AN ABSTRACT OF THE THESIS OF
DINA NG NG
for the
DOCTOR OF PHILOSOPHY
(Name)
in
(Degree)
MATHEMATICS
presented on
August 10, 1973
(Major)
Title:
(Date)
AN EFFECTIVE CRITERION FOR CONGRUENCE OF REAL
SYMMETRIC MATRIX PAIRS
Signature redacted for privacy.
Abstract approved:
Charles S. Ballantine
This paper gives a rational method of determining
the congruence of real symmetric pairs in IRrixn
(S11T1)
(ST1)
-
.
and
are nonsingular pairs, then
(S2,T2)
is congruent to
.
.
S11 T1 is similar to
S2-1 T2
-1
-1
S1f(S1 T1)k g(S1 T1)
and
for
If
(S2'T2)
over
7t
if and only if
and the signatures of
(S2 f(S-11T1 )kg(S-1T2
)
2
k = 0,1,2,,n-1 and for all
g(x)
E P,
are equal
where
is a relatively small set of real polynomials and
is a fixed polynomial.
f(x)
This result is then extended to
singular pairs using theorems on minimal indices.
several numerical examples are worked out.
Also
P
AN
EFFECTIVE CRITERION FOR CONGRUENCE
OF REAL SYMMETRIC MATRIX PAIRS
By
DINA NG NG
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Doctor of Philosophy
June 1974
APPROVED:
Signature redacted for privacy.
Professor of Mathematics
Signature redacted for privacy.
Ch
rman of Department of Mathematics
Signature redacted for privacy.
Dean of Graduate School
Date thesis is presented
Typed by Jolan Eross for
August 10, 1973
Dina Ng Ng
ACKNOWLEDGMENT
I would like to express my thanks to Dr. Charles S.
Ballantine for his valuable suggestions, guidance and encouragement during the working of this thesis.
I would
also like to thank Dr. and Mrs. James R. Brown, who hosted
my visit to Corvallis, and Dr. Thomas J. Mueller for their
support and encouragement and for making me feel at home
during my stay in Corvallis.
TABLE OF CONTENTS
Page
CHAPTER I.
INTRODUCTION AND PRELIMINARIES
1
CHAPTER II.
MAIN LEMMA
9
CHAPTER III.
MAIN RESULT
28
CHAPTER IV.
EXAMPLES
62
BIBLIOGRAPHY
82
CHAPTER I
INTRODUCTION AND PRELIMINARIES
The subject studied in this paper concerns the real
congruence of pairs of
and
(ST1)
(S2' T2
A
lar pair if
).
nxn
real symmetric matrices
We say that
is a nonsingu-
(A,B)
is nonsingular; otherwise the pair is called
Necessary and sufficient conditions are given for
singular.
example by Muth [7] for congruence of real symmetric pairs;
Uhlig [13] gives a canonical form for real symmetric nonsingular pairs; Turnbull [10], Trott [8], Ingraham and
Wegner [5] give conditions for conjunctivity of hermitian
and singular pairs.
The operations used are in general ir-
rational.
In this paper we give (using only rational opera-
tions) an effective criterion for congruence of pairs of
real symmetric matrices.
P
This criterion uses a finite set
of real polynomials whose cardinality is relatively wall.
Namely, for
T1)
nonsingular pairs
is congruent to
and only if
fixed
nxn
SilTi
f(x) E1R[x],
(S2'T2)
is similar to
(S1,111)
and
over the reals
-1
S2 T2
the signatures of
(S2,T2),
M2
if
and for some
S1f(S11T1)
kg(s11T1)
2
and
are equal for all
S2f(S1T2)kg(S-21T2)
k = 0,1,2,,n-1.
and
g(x)
An explicit construction for
E p,
f(x)
is given depending only on the similarity class of
5)
S11
T1.This result is then extended to the singular case,
using known theorems on minimal indices.
If
pair and
also if
is an
(S,T)
S-1T
f(j)
Al1A2,,An,
has a distinct simple roots
(Ai)
then there exists
g(x)
j = 1,2,, n-1,
i = 1,2,,n,
for
0
Tog(A)),
±(Sog(A),
nonsingular real symmetric
nxn
such that
E p
where
is congruent
(S,T)
is the jL1-1- deri-
f(j)(x)
vative of the characteristic polynomial,
f(x),
and
S0, T
A
is the companion matrix of
f(x)
of
0
are any
S-01T0 = A.
fixed real symmetric matrices such that
S-1T;
At
the present time we are not able to generalize this result
to any pair
where
(S,T)
f(j)(Ai)
case
(S,T)
= 0
but are able to extend it to the case
for some
is congruent to
i
(ED
i=1
(S.,T.)
and
1
S,
.
j (j
>
2).
0 T.),
where
.
1=1
is a nonsingular real symmetric pair and
is a companion matrix for
In this
-1
S. T.
i = 1,2,,m.
Now we will state without proof some known results in
matrix theory which will be used in the later chapters.
3
All capital letters denote matrices unless
Notation:
otherwise specified.
cp Ai = diag (A1, A2,
An) =
i=1
the field of complex numbers.
is the field of real numbers.
IR
nxm matrices over a field
is the set of
Fnxm
is the determinant of
det A
F.
A.
I
is the
qxq
Jq
is the
qxq matrix with l's on the super diagonal and
identity matrix.
O's elsewhere.
Eq
qxq
is the
matrix with l's on the anti-diagonal and
O's elsewhere.
C*
C
t
sigA
f(
is the complex conjugate transpose of
is the transpose of
.
is the signature of
is the
(x)
note
f(1)(x)
Definition: Let
(S1, T1)
by
5?..th
C.
A.
derivative of
by
f(x).
.LL=
be in
51, T1, S2, T2
(S2, T2),
We will also de-
f' (x).
is conjunctive (congruent) to
(S1, T1)
C.
over
C
(F)
Cnxm (Rnxn);
then
(S2, T2), denoted
if there exists
C
4
in
Cnxn (Rnxn), det C
over
C*S1C = S2
(CtS1C = S2)
C*T1C = T2
(CtT1C = S2).
A
Every
Theorem 1.1:
to a matrix
C,
m
(A.
iq
a . I
)0(0
((A
q =1
i=1
qki
a +
2
m
0
i=1 q=1 j=1
I.Iq +J q
0 (-Z.1
I
2
m
m
r1
is similar,
mxm
r2
kg.
+ J )
q
IR
where
J,
i=1 q=1 j=1
=
or
T mxm
in
00
rl
J =
iq
such that
0,
q
+ J
)
A.a
1-
,
= m,
1
i=1
kqi
q=1
the
A.
'S
are the distinct real characteristic roots, and
the
a.
'S
are the distinct roots of
1
1
This
imaginary part.
having positive
is called the Jordan canonical
and it is uniquely determined by
A
form of
J
A
numbering of the roots is determined).
A
(once the
The proof of this
well-known result can be found for example in [2].
Theorem 1.2:
det S
0
and
Let
J
as in Theorem 1.1.
S, T
be hermitian in
Tmxm
with
be the Jordan canonical form of
Then
(S,T)
(M,N)
over
T,
S-1T
where
5
m
2
e.
.
1,q,3
E
q
(F)
kl'
(:=1 q=1 j=1 E
2)
J®A
\i=1 q=1
kqi
e 6 i,c1fjEg(AiIq+Jq)
0
N =
= ±1
for all
i,q,j.
A
PRImxm
see [5].
Proof:
Every
Theorem 1.3:
J',
a matrix
m
[ai
iq
=
X.1's
T2
kqi
i=1 q=1 j=1
A' .
in
Iq Ob.1
A.' 4-J
1 q
i=1 q=1 j=1
(CD
a."L*
I
+q 12
)
lq
)
are the distinct real characteristic roots, the
A with
denotes the tensor product.
is called the real-Jordan cannnical form of
and it is uniquely determined by
of the roots is determined).
Proof:
A!
CD
_b4i)
positive imaginary part, and
J'
to
1(1.
m
are the distinct characteristic roots of
a3.+JJD.
This
is similar over DR
where
000
(r1
the
kcal
i=1 q=1 3=1 zq iq
j=1
6.1,q,
and
r2
See [6].
A
A
(once the numbering
6
Let
Theorem 1.4:
mxm
Let
pair.
(S,T)
be the real-Jordan canonical form of
J'
as in Theorem 1.3.
S-1T
be a nonsingular real symmetric
Then
(S,T) =
over
(141,1\11)
IR,
where
=
)0(6
E
C.
1,q,j q)
i=1 q=1 j=
m
r1
N'
r9m k'.
m
(r1
0 ED
=
kqi
i=1 q=1 3=1
c.
and
S
Theorem 1.5:
and
S
T
in
H
singular
S,T
Let
Definition:
Then
qi
e
ED .0
1=1 q=1 j=1
ED
E
A!
2q iq
i,q,j.
It is given by Uhlig [13].
Proof:
Then
'
for all
= ±1
.
1,q,3
Eci(Xi I+Jci)
k'.
m
2
Ei
2g)
q=1 j=1
Cnxn*
are isoconjunctive if there exists a nonCnxn, H*=H,
Let
T
be complex hermitian in
S,T
such that
be nonsingular
T = HSH.
nxn
hermitian.
are isoconjunctive if and only if
(S-11T) = (T-11S).
Proof:
See [12].
Fact 1.6 (Taylor's Theorem):
I
= I, J
=J,
then
[2]
Let
f(x)
IF[x].
If we let
,
7
n-1
(k)
f(AI+J) =
k=0
Fact 1.8: Let
Let
E IF [x] .
Then
.
f(C-1AC)
IF
= C-1f(A)C
[2].
be square matrices over
An
A1, A2'
f (x)
over
A, C
for any square matrices
.
Then
f(x) EF[x].
Let
Fact 1.7:
Jk
(A)
k!
[2]
f(C) Ai) = ED
f(Ai).
i=1
i=1
Fact1.9:LetA.be complex hermitian (real symmetric)
i = 1,2,.,n.
matrices for
sig(
ED
sig A.
Ai)
i=1
1=1
Fact 1.10:
A
Let
Then [2]
nonsingular complex her-
nxn
be an
mitian (real symmetric) matrix,
of
such that
Cnx1 aRnxi)
x E E+,
and
E
-
Fact 1.11:
and
M1, M2
-
Let
x*Ax > 0
be a maximal subspace
(xt
> 0)
for all
be a maximal subspace such that
x*Ax < 0 (xtAx < 0)
Cnxl = E+ 0 E
E+
for all
°ERnx1 = E+
V
x E E.
0 E-)
Then
[3].
be a finite-dimensional vector space
be subspaces of
V
such that
V = M1+M2.
8
Then
dim V = dim M1 + dim M2 - dim
where dim S
Fact 1.12:
(Min
denotes the dimension of the space
If
is congruent (conjunctive) to
S
M2),
S[3].
T,
then
sig S = sig T [2].
Fact 1.13:
Let
f(x)
EIF[x],
A
Then the characteristic roots of
,
of
f ( An)
A[2].
where the
X's
be any square matrix.
f(A)
are
f(X1), f(X2),
are the characteristic roots
9
CHAPTER II
MAIN LEMMA
In this chapter we aim to find a finite set
of
P
real polynomials which will be used in proving the main
theorem of this paper in Chapter III.
We first prove the
following lemmas.
Let X1,X2,-1Xr be nonzero
Lemma 2.1:
vectors over a field
[X1'X2''Xr-1]
If the
F.
has rank
r-1
nxl
matrix
nx(r-1)
and
Xr
column
can be expressed
(uniquely) as
r-1
Xr
=
X
i=1
b.X.
ii
with
for any sequence
the
2nxr
11'12,---,1r
in
matrix
X1 X2 --- Xr
A =
y1X1 y2X2---y2X2
has rank
r.
