a E Qm,
submatrix of A[wo].
Thus by Theorem 4. 3. 3,
n
and
max A(a) >
(A(w0))n1+1
aEQm,n
Cw
so that
aE
a C wo, A[a]
max A(a) > (A(
m,n
is an
mxm
48
and therefore combining this with
A(w0)
max
=
WE Q
A(w)
m+1, n
gives
_m
1
max A(a)
MY
CLE Q
1
max
>
A(a)m+1
.
aE Qm+1,n
m,n
If equality holds in II then also
max A(a) =
A(w0)m+1
aE Qm,n
and the rest of the theorem follows by applying Theorem 4. 3. 3 to
A[w°] .
Remark 4. 3. 5. The first part of Theorem 4. 3.4 coincides with Thy
orem 4. 3. 1 for positive semi-definite Hermitian matrices. That the
last parts of the two results differ follows from consideration of
A=
A
is generalized Hadarnard and has rank 3 even though all
principal minors are zero.
49
Furthermore A
1
0
1
1)
(
is generalized Hadamard and
1
max A(a) = (det A)
a.E
1,2
but A is not a multiple of the identity.
However
SA = I.
Remark 4. 3. 6.
/1
1
1)
1
1
1
A= 111
shows that (1), (2), (3), (4), and (5) are not necessary for (7).
4. 4. A General Inequality
A proof of the following theorem, which was used by Fan (1967)
in the proof of Theorem 3. 2. 2, was given by Carlson (1968a):
Theorem 4.4. 1. For any
2tk > tk-1 + tk+1
real numbers t0,t1,...,tp' for which
the quotient
p+1
1 < k < p-1,
t-t
qk
,
q-k
does not increase when
Definition 4. 4. 2. If
q
or
To = 0,
k
q
k
increases.
Theorem 4. 4. 1 gives the following
chart which will be referred to as the Fan-Carlson chart:
50
T-T
1
O
>
T2-T 0
2
1
T
>
>
-T
0
p-1
vi
vi
--T
2
1
T
>
>
Tp
p- 1
P
vi
T -T
p-1 -T 1
P
>
p-2
1
O-T
1
p-1
vi
vi
vi
T
p-1 -T p-2
vi
Tp-T p- 2
2
1
vi
T -T
P
P-1
1
Theorem 4. 4. 3. If two consecutive terms are equal,
T.-T.
i-j
jth
in the
T.
1+1
-T.3
i-j+1
row of the Fan-Carlson chart where
0 < j < i < p then
equality holds throughout the chart to the left and below the term
T.
1+1
-T j
i-j+1
'
i. e.,
T.3+1 -T.3
1
T.3+2 -T.3
-
2
II
T.
3+2
T.
-
.
1+1
-
-T.
3
i-j+1
II
-T j+1
1
Ti+1- j+1
1-3
II
51
T.-T.3
i-j
1
Proof:
implies
T.
1+1
-T.3
implies (i-j+1)(T.-T.)=
j
i-j+1
(i-j)Ti + (Ti-T.) = (i-j)T1+1
1(i-j)(T.+1
-T j
implies
-T.3 = (i-j)T.1+1 - (i-j)T.1 - T. implies
1
T1+1.
-Tj = (i-j)T.1+1 + T.1+1
- (i-j)T.i -T. implies
1
T.
1+1
-T.3
T.
1+1
i-j+1
-T.
1
Thus by transitivity,
-T
Ti +lk+1
Ti+l-Tk
i-k
i-k+1
for j < k < i
.
Setting
j=k,
T
T.
-T.3
T.
1+1
-T.3+1
i-j
i-j+1
(i-j)Ti+1- (i-j)T. = i-j)Ti+1 - (i-j)T j+1 + T i+1 - T.3+1
(i-j)Tj+1 - (i-j)T.3 + T.+1 = Ti+1 implies
(implies
implies
3
(i-j)(T.3+1 -T.j + T j.44 - T. = T.1+1
J
T.
3+1
-T.
3
3+1
1
-T.
3
T
i+1
implies
-T.
3
-T.
T
-T.
k+1
for j+1 < k < i+1.
k-j+1
k-j
T. -T
1+1
i
and the result follows by transitiv
1
Thus by transitivity,
T.
j
i-j+1
1
Therefore
-
k
52
in the chart.
