AN ABSTRACT OF THE THESIS OF
ELIZABETH LINGFOON YIP
for the
DOCTOR OF PHILOSOPHY
(Degree)
(Name)
in
November 29, 1973
presented on
MATHEMATICS
(Date)
(Major)
Title: MATRICES CONJUNCTIVE WITH THEIR ADJOINTS
Signature redacted for privacy.
Abstract approved:
C. S. Ballantine
This paper studies necessary and sufficient conditions for
a matrix to be conjunctive with its adjoint. The problem is solved
completely in the usual complex case, in which it is shown that a
matrix is conjunctive to its adjoint iff it is conjunctive to a real
matrix. The problem is extended to pairs of fields
where
[4:
2
and characteristic
2.
if a matrix is conjunctive to a matrix over
,
6,
It is shown that
it is then conjunc-
tive to its adjoint. To achieve this result, we first show a matrix
over any field
to its transpose. We
is congruent over
also show that it is sufficient to consider non-singular pencils by proving the uniqueness up to conjunctivity of the non-singular summand of
the pencil
H*
XH +
H and
conjunctivity over
where
K* = K,
)
X
when
and
1.1.
are indeterminates over
XFI + 1.1.K
is decomposed (by
into a direct sum of its minimum-indices
part and a non-singular part.
Matrices Conjunctive with Their Adjoints
by
Elizabeth Lingfoon Yip
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Doctor of Philosophy
Completed November 1973
Commencement June 1974
APPROVED:
Signature redacted for privacy.
Professor of Mathematics
in charge of major
Signature redacted for privacy.
Cha.
an of Department of Mathematics
Signature redacted for privacy.
Dean of Graduate School
Date thesis is presented
Typed by Clover Redfern for
November 29, 1973
Elizabeth Lingfoon Yip
ACKNOWLEDGMENT
I would like to express my gratitude to Professor
C. S. Ballantine for accepting me as an apprentice and for
his generosity in giving me his time and advice during the
preparation of this thesis, which, without his infinite
patience, encouragement and understanding would have
never been completed.
TABLE OF CONTENTS
page
Chapter
I.
INTRODUCTION AND PRELIMINARY
Introduction
Uniqueness of the Non-Singular Core of
*-Symmetric Pencil
II.
GENERAL CASE
Congruency of a Matrix to Its Transpose
General Results
Non-Singular *-Symmetric Pencil with All
Eigenvalues in
III. USUAL COMPLEX CASE
General Results
Geometrical Approach to the 2 x 2 Usual
Complex Case
1
1
7
17
17
28
34
59
59
65
BIBLIOGRAPHY
70
APPENDICES
Appendix I: Pictures of Matrices
Appendix II: Miscellaneous Results
71
71
73
MATRICES CONJUNCTIVE WITH THEIR ADJOINTS
I. INTRODUCTION AND PRELIMINARIES
1.
Introduction
Let
3.
(i.e.,
a subfield of index 2 in
sion 2 over
\6' be
be a field of characteristic not equal to 2 and
8
is a vector space of dimen-
Then there is a unique automorphism
).
S
whose fixed field is
*
of ,....;
:
(a+b)* = a* + b*,
(ab)*= a*b*,
and
a* = a iff a
we also have that
conjugate of
a.
E
*
If
E.
Because
J
is involuntary:
A
has dimension 2 over E....,
(a*)* = a. We call a*
is a matrix over
-.q.- ,
we denote by
the
A*
the transpose of the conjugate of A. Then we have the following
properties:
(A*)* = A,
(A+B)* = A* + B*,
(AB)* = B*A*,
(aA)* = a*A*
when A+B
S
and AB are well defined and a E-7- We shall say
is *-symmetric iff S = S*,
and say
S
is *-congruent to a
2
matrix
iff there exists a non-singular matrix
T
C*SC = T. We say a matrix
S*; we say
is of type 1 iff S is *-congruent to
S
is of type 2 iff
S
such that
C
is *-congruent to a matrix over
S
I.
We also consider our matrix
7f,
from
an n-dimensional vector space over
*-dual of
If
( 11) * =iJ
Then 5*
.
11*
into
11.
tAr and V:, are being
Elements of both
Suppose
In this context,
<x, Sy> = <y,Sx>* for all
x
E
2f= 4 n x 1
<x, Sy> = <y, S*x>;* for all
<x, f> = x*f. Then
define
E
is *-symmetric iff
S
x,y
E
if.
is *-congruent to
S
there exists a one-to-one, onto, linear transformation
<Cx, SCy> = <x, Ty> for all
such that
is a basis for
if,
the matrix of
S
a subspace of
lf
if=
renrti..
j=1
11,*
=
if* the
is a linear transformation from
written as column vectors over
x, y E
into
f(x+y) = f(x) + f(y), f( ax) = a*f(x)1. We call
1)-* = {f:
f
as a linear transformation
S
then
<e., Se.>
x, y
1)
e. Se.
1
If
E
is the
C:
{el,
,
we define
7.1..° = {x
E
If 26 is
,en}.
x*11.= 0}. Suppose
Define
J
(1)-1 e "112 e ...e Vj -1
(DV j+1
en}
entry of
(i, j)
with respect to the basis {e1,
,
iff
T
0-)r
m-1
v
m
3
,.,,
,..*
then V
acts as a 4-dual of
.
LI . n v1r*.
n 1-0
i. e.
,
j;=
m03
ti
j=1
If
and
J
3
J
_
u,
3
n=
and we partition the matrix
k.,
/
1
i=1
Sll
.
S12
51p
S21
s=
S
.
Spl
PP
...
where
is a k.1 x k.
S..
13
J
If
matrix, we write S = (S..).
13
P
, M),
and call
C
is a
we write
matrix such that C = M1 ED ... El) M ,
C = diag(M1,
C
a direct sum of M, ... ,M
1
We have the following facts which will be useful in the work
following (we shall not prove those quoted from the literature):
Fact 1.
If
S = (S..),
C = diag(Ik, -Ik,Ik, ...),
13
then P ih
P = (Pih) = CSC,
Fact 2.
P = C*SC ,
If
and
= (-1)i+hSih
C = diag(Ik, jIk, Ik,
where
. )
j* = -i ,
then
h
are odd
if
i
and
= jSih
if
i
is odd and
= -jSih
if
i
is even and h
= -j2Sih
if
i
and
Pih = Sih
h
h
is even
is odd
are even.
and
4
The proofs of Fact 1 and Fact 2 are routine computations.
Fact 3.
where A is non-singular and 0
S=Ae0
If
a zero matrix, and if
B e 0 where
is *-congruent to
S
then A
non-singular and of the same dimension as A,
*-congruent to
Proof:
Suppose
C
C11AC 1 I = B.
C=
C11 C12
(C
is also non-singular. Thus
singular. Suppose
K
C
21
accordingly. Thus
22)
C11AC I I
A and
Since
Suppose H and
A
B
0
B
o
00
-1m-1
H-PI)
0.
0
are both non-singular,
is *-congruent to
B.
are *-symmetric and K is non-
(K-1H-PI)
is nilpotent for some
be a positive integer such that (K-1H-13I)m = 0,
(K
is
is a non-singular matrix such that
C*SC = C*(AeO)C =
C11
is
B
B.
C*SC = B e 0. Partition
Thus
is
If we define an i nner product
from Mal'cev [7], we can choose a basis for
If
pE
6.
Let
but
(x, y) = x*Ky, then
so that
K-1H
block diagonal with each diagonal block a Jordan canonical block of
eigenvalue
P,
and the Gram matrix of the inner product is con-
formably block diagonal with each diagonal block of the form
is
5
1.
1
1
where
E
E
Thus we have:
s'6°
Fact 4. Suppose
indeterminants over
S = XH + p.K,
or
,
where
both
is non-singular, and
X
and
p.
are
and K are
H
is nilpotent for
*-symmetric,
K
some
Then there exists for .11 a basis such that (the
matrix of)
(K-IH-13I)
is block diagonal with each block of the form
S
X13+p.-
where
E
E
.
We define
En
as the n x n matrix with
diagonal and zero everywhere else,
Fn
I 's
on the anti-
as the matrix with
l's
on
the first super-anti-diagonal and zero everywhere else. Also define
Nn = EnFn.
Fact 5.
(K
-1
H-P)(K
-1
If
H-P*)
K
and H are *-symmetric and
is nilpotent with
p
p*,
and if
X, p.
are
6
lf such
indeterminates
over;
that XH + p.K
is block diagonal with each block of the form:
then there exists a basis for
,
z
.X.
X
1
XP+p.
XP+p.
:2. exp*i_p,
xp*+11
Fact 6 [14
If
symmetric over 6.
E,
,
then the pencil
H
is symmetric and K is skew-
and
and
X.
p.
are indeterminates over
is congruent over
XH + p.K
ing form
0
A
L
E.
0 1
1
where
z
0
A
L
0
e (XH 0+p.K 0 ),
to the follow-
7
is called the minimum-indices part of the pencil
L
=
E
and
L
IL
XH + p.K,
and
=
i
1
det(XH0 +1.1.K0
is a non-zero polynomial.
)
Fact 7. Witt's Theorem [9].
A, B1, B2
If
*-symmetric non-singular matrices over s-4
implies
*-congruent to A e B2
is
B1
are
A e Bl'
*-congruent to B2.
then
Uniqueness of the Non-Singular Core of a *-Symmetric Pencil
2.
Theorem I. 2. 1.
pencil
XH + pa,
Let H
where
is *-congruent to L
M,
and
and
X
K
be *-symmetric. Then the
are indeterminates over
p.
where
0
L=0
Lei
1:13
i=1
LE
L
is the conjugate-transpose of
is as defined in Fact 6,
LEi
i
and M =
LEi,
0
XH0 + ill<0
is *-congruent also to
is a non-singular pencil. If
XH + 1.1.K
L Ef) N where N is a non-singular pencil,
8
then M is *-congruent to
N.
Turnbull proved that a *-symmetric pencil can always be
reduced to the form L
where
is the minimum-indices
L
part and M is a non-singular pencil [11]. In this section we shall
only prove the uniqueness (up to *-congruency) of the non-singular
summand M.
Before we proceed with the proof of the theorem, we shall
discuss some of the properties of the pencil
Suppose
14)
is a subspace of 7.f
transformations from
,
L IFD M.
S, T are linear
and
V*. For our discussion in this sec-
to
tion, we need to define, for 24 c 7J (s-1T)2L= {x lf:SX alt})
E
Ell:SX
and, inductively, define
for
= 2,3,...
i
E
T(S-1T)i-1210},
E
(s-1T)°U
.
Also note that if we let
A
H + I.LK
be a pencil of
*-symmetric matrices, considered as the matrix of a pencil of
bilinear forms, then, for each co-ordinate subspace, the matrix of
this pencil of forms restricted to this subspace is the corresponding
principal submatrix of A,
Lemma 1.