IF
then
b
for which
yj
yr,
10
We will show that the columns of
Proof:
independent.
Let
a
be any sequence in
r
a2'
are linearly
A
F.
Suppose
a.X. = 0
11
(1)
a.y.X. = 0.
111
(2)
y a.X. = 0,
(3)
i =1
Then
yr.(1)
X
r
i=1
(y.-y )a.X. = 0,
(2)-(3)
r
i=1
r-1
i.e.,
ra.
(y.-y )a.X. = 0.
y
i=1
Since
are linearly independent, we have
X1,X2,'.,Xr-1
for all
ai(yr-yi) = 0
a3
But
yr
y .;
(y
j,
i =
In particular, for
r -y
r r
=-
we get
= 0.
.3)
a. = 0.
therefore
a X
r-1
X
i=1
From (1),
a.X..
11
Thus
r-1
r-1
arX
1=1
i.
b.x. = -
a.X..
i=1
11
Therefore (since
are linearly indepen-
X1,X2'...,Xr-1
dent)
a lb.
r
= -a.
1
for all
i.
Hence, in particular,
= 0.
a b. =
r 3
b
But
so
0,
ar =
0.
Therefore, by (1) again, we
have
r-1
11
a.X. = 0.
i=1
Since
a. = 0
1
for
are linearly independent, we get
r-1
Xl' X2'
X2
X1
X
X
y2X2...y r r
y1X1
has rank
So the matrix
i = 1,2,w,r-1.
r.
Lemma 2.2:
Let
IF
be any field,
Xi,X2,,xr be
r dis-
tinct nonzero column nxl vector over F. Then there exist
s
lxr
row vectors
Yj = (aiirctivr),
with Yi=(1,1,,1)suchthatforj>2,Y.is a
(componentwise) product of at most
matrix
matrix
[Xl'X2'r]
and
r-1
s < 2r-1
rows of the
and the
snxr
.
nxr
12
has rank
a11X1
12X2
a21X1
a22X2
aslX1
as2X2
alrXra2rXr
a
X
sr r
r.
r = 1,
Proof:
The proof is by induction on
r.
For
X/
implies that
has rank
1.
Clearly we can
take
0
s=1
For
Y1=(1).
and
ly independent, then
can take
[X1]
[X11X2]
s=1, Y1=(1,1).
If
if
r=2,
are linear-
X11X2
has rank
2
and again we
are distinct and non-
X11X2
zero but linearly dependent, then consider the row
(a21'a22)
of
such that
[X1'X2]
X1
a21X1
has rank
2.
X2
a22X2
Clearly we can take
which is a product of
1
row of
s=2
and
Y2 = (,21,a22)
[X1,X2].
Assume the lemma is true for
k+1
Such a row
By Lemma 2.1, the matrix
X17X2.
exists because
a21a22.
r=k.
Suppose we have
distinct nonzero vectors, X1,X2,... ,Xk+1.
duction hypothesis, there exist
Y1 = (1,1,...,1)
such that the
By the in-
Yj, j=1,2,...,s,
snx(k+1) matrix
with
13
al2X2....a1,k+1Xk+1a21X1
a22X2..."12,k+1xk+1
aslX1
as2X2....as,k+1Xk+1
[R1R2 Rk+11 =
namely with
has rank
>k,
pendent.
Also we can take
k-1
duct of at most
If
rows of
0,
11
b.R.
k+1 = i=1
then some
Now consider the row
such
thatY
11R1
j = s+1, s+2,
bi
in
bi's
F,
are not zero, say
*
of
2snx(k+1)
Define
Rk+1
Y2R2
1k+1Rk+1
= (ajl'j2'
2s-1, 2s=s1
by
x.#X k+1
matrix
R2
Y
ly0.
[X1,X2,...,Xk+1]
This row exists because
R1
k+1.
> 2.
.
(Y1,Y2,'Y k+1)
Then, by Lemma 2.1, the
has rank
j
such that
R
Rk+1
for
(X1,X2,...,Xr]
If not, then there exist unique
i=1,2,...,k,
Since
s<2](-1,andY.is
a pro_
are linearly independent, then we are
R1,R21..., Rk+1
done.
linearly inde-
R11R21...,Rk
'aj,k+1)
for
.
14
as+1,1 X1 a+12 xas+1,k+1 xk+1
..
as+2,1 X1 as+2,2 X2'. 'a s+2,k+1 xk+1
= [Y11111...'Yk+1Rk+11.
as,1 Xl, as'2 X2'
'as,k+1 Xk+1
Then
R1 R2
L1'1
Rk+1
Y2R2"k+1Rk+1
a11X1
a12X2....a1,k+1 Xk+1
a21X1
a22X2--a2,k+1 Xk+1
aslX1
as2X2....as,k+1 Xk+1
as'2X2-...as:k+1 xk+1
1X1
Clearly
5'=25<22k-1=2k
at most
k
Lemma
and each
.
p = 1,2...,k+1.
2.3:
of Lemma
Suppose, of the
2.2,
the first
linearly independent.
be chosen
<2r-h
h
r
Y,
for
I
vectors
vectors
Then the
is a product of
=y'pa.ip
a s+1,p
rows, because
i=
as+1,k+1 xk+1
"s+1,1X1 a s+1,2X2
s
Xl'X2'r
X1,X2'h
are
of the conclusion can
15
Proof:
r=h,
Then Lemma 2.3 is obviously true if
r>h).
the
fixed and use induction on
h
We consider
k
(of the proof of Lemma 2.2) be
Lemma 2.4:
and suppose
>h
With the same hypothesis as in Lemma 2.2, if
F
the characteristic of
(1,1,-4)
so let
st = 25<22k-h = 2k+l-h.1
and get as before
5<2k-h
have
(for
Then in that proof we will now
r=k.
Lemma 2.3 is true for
r
and if there exist rows,
2
(1,-1,1,,(-1)r-1), of
and
EX1,X2,,Xr],
then we can take
r-1
2
-1
if
is odd
r
s<
r-4
32
for
r>4.
Proof:
For
2-1
r = 1,2,3,
Suppose
rank 2 because if
r
if
is even
r
we can take
is even and
X1 = aX3
r>4.
for some
there exists a row (1,1)) 1=a.
So
s=1.
Then [X1,X3] has
a E/F,
X1=X3
then (since
which is a con-
tradiction to that fact that X13.Therefore we can apply
Lemma 2.3, with
x2k+1'...,xr-1
h=2,
to the
1:
vectors
Thus we can choose rows,
a2,r-1),-.. (asl'as3''as,r-1),
X1,X31...,
(a
21'a23'...'
componentwise products
16
of rows of
and hence rows,
[X1,X3'...,Xr-1]
(a21'a221" a2r);...(a slas2
'
of
[X1,X2'...,XrI,
' asr
'
),
product of rows
such that the matrix
Xr-1-
X1
X3
a21X1
a23X3
a2,r-1Xr-1
aslX1
as3X3
as,r-1 Xr-1
A =
has rank
f
choose rows
and
r _2 r-4
=2 2
s < 22
(B
B
Similarly we can
.
U3s11's'2'...'13str"
'
product of rows of
r
[X1,X2
such that
X
X4
X2
],
622X2 624X4-62rXr
B =
13
s'2X2
s 14X4
r
r
has rank 2- and
s'
<
-
22
strXr
-2
r-4
=22
.
Define
17
X1
X2
21X1
/322X2
X3
13.23X3
32rXr
,
s'lX1 (3s'2X2 ISs'3X3
X1
-X2
X3
21X1
22)(2
23)(3
S
-(32rXr
S r
s'lX1s'2X2 e.s13X3,
a21X1
a22X2
a23X3
aslX1
as2X2
as3X3
Then M has rank
r
X
r r
r
a2rXr
a
X
sr r
and hence
r-4
s' + s' + s-1 < 3.2
M has rank
To show
- 1.
we suppose the columns of
r,
are linearly independent.
IF
such that the follow-
ing equations hold:
a.X. = 0;
i=1
1
i=1
13.
ji
= 0,
M
Then there exists a sequence
not all zero in
a1,a2r,
2
j = 2,3,.--,s'
18
i-1
(6)
=0 ;
i=1
ji 1
i=1
1
i=1
ii
= 0, j = 2,3,,s' ;
(7)
k = 2,3 ,s
a.a .X. = 0,
kl
(8)
Then
(4)-(6)
11
a . X .
i=2
= 0
i even
1= 0,
(5)-(7)
i=2
i 31
j = 2,3,,s'
.
i even
for
Sincel3hasrank-/-.theriNgegeta---0
1
2'
Substitute these values in (4) and (8).
i even.
Then we have
r-1
a.X. = 0;
i=1
i odd
r-1
a.
i=1
i odd
X.
ki1
= 0,
k = 2,3,4,,s.
But.Ahasrankthereforea.=0
for
1
2'
the columns of
rank
r.
M
odd.
are linearly independent and so
Hence
M has
19
r
Similarly for
4".
odd and
r>5,
we get matrices
X1
X3
X5
Xr
a21X1
a23X3
a25 X5
--- 2r
aXr
A=
B =
13s'2X2
Xr-1
(3244
2,r-1xr-1
s'4)(4
s',r-1xr-1
r+1
and we can choose
r-1
s'
< 2
2
s < 2
2
r-1
r+1
-f- ,
are
A, B
where the ranks of
2
= 2
2
and
r-5
2
= 2
2
.
Therefore
r=3,
a1'a2
Since there exist rows
X
X ]
22_i
2-1
= 2
X1'X2'X3.
such that
X3 = a1X1 + a2X2
[X
+
consider the vectors
E F
r-1
r-3
s'+s'+s-1 = 2s'+s-1 < 2.22
there exist
respectively,
r-3
2
r-5
For
X
sr r
X4
X2
(322X2
a
as5X
aslXsl as3X
(1,1,1)
then we have
.
and
(1,-1,1)
of
.
Suppose
20
1 = a
+ a
1
-
2
1 = al - a2
.
Adding the two equations, we get
0.
So
tradiction to the fact that
X3
a1 =
1.
Therefore
has rank
[X1,X2,X3]
(1,1,,1)
there exists rows
[X1 ,X2"
X. 's
1
Therefore
X/.
Hence for
3.
This is a con-
r = 1,2,3,
we can
With the same hypothesis as in Lemma 2.2, if
Lemma 2.5:
of
X3 = X1.
and hence
I
s = 1.
take
a2 =
2a1 = 2
are in
and
(1,-1,1,,(-1)r-1)
and if all the components of the
X r]
then we can take
{1,-1}.
r-3
2
s<
2-1
is odd
r
if
r-6
3
for
r>6.
For
2.2
-1
if
r
r = 1,2,3,4,5,
is even
we can take
s=1,1,1,2,2
respectively.
Proof:
If, for
i,j,k
distinct, X.,X.,Xk
are linearly
independent, then by Lemma 2.3 and as in the proof of
Lemma 2.4, rank
= rank
A = E
2
B
for
r
even.
Hence
21
r
r-6
2
=2
s < 22
s'
;
<
-3
2-2
r-6
2
=2
thus
;
r-6
s'+s'+s-1
32
<
2-1
22;
2
r-1
S'
2
< 2
r+1 -3
22.
B =
rank
Hence
=
2
2
r-7
r-5
2
2
+ 2
Let
independent.
22
;
1
Now it remains to
-1.
distinct,
are linearly
X.1Xj ,Xk
be any scalars EJF.
Suppose
then
3
1 = a. + a.
1
7
X.