Corollary 4. 4. 4. If two consecutive terms are equal,
T.-T.
T.-T.
3
1+1
1
J
j-i
in the
j-1-1
column of the Fan-Carlson chart where
jth
1 < i < j < p,
then equality holds throughout the chart to the left and below the term
3'
T.-T.
Remark 4. 4. 5.
Let A be a pxp GH-matrix for which
for all a C {1,
S
for
0
1 < k < p,
Let f(a) = log A(a),
, 13}.
=0
5
_
Tk
f (ti 1} k.)
k
=
Sk
p
for
Fan-Carlson chart holds for
0 < k < p.
A.
k})
By Theorem 3. 2. 2, the
The first row is Szasz's inequality
and the last column is Mirsky's inequality.
Theorem 4. 4. 6.
A(4a) > 0
A(a) > 0
Let A be a pxp GH-matrix for which
for all a C {1,
,
p}.
T
Tp-T 0
p-1 -T 0
or
13-1
ppl
53
T -T
T -T
p-1
p
if and only if A = PTPt where P is a permutation matrix and
T
is a triangular matrix.
Proof: By Theorem 4. 4. 3 and Corollary 4. 4. 4,
T
-T
p-1 0
P-1
Tp-T 0
p
Or
Tp-T 0 Tp-T 1
13-1
13
if and only if equality holds throughout the chart. This occurs if and
only if Szasz's equality (3) holds which by Theorem 4. 2.12 is equivalent to (1), (2), (4), (5), and (6). Therefore the result follows by
Theorem 2. 2.5.
Remark 4.4.7. Equalities (1) through (7) and reducibility for
are not necessary for equality of two arbitrary terms in the chart.
For consider the OH-matrices
A=
1
1
0
1
and B= (2
For
A,
1)
P1 = 1, P2 = 1, P3 =
1
)
111
1
1
-2
2
2
1
TO = 0' T1 = T2 =
T = - log 2 and the corresponding chart becomes
2
54
T
1-
T
0
T
2-
T3-T0
T
2
1
T
3
T3-T1
2-T 1
3
1
T3-T2
For
T2 = log
21
B, P1 = 4, P2 = 2, P3 =
/3
,
T3 =
0
so T 0 = 0, T 1 = log 4
1
1
3
and the corresponding chart becomes
T1-T0
T 2-T 0
1
2
3
V
v
T
3
T
2-T 1
1
=
-T
0
3-T 1
2
II
T3-T22
1
In fact equality in no proper subchart of a chart with inequality
elsewhere for a
3x3
GH-matrix A implies equalities (1)
through (7) or reducibility for A.
55
V. TNN-MATRICES
5.1. A Characterization
Definition 5. 1. 1. Totally non-negative matrices will be referred to
as TNN-matrices. These are matrices for which all minors of all
orders are non-negative (Gantmacher, 1960).
Gantmacher (1960) proved:
Theorem 5. 1. 2.
If A
is totally non-negative,
A(a) = 0 and
then A(I3) = O.
a C 13
As already observed, TNN-matrices for which all principal
minors are positive are GH-matrices.
The following result has essentially been proved by Fan (1960):
is an n x n TNN-matrix, then A is a
Theorem 5. 1. 3. If A
OH-matrix.
Proof: Since A is totally non-negative,
for all a, {i}, {j}C {1,
,
A(a) > 0, A(°a j.) > 0
Therefore A is a WSS-matrix and
n}.
the conclusion follows from Theorem 5. 1. 2 and Theorem 3. 1. 5.
Koteljanskii (1950) proved:
Lemma 5.1. 4.
a. . > 0
1,1
j E {1,
for all
n}
Let A be an
i E {1,
.
.
,
}.
that a.1,j = 0.
nxn
TNN-matrix for which
Suppose for some fixed
If
1... i
1< j then A[ j...
n
56
If
i> j
then
A[ i
1... j
= O.
Theorem 5. 1. 5. Let A be a TNN-matrix for which
all
I
where
E
{1,
T
n}.
a.1,1 > 0
.
for
Equality (2) holds if and only if A = PTP-1
is triangular and P is a permutation matrix which can
be partitioned so that
is a block diagonal matrix where
1 < m < k.
P.