If(j = 1, 2,
and let
Let
and conversely.
be the *-dual of
be subspaces of
k)
If
,
let
(11 such that if= tOk
=be the corresponding *-dual direct sum
=1
j
defined in Section 1 of this chapter. If
S
and
T
are linear
9
transformations such that S, T:
TIT C
3
/I*
i _<h-i.
J
[lirm(S-1Thf.].
3
J
The proof proceeds by induction on
i = h. It is enough to show
Consider
,r
-1
r
T)h.V.)
T)h-LiCe3(U.r(S
J
k
k
i=1
x-
and
s V. C V3*,
the lemma is obviously true. Suppose it is true for
i = 0,
-1
®
j=1
3
Proof of Lemma 1.
If
k
(S-1T)i/l=
then
,
.
V*,
x.,
Suppose
(s
with
EV.. Then
x
(S-1T)11,
and
k
3
3
xE
Sx =
Sx.
T.
E
(S-1T)h-111
3
j=1
j=1
By our induction hypothesis,
T(63V. (--(S-1T1/
111-1.)
T(S-1T)1-1-11).
3
T(V. rm(S-1T)h-hr)
=
3=1
Thus
Sx
)Sx. =
yi
Therefore
(.174(
(
J
J
J
k
Thus
xE
43)
if
i=1
Lemma 2.
K = K*
(Th
Ty.
and
Sx. = Ty. E T(S-1T)h-1
3
jVel
j=1
j=1
where
Ty.
?I
.
71f.
But
Sx.
J
3
SO
X.VE
J
.
rm
::'.
3
(S-1-T)h7f.
J
3
3
(S - 1 T )h
1 ..*
q. e. d.
3
i
Let (as usual) characteristic
be linear maps of
if
into
If
.
Let
i 2, H = H* and
X, p.
be
10
indeterminates over
to
co
M00f$ M 0
L
Suppose
.
where
M1,
has no elementary divisors,
L
has elementary divisors only of the form
elementary divisors only of the form
divisors neither of the form
M = Moo EH M
Let
U.
and
y
of *-bilinear forms
A
and
x*(XH
0 e Mo ED M1,
to the
and
A
to the pencil of
is *-congruent over
0 ED M
Without loss of generality, we may
Proof of Lemma 2:
=L
2/
1110y
restricted to 26 is *-congruent over
*-bilinear forms whose matrix is
as sume XH +
p.q
has elementary
745.= (H-1K)ng. (Here n = dim
pencil of *-bilinear forms whose matrix is
restricted to
M1
has
M0
is determined as follows:
as usual.) Then the pencil
x
and
Xcl,
nor of the form
Xq
(K-1H)nif
=
and
is a non-singular pencil. Then the *-congruency
M1
0
type of M over g.
with
is *-congruent over
X.1-1 + iK
Mec
Mo
0
i=1
L
0
k'
L=
and
M1
63
L'E.
)1/4
0
1
Let 11=
Irooe
Here
denote the co-ordinate subspaces corresponding to
respectively. Denote by
= span {x1,
L, Moo, Mo, M1,
'xzrnfl }, m > 0,
one of
11
the co-ordinate subspaces corresponding to one of the direct summands in L,
Let
i(R
of
where x
= span{x 1 ,
thus
XH + 1.1.K
,5Z2m+1}
be the corresponding subspace of
acts as a *-dual of 6
restricted to (;'i
co-ordinate vector.
jth
denotes the
Note that the matrix
.
is
m+1
IL
X
11,
X
P.
P.
m+1
For simplicity in notation we drop the subscript
Hx1 = xm+2
Kx1
Hx 2.= x m+3
Kx2
Hx
x
m
Hxm+1 =
Hx
m+2
Kx
2m+1
0
,
=x l
,
P.
Hx2m = xm-1
Hx2 m+1 = x m
in
=0
=
.
xm+2
x
m+3
Kxm+1 = X2m+1
Hxm+3 7 x2
A
3
i
Kx
Kx
m+2
-x2
A
m+3
x3
Kx2m =xm
Kx
2m+1
X
m+1
Thus
12
Thus
_
= span {x1, x2, .
(K1 H)
span {xi x2,
grm (K-1H)2R
orm
-1H)rn+i
.
(This
oo
mapping on
V-00.
;
K-1H 7f (-
so
f"
oo
E
-
1H)V0
Then
x E (K
Thus
li
0
-1
=
H)k
if 0.
l(../r
Suppose
iff
0
(Th (K 1 Mil Li-0 =
Kx E
is
0-
H is a nilpotent
K1
and
(K-1H)n iloo = 0.
Thus
(K-1H)1/C
K-1Y-*=
{x:Kx if*1
0
0
0
(K
for all k > m+1
hence K LLoo = 15*co and
is non-singular on 1/0. Thus,
Thus
x2m-1}
here.)
m+1
is non-singular on 1j."co
C /f*
'
restricted to 0'? rm(K-1H)110';)
A
matrix is of order
0
x2m}
- span {x1, x2, ... ,xm+1}
0:?rm(K-1H)ko-,z
Thus the matrix of
,
=
H 7).0" =
,*
() 0.
170.
Ifo, since K( /f)C
0
(K-1H)1 270 = Vo
Elm-le-1)Vr0
for
i <k-1
....
- 0Vr0 - -)Lir0 .
. 1- TO.
and H restricted to
-1X1
are non-singular. Thus
(K-lipnLi)r
Therefore
26= (K-1H)11Y,
x*Ay
x*(XH+p.K)y,
with
x, y
restricted to
is *-congruent to a pencil of *-bilinear forms
whose matrix is 0 ®
M0
M1-
Similarly, we can show that x*Ay, with x, y restricted to
13
741-= (F1-1K)nV is *-congruent to a pencil of *-bilinear forms
whose matrix is
0 ® M 00ED M1.
1
Write
Proof of Theorem 1.2. 1.
=A,
L e M = XH +
L ED N = XH1 + F.LK1 = Al
.
Without loss of generality, we may assume M = Moo]. El) mo1 e mu
N = M002 0 MO2 ED M12, where Moo 1 and Mon2 have elementary
divisors only of the form
flq,
divisors only of the form
Xq,
and
M01
and
M11
have elementary
MO2
and
M12
nor of the form
elementary divisors of the form
recall that M and N are non-singular pencils.
singular matrix over
map of
then
21
such that
into
and
C*FIC = H1
By Lemma 2,
(K-1H)n U. and
u*(0 ED
(we also consider
,
x*Ay,
and
u,v E (K-1H)nir and
C
and
is a non-
as a non-singular
C
C *AC = C*(L
and
x*Aiy,
M)C = L (14 N,
with
x, y
restricted to
y respectively, is equal to
u*(0 ®M
)n?r,
(K-1H1
1
ED M12)v,
respectively, with
respectively.
We want to show by induction on i
-1
- 1H) ic 11.
[(C *KC) -1(C *HC)]7f=
X.q,
C*KC = Kl.
(K1-1H1)n
MO1 ED M11)v
If
have no
that
14
i = 1:
For
xE
[(C*KC)-1C*HC)]?,7
iff
C*KCX E C*14C
iff
KCx E
iff
Cx
K 114C 2.7
iff
xE
C-1(K 1H)C
HCV
Suppose
[(c*Kc)-1(c*Hc)]k-12i.
Then
1(C*FIC)PV
X E [(C*I<C)
iff
(C*KC)x E C*1-1C[(C*KC)-1(C*FIC)P-12/
iff
KCx E FIC[(C*KC)-1(C*HC)11C-1Y.
(by our induction hypothesis)
iff
=
HCC-1(K-1H)k-101-
=
H(K-1H)1(-1Clr
Cx E K
iff
XE
-1
H(K-1 H)k-1 C v
C-1(K-1H)kC
Letbe a'xm}
basis for
(K-1H)91 Then
Lx1' x2'
{C
x1,C
x
,C
xm} is a basis for
C(K-1H)ny'
C-1(K-1
-1
= [(C*KC)
n
(C*HC1f1
15
Therefore
'x.)
x.)*A (C
(C
_1
A C- x.
x.) = x. (C*)
1
1
-
*
= x. (C*)1 C*ACC 7 lx
= x. Ax.3 .
1
Thus
x*Ay
with
°
IT'
3,
E
is *-congruent to
Hi)
Thus
is *-congruent to 0 El) MO2 e M12. Thus by Fact
/) Ml l
MO1
-1
with
u 1(OA) MO2 ® M
U
x,y restricted to
M01* M11
is *-congruent to
M.. = XH.. +i = 0,1,
MO2
j = 1,2,
13
e M12. Let
where
and
H..
*-symmetric. Then
C1(H01 ED H11)C1 = H02
Cl(K01
A)
H12
K11)C1 = K02 ED K12
Thus
-1
-1
-1
C1 (H01 K01
H11K11)C1
= [C 1(H01 ® H11)C11
-1
H02 K02
Partition
-
C1
B..
= H..1K...
13
13
1j
*
[C1 (K01
K11 )C1]
-1
e H12
K12 .
accordingly:
Then
-1
C1 =
C11
C12
C21
C22
(
)
.
Let
K..13
are
16
= C1(B02 ® B12)
(B01 e B11)C
-
_
B01C12
B01C11
IMli
C11B02
C12B12
C21B02
C22B22
=
B11C22
B11C21
_
-
B11 C21 = C21 B02 .
Since
[12]
B11
and B02
Similarly,
have no eigenvalues in common,
C11M01C11 - MO2'
*
C22M11C22 = M12 .
thus
C11
and
and
M01, MO2, M11
C22
0
*
Thus
C21 = 0.
C12 =
are non-singular.
are non-singular,
M12
So
M01
is *-congruent
MO2 andM11 is *-congruent to M12. Likewise, L ® M
restricted to (H - le?)-- is *-congruent to 0 0 M001 0 Mll
to
L ED N
)n-lf
1
(FI-1K1
restricted to
o e Moo2 e M12,
where
Moo
1
is *-congruent to
and
M002
have elementary
divisors of the form ilq. With an approach similar to that of our
consideration on
Mc02.
?/../
,
we can show that
M
oo 1
Since
MM = O1
® A4
.01
e
m 11
N = MO2 ED M.02 ED M12
we have that M is *-congruent to
N.
q.e.d.
is *-congruent to
17
II. GENERAL CASE
1.
Congruency of a Matrix to Its Transpose
If
p.
H and K are over
H = H', K = K',
are indeterminates over E
then
,
XH + 1.1.1c
,
and
X
and
is congruent to the
following:
L
where L
M
only of the form
Xq
00
M1
is described in Fact 6 of Chapter I,
divisors only of the form
form
M
0
nor
,
M1
and
p.q,
and
Xq,
Moo
M0
has elementary
has elementary divisors
has neither elementary divisors of the
M0 eM00
M1
is non-singular.