2
-1 = 2
3
3
Xj
and
a. = -a..
1
k
ak = a.1,
(9)
.
the components of
there exists a row
a
If
=
r-3
a.,a.
Xk = a.X. + a.X.;
such that
22
<
therefore
;
i,j,k
show that for
{1,-1},
s
r-5
r-7
3
s'+s'+s-1 < 2 -2
Since
odd, rank
r
r-1
r+1
A =
Similarly, for
.
(ak' a
j'
X1.,Xj ,Xk
of
)
are in
[Xk X.,Xj
]
So we have
= a.a. + a.a. = a.a. - a.a..
ii
3
then (since
ii
3
3
1
ai0)
(10)
1
Combine (9) and (10), we get
a=0.
Similarly, if
aOai,
then ak=-a.andhencea.=0.
Both cases contradict
1
1
that
Xk
3
' X.,X.
are distinct.
1
22
13
element is
Let
f(x)
f(x).
of
whose (i,j)--
3
sgn a =
Let
Fnxm
a1..
in U [x],
then
R,
be any number in
a
th
is a matrix in
A = (a..)
Notation:
1
if
a > 0
0
if
a = 0
if
a < 0
then
.
.th
is the
f(j) (x)
3
-- derivative
We will denote
f
(1)
(x)
by
fl(x).
a. = a1a2a3an
11
1i=1
Definition:
LetA=(a..)be
in
13
is the matrix
B = (b..)
all
b.. = sgn a..
such that
13
then
Rnxm;
13
13
f(x) =
Let
R[x],
(x-Ai)
II
where
i=1
real for
Assume
i =
for
i
Lemma 2.6:
Let
sgn A
= 1,21,r,
X1>A2>...Ar
nonreal for
and consider
0
1
Then the
i = r+1,
f(X)n_.
f(x+t) =
j=0
X. = sgn(f (A.),f 1(A.),...,f
1
n-1
1
is
A.1
Xi's
(A..))t
for
are distinct, and hence
are pairwise linearly independent.
23
We will use the following theorems in the theory
Proof:
of equations without proof.
degree
f(a) 0, f(b) O.
and
n
a,b E R,
Let
Budan's Theorem:
f(x)
E R[x]
Let
V(a)
the number of variations of signs of
f(2)(x),,fn(x)
have been discarded.
f(x)=0
to
V(a)-V(b)
for
between
X1
f(x), f'(x),
vanishing
terms
and
b is less than or equal
by a nonnegative even integer [1].
Rolle's Theorem:
f(Xi)=0
denote
Then the number of real roots
a
between
of
after
x=a,
for
of
Let
i=1,2.
and
E R[x],
f(x)
X11X2 E IR
Then there exists
c
c E IR,
f'(c)=0 [1].
such that
X2'
and
First note that sgn (f0(Ai"fi(Ai)''fn-1(Ai))
= sgn (f(n)(X.),
f(n-1)(Xi ),,f(2)(Xi ),f'(Xi )). Since
1
are distinct for i=1,2,.,r, then f'(Xi) 0
the Xi 's
i=1,2,---,r.
for
f(A) = f(Xci) = 0,
Clearly
X
<
P
Consider
such that
<
X
,
P
X
,
where
r>p>q>1.
q
hence there exists
f' (.) = 0
li
E IR,
(Rolle's Theorem).
Now
Xq
apply Budan's Theorem to
f' (x)
and consider the following
sequences,
f'(X
)1 f(2)(A )1ff(n) (A
)
;
24
f(2)(X ),, f(n)(X
fi(X ),
g
Then
V(A )-V(A )#0,
existence of
-
)
otherwise it will contradict the
Hence we have sgn(f1(A ),f(2)(A
13.
f(n) (A ))#sgn(fy(A ),f(2)(X ),...,f(n) (A )), for any
p,q, 1 < p,q < r
If there exists
and
a E IR
v i.
nearly independent for
f0
Therefore
XP
=aX
such that
#X
for
,
then
and
Xq
p#(1.
a=1 be-
XP
cause sgn f0(A1)=1
Remark:
p#q.
Therefore
p#q.
Xp
are li-
1
(x) = 1.
Let
Proposition 2.7:
f1 (x)
fo(x), f1 (x),
(x)
be as
n-1
in Lemma 2.6.
Let
P = {g(x)Ig(x) =
II
f1(x)ii,
i=1
ji E {0,1,2}1.
g(x)
Then there exist
gi(x), g2(x),...,
such that the matrix
P
gi(xl) gl(x2)
g2(2'1) g2(A2)
gi(xr)
g2(A)
sgn
gr(A1) gr(A2)
gr(Ar)
is nonsingular.
Proof:
Let
Xi = sgn(f0(Xi), f1(Xi".**'fn-1(Ai))t*
Then
25
the
X.1's
are pairwise linearly independent (Lemma 2.6).
f (X.)=1
1
0
Since
and
f
n-1
the first row
(X.)=(-1)i-1,
and the last row of
[X1,X2,...,Xr]
(1,-1,1,...,(-1)r-l)
respectively.
are
and
(1,1,...,1)
So (by Lemma 2.4)
r-1
there exists positive integer
2-1
s < 2
if
is odd;
r
r-4
s <
3.22_l
for
h.. E (0,1,21
r is even; and numbers
if
i E {0,1,,n-1
and
13
E {1,2,,s}
j
such that,
with
n-1
.
n-1
h.
=
sgn fi(Xk)
j
X,
E {0,1,2,,s}
has rank
r.
and
k E {1,2,...,r},
a22X2
a2rXr
a51X1
as2X2
a
r
3
where
t.,1
the matrix
(12)
X
sr r
linearly independent rows of the mat-
rix (12) and denote them by Al,A2,,Ar
A, = sgn(a
13
k
)
X2
a21X1
Pick
sgn f.(X
1
i=1
i=0
for
h..
lj .
f
q(X 1 ),a t.,2 f
1<t.<s, 0<q.<n-1
(X
.
.
3
q3
for
2
),...,at.,rfq.(Xr))
3
3
j=1,2,...,r.
n-1
a
f
(X ) =( II
sgn f. (X )
m
t.,m q.
1 m
i=1
3
3
Then
So we have
hi,t.
3)fci.
3
(Xm
)
26
j=1,2,,r.
for
f0(Am)=1 V m)
Hence (since
sgn ati,mfgj(Am) =
sgn f.(A
II
1
i=1
where
h!.
E {0,1,2}.
13
.
n-1
.00 =
g3
i=1
m
13
)
Let
h!.
f.(x) 13
1
.th
Then the 3 row of the matrix (11) is
sgn(gj(A1),gi(A2),...,gi(Ar))
h!
n-1
n-1
= sgn( R f. (A1)1j, R f
1
i=1
.(A2
)
n-1
h!.
R fi(Ar) 17)
i=1
h!.
13
i=1
= sgn(at.,ifg. (Al"at. 2fg (A2)'...'Xt
rfq.(Xr"
3,
7
= A..
Therefore matrix (11) =[A A
A ]t
and hence non-
singular.
Corollary 2.8:
Let
fo(x),f1(x),,fn_1(x)
n-1
Lemma 2.6.
Let
P' = {g(x)g(x) =
E {0,1}1.
all
iE{0,1,2,...,n-1}
polynomials
If
and all
P'
in
ji
f.(x)
i=1
1
11
fi(Xj)
with
is nonzero for
jE{1,2,,r1,
gi(x),g2(x),,gr(x)
be chosen from
be as
then the
of Proposition 2.7 can
(which is a subset of
P).
27
=sgn f.(X.)
Here sgn f.1 (X.)2=1
3
3
Proof:
and all
i E {0,1,2,---,n-1}
n-1
g(x) =
h(x)
ji=0
f.(x)
i=1 1
j.1
63,
IT
n-1
n
i=1
=
E
j
PIF
then there is an
where
Let
correspond in this way to the
is sgn hi(xj).
Thus if
ji=ji
if
ji=1
and
such that sgn g(A) = sgn h(X)
E 11,2,,r1.
of Proposition 2.7.
for all
{1,2,,r}.
344
ji E {0,2},
if
for all
f400
j
0
hi(x), h2(x),,hr(x)
gi(x), g2(x,
g(x)
The (i,j)thentry of the matrix (11)
28
CHAPTER III
MAIN RESULT
(S21T2)
Consider two nonsingular pairs, (S11T1) and
_
of symmetric matrices in
lar to
-1
S2 T2.
nomial.
Let
Fluxm
such that
S11T1
is simi-
Then they have the same characteristic polybe this characteristic polynomial and
p(x)
be the greatest common divisor of
gcd (p(x),p1(x))
Define
and p' (x).
f( x) -
Then we have
f(x) =
p(x)
p(x)
gcd (p (x) ,p ' (x) )
II
(x -Xi),
where the
X.'s are the
i=1
distinct real characteristic roots of
-1
S1 T1and S2 T2
for i=1,2,,r and are the distinct nonreal roots for
Let us assume that X1>X2>>Xr and let
i=r+1,r+2,,n.
fo(x),f1(x),,fn_1(x)
and the polynomials
j=1,2,,r.
nomials
fk(x)
Then
and
be the polynomials as in Lemma 2.6
g(x)
g(x)
as in Proposition 2.9 for
is a finite product of the poly-
29
g1(X1)
g1(A2)
g1(Xr)
g2(A1)
g2(X2)
g2
(A
sgn
g(A2)
2r(A1)
*** gr(Ar),_
is nonsingular (by Proposition 2.8).
Now we will state an effective criterion for congruence of nonsingular real symmetric pairs in the following theorem.
be
(S11T1), (S2,T2)
Let
Theorem 3.1:
real symmetric (or hermitian) pairs.
n-1
= {g(x)g(x) =
f(x)
II
i=1
Then ST1)
to
-1
(S2'T2)
j4
mxm
nonsingular
Let
ji
10,1,21}.
1
if and only if
S11T1
is similar
and
S2 T2
-1
sig Sif(S, Tl)k g(Si-1 T1)=sigS2f(S2-1 T2)k g(S2-1 T2)
for all
g(x)
and
P
k=0,1,2,,m-1.
We will prove this theorem for the real symmetric
case only.
For the hermitian case, the proof is similar to
the real case.
The following lemmas will be used in the
proof of this theorem.
Lemma 3.2:
Let
(S,T)
hermitian) pair and
be a nonsingular real symmetric (or
f(x)
E R[x].
-1
Then Sf(ST)
is real
30
symmetric (or hermitian).
Let
Proof:
E R[x].
f(x)
Then
f(x) =
k=0
a's
the
=
R.
Let
Let
ak(At)k.
k=0
A
Rmxm
be in
then
,
akxk,
with
f(A)
be any natural number; then
k
(S(S-1T)k = (TS-1TS-1.TS-1T)t
=
Tt(S-1)tTt.--(S-1)tTt(S-1)tTt
= TS 1T...S-1TS-1T
= S(S -1T)k.
Thus
(Sf(S-1T))t=(1
a S(S-1T)k
al,S(S-1T)k)t= 1 ak(S(S-1T)k)t=
k=0 k
k=0
k=0 j"
= sf(s-1T).
Lemma 3.3:
E+
Let
Let
A
A
hermitian matrix, det AA.
be a maximal subspace of
positive definite on
that
be an nxn
E+, E-
x*Ax=0
such that
A
is
be a maximal subspace such
is negative definite on
subspace such that
Cnxl
E-
for all
and
N
x E N.
be a maximal
Then
dim N < minimum (dim E+, dim E-)
where
Proof:
E+
means the dimension of
dim S
Clearly
+NCCl,
nx
S
for any space
E+nN=E nN= E+nE = 0.