1
is a permutation matrix for
P.
is either the identity matrix or P.
has ones in
1
the secondary diagonal and zeros elsewhere, I. e.,
01
1
\I
If
is of either form for
P.
1
0)
1 < m < n-1
m
remaining form.
then
is of the
P.
m+1
Proof: Suppose A(a.v13)A(a(m p) = A(a)A(13) for all
a, PC {l,
,
n}.
For any i,j
a.1,1.a.j,j - a.1,3.a.j,i
E
{1,
A({i} j {j})A({i}
=a1,ia...
3,i
,n},
{j}) = A(0.1)A({j})
57
Therefore
a.1,3.= 0 or a..
= 0.
3,1
In particular either
generality suppose
a1,2 = 0 or
and let
a1,2 = 0
< n-1 such .thata .
i,i+1
a2,1
Without loss of
= 0.
be the greatest index
i1
= 0 for 1 < i < i1' Then A. = A[{1,...,
1
is lower triangular. If i1 = n-1 then A is lower triangular so
P = I, T = A and the proof is finished. If
i1 <
n-1 let
B1 =1+1,...,n}]. Then by Lemma 5. 1.4, A has the partitioned
form
0
A=
(
B1
Sincea .ii+1,i1+2 > 0, a.11+2,i1+1
index i <
_ n-I
B1
such that
.
ai+1,i
= 0. Let
i
be the greatest
2
= 0 for i 1+1 < i < i -1. If
is upper triangular. If i2 < n-1 then A.
=
Athen
i
= n- 1
2
[Ii1+1,...,i21]
12
is upper triangular. Let B2 = A[{i2+1,... , n}]. Since
ai. +1,i > 0
2
then
ai2,i2+1=
2
Therefore by Lemma 5.1.4, A has the parti-
tioned form
'A.
0
0
Ai
0
11
A=
X1
\O
2
B21
X2
/
There exists an integer m < n such that
A.
0
11
A=
\X
\
A.
1
58
where
Ai
for
1 < k < m,
ik} I
= AE
i = 0, i
= n.
A.
is lower triangular if k is
lk
odd and upper triangular if k is even. Let ak be the order of
A.
Let
ik
is the a k x a k identity matrix if k is odd and P.
lk
is an ak x ak matrix of the form
where
P.
1
if k
is even.
P.
P
Therefore
P
kk ik
A.
P.
11
A.
11
is lower triangular for each k.
P.
11
PAP =
is lower triangular.. Since P is a permutation matrix P-1
=
Pt
59
and by definition
Thus A = PTP -1 .
P = Pt.
Conversely, suppose A = PTP -1 with T, P as in the hypothesis.
Let a, p c {l , .n} and let Q be the permutation on
the columns of the
n x n identity matrix which produces
P.
Then
by Lemmata 2. 2. 2 and 2. 2. 3,
A(ak../P)A(ar) 13) = T(Q(a)k..)Q(P)Yr(Q(a)r\Q(13))
= T(Q(a))T(Q(P))
A(a)A(p) .
Remark 5. 1. 6. The results stated in Theorems 5. 1.4 and 5. 1. 5 do
not hold for GH-matrices in general as consideration of
Az
shows.
More directly, to show that Theorem 5. 1. 5 does not hold, let
111
001
A= (0
1
1
and Q
QAQ1 is a GH-matrix and P = Q-1
o i\
(o1 0 0
0
1
0/
is the unique permutation ma-
trix such that P(QAQ-1)P-1 is triangular. However P is not of
the form described in the conclusion of Theorem 5. 1. 5.
60
5. 2. A Canonical Form
Definition 5. 2. 1. A TNN-matrix A
is called oscillatory (Gant-.
macher, 1960) if some integral power of A is totally positive.
Remark 5. 2. 2. Some typographical errors appeared in the transla-
tion of (Koteljanskii, 1950) in Lemma II and Theorem II which are
restated below as Lemma 5. 2. 3 and Theorem 5. 2. 4.
Lemma 5. 2. 3. If a matrix is totally non-negative and one of its
, s+m) is zero, then the rank
minors of order m+1, A( s,t, s+1,
t+1,
t+m
of at least one of the four matrices
s, s+1,
I.
AI
1, 2,
s+m
...
Is, s+1,
t, t+1, ... , t+m
[1,
,n
III. A
IV.