In this section, we shall provet
Theorem II. 1.1.
Every matrix over
E
is congruent over
to its transpose.
Lemma 3. Let ?./ be finite dimensional over
always)
characteristic
g
K: 'if' -.f be skew, and
2.
Let H:
B = KH:
with (as
71' be symmetric,
be non-singular. Let
q(t) be irreducible in
[t]
polynomial for
VT(Here R1 is the (ordinary) dual of Y.)
B
on
and let q(t2)ni be the minimum
Then there is a maximum B-cyclic subspace
21 of 21 such that
18
lLm
(H
ni = 0.
U)0
Proof of Lemma: Let
thus there exists
e
H1 = Hq(B2)m-1.
e'Hle 1 0. Let
such that
E
f(t)
[t],
E
,-,(H it
(f(B)e)tH
q(t2)4(t). Let r(t)
it is sufficient to show that for
= 0,
f(t) = [q(t2)]ip(t)
Write
= 0.
g. c. d (p(t), p(-t)).
and some
k > 0,
some
s(t)
E
g [t],
q(t2)
tn(t2).
r(t)
for some
/ >0
m>0
and some
h(t)
E
E [t]
m(t) = n(t2). Thus
SO
which in turn implies
.
n(t) = 1.
Thus
r(-B)
Thus
so
1
or
B
E
t*s(t2). We
6[t],
such that
q(t2)= t2mh(t2) for some
Thus
til,'h(t2 ).
is non-singular
which implies
q(t2)1 p(t),
k = 0,
n(t) q(t),
n(t) = q(t),
n(t) = q(t). If
then
which is a contradiction.
is non-singular, so r(-B)2t =2t
,
for
m(t) = 1. Since g. c. d (r(-t), q(t2)) = m( -t) = 1.
r(t) = a(t)p(t) + b(t)p(t),
0 = (f(B)e)TI-fit
n(t)
such that
n(t2) g(t2),
n(t)
n(t2) = m(t)Ir(t),
q(t2)
and
since
q(t) = tmh(t). But q(0) 1 0,
tks(t2),
m(t) = g. c. d (r(t), q(t2 ))
It follows that
Celq(t2).
Thus
where
is even or odd, then
is even and
m(t) = tn(t2)
r(t)
Then
such that
claim that g. c. d (r(t), q(t2)) = 1. Let
Since
be the
= 0 => q(t2)11.1 f(t).
(f(Ble) TH
Suppose
)0
?./..."
0,
21 is maximum
B-cyclic subspace generated by e. To show that
cyclic and that
HI1 = H1
Then
for some
a(t), b(t)
which implies for all
E
i > 0,
.
Let
[t]. Now,
that
19
0 = (f(B)e)11-1Bie
[II. 1. 2]
Since
=
[q(B2)ep-(B)ePHB1e
K = HB and H' = H, B'H = (HB) ' = K' = -K = -HB. Induc-
tively, we can show HBi
= (
Thus,
-Bi) 'H.
[II. 1.2] = ((-B)ie)'H[q(B2re p( -B)e]
=
[((-B)ie)1H[q(B2)113(-13)e111
=
[q(B2) ep(-B)ePH(-B)ie
.
Thus
[q(B2)ep(-B)ePH 11 = 0 = [q(B2)1p(B)ePH 2L,
SO
0 = [q(B2)ep(-B)e1'H ../Z D [q(B2)1p(-B)e]'Hb(-B)7t
=
[q(B2)ep(-B)b(B)ePH21.-
and
o=
[q(B2)p(B)epa. D [q(B2)e p(B)e]'Ha(-B)a
=
[q(132rep(B)a(B)ePH ??..,
Thus
0 = [q(B2)e[p(B)a(B)+p(-B)b(B)]ePHit
=
[q(B2)1r(B)ePH/L
=
[q(B2)1 e] 1Hr( -B)1
=
[q(B2)1e]'H2t
20
[q(B2)1e]' H q(B2)ie
e'Hq(B)m-le
=
for all
e'Hq(B2)1fie = 0
>m. Thus q(t2)mlf(t).
0,
q. e. d.
Without loss of generality we may
Proof of Theorem 11.1.1.
consider the pencil [II. 1. 1]
i > m. Since
L
Mo KB M
described in the
M1
first paragraph of this section; and we replace the indeterminates
X.,
with 1.
11
We shall show separately L, Mo, Moo, and M1
are congruent over
Con
L,
8. to their respective transposes.
L. For each non-zero block (of order 2c.+
multiply on the right and left by
IEi+1
6 E E.
.
of
)
lsider
Thus in L,
the
E.
first column becomes the last column, the second column becomes
A
the second to the last column and etc. ; in L
the first row
,
Ei
becomes the last row, and the second row becomes the second to the
last row and etc. Thus, we have for each non zero block of
L
z
o
[II. 1.3]
2
E.
1
where
ii
1
'el
1
Z
4.
E.
=
1
1
1
1
61
Multiply [II. 1.3] on the right and left by
diag(1, -1,1, -1,
, 1);
21
we get the transpose of [11.1.3]. Thus
Next consider
corresponding to the direct summand M0
S=H+K=L ED Mx *Fa
0
is congruent to L'.
be the co-ordinate subspace
Let
Mo.
L
in
H is non-singular and B = (H-1K)
e)N41.
is nilpotent on
Vo. Let m be a positive integer such that
B rn
Br11-1
= 0,
and
We shall consider the cases m even and
0.
m odd separately. We shall prove the following:
is odd, there exists a maximum B-cyclic subspace
m
if
k(Th (HiL)° = 0;
?I. such that
if
m
is even, there exist two maximum B-cyclic sub-
U and 24j such that
spaces
71/
0=
)°
=
riff (H2L)° =(1
=
Suppose m is odd.
eE
ir
such that
(HBm-1)1
e'HBrn-le
0.
space generated by e. Suppose
p(t)
E
nomial of
p(-B)W.
Thus
such that
E,,[t]
B
=
on
M.. Thus
t.i(p(t).
p(-B)
(BW))°
74jc (H26) °.
and
11. C (HU )°,
(H (.
)
=
HBm-1
There exists
Let W./ be the B-cyclic sub(131p(B)e)lHa..= 0,
Since
tm
where
is the minimum poly-
is non-singular and hence
0 = (131p(B)e)'HU,
=
(Ble)'Hp(-B)2t= (Ble)1H2L
> m. We therefore have completed the proof of (i).
The following discussion is not needed in the proofs of (i) and
22
(ii), but useful for later purposes. We note that B UC
_ U. because
is B-cyclic.
21....
B(H M
)o C (Hit)o because if
which implies
x'Plit. D x'FIBU, = -(B,o'HU.,
m-1
a e ; we can choose al ' a2 '
Let e1 =e
I
0
x
Bx
,
,--
i.B
*
i=1
k <m-Z. Let
for
e'1HBke1 = 0
2.1.,1
*
.
(H76)°, then
E
E
(H46)0.
so that
' a m-1
be the B-cyclic subspace
generated by el. 21.1 has all the properties proved for 21..
restricted to 21.1 is the Jordan canonical block of eigenvalue
H restricted to
is a matrix
./4,1
H0
K0
0,
with non-zero elements on
the anti-diagonal and zero everywhere else. Thus
restricted to 261 is a matrix
B
K = HB
with non-zero entries on the
first super-anti-diagonal and zero everywhere else. Multiply
on the right and left by diag(1, -1, ... ,
H0 + K0
1).
Since m is
odd, by Fact 1, we have multiplied the first super-anti-diagonal by
and have not changed the anti-diagonal. Thus we obtain the
-1,
transpose of Ho + Ko.
In the next paragraph, we shall prove (ii).
Suppose m is even; since HBm-1 i 0 there exist
e,f Y"
such that
E
e1l-IBm-1f
e'HBITI-lf = 1. Let U.
f
and
If
xEUt--N Zti
E
E
[t]
,
e, f
so that
Zti be the B-cyclic subspaces generated by
respectively. First, we want to show
e
p(t),q(t)
,
we can choose
0;
then
x = Bkp(B)e
such that
0(0),
=
B1q(B)f,
Orq(t).
7/4. (ThW = 0.
for some
Without loss of generality,
23
assume k
If
/ > m,
then
x = O.
Therefore assume I < m.
Note, since to(t), tirq(t), p(B) and q(B)
are non-singular,
r(B) = p(B)q(B)-1
is non-singular and we can even take r(t) E 6[t]
here, and r(-B)
is also non-singular. x = Bkp(B)e
=
Blq(B)f
and
k >/. Therefore
r
0 = Bi Lq(B)f-Bk-/p(B)e]
=
q(B)Bi (f -Bkr(B)e)
0 = B (f-Bk-/r(B)e)
since
is non-singular. 131(f-Bk-Ir(B)e)
q(B)
0=
=
If
implies
=0
&FIBm-1(f-Bk-/r(B)e)
etHBm-1f
-
e'HBm-1+k4r(B)e
.
k >1+1, &FIBr(B)e = e'FIBmBk-4-1 r(B)e = 0;
0 = &FIBm-1f
e'HBIn-lr(B)e
because m-1
which is a contradiction. If k =
e'HBril-lr(0)e
=
=
= -eHB
e'HB111-lf ,
is odd (and hence
= 0 = e,HBm-lf
eIm-le
HB
which impliex
thus
HBm-1
is skew). Thus
which is a contradi ction. Thus
x = 0. Thus
r1,6
, 2ii
m-
'
>m,
= O.
Secondly, we want to show ?A-(--) (H741)0 = O. Suppose
(Bkp(B)e)11-1111= 0
p(B)
and
p(-B)
for some
p(t) E 6 [t]
such that
are non-singular, so p(-B)t,
=
0(13(t).
Thus
24
0 = (Bkp(B)e)II-124).= (Bke')Hp(-B)241
= eiH(-B)
e'H(-B)k+ilif for all i >
k >m Thus 7A.
Thus
7)7, (H)0
(H
Zii)° = 0. Likewise we can show that
= 0.
Thirdly, we want to show
If
xE
(
where
.1103, W)
rTh
assume
(Hlle) W))°,
[t]
p(t) , q(t) E
> k. Suppose
0=
11.1)
and
t
then
r (FT( U et ?Ai))° = o.
x = Bkp(B)e + Blq(B)f
p(t),q(t). Without loss of generality,
k <m-i. Then
[Bkp(B)e+Blq(B)f]'H(Zi e141)
03W)
[Bkp(B)e+Blq(B)f]tH(-B)m-1-k(
=
[Bm-1p(B)e+Bl+m-1-kq(B)f]'H(72. fdAj)
[Bm-1 p(B)e+B
=
[Bni-lp(B)el'H7f1+ [131+m-l-kq(B)f]TH241
p(0)[Bm- 1 ePHf + [B
If
1>k,
then
B
+m -1 -k
m- 1 - k
= 0. If
,
,B)f
clkJ'Hf.