S.
Since
we have
+
+
n > dim(E +N) = dim E+ dim N
(by Fact 1.11)
31
By Fact 1.10 and 1.11, we
Similarly, dim E-+ dim N < n.
dim E+ + dim E
have
= n.
Thus
dim N < n-dim E+ = dim E
= dim E+
dim N < n-dim E
Hence
dim N < minimum (dim E+, dim E).
Lemma 3.4:
;
.
I
Let
(:::-/) a 23.1
a2
1
A=
a2
al:- - a2... -an
a. ERyi and a10.
Then
sgn al
if
n
is odd
if
n
is even.
sig A=
0
Suppose
Proof:
of
Rnxl
Note that
1
A
et
ek+1 = a1.
k+1
If
be the subspace
,
where
.th
for
j=1,2,...,k
and
7
a1
> 0,
then
Ek+1
n E
= O.
dim E- < n-dim Ek+1 = 2k+1-(k+1) = k.
Also by Lemma 3.3, we have
2k+1-dim
e.
on the j-- component and O's else-
e.Ae.=0
3
E
e11e2...e
spanned by the vectors
is the vector with
where.
Let
is odd.
n=2k+1
= dim E- > dim N > dim Ek = k.
Hence
32
Thus dim E
= k
dim E+ = k+1.
and
Therefore
sig A = k+1-k = 1 = sgn al.
Similarly, if
Hence
al < 0,
then dim E- = k+1; dim E+ = k.
sig A = -1 = sgn al.
Suppose
n = 2k
dim E
dim E
dim E- = 2k,
then dim E
Thus dim E+ = dim E
< k.
dim E
and
fore by Lemma 3.3, dim E+ > k
There-
Then dim N > k.
is even.
Since
> k.
2k-k = k
and
and so sig A =
= k
0.1
The above result is also proved by Haynsworth and
Ostrowski [4].
Corollary 3.5:
Let
a'
1
'
':1
,
A =
a2
al
and let
2
a:
a'
2
be the first nonzero number of the sequence
ak
a11a2n.
Then
sgn ak
if
n-k+1
is odd
if
n-k+1
is even.
sig A =
0
33
Proof:
Write
1
A=
.ak
1
a'
k
1
1
ak+1
.
1
.ak+1.
.
1
..
ak---ak+1.
i
n-k+1.
a
n-
and Fact 1.9,
Then by Lemma 3.4
k
k
..a..
sig A = sig
an
sgn ak
0
f(X) = 0
is odd
if
n-k+1
is even .
k > q
for
(since
Define
Jc1=0
J
and
X
e R
with
and
sig Ef(XI+J)q-ig(XI+J)
Proof:
and
Let E=E,J=J,I=I. Then
f'(X).
f(XI+J)k = 0
n-k+1
f(x), g(x) E R[x]
Let
Lemma 3.6:
if
0
= I.
= sgn f"(X)q-ig(X).
By Taylor's Theorem, we have
f(XI) = 0)
34
q-1 f(9)0,I1
" J
f(XI+J) =
9!
9=1
q-1 f
( 9, )
(X)
JR.
2,=1
q-2
f(k) (A)
/
(9,+l) If' (A)
= jf (A)
k=0
Jk
q-2
= J
c J
(X)
9,=0
where
f(X)
co =
and
ct - (9+1)1f'(X)
q-2
f(XI+J)
= J ft (X)
c J )
(
Hence
1.
k
.
9=0
Since
all
Jk = 0
k > q.
for all
Let
we have
k > q,
for
Then
E R[x].
g(x)
f(XI+J)k = 0
Ef(XI+J)g-ig(XI+J)
= EJg-ifV(X)
q-1 g-2
(
q-2
CJ 9)-a- 1(g(X)I+q)I
(p+1)
g
p=0
9=0
(A)
(p+1)!
q-2
=
EJq-1V(X)q-1g(X)
c
(
J9)q-1
9=0
q-1(17-2
EJgft (A)
c J
2,q-1,c1-2 g(P+1) (A)
)
p=0
9=0
The second term drops out because
Jg=0
Ef(XI+J)g-ig(XI+J)
q-2
=
EJc1-1V(X)q-19.00(
c Jk)
9=0
JP)
(p+1)!
q-1
Thus
.
JP)
35
= EJ(4-if'(X)q-ig(X)crlIci-1
f,(x)q-lgmajq-1.
yxq
The matrix in (13) is a
is
(13)
matrix whose
(q,q). entry
and whose other entries are all zero.
f'(X)q-ig(A)
Hence
sig Ef(XI+J)q-ig(AI+J) = sgnf'(X)q-ig(A).
Lemma 3.7:
(S11T1), (S21T2) be
pairs.
(S11211) = (S2,T2),
If
junctive to
Proof:
nxn
then
for all f(x)
S2 f(S2-1T2 )
(S11T1) ; (S2T2),
Since
C Enxn such that
S2 = C*S1C
nonsingular hermitian
-1
Sif(Si TI)
is con-
E R[x].
there exists nonsingular
and
T2 = C*T1C.
Hence
-
S2f(S21T2)
1C*T1 C)
= C*S1 Cf(C-1S-1(C*)
1
= C*S1CC-1f(ST1)C
-1
= C*S1f(S1 T1)C.
Lemma 3.8:
If
pair such that
p(x)
E R[x],
Proof:
(S,T)
S-1T
is an
nxn
nonsingular hermitian
has no real characteristic roots, and
then sig Sp(S-1T) = 0.
Without loss of generality (by Fact 1.9) we may
assume that
S-1T
has only one pair of complex conjugate
36
S
1T
(XI +J
q q
is similar to
(S,T)
By Theorem 1.2,
)
+ (TI
(A,B),
=4-
(x-T)q.
and
(x-X)q
nonreal elementary divisors,
+J ),
where
Then
2q = n.
q
q
where
E
A =
0
E
woo,
Eq(TIq+Jq)
B=
Let
3.2)
(XIq +Jq
Sp(5-1T)
Then
E R[x].
p(x)
0
)
[0Eq
is hermitian (by Lemma
and by Lemma 3.7
rQE
Sp(S 1T) = C*Ap(A 1B)C
qp(TI q
+J
)
q
= C* [EP (XI +J )
q q
q
Then
K* = E p(TI +J
q
in
C
for some nonsingular
Tnxn.
Hence
Let
Sp(S 1T)
K=E p(XI +J
q
is conjunctive
q).
to
M =
Also we have
K*
0-
1
0
0
K*
I
[0
-I
K
0
0 -I
q).
K
0
37
Hence
is conjunctive to
Sp(S-1T)
Corollary 3.9:
pair.
If
Let
(S,T) =
(S,T)
(Ale
A2,
are matrices such that
-1
A2 B2
Thus sig M=0.
sig M = sig(-M) = -sig M.
we have
M,
nxn
be
sig Sp(S-1T)=0.I
nonsingular hermitian
Bap B2),
where
A1,A2,1311B2
has only real roots and
A-11B1
are only nonreal roots, then
-1
sig Alf(Al Bl)
Since
sig Sf(S-1T) =
E R[x].
for all f(x)
sig S f(S 1T)
Proof:
= sig(Ai
A2) fc (A-11
0
A-21) (Bi
0 B2) ) (by Fact 1.12)
1
= sig A1f(A1 B1)+sig A2f(A2-1 B2)
1
= sig Alf(Al Bl)
(by Fact 1.8, 1.9)
(by Lemma 3.8)
.
Proof of Theorem 3.1:
Suppose
for some nonsingular
C E Rmxm.
CtS1 C = S2
and
I
CtT1 C = T2
Then
S21T2 = (CtS1 C)-1CtT1 C
t
(C
=C -1S-1
1
)
-1 CtTC
1
-
= C1 S1-1 T1C
so
S1-1 T1
and
S2-1 T2
are similar.
Let
f(x), g(x)
be
as in Theorem 3.1; then (by Lemma 3.7) S1f(ST1)kg(ST1)
38
S2f(S2-1 T2)k g(S2-1 T2)
is congruent to
and for all
Hence
P.
g(x)
k=0,1,2,.,m-1
for
sig S1f(ST1T1)kg(S11T1)=sig S2f(S-21T2)kg(S-21T2).
Conversely, if
sig
SilTi
Sif(Si-1 TI)k g(Si-1 T1)=sig
k=0,1,2,,m-1 and all
-1
-
S21T2
and
are similar and
S2f(S2-1 T2)k g(S2-1 T2)
P,
g(x)
for
SI1T1
then
and
have the same real-Jordan canonical form (by Theo-
S2 T2
rem 3.1) ,
k.
q
m
r
A.I
EDe
i=1 q=1 j=1 1 a
M.
+
ED
It is clear from Theorem 1.4 that
(ST1)
=
(S2'T2)
= (A'
A2' B1
9 B2)
B1
BI)
2
A''
1
1
have only nonreal roots.
(A1)-1131
2
2
2
'
have only real roots and
1B
A1 1B1, (A')
1
1
where
ED
(A1
)
Let (A1 ,B1'
A21B2'
B1)
1
1
(A'1
be congruent to
r
m
k.
qi
k.
ql
m
r
9
.E
e j=1
e q=1
9 q=1 j=, iqjqi=1
i=1
,
( r
m
k.
qi
$
ED
i=1 q=1 j=1
r
c!.
iqj E q
0
,
m
k.
qi
1=1 q=1 j=1
.
E.
.E
cliqq
(X .I +J )
iqjqiqq
.
respectively, whereE
ci qj' cliqj{1,-1}
E (A .I +J
for
)
v i,q,j
39
Let
(by Theorem 1.4).
g1(x),g2(x),,gr(x)
be in
6'
such that
g1(A1)
g1(x2)
g2(A1)
g2(X2)
g(X1)
gr(A2)
g1(X)
g2(X)
sgn
gr(Ar)
Then
is nonsingular (by Proposition 2.7).
-1
m-1
-1
gs(Si T1)
sig Sif(Si Tl)
(by Corollary 3.9)
= sig A1f(A11B1)m-1g5(A11B1)
m
®E
f(X.I
+J
lqg
Jci
i=1 q=1 j=1
kqi
r
= sig( ee e
E.
1c1
)m-1gs(XiIq+Jc1))
(by Fact 1.12)
kqi
m
r
C.
y
i=1 q=1 j=1
=
1'43
V.
k
sig[y(xiiqi-Jq)m-1gso.iyiy]
.
(by Fact 1.9)
.
TiE i mj
sgn(f1(Xi)m-1qs(Xi))
.
(by Lemma 3.6)
i=1 j=1
-1
Similarly, sig S2f(S2 T2)
-1
m-1 g5(S2
T2)
k.
mi
16'.
i=1 j=1
r
.
im3
Define
y'
G
sp
by
sgn(fl(xi)m-1gs(x
)).
40
k51
.
y.
lq
=
E.
igj
j=1
kqi
Ye
=
1q3
j=1
"I
-1
a
(S,
= sig S1 f(S1-1T1 )p-1gs1
sp
Ti)
,p-1
(s-1T ,.
-1
= sig S f(S.., T2)
gs' 2 2)
2
Hence we have
z
r
sgn(f1(Xi)m-1gs(Xi)) = a SM
y.
1
1m
i=1
The coefiicient matrix of this system
for s=1,2,,r.
of equations is
.1
Xl)
g1(X1)
ft"2)
g2(X1)
ft "2)
sgn
gr(X1)
ft "2)
Since
f'(X.) 0 V i,
1
by Proposition 2.7.
g2(Ar)
y'
1
Hence
Yl,m-1
Similarly we have
sgn(f'(Xi)m-1gs(Xi) = asm
yim = yim
for i=1,2,,r.