2,
... , s+m
A
Lt, t+1,
,
is less than or equal to m.
Theorem 5. 2.4. Every n x n,
non-singular, totally non-negative
matrix A is either oscillating, or is of the form (18):
Av
Cy
the
are oscillating matrices, and from every pair of matrices
Bv,
at least one has rank zero.
Remark 5. 2. 5. A typographical error occurred in III, which has been
corrected above, and in the chart accompanying the proof of Lemma
61
5. Z. 3 where the labels of matrices III and IV are reversed. Since
Lemma 5. 2. 3 was used in the proof of Theorem 5. 2. 4, the same er-
ror was made in identifying matrices III and IV there also. In the
statement of Theorem 5. 2. 4 the word "order" should have been
"rank" as in the statement above. Since there are at most
Av
n
of the
mentioned in the statement of Theorem 5. 2. 4, the largest sub-
script appearing in the chart which accompanied the proof should not
have been
n.
Finally as noted by the translators the definitions of
Bv
and
Cv were omitted from the original Russian text. In the proof of Theorem 5. 2. 4 the matrices
Bv
and
Cv
are referred to as being
"in the extensions of A in the direction of larger indices. ft The
extensions of A in the direction of greater subscripts were defined
by Koteljanskii, (1950) in the statement of Lemma 1 to be the follow-
ing matrices where
Av= A[iv, iv+1,
,
A
v +1,
iv 11
and
, iv+1]:
A
iv, .
,n
v v +1, ... iv+j
Therefore, it appears that the matrices Bv and Cy were
intended to be subsets of these matrices so that:
62
i
B
=A
[iv+1,... ,n
+1 ..." iv
v-1
,
iv+1,
iv-1+1, .
. . ,
i
Based on (18) and the corrections mentioned above, the rest of the
proof of Theorem 5. 2. 4 holds mutatis mutandis.
Koteljanskiits proof of Theorem 5. 2. 4 also proves the following
theorem which could have been what he intended in Theorem 5. 2. 4.
non- singular, TNN-matrix A is
Theorem 5. 2. 6. Every n x n,
either oscillating, or is of the form (19) below where the Av are
B,
v
oscillating matrices, and from every pair of matrices
least one has rank zero.
Cv
at
(19): A can be partitioned so that Av
is defined as above but define
1, 2,
B=A
,
,
,n
iv+1,
Proof: Since det A > 0,
a..
> 0 for all
1,1
i.
C=A
v
iv+1,
1, 2,
,
iv
Theorem 5. 1. 2 implies that
Necessary and sufficient conditions for a totally
non-negative matrix to be oscillating were given by Gantmacher and
Krein, (1950)tobethat a.1,1+1 > 0,
.
a.
>0
1+1,i
for all
i
such that
1< i< n-1.
Let some
a.
1,1+1
= 0.
Of the four matrices in Lemma 5. 2. 3
of which one must have rank zero, I, II, III have diagonal elements.
Thus IV has rank zero. Similarly if someai+1 ,1
0,
then the
63
associated matrix III must have rank zero.
Let i1 <
be all those values of i E {1, ... , n} for which at least one of the
elements
a.
and 1+1,i
a.is zero.
.
1,1+1
Av= A[iv- 1 +1, ... , iv], where i0 =
< v < k, and A[ik+1,..., n]
0,
are oscillating. Since the matrices
and
Bv
IV and III respectively, at least one of
each
Then the matrices
B
C
Cy
are the matrices
v has rank zero for
v.
The proof of the following theorem is similar to that of Theorem 5. 1. 5.
Theorem 5. 2.7. Every n x n,
non-singular, TNN-matrix A
has the form A = PTP1
where P is a permutation matrix and
P
and T can be partitioned with the same partition so that
T=
( Al, 1..
x
At,t
is triangular where Av= A v,v is oscillating for each
v
with
be the order of Ay and let Iv be the
pv x pv identity matrix. Then, the partitioned form S of P has
exactly one non-zero element in each row and each column. The non1 < v < t.
Let
pv
zero element in the vth row is Iv.
S
can be partitioned so that
64
S=
S.
0
has exactly one non-zero block in each row and column and is a block
diagonal matrix where for each m with 1 < m < k,
S.
has one
of the forms
Ia
0
(I a.im-1
im- 1
0
a.
ai
If for any m with
Si
rn+1
,
1 < m < k,
1
has one of these forms, then
S.
has the remaining form.