= k,
[Bm-lq(B)fillif
[B/ +m- 1 -1(q(B)fi
=
q(0)[Bm-lf]'Hf.
25
(-1)111-1VHBm-if. Since
=
is skew. Thus
Both
fIFIBm-1f = 0.
is odd, then HBm-1
m-1
p(0)
>m-1. Thus
nonzero. Thus we have a contradiction. Therefore
we have shown
(U el())
,Th
are
(Bm-1e)Ilif
and
[H(11 e 145)]0 = o.
Fourthly, let p(t) =
and let
a.t2j
j>0
f = f + Bp(B)e
= f+
.B
)e
j> 0
be the B-cyclic subspace generated by f. We can choose
Let
a ,a
0
1
such that
C
H
fa°-
-A
3
3
j>0
j>0
=
Thus we can solve for
VI-1Bm-2f
a
0
.
a.B2j+le)
a.B2j+le),HBm-2(f +
0 = fIFIBm 2f = (f+
-
2a0e'HBm-1f,
If we consider
f'HB
m-4A
we will
f = 0,
get an expression in terms of f'HBm-4f, a0, f'HBm-3e, e'HBm-2f
and
and solve for
e'HBm-lf,
by considering f'HB
m-2(k+1)A
f.
al. Likewise, we can solve for ak
A
In like manner, let e = e + Bq(B)f
A
where
q(t) =
p
j>0
2.tj.
We can pick
rf01
.
.
so that
(2/6
,
the
26
satisfies the property H
B-cyclic subspace generated by
A
Zt, and kr obviously satisfy the first 3 properties we proved for
Thus we have completed the proof of (ii). Further,
21- and
A
let f = f +
fJ
a
a2,a3, ...
we can choose
j.B -If
so that
A
A
for
e HBkA
f = 0 j-2
A,
k < m. Let
7/tY
be the cyclic subspace generated
A
by
f;
then the pair 7.6
also satisfies the properties we have
,
A
It
proved for the pair
B restricted to 'IL 63, w is the direct
sum of 2 m-dimensional Jordan blocks of eigenvalues
0.
A
H,
restricted to
Ho =
1.)
®
0
-S
[S
0
,
is
where
S=
as.
A
and
A
A
K, restricted to 'ILEB"lif
,
K0,
is
NM,
Ko =
r-T
where
T=
o
0
0
.0
Multiply by
r
I
-H0 -K0 .
diag(1, -1,
Since
, -1)
I
the right and left of Ho + Ko we get
0
H0
is of even dimension, if we multiply by
on the right and left of
the anti-diagonal has been multiplied by
-1,
-H0 -K0'
then by Fact 1,
and the first
27
super-anti-diagonal is not changed. Thus we get
In case m is odd,
V--0
ZIED (Hit )°,
=
H(HU.)° C /C), KU, = HBll C
0
),0 ;
1, 0
), 0
K(Htt..) = HB(H GC) C H(Hit) C
1-10-K0.
ITO = H
(Es
and
process with
0
thus we can repeat the above
-)/
Likewise, for m even,
(H)°.
replaced by
we can repeat the above process with
replaced by
0
j)))°. Thus, after a finite number of steps, we show that
(H( U
is congruent to its transpose.
0
The process of showing M
to
in this case, instead of HK, we consider
Now we consider
N/Icc
K-1H.
(and thus we assume H and
MI
non-singular here). Suppose
1
I
K.i = 2(N.-N!),
+NI),
that if H.i = (Ni
i
i
i
2
1
of
Note that
K(K1 H)K-1 = -HK
nomial of
K-1H
-
xolynomial
=
are
)
f.(x)
and
i
i=1
such
N
is the minimum
is prime to f.(x) for
j.
so the minimum poly-
-(K-1H)',
is even; thus we place another restriction on the
f.(x); f.(
p
f.(
then
p
K
is a non-singular matrix such that
C2
C'2 M1C 2 = M. Without loss of generality, write M =
K.-1H.,
is similar;
=
xolynomials
[q.(x2dm where
q(x)
is irreducible
x)rq(x)). Without loss of generality, we may assume
(and
M1 = HI
+ K1. Lemma 3 (proved in this section) assures us that we
can find a basis for
1
(Mil+M'il)
_Jr
,
such that
and K il
1
M1
=
P
el
i=1
M.1,
il'
if
il -M' ) then Kil-1 Hu. is
similar to a companion matrix. Thus it suffices to prove the following:
H
H.
=
2
=
2
(M
28
Lemma 4.
and K as defined above are non-singular,
H
Let m be the degree of the minimum polynomial of
Proof.
Thus, for some
A = K-1H.
U.
basis for
is congruent to S'.
is non-derogatory, then S = H + K
K-1H
and
If
Write
,Am le} is a
e E V--, {e, Ae, A2e,
e. = Ai-le. Recall
is skew when
KAi
i
3
is even, is symmetric when
i
e!Ke. = (-1)je1KAi+j-Ze1 = 0
for
i+j
even,
for
i+
odd. Let
3
is odd. Thus
1
e!He.
= (-1)je1HAi+j-2e1 = 0
3
1
Ji
C:
11-- 2r such
that Ce. = (-1)i-1e.. Then
e
(1.C'FICe. Ce.)'H(Ce.) =
1
because if
i+j
1
3
is odd,
(-1)1+3e!He.1
3
=
3
1
e!He.1 = 0. Also
3
= -e.'Ke.
el.C1KCe. = (Ce.)1K(Ce.)1 = (-1)i+je!Ke.
3
because if
2.
1
1
3
i+j
is even,
e!Ke.
= 0. Thus
3
3
C 'SC = S
1.
q. e. d.
1
General Results
In this section, we shall prove some simple results, some of
which lay the ground work for the following section.
Proposition 11.2.1. Let
S
be a non-singular matrix; then:
29
if
is of type 1, then det 5* = ±det S.
S
if
if
is of type 2,
S
S
E.
det
then
is of type 2, then
is *-congruent to 5, where
S
is the conjugate matrix of S.
if
Proof.
S
(i) If
(C*)2 S*(C2 ) = S*.
If
is of type 2, then it is of type 1.
Therefore
C*SC = T
det S = (det T)(det C
If
then
C*SC = S*,
-1
S
is over
)(det
C*SC = T
det(CC*)
)*,
C1
so we have
C*S*C = S,
Thus det S* =
±1.
det S.
then
,
which is in
is over
,
C*S*C = T*= T' = C'S'C. Therefore
then
(CI)
C*S*C(C)
-1
= 51.
(C1)-1C4SC(E)-1 = (S')*=
If
C*SC = T
6,
is over
and from (iii) of Proposition 11.2.1,
S
is *-congruent to
Proposition II. 2. 2.
where
j* = -j,
and let
is similar to
A
If
If
K
H
T,
from Theorem
(hence *-congruent over
I. 2.1, is congruent over
then
then
S
to
)
is *-congruent to 5;
(5)' = 5*.
Let
A=
(A
H=
1
(5-S*),
and let K
1
2j
51S*. Then
-1
)*
is non-singular,
is non-singular,
B = K-1H
B1 = H-IK
is similar to B*.
is similar to
B.
TI,
30
If
is of type I (hence also if
S
is of type 2),
S
is
A
similar to A*.
If
is of type 1, i.e.,
S
singular, and
is non-singular, then C-1K
K
Proof. (i) SAS'
KBK
-1
-
S(S-1 S*)S1
=
K(K-1H)K-1
=
=
If
C
-1
is also similar to
C*HC = H
=
= -K
-1
(A*) 1,
-1
-1
1-1HC
H.
= B.*
1
(C*S*C) = (S*)-1S
thus
A
=
then
A-1. Since A
is similar to A*.
is non-singular such that C*SC = S*,
C
-
(A*)'.
is non-singular such that C*SC = S*,
C
AC = C -1(S -1S*)C = (C*SC)
If
-
S*S1
is non-
C
HK-1 = B*.
-1-1
= H(HK)H= KH
=
-1
HB1H
where
C*SC = S*
then
and C*KC = -K. Thus
,-1
C-1(K-1H)C = [C*KCJ (C*HC)
= -K-1H
Proposition II. 2. 3.
S=S0 OS co
A
oo
=
S'S *co
oo
(1) S
1
where
(i)
-1 *
A0 = S0
S
0
has eigenvalues only at
eigenvalues neither at
I
nor at -1.
matrix such that C*SC = S*,
o C
is such that CSC
000 = S,
CI
Suppose
is such that CISIC, = Si
S
q. e. d.
is non-singular, and
has eigenvalues only at
-1,
If
and
C
-
*
1,
has
Al = S11 S1
is a non-singular
where Co
1,
C=C0 f9C co
is such that C00S0000
Coo = S oo and
then
,
31
(ii) Suppose H =
singular. Let
1
K-1H
B
1
where the
B.,
(f)
j=1
1
some
qi(t) E
e[t],
=
and g. c. d (f.(t), f.(t))
j
are such that if
B.1
f.(t) = qi(t2)
then
is the minimum polynomial of B.,
f.(t)
and non-
and K = -2- (S-s*)
(S+S*)
iij Suppose
when
1
for
1
S = ED S.
.
1=1
where if
H.
1
=12 (S.+S.)
1
1
1
K. = --(S.-S.
)
2j
then
B. = K.1H.
1
1
If
.
1
C*SC = S*,
then
C
is a non-singular matrix such that
C = e C. such that
C.1 S.C.1 = S.
.
1
1
Proof. (i) Since
C'
AC AC = C -1(S
=
[C*SC]i[C*SC]
=
(A*)-1
AC = C(A*)
Partition C = (Cih)
- 1S *)C
conformably to
A.
-1
.
i = 0,00,1
and
h = 0, 00, 1.
Then AC = C(A*)-1
A.
and
(A*) - I
implies
*-
Ai Cih = Cih
(A,
)1
If i
h,
have no eigenvalues in common and thus C ih = 0.
32
We have shown that the off-diagonal blocks of
C
are
Thus (i)
0.
is proved.
(ii) Since
C-1BC = (C *KC)-1(C*HC)
= -B
Thus
BC = -CB. Partition
B.C.. = -C..B., i,j
1.3
13
C
1, 2, ... ,p.
conformably to
B.
Thus
j,
B.
and
If
i
then
-B.
J
= 0. We have shown that
have no eigenvalues in common, thus
the off-diagonal blocks of C are
0,
thus we have proved (ii).
The next result is based on Taussky-Zassenhaus' result [10].
Proposition 11.1.4.