= V!
we first observe that
as,m-i ,
gr(Ar)
then the matrix (14) is nonsingular
lm
'1,m-1,
(14)
m-1
gr(A2)ft (Ar)
r
i=1
g(A)
m-1
g2(X2)ft(Ar)
m-1
m-1
ft (X1)
gl(X2)wft(Xr)
m-1
m-1
ft(A1)
m-1
m-1
m-1
f'
To show that
= sig Sif(S11T1)m-2g5(ST1)
=c.
m
r
41
kqi
sig(E f(X.I +J )m-2g (X.I +J ))
4
q
IqJ
i=1 q=m-1 j=1
r km-1,i
ci,m_1,j sig(Em_if(XiIm_i+Jm_i)m-2gs(AiIm_i+Jm_i))
=
i=1 j=1
k.
r
+
sig E f(X.I +J )
1 m m
m
Eimj
1
X
i=1 j=1
Define
m-2
g (X.I
+J
1 m m
s
by
a.
iqps
a.
= sig Ecif(XiIci+Jci)p-1gs(XiIci+Jci).
3.6)
a.
iqps
ipps
)
and we have
= sgn f'(Xi)p-1gs(Ai)
r
as,m-1 =
km-1,i
13 1-1,s
c.,m-1, .
j=1
i=1
r
Then (by Lemma
a.,m-1,m
k.
ma.
+
X
i=1 j=1
a.
c.
1 ,m,m-1,s
lmj
r
=
sgn(f'(A )m-2g (A )) +
y.
A
i=1
=1
i=i
im
a.
1,m,m-1,s
Hence
as,m-1
=
i=1
Yimai,m,m-1,s 1=1
.
Here,
s=1,2,...,r.
for
y!
as,m-1 i=1
Since
yim = Yim,
y!
a.
'
we have
'
is fixed.
rn
im 1,m-1 s
v.
'1 m-1sgn(f1(A.i)m-2gs(Xi))
.
1=1
1,m-1
Similarly
Si
sgn(f1(A.)m-2g
(A)).
1
42
i=1
for
Yi,m-1sgn(fy(Ai)m-2gs(Ai) =y1,m-1sgn(f1(Ai)m-2gs(Xi))
The coefficient matrix (whose (i,j)th
s=1,2,...,r.
entry is
W(A.)m-2gi(Aj)))
is the same as the matrix
7
obtained from (14) by replacing
f'
(A.)"1
by
f'
(Xi)m-2,
HenceY .i,m-1 = y!
11,m-1.
and thus it is nonsingular.
In
general
as,m-k =
i= 1 Yi
sgn(f' Uti)m-k-1 gs(Xi"
m-k
'
r
m
i=1
q=m-k+1
+
a.
=
y!
l'm-k
i=1
Y.lq
sgn(f'(Ai)m-k-1g s(A.))
M
r
+
1,q,m-k,s
y
a .
q=m_k4.1 lfq,m-k,s
i=1
for s=1,2,,r.
Since the matrix
m-k-1
f'(A1)m-k-1gi(xl)
tx
ft(A 1 )m-k-1 2' 11
2
f' (x2)m-k-1 gl (A)
r
(A)
gl " r)
Xr)
(Ar)
f,tx Im-k-1
'
2'
g2 X2)
I
sgn
-f'(A 1
)m-k-1 gr(X1) fl(X2)m-k-1 gr(A2)
(A )m-k-lgr(Ar)
is nonsingular (by Proposition 2.7), we get
Yi,m-k = Yi,m-k
successively for
(S11T1) ; (S21T2).
k =
Therefore
43
Then
Cnxn.
S
and
(T1
,S)
(S-1,T)
S, T
Let
Corollary 3.10:
be nonsingular hermitian in
are isoconjunctive if and only if
T
if and only if
sig S-if(ST)kg(ST) = sig T-if(TS)kg(TS)
and
k = 0,1,2,,n-1 and for all
and
g(x)
TS
for
(Here
E P.
are defined as above except that
P
f(x)
is replaced
S
-1
S1.)
by
Let
A
...
A
l' 2" n'
nxn
be an
(S,T)
pair such that
A
is similar to
ST
nonsingular real symmetric
S-1T
has
n
where
the
A.'s are real for
and nonreal for
distinct characteristic roots,
i = r+1,
Let
n.
p(x) =
(x-Ai).
i=1
Then
A
p(x)
is the characteristic polynomial of
be the companion matrix of
lar to
A.
such that
Let
Also let
0
be as in Lemma 2.6.
Then
S-1T
Let
is simi-
be any symmetric matrices such
S0' T0
S 1T0 = A.
p(x).
S-1T.
Suppose
f0
(x)
f(A1) 0
f1 "
(x)
'
v i,j;
fn-1
(x)
then proceed
as in Corollary 2.8 to obtain the set
n-1
P' = {g(x) Ig(x) =
0
v
g(x)
E P',
Clearly
1
1=1
g(X.)
j.
f. (x) 1,j. E {0,1}}.
II
v
i.
Also, since the characteristic
44
roots of
then
(Sog(A))-
E P',
every
g(x)
= g(A)
lAg(A)
= S-1T
= A
for all
Lemma 2.7.
If
(Socog(A),
-1
f.(X.)
(Tog(A))
(Sog(A))
be a symmetric nonsingular
(S,T)
fo(x), fi(x),
Rnxn;
g(A)-1(s-01T0)g(A)
E P'.
g(x)
Let
Thus for
E P'.
g(x)
1 (Tog(A))
Therefore
= A.
Proposition 3.11:
to
(by Fact 1.13), i=1,2,,n,
0
is nonsingular for all
g(A)
pair in
g(Ai)f
are
g(A)
be as in
fn-1(x)
then
0 V j,i,
(S,T)
is congruent
3
over
Tocog(A)
IR
for some
g(x)
P',
We will first prove the following lemmas.
al3.=-1
and
V j.
al.
13
Then
A'
be in
Let A = (a..)
13
Lemma 3.12:
=
13
(1
if
to
if=
A
a..=±1
13
a..
=-1
13
.
13
is nonsingular.
A
is nonsingular.
A
are linearly independent.
R
and suppose
a.a!
j=1
with
where
A' = (at.),
Let
is nonsingular if
Proof: Suppose
Rnxn,
3
.
13
Let
= 0
Then the columns of
al, a2,...,an
be in
45
for
Then let
i =
= 1}
K. = {j/a!
1
So for
j
Ki,
1
a.a!
a! .=0.
13
Hence we have
n
i
j
EK1
ii
3
a.a
1
3
I
=
a. = 0
j=1
I
=
2j
a. = 0
jEK2
EK2
I
=
3
(13a)
3
a. = 0
3
j
EKn
EEC
Now multiply each equation in (13a) by
We get
-1.
ctO.
(14a)
j EKi
But
a..=
-1
13
I
so (14a) becomes
E K.,
1
j
(15)
a. .a. = 0.
1J
4 _az
J
for
3
-,'"i_
Subtract each equation in (13) from
Y,
j=1
a. = 0.
Then
3
we have
I
a. -
j=1
3
IsKi
We know that if
for
j
4 Ki.
I
j
a. =
3
Ki,
Y
xi
then
a. = 0.
(16)
3
al. = 0.
13
Therefore (16) becomes
So
a.. = 1
13
46
a..a. =
13 3
(15) + (17)
jK.
3
i = 1,2,,n.
a. = 0
0
A
Since
j = 1,2,, n.
for
a. .a. = 0
13 3
3 13
j=1
for
V
j K.
a.a.. =
(17)
.
+
a. .a.
13
0
we have
is nonsingular,
Therefore
is also
A'
3
nonsingular.
Lemma 3.13:
F = {1,-1}, F' = {1,0}.
Let
Then the fol-
lowing are true.
are groups with respect to multiplication and
F,F'
1+1 = 0.
addition respectively in which we define
Define
f: F
F'
f(1) = 0; f(-1) = 1.
by
Then
f
is a group isomorphism.
Define
f*: Fr
by
(F')r
f*(xl,x2,...,xr) = (f(x1),f(x2),,f(xr))
where
F
Fr
(resp Fir)
(resp F').
Then
is the product of
f*
copies of
is a group isomorphism,
where multiplication (resp addition) in
(F')r) is componentwise.
Proof:
n
Immediate by the definitions.
Fr
.(resp
47
Proof of Proposition 3.11:
gruent to
(D0' J0
We know that
)
(S0 ,T0
is con-
where
)
e
Do =
E2)
i=1
Jo = diag
1A1iE2A2I
e
1ErAr
E2Bi)
i=1
and
c!
By proposition 2.8,
E {1,-1} y i.
g1 (x) = f0(x),g2(x),...,gr(x)
G = sgn
is nonsingular.
E Pi
such that
g1(A1)
g1(x2) "' gi(xr)
g2(x1)
g2(x2)
r(A1)
gr(A2)
Then
g2(A)
g r(A r
G' = (gli),
for
g!. = f(sgn g.(A.))
there exist
)
where
i > 2,
gli = f(-sgn gi(Xj)) = f(-1) = 1
is nonsingular (by Lemma 3.10).
g(x)
Define
for
i > 2
for
i = 1.
hi(x) =
If we consider
the rows of
a. s E F',
1
G'
(FI)r
as a vector space over
form a basis for
we have
f*(xx2," Xr)
(Fl)r.
F',
For some
then
48
f(xr))
= (f(x1),
=
1=1
ai(gi1'g12'...1gir)
= al(f(-sgn g1(X1)),f(-sgn g1(X2)),..-,f(-sgn gl(Xr)))
ai(f(sgn gi(X1)),f(sgn gi(A2)),,f(sgn gi
+
) ) )
i=2
= al f*(-sgn g1(A1), -sgn g1(X2),..., -sgn gl(Ar))
ai f*(sgn g1(X1), sgn gi(X2),-..,sgn gi(Ar))
+
i=2
=1
f*(sgn(hi(X1), hi(X2),..., hi(Ar)))
1=1
a.
= f*(
H
sgn(hi(X1), hi(A2),..-,hi(Ar)) 1
.
1=1
Therefore (by Lemma 3.11 (c))
a.
(x1,x2n) =
11
(h. (A ),h.(A
II
1=1
1
1
2
r
))
ai
Let
q(x) =
H
1=1
hi (x).
Then
q(x) = (-gi(x))
ai
r
al
gi(x)
H
i=2
al
= f-f°'
fxl`
a
= (-1)
"
1
n-1
r
(
"L.
i=2
1.1
j=1
4
f
(x)
j
84.a.
n-1
f.(x) J1 1
11
n
i=2 j=1
r
jj")
a-;
-1"
49
where
Let
E {0,1}.
31
=
y.
3
X
i=2
Then
31
al n-1
q(x) = (-1)
f.(x)
II
j=1
3
and
sgn q(Ai) = (-1)
al n-1
II
sgn f.(A
j=1
= (-1)
y. mod 2.
y!
h(x) =
Then
Yj mod 2
sgn
(-1)a1 n1711 f.(x)
h(x) = ±g(x),
= sgn q(Ai).
fj(X)
Let
E {0,1}.
yi
j=1
Then
)
n-1
a1
j1
:
Let
i
.
3
where
g(x)
Also sgn h(Xi)
E P'.
Thus
(xl,x21.--,xr)
=
sgn(hi(X1),hi(X2),---,hip.r))
II
a.1
i=1
a.
= sgn
1,,
(
1=1
r
i1
=
r
a.
h.(A
1
h.(X
)
2
a.