A has the partitioned form
Proof: By Theorem 5. 2. 6,
/A1
.
.
.
A1,t
1
M=
At,t
At,1
where AV= A v,v is oscillating for all
least one of the matrices
Cy
1 ,
,v
Mr 1,
v+1,
t rank zero.
v+1
= M[
" has
v
,t
with
1<v<t
and
Suppose, without loss of
and at
65
generality that
i < t-1
has rank zero and let
B1
such that
B.
1
has rank zero for
be the greatest index
i1
1<i<i
Then
.
= A[1, ...' i 1 ] is lower triangular. If i1 = t-1 then M is
il
lower triangular. If il < t- 1 let N = M[ii+ 1, . . . , t]
Then M
l`M.
.
has the partitioned form
M.
0
X
N
M
Since
11+1
zero for
i1
+1 <
= t-1
2
then
< i2 -1.
M.
12
i2 < t-1.
pose
has rank zero.
Ci
1+1
be the greatest index i < t-1
Let i
i
has rank greater than zero,
B.
=
i2 = t-1
If
MIf
such that
1+1,
...
,
i
then
2]
N
C.
has rank
is upper triangular.
is upper triangular. Sup-
There exists an integer in < t such that
/M.11
\
M=
\x
where
M.
ik
i m = t.
is k
= M[i+1,
... , i]
k
k- 1
irn
1 < k < m,
for
is lower triangular if
M.
M.
k
i
0
= 0,
and
is odd and upper triangular
ik
is even.
Let
a
be the order of
M.
for all k with 1 < k < m
66
the order of Ay for all v with
and pv
1 < v < t.
S.
11
s=
S.
0
where
S.
k
is thex
ak a
)
matrix of the form
/I.
0
'k- 1+1
S.
=
S.
=
is odd and
if k
lk
0
k
is even where
if
k
S.
M.
lk
S.
Iv
p x pv
is the
identity matrix.
is lower triangular for each k.
M.
S.
0
S MS =
=T
X
is lower triangular,
Therefore,
M
S.
1
m
i
S.
mm
1
Let
67
Since
St
M = STS-1.
=
Let
P, B represent S, T
Then P is a permuta-
respectively without regard to a partition.
tion matrix.
Since M is the partitioned form of A, A = PBP-1 which
was to be shown.
Determinantal Equalities
5. 3.
Lemma 5. 3. 1.
for all i e {1,
aC
{1
,
n}
Let A be an n x n, TNN-matrix with
, n}
where n> 2.
If
a.
.
1,1
SA[a] = dia A[a]
>0
for all
then SA = dia A.
Proof: A has no non-zero diagonal corresponding to a noncyclic, non-identity permutation
such a
a-
o-
of
{1,
,n}
since a factor of
corresponds to a non-principal diagonal of a principal
submatrix of A which is zero by hypothesis. By hypothesis and
Lemma 2. 1. 5 A has at most one non-zero diagonal corresponding
to a cyclic
a-
of length n.
Suppose A has one such non-zero
diagonal
10-(1)' aThere
an,a-(nr is some
that a-(j) = 1.
o-(1)
j
j,
1<j<n
since by hypothesis AL1 j] has diagonal
support.
A(
1j
1 a-(1)
such
)=a
a.
a
1,1 bo-(1) - a10-(1) j, > 0
68
and
a1(1 ) > 0, a1,1 > 0,
fore
aj, 1
imply that
>0
a2,0-(2)," " "'aj-1,0-0-1Y ai,Gr(
a.
(1)
There-
> 0.
is a
ai+1,0-(i+1),
non-zero, non-principal diagonal of A(2... n) which is a contradiction to the hypothesis. Therefore SA = dia A.
Corollary 5. 3. 2.
where n> 2.
If
Let A be an n x n, TNN-matrix with
for all a C
SA[a] = dia Ala]
{1,
,
det A > 0,
then
n}
SA = dia A.
Proof: A totally non-negative implies that A is generalized
Hadamard.
Thus
det A> 0 implies a> > 0 for all i E {1,...,
1,1
Let A be an n x n TNN-matrix where n> 2.
is at most dia Ala] for all a C {1, ... , n} then
SA
#
Theorem 5. 3. 3,
If
SAfai
is at most dia A.