If
S
is non-singular and S-15* is
non-derogatory, and C*SC = S* for
C
is *-congruent to a symmetric matrix over
non-singular, then
.......4
C*S
.
Proof: First assume A = S-15* is a companion matrix.
is similar to A* hence similar to a
is of type 1 therefore
A
matrix over
since
matrix over
[2];
."-8
A
is a companion matrix,
A
is a
33
Since
C*S(A)S-1(C*)-1
C*S(S-1S*)S-1(C*)-1
=
-1
C *(S
) (C *)
=
(C*S*C)(C*SC)-1
=
S(S*)-1
= A* = A' .
Since
A
is a companion matrix, by the Taussky-Zassenhaus result,
is symmetric.
C*S
Now suppose
A = SS* is not a companion matrix, but
is non-derogatory. There exists
A0
D
such that
D-1AD = Ao, where
is a companion matrix. Let T = D*SD. Then
S = (D*)
- 1 TD-1,
5* = (D*)
-1
T *D-
1
.
C*SC = C*(D*)-1TD-1C
= S*
=
Thus
(D*)-1T*D-1
(D1 CD)*TD-1 CD = T*. Note
T-1T*=
D-15-1(D*)-1D*S*D
=
D-1AD = A0
is a companion matrix. From what we have shown for the case
S-15*
being a companion matrix,
[D-1CD]*T
A
is a symmetric
34
matrix over
[D-1CD]*T
(D*)C*(D*)-1D*SD
= D*C*SD
Thus
3.
is *-congruent to a symmetric matrix.
C*S
Non-Singular *-Symmetric Pencils with All Eigenvalues in
Let
and K be *-symmetric. Then Theorem I. 2. 1
H
asserts that the pencil XH + pa,
where
is *-congruent to
terminates over
and
X
L
1.1.
are inde-
M where
is the
L
minimum-indices part as defined in Theorem I. 2. 1 and M is the
non-singular core, unique up to *-congruency. If T = H1+ jKl,
where
matrix over
the matrix
pencil
and
H1
K1
then H1- jK1 T*=T1 = HI1 +jK 1 Thus proving that
S = H + jK is of type 2 is equivalent to proving that the
XH + H.K
is *-congruent over
*
Hi = Hi = Hi ,
jKi = -(jKi)
are *-symmetric and j* = -j , and jKi is a
E
*-congruent to
-K1 =
.t
K' = K,i
1
e.
to
X.H1
+1 where
rxwl.
and
H1= H1
We say a pencil is of type 2 if XH + pa
X.H1 +1 where
H1
above. We say a pencil is of type 1 if
XH - pa.
First we want to show that:
and
XH +
K1
is
are as described
is *-congruent to
35
where H and K are
Proposition II. 3. 1. If S = H + jK
*-symmetric and j*= -j,
core of the pencil
then
is of type 2.
XH + p.K
M where
XH + i.K = L
Proof: Write
is of type 2 iff the non-singular
S
is the minimum-
L
indices part and M is the non-singular core.
("only if")
*-congruent to
If
XH + p.K
XH1 + p-K1
.
1K
(51K1)* = -3-
thus
Ko = j -1K
=
*
-KI1 = K1 =
k
to
K
1
Let
is a matrix over
3-1K
is
XH + p.K
H1 = H1 = H1
where
is congruent over
XH1 + p.K0
.
is of type 2, then
L0
41) M1 where
L
0
1
L0 = 0
13)
A
i= I L
0
1.
A
are as defined in Fact 6 and M1 = XH1 + p.K0 is a
E.
Lei
A
A
A
A
and HI1 = H1,
non-singular pencil and HI, KO are over
1
and
L
1
A
A
(Ko)' = -K0.
A
Thus
A
L 1e X.H1 + iljKo,
kl-11 + 1.1K1 = XHi + jp.K0
where
kl
1
where
WEi
= 0 ED
L
i=1
kV*
0
is *-congruent to
36
E.
1
W
0
i
E
Because of the uniqueness of the minimum indices,
W
0 Li
*-congruent to
*
-
Since
0
kH1 + pK1
and
1
*
is
0
E.
1
kH +
.1.1K
have
LEi
Ll is *-congruent to L [11].
(Recall in the beginning of the proof, we assume kH + 1.1K = L Eli M,
the same minimum indices,
where
L
is the minimum-indices part and M is the non-singular
core.) Thus kH + i.LK is *-congruent to
*-congruent to
L
A
= jKl; kH1 +
(jKI)* =
H1 = HI = H1,
A
L1
A
(j11)* =
A
)* = H?1 = H1.
(H1
A*
A
L EDkH1 + FIKl
A
where
p.K1
in turn is
which in turn is *-congruent to
A
A
singular core of pencils
before
+
Ki = jKo Thus kH + iK= L
+ pjKo. Let
*-congruent to
.A
kH1 + 1.1.3K0,
.A
kH1
A
5K1 =
A.2A,
-jKi = --J(JK0)=,' = J K o -
is
By the uniqueness of the non.2 A
Ko
A
A
Hal. We said
and
is a matrix over
is *-congruent to
M
M
kH1
+
A
-J.2AKo - -3K1
Thus M is of type
2.
(UHT?)
We only need to prove the minimum-indices part
L
of
37
the pencil is of type 2. Therefore without loss of generality, assume
A = X.H + 1.1.1C
IMO
where
L
E.
is defined in Fact 6 of Section 1 of Chapter I. Multiply
on the right and left of A by
By a calculation similar to
I
EEi.
a calculation appearing in the proof of Theorem II. 1. 1, we can see
that
A becomes
[iii. 1. 1
A2
where
Multiply on the right of [III. 1. 1] by C1 = diag(1,j,
,j, 1)
and the
left by Cl. Since [III. 1. 1] is of odd order, by Fact 2 of Section 1,
Chapter I, we have multiplied the entries on the anti-diagonal
38
alternately by
j
alternately by
that
E
{1 ,
1
and
-j
and
-3
.2
Thus
.
6
is a matrix over
H2
entries are
E
{j, -j}
is such
C 1A2C1 = XH2 + H.K2
because its non-zero entries are
E because
is a matrix over
and jK2
-j2}
and the entries on the anti-diagonal
K2
Is
non-zero
q. e. d.
.
In this section, we shall consider 5 = H + jK where H, K
are *-symmetric and j*= -j,
such that the pencil
XH + p.K
is
non-singular, and such that all the eigenvalues of the pencil are in
(Note that
5-1S*
=
is non-singular,
(H+jK)-1(H-jK) If H
then
[H(I+jF1-1K)I1H(I-jH-1K)
S-1S* having eigenvalue at
Therefore
being nilpotent. Likewise,
equivalent to
Suppose
and
and
-
HI1 K1
K.
I
-1(I-jFI-1K)
=
is equivalent to H-1K
1
having eigenvalues at
5-15*
-1
is
K-1H being nilpotent. )
S=
P
ED
i= 1
S.
and Hi =
*
1
*
1
and Ki = Ti (Si-Si)
(Si+Si)
1
is nilpotent,
-
is nilpotent, and for i > 3,
K21 H2
are non-singular, and
K.-1H.
1
1
is similar to
-K-1.IH.1
H.
1
and
,,
has eigenvalues only at Pi,
the minimum polynomial of
13i
-13.
,
1
K. 1H
1
1
and
for
R
"i
'
i > 3,
an if
f.(
i
td
)
is
then g. c. d.
S
(f.(t),f.(t))
j = 1 for i i j. Then Proposition II. 2.3 asserts that
is Of type 1 iff each Of the S. is of type 1. Therefore, without loss
1
1
39
of generality, we consider the following cases separately:
K-1H has eigenvalue only at
is similar to
K-1H
and
p
±3,
K-1H
and
is similar to
is similar to
H-IK
where
p, 1, -R and
P
=
P
-I3
-K-1H.
K-1H has eigenvalues only at
and K-1H
-p
-K-1H.
K-1H has eigenvalues only at
p
and
p,
-13
where
p
-K-1H.
is nilpotent,
H is nilpotent.
K1
Theorem II. 3. 2. If
has eigenvalues only at
K-1H
then
S
P'
p
-p , -p
-K- 1H,
where
and
P
is of type 2.
Proof:
chapter,
is similar to
K- 1H
S
First consider
p
=
-p
By Fact 5, in the introductory
is *-congruent to a block diagonal matrix whose
diagonal blocks are of the form:
40
1
13+i
1
-a+i
-1j
1
1
-13+J
Icrote:
j=
(R+j)* =
(-13+j)* = R - j =
Therefore j(p+j)
and
j(-i3+j)
are elements in
6.
We shall
Therefore
show that each block is *-congruent to a matrix over 6
without loss of generality, we may assume S = [ H.3.1]. Multiply
.
by
C = diag(11 j, 1,j, ...) to the right and by
C*
on the left of
then by Fact 2, the entries on the first super-anti-diagonal of
.2
or -3 or (in case of the middle entry)
become either
1
the entries on the anti-diagonal are multiplied alternately by
-j. Thus
S
S;
S
0,
j
and
and
is of type 2.
Next consider
13*
±p. By Fact 4,
matrix with diagonal blocks of the form:
S
is 4-congruent to a
1+d-
1-F9-
.
f+d-
1+44d-
..
-
f+*d_
'NI e JAI =
1
1
f+dI
.
I
1
f+*d
.
'
f+,:cd
. 1+,4
I
1
I
f+d
.
,
f+d
1
,
I
f+d
1
42
where M is the upper left 2m x 2m block and N is the lower
diag(1,j, 1,...,
right 2m x 2m block. Let C = diag(1,
2m
Consider
N)C = P e Q. p.. = q.. = 0
C*(M
and also
2m,
p
m, 171
q
2
Pi,2m-i
2m-i
-j
=1
i-1
2m+1 -
q.(
= (-1) i-1j(P+i)
2m+1- i
2m+1
when
i
is even
when
i
is odd.
=
(-1)i-l[j13*+2j2]
i<m
=
(-1)i-i[jP-Fj2 ]
i > m.
-1)i - 1( -j P+j 2)
=(1)i-1(-j*+j2)
=
i+j
and
=0
M, M
for
i<m
i > m.
Consider
[I
I
PlI
1:-)
P+1-5
P2+P2
P*P+PIP
P*P2 +PI2P
too.
which is over
6
Note that
P*
PP
15
P
and
43
= det(P -P)
det
±i2m(p+p*)2m V
because
Therefore M
0.
±j(13±P*)
N
0
is of type 2. Thus
is of type 2.
From this point let
and
B.
denote non-singular diagonal
D.
matrices over
Theorem 11.3.3.
basis for
such that
11
is nilpotent, then there is a
H-1K
If
0
0
where
ED T.
i=1 1
S
T. = (E
1
+jF
n.1
Bi
)
,
ni
and n. i n. for i i j.