'
i=1
1
r
)
1)
= sgn (q(X1), q(A2),-..,q(Ar))
= sgn (h(A1), h(A2),-.., h(Ar))
cog(Ar)),
= sgn (c0g(X1), cg(X2),
where
c
0
=
(-1)al E {1,-1}.
Let
(S,T)
be congruent to
50
(D,J)
over
where (by Theorem 1.4)
R,
t
D = diag(c1,621-..,cr,
et E2)
j=1
t
J = diag (c1X1, c2X2,...,crArai. E2Bi)
,
-b.-
-a,
3
]
B. =
1
3
a.
b.
- 3
3-
e the non1
3
real roots of
xi,x21,xr
S 1T
for
j=1,2,--,t
3
and
r+2t = n.
Let
such that
E F
(61'62'...'cr) = (El'62'...'q )(xl'x2'...'xr)*
Then
,Er).= (662'...'q)sgn(60g(Xl"E0g(X2"
-
.-.,cog(Ar)).
Hence we havec.=c!csgng(A.)1/
i < r.
1
0
1
1
Now consider,
for
k=1,2,-..,r,
sig(c0S0g(A))gk((c0S0g(A))-1(c0T0g(A)))
= sig coSog(A)gk(A)
-1
= co sig Dog(D-01Jo)gk(Do Jo)
(by Lemma 3.7)
= cosig(diag(c1,,cpdiag(g(X1),g(A2),..-,
g(Ar))diag(gk(A1),...,gk(Xr)))
51
=E
0
!
E.
i1
=
=
sgn(g(X.)g k(?"))
sgn(cocig(Xi)gk(Xi))
i=1
=
!
E. sgn g(Ai)
i=1
sig(diag(el,c2,,cr)diag(gk(A1),gk(A2),...,gk(Ar)))
=
(ST)
= sig S
Therefore.
Therefore (by Theorem 3.1)
(coSog(A), coTog(A)) = (S,T)
Proposition 3.12:
Let
S-1T
If
real symmetric pair.
nxn
be an
(S,T)
n
has
nonsingular
distinct roots,
then
(S,T) =
where
-
S.1T.
3
ED
S.,
$
T.)
j=1
3
i=1
3
is a nonsingular real symmetric pair and
(S.,T.)
7
(
7
is a companion matrix for 3=1,2,,m.
3
Proof:
Let
f(x) =
(x-Ai)
II
be the characteristic
p0-
i=1
lynomialofS-1T,wheretheX.'s
i=1,2,,r and nonreal for
gcd (f(x), f(j)(x)) = 1
= f
.(X.)
n-3
for
0 V i
are real for
If
i=r+1, r+2,
j=1,2,,n-1,
then
and therefore the proposition
52
such that gcd (f(x), f(j)(x))
j, 1<j<n-1
exists
Suppose there
m=1) by Proposition 3.11.
is proved (with
thereexistsiforwhichf(j)(A.)= 0.
g1
and let
(x) = gcd(f(c), f(j)(x))
because
f(x)
gcd(g1(x),
has only simple roots.
g1(j)(x)) = 1
is the degree of
not, let
for
such
al[x]
Now if
where
j=1,2,,d1-1,
1 <
Hence, eventually in this way,
pi(x).
such that
gcd(hi(x), hi(x)(j)) = 1
< d.-1,
j
and
d.
f(x) =
h(x)
II
1
v
j, i,
where
is the degree of
hi(x).
Let
be any nonsingular real symmetric pair such that
(M11N.)
3
-1
M. N. = A.
7
is the companion matrix of
Let
h. (x).
3
(MN) be congruent to
where
(W.,V.)
3
3
If
Apply
i=1
and
d1
g1 (x).
and so on.
g2(x) = gcd(gi(x), gi(j)(x))
h.(x), i=1,2,...,m
we get
e
then we are done with
g1 (x),
the same process to
Then let
Clearly gcd(gi(x), pi(x)) = 1
f(x) = gi(x)pi(x).
that
p1(x)
1, i.e.
3
t.
D! = diag(c!
,E1
,,61.3
31
j2
3
,
Pj
e
k=1
E 2 B 2,3
'
t.
,Et X.
KI = diag(0.X.
31 31 32 32
X.
3p. 3p.
,
e
x-1
x3
53
atj
B.=
azt
c'
jk
E
and the numbers
{1,-1) V j,k
ncmreal moots of
Let
A-.
(S,T)
(a
.)
kj ±ib23
are the
be congruent to
(D,J),
7
where
D = diag(61.62,,6r, R,151 E2)
0
J = diag(c1A1re2X2,.,crXr1 k=1
E
2B
)
.411,
-b
a
BR,
a2-
Then
{1,-1} V i.
and
1B1,B2,-..,Btl
Bzi
Ajz E
{Al1A21..-,Xr1
y 2,j.
By Proposition 3.11, there
exist
d.-1
k.
1
r.(x) = IIh(.c1/(x) 3c1
3
q=1
k.
and
c.1
E
7c1
0,11
such that
(D.,K.)
3
±(M.r.(A.), N.r.(A.)),
3
3
3
3
3
3
where
3
t.
D.
31
32
E
3
Pj
2=1
)
2
54
,.32
diag(Ej6
1
K.=
and
,,6.
where
TrEoNiri(Ai),
m
e
3
Then
E {1,-1}.
co
and
Si=60Mirj(Ai)
Let
j,k.
e
T.)
j=1
3
e
6 M.r.(A.),
(
j=i °
=
z=1 E 2B 2,3.),
,
in
S.,
j=1
=
7
3pi
E fel,E2,...,Erl
E.
t.
e
3
DA
j=1
c M.r.(A.))
0
j=1
3
3
e
K.)
j=1
3
3
3
3
= (D, J)
(S, T).
Minimal indices:
I
(ST1)
=
(S2,T2)
mogeneous parameters
p,A
over
R.
and consider two families
R
in
First we introduce ho-
of matrices, called pencil of matrices,
11S2+ T2.
If det (pS1+XT1)
real symmetric
We will find conditions for
singular pairs (S11T1),(S2T2).
which
nxn
Consider pairs of
and
and
1151-1-ATI
det (uS2+XT2)
are both
not identically equal to zero, then we transform the parameters by the formulas,
A =
a1X' + a2p'
P = 13,1A
where
ct1,133, a2,
Now consider
132
are in
+ 132PI
R
and
a02 - a2131
0.
55
pS
=
1
l
+ AT
1
S1+(a1A'+a211')T1
X'
= P'(2S1+a2T1)
X'(1S1-FalT1)'
Similarly,
pS2 + AT2 = p'(32S2+a2T2)+A'(1S2+aiT2).
We can choose
and
a2132
det(P.2S2 a2T2)
Si
det (2S1+a2T1)
such that
0.
0
Let
-= 2 (3s+aT
T'1
2 1
1
=
1
S'2= 132S2 + a2T2
Thus we can consider the pairs
T
;
P.S+
1 1
=
(ST)
a1T1
'
1S2 + a1T2.
and
which are already considered in Theorem 3.1.
(S,T)
The above
discussion can be found for example in [2, Vol. II, p.27].
It can be easily proved that
(ST1)
(S2'T2)
(S''
l T')
1
(S''
2
2 T').
if and only if
Now we will only consider the case where
det (pS1+AT1)
equal to zero.
)
(S2' T2'
and
det (pS2+XT2)
Obviously, if
then det1+AT1)
are both identically
(Si,Ti) is congruent to
and
det (pS2+AT2)have t3be
56
both zero or both nonzero.
Let
u
nomials in
be
lxn
a
vector whose components are poly-
Define
X,p.
u.(11,2)1
deg u = max
{degree of
1<i<n
where the
are the components of
u.'s
nxn
real symmetric matrices such that
all
X,p E
R.
L t
A,B
det(pA+XB)
Then there exists a nonzero vector
components are polynomials in
and hence
u.
X,p,
such that
E
u,
be
0
for
whose
u(pA+XB)=0
Let
(1.1A+AB)ut = 0.
P = {ulu(pA+AB) = 0}
m1
m1
=
minimum {deg u}
uEP
is defined because
deg ul = ml,
where
m2
where
Define inductively,
and
u1
E P.
Then let
and
u
R[A,P]
is linearly independent over
with
111}.
mk = minimum {deg 1}
uEPk-1
Plu
Pk-1 = {u
span, over the quotient field of
{ul,u2,...,uk_111;
Suppose
uEP 1
P1 = {ulu G P
5'o = P
is not empty.
= minimum {deg u},
the quotient field of
where
.
uk E P k_l
and
is not in the linear
R[X,p],
of
deg uk = mk.
Clearly
57
m1 <m2x-<..-
etc.
This process will end, say with
We call
vectors.
the minimal in-
1<s<n,
mm2'-..,m5,
Define for each natural
pA+AB.
dices of the pencil
number
linearly independent
n
because we have at most
k=s,
m,
m+1;
Lm =
pA+xB
is
onjunctive over
0
P
e
L
Lm)
\i=1
m.
m.
C
to
GM,
1
0
(_.;+1(1.1'
= [0] (1x1).
A,B are complex hermi-
In [10],Turnbull has shown that if
tian, then
L0
)
1
wheremisnonsingularandthemi's(m.=0
m.>0
for
call
M
p<i<k)
are the minimal indices of
a nonsingular core of
pA+AB.
for
i<p,
pA+XB.
We
Since the method
he used for the above reduction is rational, then it is also
applicable to real symmetric matrices and congruence over
R.
He also has given [9] a method of determining the
smallest minimal index as follows:
58
A
M1
M2
B
B
0
A
B
B
0
0
0
A
B
0
0
0
A
B
0
M3
I
etc.
p. = rank of
1
M.
1
11.=in-p.(i.e.,11.is the nullity [in right hand
notation]ofIC).
1
pm+1
Theorem.
If
sequence
111,1121,
minimal index while
which are equal to
Fact 3.13
11.A+XB
mation of
m
then
pm+1
m.
is the value of the smallest
is the number of such indices
[9]
Minimal indices are invariant under two kinds of
transformation,
pencil
is the first nonzero integer in the
A,p
(i) equivalent transformation of the matrix
(ii) homogeneous nonsingular linear transforto
p,a,
i.e., change of basis [11].
Turnbull [10] claims (without offering proof) that
the following holds:
Assertion 3.14:
Let
'IS1+XT
pS2+XT2
be hermitian (real
59
Then if
symmetric) pencil.
'IS2+AT2'
is congruent to
11S1+XT1
then the corresponding nonsingular cores of the
two pencils are congruent.
We shall use Assertion 3.14 also without proof.
(There does not seem to be any routine proof of Assertion
3.14 but it certainly seems plausible enough.)
11S1+AT1
and
Then the pencils
(S11T1) = (S2,T2).
Now suppose
pS2+AT2
are congruent over
R,
have the same minimal indices (by Fact 3.13).
PS2+AT2
R
be congruent over
A1 =
i=1
Lc))
A2 =
1-1
L )
o
Lm
0
i=p+1
11241
0
Lt.
Lm.
0
e
.=
1
(61.I
m9
)
congruence is an equivalence relation, then
congruent over
R.
Thus
-
are nonsingular cores.
Mi,M2
respectively, where
pSi+XT1,
9
(
k
(
Let
to
k
9
hence they
Mi
is congruent to
Al' A2
M2
Since
are
(by
Assertion 3.14).
Conversely, if
M1
and
M2
are congruent over
and the two pencils have the same minimal indices, then
R
60
PS1 + AT1
rp
= p t 0 I,
1=1
/k
\
®e
i=p+1
°,
J
0
Lm.
1
Lt
mi
/0
Lt
mi
S L )S
=pt
where
L
\
m.
0
G C M
.