Proof: By hypothesis and Lemma 2. 1. 5 A has at most one
non-zero, non-principal diagonal
{a
1,cr(1)'
an,o-(n)} and
cr
must be cyclic of length n.
If
tion
SA
A has no non-zero, non-principal diagonal then by defini-
is at most dia A.
Suppose A has one such non-zero, non-principal diagonal.
We will show that A then has
a
For an arbitrary
there exists an i such that
o(i) = x.
x, 1 _< x _
< n,
We may assume that
x,x > 0 for all
i < x.
x E {1
.
Otherwise we could
.
n}.
69
consider At.
First suppose that cr(x)< x then
A(
or(x) x
By hypothesis
non-negative,
Suppose
o-(y)< x
x
i
) = a.(x) a -aX,0-(X)
a. 1,X
>0.
-
ax,cr(x)>
a
X,X
1,Cr
and a.
0
13X
>0
and since
A
is totally
x,x > 0.
There exists y, y> x such that
x < o-(x).
since otherwise
cyclic implies that A contains a
a-
non-zero element in each row and each column so that A[1,
,
x]
would also contain a non-zero element not on the principal diagonal in
each row and each column contrary to the hypothesis that
is at most dia A[1,
A(
0.(y)
SA[1,...,
.,x]. Therefore,
a
xy ) = a.
1,a(y) y,x
- a.1,x a y,cr(y)> 0.
ai,x > 0, ay,o_ (y) > 0 and A is totally non-negative, it follows that a
y,x > 0. Therefore
Since
A(
ax,o- (x)
x ycr(x)) = a
x
> 0, a y,x > 0,
a
x,x y,o-(x)
-a a >0
X,Cr (X)
y,x
and A totally non-negative imply that
Since A has positive principal diagonal, Lemma 5. 3. 1
70
implies that SA = dia A.
Remark 5. 3. 4. Even though the proof of the following theorem fol-
lows from Theorem 4. 2. 8, it can be established in a different way
using Lemma 5, 3. 1.
Theorem 5. 3. 5.
for all
i E {1
.
Let A be an
n}
TNN-matrix with
n x n,
a.
.
1,1
>0
then (3) implies (1).
Proof: (1) and (2) are equivalent by Theorem 2. 3. 1.
The proof is by induction on n.
vial.
If
n=2
If
n=1
the result is tri-
let
a1,2
a2,1
A = (a1,1
then (3) implies
a22)
A(1)A(2) = a1 1a22 = A(1,2) = a1 1a22 - a1 2a21
Therefore a12
= 0 or
= 0 so that (1) holds.
,
a2,1
Suppose the theorem holds for k = n-1 where n> 3.
(2) and Theorem 2. 2. 5 imply that SA[a] = dia A[a]
a C {1,
,
n}.
Therefore by Lemma 5. 3.1
SA =
Then
for all
dia A and again
by Theorem 2. 2. 5, (2) holds for A and therefore (1) holds for A.
Remark 5. 3. 6. The following three general results (Marcus and
Min.c, 1965a) are needed for the last two results.
71
Theorem 5. 3. 7. A matrix A is normal, (i. e. ,
AA* = A*A),
if
and only if it is unitarily similar to a diagonal matrix.
Lemma 5. 3. 8. Let T
TT* = T*T
if and only if T
Xt,
t
E
{1 ,
.
.
n}
triangular matrix.
n x n,
is a diagonal matrix.
If A is an
Theorem 5. 3. 9.
values
be an
n x n matrix with characteristic
then
a. .12
1,
with equality if an only if A is normal, (i. e.
Let A be an
Theorem 5. 3. 10.
holds.
n x n,
is normal if an only if
A
A
,
AA* = A*A).
TNN-matrix for which (2)
is a diagonal matrix.
Proof: The sufficiency is obvious. By hypothesis and Theorem
5.1. 5,
A = PTP-1 where
{t.1
Pt
=
P-1 = P and T
P
is a triangular matrix with
.11 < i < n} = {a. .111 < i < n}.
1
=
P(TtT)P
=
AtA = A*A
Lemma 5. 3. 8,
Theorem 5. 3. 11.