3
1
that
If
h
S=
Ett
i=1
co
p,
and
0,
p
1=1
.
1
1
,
E
1
U.- such
n. i n. for i r/j.
1
3
-
-K11H and has eigenvalues only
E, then there exists a basis for
where
1-
= F n. CS) ( B AD D.) + En.
1
1
i
is similar to
K-1H
-p,
= (jE+F)
,0 B.
n. n.
T.
1
such that S = @ T.
P.
cc
where
T.
If
at
is nilpotent, then there is a basis for
K-1H
[(13+j)B. ED
1
i
1
(T
-P+j)D.1
1
(Pictures of blocks of Tr, z = 0, .0 , p where
n.
1
13 /0,
n., for
pE
Appendix I).
Proof:
By Fact 4,
(i)
is *-congruent to a matrix in
S
diagonal blocks with each block of the form:
WO
E, n
= E(E
+F
)
E /n,
EE _E
i
j.
3
are in
44
Thus
rk.
S=
Wo
E., n.
j=1 1=1
3
1
)
k1
where.n. / n.i
i / j. Suppose S = 0 W:3, n
if
and
B = H-1K,
{el '
Be
1'
Write
n = n1,
n-1
is a
1
k = kr Then
... ,Bn- le
elifor
basis for
i
i=1
J
1
, e2, Be2, .
some
.
,B
ek ,
le2,
el, ... ,ek
. .
,B
ek}
If' Rearrange the basis
E
elements in the following manner:
{e1,e21.. ,ek,Bei, Be2, .. ,Bek, . .,Bn-1er
respect to this new basis
S
becomes
To
.,Bn-1ek}. With
8
=
1
(En1+jFnl)
B1
B1 = diag(er . . ,Ek). Repeat the above process for
where
j = 2, 3 ,
0
h
. . .
Thus if
, h.
S=
h ki
Q3
3-1i1
is *-congruent to
S
W
Ei' nj
ED T
j=1
nj
If
K-1H
is nilpotent, by Fact 4
is *-congruent to a
S
matrix in diagonal blocks with each block of the form:
W
E$
E / 0,
n = E(jEn+Fn)
H-IK being nilpotent,
Going through a similar process as in the case
we can show
S
is *-congruent to
Eft
T.
j=1
6.
EE
.
3
(ii) When K-1H has eigenvalues only at ±p
by Fact 4, we have
S
o,
*-congruent to a direct sum of blocks with
these blocks paired off in the form U
13
E
,n
et U
13
,
E
$n
where
45
n = E(Fn+(3+j)En),
/0,
E
5 7' 0,
6, n = 6(Fn+(-13+j)En),
Without loss of generality, assume
k.
s= e [( e u
[II. 3 . 2]
j=1
1=1
e(
ED V
E'i n j
1-1
6,nj
)]
1
First consider
;1(
[H.3.31
S=
k
3
U
ED
E
i=1
n = n1,
Write
n-1
manner
-
. ,
i
..
-
1
-
H-3I), A1 = (K1 H+PI). A
An1I dk}
.
e2,...,A1n-1 ek,d1,Aidi,...,A1n-1di,
Rearrange this basis in the following
{ei,e2,...,ek,di,d2, ...,dk,A0ei,
n- 1
n- 1
Ao
new basis
,A0
, Ai
Then in this
dkl.
is of the form
S
(-13+j)D1]
where
n1
n1
B1 = diag(E
,A0ek,Aidi,
n-1
n-1
ek, Ai di,
(303 D1) + E__ 0 [(P-1-j)B1
F
S
'=\\'i
is of the form [II. 3. 3] is of the form:
S
{ei,A0e1,...,A0n-1
2
,Aidk,Aidk,
.
,(13 e v
Let A0 =
k1 = k.
If for which
basis of
i
n,
-\\
3
,E
Ek),
,
is of [II. 3. 2]
S
D1 = dial( 5
is *-congruent to
6k)
.
d
k
ED
T'.
.
E
Thus if
q. e. d.
j=1
In the following theorem we let
z
Tz = T1,
where
z = 00, 0, p
46
and where
13 / 0
13E
,
B = B,
E = E,n
l
(i) (a) For
Theorem II. 3. 4.
hence of type 1. For
(ii) For
(i.e., iff
even,
n
odd,
n
n
T0
(b)
jB
odd,
is of type 2 and
T0
is of type 1 iff
is
B
is of
is of type 1), and (c) T0
is of type 2.
type 2 iff jB
For
even,
n
(i.e., iff
-B
n = nl, k = k1,
F=F
and
*-congruent to
we also write
;
T
on
T
is of type 2 and hence of type 1.
is of type 1 iff B is *-congruent to
is of type 1),
jB
oc
and
-B
of type 2 iff
T
is of
jB
type 2.
Proof: Suppose
Recall
is odd. Let
n
T0 = (E+jF) 0 B.
.
,
j,1).
Then
)*T°(C(xj Ik) = (C
(C
C = diag(1, j, 1,
Ik)*((E+jF)
= C*(E+jF)C
B)(C
T.k)
B
By Fact 2, the entries on the anti-diagonal of C*(E+jF)C are
alternately
or
-j2
alternately ±j(j) = ±j2,
zeroes. Thus
-B. Suppose
P
the rest of the entries of C*(E+jF)C* are
is of type 2, and hence is of type 1.
To
(b) Suppose
those on the first super anti-diagonal are
1,
n
is even, and suppose
B
is *-congruent to
is non-singular such that P*BP = -B. Let
C = diag( 1 , -1,1.. . );
then
47
P)*[(E+jF)
(C
P*BP
P) = C*(E+jF)C
13](C
= (-E+jF) 0 -B
= (E-jF) 0 B.
Conversely suppose
Let
H=
1
is non-singular such that
CI
*
C1H(H1 K)
H(H
-1
*
1
and K = 7(T0 -(T0
(T0 +(T0 ) ),
K)
n- 1
= [E
C1 =
=0
(T 0)*
C1:',E
Then
).
-)
-H(H1 K)n1
131[E F 45)
n-1
Ik]
= B Eft 0
Thus
This computation will be used again in the proof of part (c).
0. Thus by Fact 3,
0)C1 = -B
CI(B
(c) Now suppose
exists a non-singular
C2 = diag( I, j,
, j).
is even and jB
n
is *-congruent to
B
is of type 2, i.e. , there
*
such that jC013C0
C0
-B.
is over
Let
Then
(C2 0 CO) T°(C2 0 CO) = [C2 0 C0][(E+jF) 0 B](C2 0 CO)
[II. 3.4]
= C2(E+jF)C2
C0BC0
The anti-diagonal entries of C2(E+JF)C2 are alternately
+j
and
the first super-anti-diagonal entries are alternately -j3 and j.
Thus
*
j1- C2(E+jF)C2
factor by j-I
we get that
is over
6.
Thus (if we multiply the first
and the second factor by
[II. 3. 4] = j
-1
j
in the tensor product)
.
C2(E+3F)C2
0 jC013C0
is over
E
Thus
48
is of type 2.
To
*
Conversely suppose
1
1
(w+w*), K1
=
C3T0 C3 =
(W-W*)
so jici
is over
C3
Since
matrix
matrix,
--
of
D
1
1
C3R
-1
jR*H1(H1 K)
n1-1
R
is a matrix
and the first principal sub-
is non-singular. Hence jD*BD is the upper
left block of j(C3R)*(B&O)C3R=j(C3R*(H(H-1K)n-1)C3R which is
over
Thus
.
(ii)
jB
is of type 2.
The proof of (ii) is similar to that of (i).
For the following discussion, write
p
0,
and
1
= jR*H1(H1-1K11n-1R .
Recall H(H-1K)n-1 = B e 0
matrix
6. Likewise H = H * = HI
has its first k x k principal sub-
is a permutation matrix,
over
Thus
non-singular. Thus
D,
R
K1 = -K1 .
1
is over
*
jR*C3(H(H1
K)n1 )C
Since
Let
is non-singular, there exists a permutation
such that C3R
R
is over
then W = H1 + jK1,
W* = H1 - jK1 = H'1 + jKl1 . Thus
(jKI)* = -jKi = jKi,
W
pE
,
T
q.e.d.
as
T.
Recall that
ni= n. (Refer to picture in Appendix I.)
Theorem II. 3. 5.
T is of type 1.
T is of type 2.
The following are equivalent.
49
(3) B is *-congruent to (-1)nD.
M=
(
Lemma 5.
If
C
C*
then
C*)
,
C
such that
cE
c* 1 ±c, C = cI,
E
E
(c*)2 A+Ec2 A
[A
M=
M*
0
c2A+E(c*)2A* cc*(A+EA*)
[cc*(A+EA*)
EA*
The proof of Lemma 5 is routine computation.
Lemma 6.
Suppose
C, B, D are non-singular.
C*(B
(-1)nD)C = -(B El) (1)D)
C):c(B ED
(l)n1 -1D)C = B
If
and
( - 1)n -1D
then
is *-congruent to
B
(-1)n-1D.
Proof of Lemma 6. Partition
C=
C11
C12
conformably:
then
C22);
(C21
[B
*
0
C*
C=
0
where
C
E
ED
1
C 11BC 11+EC 21DC21
*
*
*
C 11BC 12+EC 12DC22
*
C12BC 11+EC22DC21
*
C
12BC 12+EC 22DC22
.
Referring to the first condition, we have
C11BC11
+ (-1)n C21DC21 = -B.
50
Referring to the second condition we have
-1
+ (-1)n
Cl 1BC1 1
*
C2 1DC21 =
B.
If we add the above two equations together, we get 2C1 1BC1
n- *
B
Thus B is *-congruent to
Thus (-1)1
=
1
C2 1DC2 1
Lemma 7.
1
K =23.-(r-T*),
If
T
1
is as in Theorem 11.3.5, and H = 2(T+T*),
then
(0)n-1(B
K[(K-111)24,2i]f-1
H[(K-
(0)n- lp(B
1H)2..p2I]n- 1
Proof of Lemma 7.
(-1)n-1D)
®
1 )nD)
Recall
(-P+3)D]
T = F 0 ( B sEs D) + E 0 [(P+j)B
(Refer to
and
picture.)
F
(B
D) +E® (PB Q3 -PD)
K=E
(B
D)
H
K
Let
-1
= E 0 (8 -1
,
Q3 D
-1
)
.
k = dim B.
K-1H=EFOI 2k +I nGD(PI
N=EF
is the
nxn
matrix with
1 's
-
).
on the first sub-diagonal
51
and zero everywhere else.
1
1
H- I=N
I2k + In 0 (0k
=N
H+
+
I2k
- 2 61
)
0 (2PIk $k)
SO
=
-
H-
(K
13I)(K H+PI)
I2k + N
N2
(26
1k
ED -23I
'
k
)
= (N (8) I2k)(N 0 I2k + I 0 (2pIk 0 -23I ))
.