CtM2C\
j
K P
Rt(pS2 + AT2)R K,
P,C,R,K
are nonsingular real matrices and
K = (I
e
o
1/
I2m.+1
i=1
So
Lm
e
i=1
= Kt
Lin)
1
0 )
/k /0
\i=1
®mi
0
(S11T1)
=
(S2'T2).
e
C
P
Now we have the following con-
clusion:
If
(S11T1), (S2,T2) are
singular pairs and
all
A,p E 1R,
then
nxn
real symmetric,
det(uS1+AT1) = det(uS2+AT2)= 0
for
(S1,T1) = (S21T2) if and only if the
two pencils have the same minimal indices and the corresponding nonsingular cores
Let
M1, M2
are congruent over
IR.
61
M1 = pA1 +
AB
1
.
M2 = pA2 + XB2
Without loss of generaiity, we may assume
Then
nonsingular.
AS1 + pTl'
A11B1
(S11T1) = (S21T2)
AS2 + pT2
is similar to
k, s,
Theorem 3.7.
where
are
if and only if
have the same minimal indices and
A21B2
sig Alf(AT.1Bi)kgs(Al1Bi)
for all
Al, A2
and
-1
= sig A2f(A2
f(x), g5(x)
B2)k gs(A2-1 B2)
are defined as in
62
CHAPTER IV
Example 1:
S
1
Let
-0
5
0
-1
5
0
-1
0
0
-1
0
0
__-1
0
0
0
0
0
0
-1
0
0
-1
0
0
-1
0
0
- -1
0
0
0
=
r
,
S
=
2
Show that
-
-
-
0
0
0
5
0
-1
0
0
-1
0
- 0
-1
0
0
4
0
0
0
0
-5
0
-1
0
0
-1
0
0
-1
0
,I,
-
T2
(S2'T2)
0
=
=
-
...,
=
(S11T1)
T1
4
.
Solution:
0
0
0
-1
0
0
-1
0
0
-1
0
-5
-1
0
-5
0
.1".
-1
S1 =
S-1
2
= S2
0-
63
_
-1
0
1
0
0
0
1
0
q-lm
0001
S1 T1 =
-4
0
1
-2 "1=
-4
0
5
0100
0010
0501
The characteristic polynomial of
-1
Si Ti
0
0
and
0
-1
S2 T2 is
.
P(x) = x4-5x2+4
p' (x) =
4x3-10x =
2x(2x2-5)
(x)
-
f (x)
- p (x)
.
gcd (p (x) ,p ' (x) )
-1
Since all roots are simple,
4
f.(x)t4-j
f(x+t) =
j=0
f0
f1
S
-1
2
.
.
3
(x) = 4x
6x2-5
f3(x) = 4x3-10x
Al
is similar to
(x) = 1
f2(x) =
Let
S1 T1
> A2 > A3
f3(x) = (x- 1 )
>
(x-2)
A4
=
2x(2x2-5).
be the roots of
p(x).
Then
(x-3) + (x- 1) (x-2) (x-4)
+ (x-A2)(x-A3)(x-A4).
So
Let
Then
sgh. f(A-
31
Xi = sgn
X.'s
=
(-1)1-1.
Clearly sgn(f(Ai = 1.
(f0(A1),f1(Ai),f2(Xi),f3(Xi))t,
i =1,2,3,4.
are pairwise linearly independent (Lemma 2.5).
64
Also the matrix
(--
has rank
Hence the
4.
X1
X2
X3
X1
-X2
X3
IS"
X4
1
-X4
(in Corollary 2.8) is
S" = {g(x)Ig x) = fk(x)f3(x)j, j=0,1
But we need only the following
k=0,1,2,31.
and
g(x)'s e I"
to get the
Let
matrix (11) is Proposition 2.7.
go(x) = fo(x), g1(x)=f1(x), g2(x)=f2(x), g4(x)=f1(x)f3(x),
g5 (x) = f2(x)f3(x).
We will show that
-1
sig S2gk(S21T2) = sig S1 g k(S, TI) for
k = 0,1,2,...,5.
-1
sig Sigo(Si
TI) = sig S1 = sig
sig Sigi(SilTi) - sig
S1(4S11T1)
0
5
0
-1
5
0
-1
0
0
-1
0
0
-1
0
0
0
= 0.
= sig Ti=sig
4
0
0
0
0
5
0
-1
0
0
-1
0
0
-1
0
0
= U.
-1
sig S1g2(S1 T1)
=sig
2
t- 0
5
0
-1
r- 0
1
0
0
5
0
-1
0
0
0
1
0
0
-1
0
0
0
0
0
1
-1
0
0
0
-4
0
5
0
-51
ii
65
-1-
0
= sig
0
0
0
0
0
0
-1 n
_
0
5
0-1
0-1
- -1
0
1
0
0
0
0
1
-4
0
5
0
0
-4
0
5
.-- -5
0
6
0-
0
6
025
0
6
..-,
0-5
0
5
= sig
0
(
-1
0
0
0
0
^
-0
-24
0-24
-51
I
25
0
..
-
0
-1
0
5-
-1
0
5
0
0
+5
0
-6
- +5
0
-6
0
= sig
=0
_
,
_
0
0
1
0
0
0
0
1
-4
0
5
0
0
-4
0
5
0
-5
0
2
-1
0
-5
°
0
0
2
0
-1
0
-8
0
5
0
-1
0
0
-8
0
5 --1
0
0
0
0
5
0
-1
0
0
-1
0
-1
0
0
4
0
0
0
5
0
0
/
2
ow,
_
,-
= sig
_
0_
-
\
-
66
-20
= sig
0
8
0
0 -17
0
5
8
0
-5
0
0
5
0
-2
= -4
sigS1g4 (SilTi)=sig Sifi (SilTi) f3 (Si-1Ti) = sig 4 T1f3 (Si-1 Ti)
= sig [4T12 S-11T1(2ST1)2 - 5 I)]
)
)2-5I) =sigT1 (2 (S1-1T1 )2-SI) (S-1T1
1
=sigT1 S1-1T1 (2 (S-11T1
= sig
= sig
4
0
0
0
-5
0
2
0
5
0
-1
0
-5
0
0
-1
0
-8
0
-1
0
0
0
-20
0
8
0
0
1
0
0
17
0
5
0
0
1
8
0
-5
0
0
0
0
0
5
0
-2
_ -4
0
5
/--20 0
0-17
= sig
8
0-
0
5
8
0
-5
0
0
5
0
-2
0
1
0
0
0
0
1
0
0
5
0
0
0
1
-8
0
-4
0
5
0_i
-5
0
0-20
=sig -20
0
8
0
8
0 -5
8
0
1
6
0
0
6
0
25
0
0 -24
0
0
0-5
-24
0
25-)
-5
=0
0
67
= sig
-92
0
80
0
-35
0
23
80
0
-77
0
0
23
0
= -4
-20_
sig S2g0(S2-1T2) = sig S2 = 0.
sig S2g1(S2-1T2) = sig 4 T2 = 0.
sig
)
2
S2g-z (S-1T2
-1
= sig S2(6(S2 T2)
2
-5 I)
0
0
0
-1
0
1
0
0-
2
0
0
-1
0
0
0
1
0
0
-1
-5
0
0
0
5
0
-1
0
0
0
-4
0
0
=sig
1
0
0
0-\
0
1
0
0
1
0
0
1
0
0
0
0
0
1.1
-0
0
0
0
0
-1
0
0
-1
0
0
_-1
0
0
0_
0
0
0
0
0
-1
0
-1
-1
0
=sig
=sig
dm.
-1-
-1-
-0
0
1
0-
0
5
0
1
-4
0
5
0
0
-4
0
0
6
-
o oio
o
-5
Nam
,-
-5
0
6
o7
0
0
25
0
6
0
0
-24
0
25
0
0
0
0-24
0
-5
=sig
-
0
24
0
5
24
0
-25
0
0-25
0
-6
-6
0
5
0
=0
-1
-1
2
sigS2 g3(S2iT2)=sigS2 .2S 2 T2(2(S2 T2) -5I)=sigT 2(2(S 1T2) 2-5I)
68
-4
0
0
0
-5
0
-1
0
0
-1
0
0
-1
0
0
=sig
-
,
^
-I
0/
0
0
1
0
5
0
-4
0
5
0
-4
0
2.
\
_
-
0
0
- -5
0
2
0-
-5
0
-1
0
5
0
2
o
0
-1
0
-8
0
5
0
o
-1
0
0
0
-8
0
-5
-
0
0
0-
0
1
0
0
0
0
1
0
_0
0
0
1-,/
-5
0
=sig
1
--20
=sig
-
-
0
8
0
0 -17
0
-5
8
0
-5
0
0
-5
0
-2
sigS2g5(S2-1T2)=sigS2f0(S;1T2)f2(S-12 T2)=sigS2g3(S-23T2)g2
= -4
(S-21T2)
_
-20
0
8
0
-5
0
6
0
0
-17
0
-5
0
25
0
6
8
0
-5
0
-24
0
25
0
0
-5
0
0
-24
0
= sig
_
=sig
_
- -92
0
80
0
-305
0
-77
80
0
-77
0
0
-77
0
-20
= sig
-20
-2_
-
0 -
= -4
-
-
0
8
0
0
1
0
0-
0 -17
0
-5
0
0
1
0
8
0
-5
0
0
5
0
1
0
-5
0
-2
-4
0
0
0
.. (S,,T1),---; (S2,T2)
-5_
-
0
20
1
0
8
20
0 -17 0
=sig 0 -17 0 -5
8
0
-5
0
=0
-
69
Note that if
c=
then
r_1
0
0
0
0
1
0
5
0
0
1
0
0
0
0
1
CS1C
e-
00
5
0
-1
1
0
0
0
5
0
-1
0
0
1
0
0
0
0
-1
0
0
0
0
1
0
0
1
-1
0
0
0
0
5
0
1
5
0
-1
1
0
0
0
0
0
0
-1-s
0
0
-1
0
0
1
0
0
0
0
-1
0
0
-1
0
0
0
0
1
0
0
-1
0
0
-1
0
0
0
0
5
0
1
0
0
0
-4
0
0
0-
1
0
0
0-
0
0
0
1
0
5
0
0
1
0
0
0
s_
=S2.
CC
T1
-
1
0
0
0-
0
1
0
5
0
5
0
-1
0
1
0
0
0
0
1
0
0
0
-1
0
0
0
1
0
0
0
0
1
0
-1
0
0
0
5
0
1
4
0
0
0
1
0
0
0
4
0
0
0
0
0
0
-1
0
1
0
0
0
-5
0
-1
0
0
-1
0
0
0
1
0
0
0
-1
0
0
-1
0
0
0
5
0
1
0
-1
0
0_,
=T2
70
Let
Example 2:
r."
- -1
0
0
0
0
2
0
0
0
0
-1
1
-1-
1
0
0
0
0
-1
0
0
-2
0
0
-1
-1
0
0
0
0
1
-1
0
-I
0
0
0
0
0
-1
0
0
0
0
0
0
0
-1
0
0
0_
1
0
0
0
0
1
0
0
0-
1
0
0
0
0
1
1
0
0
0
0
0
0
0
-1
0
0
0
0
1
0
0
0
-1
0
0
0
0
1
-1
0
0
-1
0
0_
0
0
1
-1
0
1
0
0
0
0
1
0
0
0
0
0
-1
1
0
0
0
0
-1
1
0
0
0
0
2
Show
(S11T1)
T1=
T2=
is not congruent to
(S2,T2).