T
is a permutation matrix for which
implies
AA* = AAt = (PTP)(PT Pt = P(TTt)P
TT* = TTt = TtT = T*T.
Thus by
is diagonal and therefore so is A = PTP-1.
Let A be an
n x n,
TNN-matrix for which
(2) holds. Shurfs equality holds, i. e., (9) holds for
A
if and only
72
if A is diagonal.
Proof: This follows directly from Theorems 5. 3. 9 and 5. 3. 10.
73
BIBLIOGRAPHY
Aitken, A. C. 1956. Determinants and matrices. 9th ed. Edinburgh,
Oliver and Boyd. 144 p.
Arnold, B. H. 1962. Logic and boolean algebra. Englewood Cliffs,
Prentice-Hall. 142 p.
Beckenbach, Edwin F. and Richard Bellman.
Berlin, Springer-Verlag. 198 p.
1965.
Inequalities.
Birkhoff, Garrett. 1948. Lattice theory. New York. 283 p.
(American Mathematical Society. Colloquim Publications. Vol.
25. Rev. ed. )
Brualdi, Richard A. 1966. Permanent of the direct product of
matrices. Pacific Journal of Mathematics 16:471-482.
Carlson, David. 1967. Weakly sign-symmetric matrices and some
determinantal inequalities. Colloquium Mathematicum 27:123129.
1968a. On extensions of Szasz's inequality.
Archly der Mathematik 19:167-170.
1968b. On some determinantal inequalities. Proceedings of the American Mathematical Society 19:462-466.
Fan, Ky. 1960. Note on M-matrices. Quarterly Journal of Mathematics, Oxford Ser. 2, 11:43-49.
1967. Subadditive functions on a distributive lattice
and on extensions of Szasz's inequality. Journal of Mathematical Analysis and Applications 18:262-268.
Fiedler, Miroslav and Vlastimil Ptak. 1966. Some generalizations
of positive definiteness and monotonicity. Numerische Mathematik 9:163-172.
Gantmacher, F. R. 1960. The theory of matrices, tr. by K. A.
Hirsch. New York, Chelsea. 2 vols.
74
Gantmacher, F. R. and M. G. Krein. 1960. Oszillations matrizen,
Oszillationskerne und kleine Schwingungen mechanischer systeme, Wissenschaftliche Bearbeitung der deutschen Ausgabe:
Alfred Stohr. Berlin, Akademie-Verlag. 359 p. (Mathematische Lehrbucher und Monographien. Abteilung I. Bd. 5)
Hadarnard, J. 1893. Resolution d' une question relative aux determinants. Bulletin des Sciences Mathematiques, Ser. 2, 17
(Part I):241- 246.
Koteljanskii, D. M. 1950. The theory of non-negative and oscillating
matrices. Ukrainian Mathematical Journal 2(2):94-101.
(American Mathematical Society Translations, Ser. 2, 27:1-8.
1963)
Marcus, Marvin and Henry Minc. 1964. A survey of matrix theory
and matrix inequalities. Boston, Allyn and Bacon. 180 p.
1965a. Introduction to linear algebra. New York,
Macmillan. 261 p.
1965b. A subdeterminant inequality. Pacific
Journal of Mathematics 15:921-923.
Markham, Thomas Lowell. 1967. On oscillatory matrices. Ph. D.
thesis. Auburn, Auburn University. 58 numb. leaves.
(Micro-film)
Mirsky, L. 1957. On a generalization of Hadamard's determinantal
inequality due to Szasz. Archiv der Mathematik 8:274-275.
Ostrowski, A. M. 1937. Uber die Determinanten mit uberwiegender
Hauptdiagonale. Commentarii Mathematici Helvetici 10: 6 9 - 96.
Sinkhorn, Richard and Paul Knopp. 1967. Concerning non-negative
matrices and doubly stochastic matrices. Pacific Journal of
Mathematics 21:342-348.
Szasz, 0.
Uber eine Verallgemeinerung des Hadamardschen
Determinantensatzes. Monatshefte fur Mathematik und Physik
1917.
28:253-257.
Taus sky, Olga. 1958. Research problem. Bulletin of the American
Mathematical Society 64:124.
75
Turnbull, H. W. and A. C. Aitken.
1961. An introduction to the
theory of canonical matrices. New Dover ed. New York,
Dover.
200 p.