Thus
,
-1
1.(K
H)2 -p2ij,n-1
M = (N
is a
blocks
= (N
I2k
n-1
- 2PIk))n
® I2k)(N® I2k+10 (213Ik
+I
(
-1
-26Ik))
261k
2kn x 2kn block lower triangular matrix with the diagonal
=
(2)n '1k(Er (-23)n-1I
)
I
,n
1
.
Thus
[E (8) (B ED D)][Nn-1 0 I2k]1\.4
=
((213)n-1B
(13
(-23)n-1D)
(213)n-1(8
ED
(-1)n-1D)
ED 0
eo
Similarly,
H[K-1H)2-p2]n-1
=
(2P)n- 1PB e (2p)n-1(-3)D
=
(0)n-113(B
(-1)nD) e 0
0
q. e. d.
52
Proof of Theorem 11.3.5. We shall prove (3) => (2) => (1) => (3).
(3) => (2). Suppose
Q
is non-singular k x k
such that
Let
C = In 0 (Ik
e Q). Let p = C*TC. Then
P = [Ina) (Ik Ef31Q)]*[F
(B e D)+E
[(P+j)Bek (-P+j)Dfi[InCi0 (Ik
Q*DQ = (-1)2IB.
Q)]
= F 6 (B 6 (:)*DQ) + E 0 [(p+i)B e (--P-I-j)Q*DQ]
(B 9 (-1)nB) + E
=F
Let
such that
cE
[(P+j)B e (--P+j)(-1)nB]
c*
±c,
M=
.
and
cIk
c*Ik
By Lemma 5
R0 = M*(B
Let
1.1* = p_j
p. = p+i,
R
(c*)2B+(-1)nc2B
c2B+(-1)n(c*)2B
cc*(B+(-1)B
Thus
= M*(p.B e
-P+j = -p.*.
Then again by Lemma 5,
np.*B)m
(-1)1
n-1 LB)
n-1
2
[cc*(p.B+(-1)
c2p.B+(-1)
(c*) 1.1*B
Note
i
cc*(B+(-1)nB)
(-1)nB)M =
(c*)2p.B+(-1)
n-1c2p.*B
cc*(B+(-1)n-1B)
53
(In 0 M*)P(In 0
= (In® M*)(F
=F
M)
(-1)nB) + E 0 (p.B+(-1)11-14*Bl(In 0 M)
(B
(M*(B e (-1)nB)M) + E 0 M*(p.13+(-1)*B)M
=FOR0 +FOR1.
Note that if
if
n
n
is odd,
is even,
jR0
= diag(1, j, 1, j,
and
).
and jR,
R0
are matrices over
are matrices over
R1
i. Thus let
Then
(UEfiI2k)*(FOR0 +EOR1)(U$I2k)=U*FU 61)Ro+ U*EU OR1,
by Fact 2, is a matrix over
6
This
T
is of type 2.
Proposition 11. 2. 1 (iv) asserts that (2) => (1).
C*TC = T*. Recall H =
, ,n-1
1
c = -KL(K H) -p
K = -23(T-T*), C*KL(K1 H)2 -(32 I]
(1) => (3). Suppose
C*H[(K-1H)2-132I]-1C
C*(H+jK)[(K
-1
H[(K-1H)2-132P-1.
1
(T+T*),
,
Thus
H)2 -p 21]n-1 c = (H-jK)[(K -1 H)2 -p 2 n-1
Lemma 7 asserts that
H[(K-1H)2I]n-1
(2)n-l(B e (_i)no)
K[(K-1H)2-p2Ir-1
(mn-1(B
(-1)n-1D)
0
.
54
C*[(2P)11-113(B ED (-1)nD
C*(H+jK)[(K-1H)2-I32I]n-1C
i(o)n-1(B
[(2p)n-l(B
_
1)11-1D)4) 01C
Ela
(..unD)
j(2p)n-1(Be (_un-1D]
By Fact 3, there exists
C11
C11(B (134
non-singular, such that
(-1)nD
( -1)nD)C 1 1 = B
C11(Be (-1)n-1D)C11 = B e (-1)n-1D
Thus by Lemma 6,
Theorem II. 3.6.
i
j,
where
is *-congruent to (-1)'D.
B
z = co
(i) If
or
0,
S
Tz.
=
z
i=1
then
S
,
with
q. e. d.
for
n.
n.
1
is of type 1
iff each
Tz.
is of type 1.
(ii)
Let H =
1
2
(Sz +Sz ),
K
j (S-S) .
1
2
z
z
Then if z = 0,
Sz
is of type 1 iff H(H-1K)2m+1
is *-congruent to -H(H-1K)2m+1 for
m = 0, 1,
is of type 1 iff K(K-1H)2m is
;
*-congruent to
Proof:
if
z = co,
Sz
-K(K-1H)2m.
The "if" parts of both (i) and (ii) are trivial. To prove
the "only if" part of (i) for
z
it is enough to prove
0
type 1 for n.%even
0
(see Theorem 11.3.4(i) ). Assume
n. >
of
for
55
i < h. Write
If
as
S0
is even, then
n1
by Fact 3
is *-congruent to
0
1
K.1 = 2j (T. -(T.0 )* )
-1
H(H
is of type 1.
1
is
H(H
so
-B1. Hence by Theorem 11.3. 4(i)
is of type I also ifn1 is even. Let
1
0
-1K) n1-1-1K)n1-1
= B1 EP 0 ,
Also H(H
-H(H
TO
T
is of type 1 implies
S
*-congruent to
B1
is odd then
If n1
S.
H. =
0
0
1
(T. +(T. )*)
2
I
then
;
K)
n2-1
n2-1
-1
n2-1
-1
H2(H2 K2)
= H1(H1 K1)
ED 0.
Also
n2-1
-1
H2(H2 K2)
and (if n2 is even then) H(H
-1n2-1
-H(H K)
*congruent to
*-congruent to
Thus (if n2
1K)
n2-1
-H1(H1 K1)
-1
n2-1
is *-congruent to
n2-1
_1
Thus by Fact 3
_1
= B2 ® 0,
H1(H1 K1)
then
-H1(H1 K1)
is
H1(H1 K1)
n -1 is
is *-congruent
B2
is even and hence in any case)
B2
2-1
e -B2* Since
n2-1
e
T2
-B2.
is of type 1. If
we repeat the same process by considering in like manner
ni- 1
_1
H(H
K)
for
i = 3, 4, ... , h,
type 1. Thus we have shown for
type 1 if
So
then we can show each
i
1,
,h,
that
is of type 1. Thus we have proved (ii)
The proof for
z = co
T.0
for
0
T.
is of
is of
z = 0.
for both (i) and (ii) are similar to the
above discussion. In this case we would only replace "even" by "odd"
n -1 by
T° and H(H K)
(and vice versa) and T0. by T.
i-1
56
n.- 1
K(K-1H)
Theorem II. 3. 7.
If
S=
(13
,
i=1
00/p/o, 13E6,
then
the following are equivalent.
S
is of type 1;
S
is of type 2;
is of type 1 for each
T1
i;
is of type 2 for each
T1
Proof: Theorem 11.3. 5 asserts (3) <=> (4) and Proposition
11. 2. 1 asserts (2) 3(1). Thus the following implications are obvious:
=> (1)
(2)
A
A
II
I
4 <=>
3
Thus to complete the proof we have only to show (1) => (3).
Let
K. =
Let
H=
2j
1
(s+s*),
(TP)-(T)*)
,
P. = KL(K
= -P.,
n.i >nh for
-1
K=
.
H)
and
2
1(S-S*).
Suppose
2 , n.- 1
C*SC = S*
,
C-*Q.0 = Q..
(-1)
n1
Q1 =
and
Q1 - jP1 = C*(Q1+jP1)C :
such that
13(B1 ED
13
13
(T i+(T. )*),
is non-singular.
where
C
2
2 ni-1
;
then
Without loss of generality, assume
i < h. From Lemma 7
n1-1
1
2
Q. = Fl[(K H) -p 1]
and
(213)
H. =
P 1 = (23)
n1
ni-1
-1
(B e(-1)
1
D1)450'
0. Apply Fact 3 to Q1 + jP1
there exists a non-singular C1
01) EP
57
n1-1
CI(BI ED (n1-1-1)D1)C1 = -(B1 ® (-1)
D1)
and
n1
( -1)1DI)C1 = B1 ED (-1) D1
CI(B1
Then by Lemma 6,
is *-congruent to
B1
(-1)nD1. Thus
TR
1
is of type 1 (see Theorem 11.3.5). Now,
2
1
P2 = K1[(K1 H1)
-2P I]
2
1
2
Q 2 = H1 [(K1 H1 ) -(3 I]
n2-1
n2-1
n2-1
D2) e o,
e
(-1)
(B2
n2-1
P(B2 e (-1) 2D2) e o,
El) (2P)
n2-1
ED (20)
and
C*(1P2+Q2)C = -017)2 + Q2'
Applying Fact 3 to this last *-congruency, we get that
-1
2
PK1[(K1 H1)
2
2
I
2
n2-1 + HI[(K1 HI) -p I]
n2-1
e (2P)
n2
(B2 ED
0)
is *-congruent to
-1
-PICI[(K-11HI)-pI]n2
Also, since
T1
-
2
+ H1[(K11 H1)2 -0 I]
is of type 1, we have that
n2-1
(0) n2 (0 013
n2
(-1)
D2).
58
is *-congruent to
-1
2
2
-PK1[(K1 H1) -p I]
Thus by Witt's Theorem,
B2
n2-1
eil 0
-
n -1
+ H1[(K11 H1)2 -p2 1]2
n2
is *-congruent to 0 $ (-1)
n2
(-1) D2. Hence T2P is of type 1.
is
*-congruent
to
B2
r
If we repeat the above process by considering P. and Q.
so
i = 3, 4, .
then we can show
T.P
1
is of type 1 for each
i.
D2
59
III.
1.
THE USUAL COMPLEX CASE
j
General Results
In this section, we take
to be the complex field and
Ee
to be the real field and we consider the problem from the view point of
a criterion Dina Ng proved in her thesis [3].
Proposition III. 1. 1.
X1, X2,..
real roots
,n
and
j2
r. (2)
{[h(1)(y)?1 Lh
(y)] ,
H=
(S -S *)
1
Suppose
(s+s*).
-12m+1
is of signature
K)
if
have that
= 0, 1, 2}
such that
grg2'..gn
is non-singular.
S
is an n x n matrix and
Then
H1
K is nilpotent, we have that
if
H(H
[h -"(y)]
Refer to Dina Ng's thesis [3].