Solution:
- Si
- 0
0
0
0
0
0
0
-1
0
0
0
-1
0
0
0
-1
0
-2
0
_-1
0
0
0
1
0
1
0
0
0
0
0
1
0
0
0
0
-1
1
0
0
2
2
0
1
....-1
-1
0
0
0_,
=
-1-
S-2.1 =
S
2
-
S-1T
1
1
=
S1-1T1
matrix.
and
--,
S21T2
S-1T2
=
2
-
-1_
are both similar to the same companion
Hence they are similar.
71
= p(x)
Characteristic polynomial = minimal polynomial
= x5+x4-2x3-2x2+x+1 =
(x-1)2 (x+1)3
Pl(x) = 5x4+4x3-6x2-4x+1 = (x-1)(x+1)2(5x-1)
g.c.d (p(x), p'(x)) = x2-1
f(x) -
p(x)- x2-1
g.c.d(p(x),p'(x))
= (x-1)(x+1)
2
f(x+t) =
f.(x)t2-j
1
j=0
f
0
(x) = 1
J
f1 (x)= 2x
;
-1
Note that
f2 (x) =
;
f (x)
-
Therefore
f(S21T2) = 0.
f(S1 T1)3 =
1
f.(S.-1 T.) = 0
sig S.f(S. T.)k
1
1
1
3
1
1
for
k > 3,
i,j.
sig Sif(Si1 TI)2 fo(Si-1 Ti)
e..
/ _-1
0
0
0
0
2
0
-1
0
0
-1
0
-1
....-1
= sig
.
1
2
1
-1
1
0
-1
-2
-1
1
-1
0
0
1
2
1
-1
1
0
0
0
0
0
0
0
0
0
0
0
0_
0
0
0
0
0-
1
-2
-1
1
-1-
2
-4
-2
2
-2
1
-2
-1
1
-1
1
2
1
-1
1
1
-2
-1
1
-1-
L_
r'..'
= sig
-
-1_,
= -1
72
-
=sig2
= sig
1
0
0
0
0
0
-2
0
0
-1
0
0
0
0
-1
_O
-1
1
-1
2
1
-1
-1 -2
-1
1
1- \
-1
0
1
2
1-1
0
0
0
0
0
0
0
0
0
0._
0
0
0
0
0,_/
2
1
-1
1
2
4
2
-2
2
1
2
1
-1
1
1
-2
-1
1
-1
1
2
1
-1
1
1-1
-
1
i
=1
sig Sif (Si-1Ti) fo (Si-1 Ti) = sig Sif (Si-1 Ti)
= sig
0
1
0
0
0
0
-1
-1
1
0
0
0
0
2
2
-1
1
0
0
0
-1
0
1
0
0
0
0
0
-1
-1
-1
0
1
0
0
1
-2
-2
1
0
0
-1
-2
1
-1
0
1
1
-1
0
1
0
-1
0
0
0
0
0
0
2
0
-1
0
0
-1
0
-1
1
1
= sig
-1-
-1
1
=sig 2
sig Sif (SilTi) f1 (Sily
\
=-1
--1
0
1
0
0
0
3
3
1
1
3
2
-2
-2
0
-2
-2
1
-1
0
1
1
-1
0
1
-1
73
-1
sig S1f0(S1
T1) = sig
sig S1f1(S1-1 T
)
S1
= -1
= sig 2T1 = sig T1 = 1
sig S2f(S2-1T2)2 f0(S21T2)
= sig
1
0
0
0
0
0
0
0
0
-4
=1
= sig
= sig
_.,
0
0
0
2
0
0
0
0
0
0
0
0
4
=1
0
0
0
0
0
2
0
2
-1
74
-
-1
sig S2f(S21 T2)fl(S2 T2)
= sig 2 T2f(S21
T2)
,
=1
= sig
0
0
0
0
0
-2
0
-2
3_
-1
sig S2f0(S2 T2) = sig S2 =
-1
sig S2f1(S2 T2) = sig 2 T2 = 1
So
to
(S11T1)
(S31T3),
.
is not congruent to
but is congruent
(S2'T2)
where
S3 =
T3
Iwo
Examples
S=
3:
0
0
-1
0
0
1
0
-1
0
0
1
-1
-1
0
0-,
1
-1
0
Let
-o
0
0
-1
0
1
1
0
1
__3
I
T =
0
-2 --'
0
0
0
-2
0
-2
.1
75
r- -1
o
0
-1
1
0
S-1 =
_
T
S
is similar to
10 -
1
0
0
0
1
0
2
2
-1
-1
-1
-1
0
--1
0
0
-1
S0 =
S
0 -
0
0
0
--3
0
-2_
' 0
0
0 -
0
-1
_0
-1
where
A,
-
-1 -
-
_
Let
0
S-1T=
0
- 0
A =
1
T0 =
0-
S-1T
1
0
0
-1
0_
= A
t_ -1
Characteristic polynomial = s3+x2-2x = x(x-1)(x+2)
3
f(x+t) =
f.(x)t3-j
j=0
f0
(x) = 1
;
f0(x) = f (x)
3
f1 (x) = 3x + 1
;
f2(x) = 3x2+2x - 2
.
sig So = -1
sig SoyA) = sig
2
-1
-1'
-1
-4
-3
- -1
-3
0 -
= 1
76
sig S0f2(A) = sig
Let (ST1) = (S0 f1 (A) '
s ig
2-
-4
2
2
-6
-2
2
-2
-3_
=-3
T0 f1 (A))
Sifl(S 11 TI) = sig Sofi(A) 2
=
-1 -25
-6
-6
-9
- -4
2
-1
sig S1f2(S,-1T1)=sig S0f2(A)f1(A)= sig
Let (S2'T2) = (S0 f2 (A)
sig
S2f1(S2-1 T2)
2-
-
2 -12 -14
2-14 0 -
-1
.
F
8
= -1
' T0 f2 (A)).
= sig S0f2(A)f1(A)
=
2
sig S2f2(S21T2) = sig S0f2(A) = sig
-4
-4 -16
-4
Let
= -1
f 1 (A) (A)
t2'
2
-4 2
-1
-7 -
T0 f1 (A)f2(A))
sig S3f1(S-13 T3)=sig S0f1(A)2f2(A)=sig
2 -90
2
-8
= -3
-8 -42
sig S3f2(S-31Tr) = sig S0f2(A)2f1(A) = -4 -16 -52
-4 -52 20
=1
77
sig S = sig
sig
-1T)
(S
Sf1
sig S
f2
-0
0
0
-1
_1
0
= sig
(S-1T)
= sig
(sig S
N
.
,
sotation:
1= -1
0
10
-1
3
0
6-
0
2
0
_6
0
6....
0
ig S 3.f1(S 3)T 3)
3
= 1
,
- .) )
sig S f,j2(S .1T
3
3
=(
)sj
We have
(-1, 1, -3)
so
(1, -1, -1) Si
(-3,-1,
-1)s2
(-1,-3,1)
s3
(1, -1, 3)
(-1,1,1)
_so
_Si
(3,1,1)-s2
(1,3,-1)
_so
(-1,-3,1)
.
(S ,T)
(S3, T3) =
(Sofi (A) f2 (A) , Tofi (A) f2 (A) )
.
78
Let
Example 4:
,--
- 2
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
1
0
0
0
0
0
-1
0
0
0
0
-1
1
0
0
0
0
0
1
0
0
0
0
0
-1
0
0..
0-1
0
-1
S1=
0
A
0
p
0
0
p
0
0
0
0
0
p
0
0
0
u1
0 -A-p
-
0
= 1.5 - p1 = 0
T1
1
0
=
rank of
-
S
0
1
T
n2
=9
1
= 2.5 - p2 = 1
1
(S1,211)
is congruent to
E
0
(S11T1) = pl = 5
S
u2
-A-p
u. = i.5 - p.
1
1
Notation:
rank of
4...0
p
F2A
det (ASi+pTi) = det
T1=
0
o
//
ex
o o oA
0o
o
0 o o
0 o o
Ao
o
o
e`
ci)
o
)?0
,e.
/r
x
s,
,..)
0i.
7ie
1?"
Xti_ca
/"\
/7)
)?o
..
e\
.
.2
I'
cP.
.
e`
e.
.,
A
c'÷
rti
Q.
o o o o0
0
A o0
4 o0
0
'es
<,.
s%
o 0 Ao
80
Similarly
is congruent to
AS2+pT2
o
L
o
L1
t
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
0
_1
0
0
Let C2 =
Then
C2 (XS2 +1..tT2
_
M2
4...
^
0
A
0
0
p
A
p
0
(Ct2 =
0
-A-P
...00.
0
Let M2 =
_
--p
A
=A
Let
-A-p
11
-,
_
0
-1
-1
0
0
A2
=
A-1
=
2
01 '
-1
[
-1
B
2
=
[ -1
-1 I
1
--A-p
P
81
A-1B
2
F
=
2
L
0
Characteristic polynomial of
Test whether
f(x) -
sig
is congruent to
M1
(x-1)2
Clearly
M2.
p(x)
- x-1
g.c.a(p(x),p'(x))
(x) = 1
= sig Al = 0
A1f0(A1-1 Bl)
-1
sig Aif (Ai Bi(fo
1r
-1
-1
0J L
-1
11
F
-1
1
L
1
-1
r
0
-11
r
0
-11
0J L
0
0J
r
2
L
-1
= sig
(Ai-1 Bi)
= sig
-
sig A2f0(A21 B2)
sig
A1-1B1
A21 B2.
similar to
f0
A2IB2 =
= sig A2 = 0
A2 f(A2 IB2 )f0 (A2-1B-) = sig
L-1
= sig
10
Lo
(Al,B1)
is not congruent to
M2
(ST1)
J
(S2'T2)
.
0
(A2,B2)
=1
is
82
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Theory of Equations.
Dickson, L.E.
Wiley and Sons, Inc., 1948.
New York, John
Gantmacher, F. R. The Theory of Matrices. Vol. I &
II, Chelsea Publishing Company, New York, 1964.
Linear Algebra, 3rd edition.
Greub, W. H.
Springer-Verlag 1967.
New York,
On the Inertia
and Ostrowski, A. M.
Haynsworth, E. V.
Linear Alof Some Classes of Partitioned Matrices.
Vol.
1,
pp.
299-316,
1968.
gebra and its Application.
The Equivalence
Ingraham, M. H. and Wegner, K. W.
of Pairs of Hermitian Matrices. Trans. Amer. Math.
Soc., Vol. 38, pp. 145-162, 1935.
Kowalski, H. J., Lineare Algebra.
bUcherei, 1963, 27, Berlin.
GOschens Lehr-
Muth, P. H. Uber reelle Aquivalenz von Scharen reeller
quadratischer Formen. J. fUr Math., Vol. 128,
pp. 302-21, 1905.
Trott, G. R. On the Canonical Form of a Nonsingular
Pencil of Hermitian Matrices. Am. J. Math., 1934,
Vol. 56, pp. 359-71.
Turnbull, H. W. On the Reduction of Singular Matrix
Proc. Edinburgh Math. Soc., 1935, Vol. 4,
Pencils.
pp. 67-76.
On the Equivalence of Pencils of Hermitian Forms. Proc. London Math. Soc. (2), 1935,
Vol. 39, pp. 232-48.
Turnbull, H. W. and Aitken, A. C. An Introduction to
the Theory of Canonical Matrices, 4th edition, New
York, Dover, 1961.
Vander Beek, J. W. Isoconjunctivity of Hermitian
Matrices. Ph. D. thesis, Oregon State University,
1971.
83
13.
A Study of the Canonical Form for a
Uhlig, D. F.
Pair of Real Symmetric Matrices and Applications to
Pencils and to Pairs of Quadratic Forms. Ph.D.
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