Proposition IlL. 1.2.
1
,
matrix P(i,j) = (sgn gj(X.i))
nxn
Proof:
K=
..
then there exist
and n = deg h(y),
the
is a real polynomial of simple
h(y)
If
S
m = 0, 1, 2, ...
H
0
S
is of type 1 iff
for m = 0, 1, 2, ...;
and K are as in (i) and
K-1H
is nilpotent, we
is of type 1 iff K(K-1H)2m has signature
0
for
60
(i) From Theorem II.3.6(ii),
Proof:
is *-congruent to
H(H-1K)2m+1
is of type 1 iff
S
(for m = 0,1,2,...).
-H(H-1K)2m+1
In the usual complex case, this is equivalent to saying that
has signature
H(H-1K)2m+1
0.
q. e. d.
(ii) The proof follows in a similar manner as that of (i).
Theorem III. 1.3. Suppose
and K are non-singular, and
1
1
H = -2- (S+S*), H
K = Ti- cs -s *) ,
K-1H is similar to -K -1 H.
H and f(x) =
nomial of K1
is an n x n matrix,
S
If
p(x)
is the characteristic poly-
p(x)then
g. c. d. (p(x), p T(x))
'
f(x) = h(x2)
for some real polynomial h(y). Let
{[h(1)(ydji[h(2) (y)]
r = deg h(y).
where
has signature
of
and all
5)
Proof:
for all even polynomials
0, 1, 2, ... ,n-1
Suppose
If
we can assume S =
,
n
is the order
S1
ED
'
j
= 1, 2'
52,
sig Kq(K1 H) = 0
[x].
q(x) E
has signature zero
K[f(K- 1H)mg((K-1H)2)]
*
-Si )
Kj = T (S.-S.)
(where
is of type 1 then
S
for m = 0,1, 2, ... , n-1 and all
1
0, 1, 2}
j.
gE
("only if")
("if ")
ir
is of type 1 iff K(f(K-1H))m g((K-1H)2)
S
for m
0
(r)
[h(r)
j2
g
Without loss of generality,
*
3233
where if H. =
1
(S,+5.)
and
-1
such that all eigenvalues of K2 H2
are
61
non-real, and all eigenvalues of
are real. We have shown in
Ki-1H1
Chapter II that 52 is of type 1. Therefore without loss of generality,
as sume S = SI. There exists a basis for Irsuch that S of
the form
r
®T
S
131
k=1 1=1
where
P1
Tk
and
B
As sume
k
=F
nk
and
n. >nh
(Bike D/k) +
(-p +OD/
[(p+i)Bik
Enk
are real diagonal matrices, with entries
Dik
i < h . Note
if
2
2
(x -Po),
f(x) =
1=1
and h(y) =
(y-i32
m=1
'L
)
m
From our hypothesis,
1. 1]
0 = sig(Kf(K
= sig[K
-
1
11
H)
n1-1 g[(K -1H)2 ])
[(K
-1
H)
m=1
Let
K
1
=
R
R
(T1 -(T1
*
)
)
,
2
2
-,111-1
-BmIl
g[(K-1H)2]
,
±1.
62
[III. 1. 1] equals to
sig Et) ly 11
= sig
mil
1=1
X ((K
-1
Z
n1-1
2
-pmin
-1
2
g(aCji
)
-1
-1
n1-1
2
2
]gUK1
Ki -1
Hi) -f3mI)
)
2 2I
(p.f)
2
)
n -1
2
2
H ) -p m1)
/
2
(K./ H1)
-1
m=1
/ =1
Note
((K
1
+ M where M is strictly lower triangular.
Thus
-
2
g((K/1)2 ) = g(P1)I + M1,
mil
(because
H
((K-1H1
./
(132-I32
m
mil
)
=
)-62m I) =)I + M2
where
hl(P2)),
M1
and
M2
lower triangular.
-1
(K,e
2
- p 2 = Nn1
H./)
M.(N
n1-1
nl
Ci) I2k)
I2k
=0
for
j
=
1, 2.
Thus
,
[III. 1. 1] = sig
2
-,111-1
1(1[11'(1)n
1=1
n1-1
,
guytN_ it) 12k)
are strictly
63
Recall
K
[III. 1 .
=E
1
n1
0 (B1 1 et (-1)
2
1]= sig 0 h
)
1=1
n1-1
n1-1
DI 1)
.
n1-1
2
(B1 ED (- 1 )
g(P1 )[En N
n1-1
DI 1)]
""
2
sig[hI(131)n1-
1
2
g(1:3,e)(131
-
e (-1) n1 1Di 1)]
1
1=1
7 n -1
n1-1
sgn[h'(r) 1 g(rdsig(B1 e (-1)
DI 1)
=
1
1=1
p(j,i) = (sgn
such that
,g
g (successively) = g 1, g2,
Choose
matrix. (See Propo-
is a non-singular r x r
gj(1312))
2
sition III. 1. 1.) Then Q(j,i) = P(j,fl(sgn(111(I31))
singular
r x r matrix. Thus sig(Bi e (-1)
each
Thus
1.
sig
1=
-sig(-1)
n1-1
= sig(-1)
Thus
B1
1
is *-congruent to
(-1)
Di
1
n1- 1
is also a non-
Di 1) = 0
1
for each 1. Thus
is of type 1 for each 1.
Write
r
S0 =
1=1
r
p
R
T11,
for
Dil
DI
1
n1-1
and let S1 =
ED
(
131
ED T.
j=2 1=1
)
Pi
T1
64
1
1
H0 =
),
2 (S0 +S 0
K0 = 2i (S0 -5 0 )
1
1
H1 =
K1 = 2i (S1 -S 1)
(51+51),
.
Then
-
0 = sig K(f(K1 H)
n2-1
-1
= sig Ko(f(Ko H0)
g((K-1H)2))
n2-1
-1
+ sig Ki(f(Ki H1)
Because S 0 is of type 1 and
f(x)
-1
g((K0 H0)
n2-1
n2-1
2
))
g((lC11HI)2)),
2
g(x)
is an even polynomial,
repeating the same process as above, by replacing
withn2,
we can show
K
with
K1,
is *-congruent
H
with
to
(-1)
/.
The proof of the nif" part is completed by similarly considering
n2
HI,
n1
Di 2
,
for each /;
PI
thus
T2
B./ 2
is of type 1 for each
the fact that
n -1
0 = sig K(f(K - 1 H)k
for
-1
H)2 ])
k = 3, ... , p.
Proposition III. 1.4.
1 iff
r
gL(K
S
S
is of type
is of type 2.
Proof:
implies
In the usual complex case,
S
It is only necessary to prove that
is of type 2.
S
is of type 1
65
Suppose
is of type 1. Here all roots of K-1H are in
S
so, in view of Chapter II, where we have shown that when K and
H
are non-singular, type 1 is equivalent to type 2, it is only neces-
sary to consider the cases K-1H nilpotent and H-1K nilpotent.
or
Without loss of generality, assume S = Tz1 where z = 0
In view of Theorem II.3.4, it follows that showing
implies
S
matrix
B
Let
c
E
of type 1
S
of type 2 is equivalent to showing that, if a diagonal
with entries
type 2. Suppose
signature
00.
then
is a diagonal matrix with entries
such that
c*I
cI
cI
c* 4 *c . Let M = (
iB
is of
and of
±1
B = I ® -I.
then without loss of generality, write
0,
C
B
is of signature 0,
±1
)
then by
Lemma 5 of Chapter II
0
iM*BM = i
2
[c - (c *)
Thus
2.
iB
is of type 2.
2
III
0
q. e. d.
Geometrical Approach to 2 x 2 Usual Complex Case
In the
2x2
case, using geometric ideas from Ballantine's
work on Positive Definite Matrices {l}, we can characterize the prob-
lem geometrically.
Define
F(S)
r(s) = {x*sx Ix
is a line;
S
E
C2x1}. We say
is contradefinite iff r(s)
S
is bidefinite iff
is the whole
66
complex plane.
If
S
is non-singular; let sgn det S = e2iP
S
[a bdb
2
-
1 det s
If
such that
y>0
define
Lc
det S
-y
ac71 +
=
-
-
a-
21 det S 1
If
S
is non-singular,
S
is *-congruent to one of the follow-
ing forms:
0]
[i
S
SI is bidefinite;
If
S
=
1
eiP
0
-i
S2 = eiP
is contradefinite if y > I.
S2
is singular and non-zero,
S
is *-congruent to one of
the following forms
0]
[0 0]
S3 =
20
S4
00
where here
eiP
=
Proposition 111. 2. I.
S
S
is type 2,
is type 1,
trace S = sgn tr S
'trace S1
The following are equivalent:
67
(iii) if
either
is positive or
det S
if
is non-singular then
S
S
is bidefinite or contradefinite;
is singular, then either
S
has real determinant, and
S
is contradefinite or
S
trace S
is real.
In view of Proposition 111.1.4, it is enough to prove
Proof:
(ii) => (iii) and conversely. We consider the non-singular cases first.
(ii) => (iii) Suppose
such that
then
C,',(SC = S*,
is of type 1, then these exists
det(CC*)det S = det S*. Also
det S*=
det S
1
det(CC*)
det CC* = +1
Since
det CC* = (det C)(detC)* > 0,
det S >0;
det S = det S*. Suppose
Without loss of generality, assume
to
0
1
SZ =
(
Zy
1)
,
det S < 0, eiP = +i
If
S
S
=
det CC*
1
.
det CC* = 1. Thus
det S = ±1.
then e iP 1 det S I
j
e
iP
= 1.
If
S
,
is *-congruent
is of type 2, hence is of type 1. Now suppose
ip
Without loss of generality, assume e = i.
is *-congruent to
*-congruent to
C
det(CC*)det 5* = det S. Thus
thus
C*S*C = S
S
S
2
= i(
1
0
Zy
1
) ,
S
is of type 1
iS* = - (iS)*. Let
L = iS + (iS)* = iS - iS* =
-2y
-2
if
iS
is
68
implies
is of type 1
S
4-y2
4(1-y2) < 0
=
contradefinite. If
iff
1 < y2
iff
is *-congruent to
S
det L < 0;
which in turn implies
iS* - is = -(iS)* - iS = -L,
det L = 4 -
is *-congruent to
L
iff
y>1
and
is
S
then in all cases,
S1
S
is bidefinite.
(iii) => (ii)
e
i_/ det S
If
det S < 0,
then
C=
det S < 0,
(
01
0
1
and is real. If det S > 0,
0
,
k
)
i
0
,
S
0
is bidefinite, if det S > 0;
then
Thus
C*SC = 5*.
is *-congruent to
i(
is of type 2 hence, is of type 1. Suppose
det 5 > 0,
is *-congruent to
S
(
hence is of type 1. If det S < 0,
By Lemma 4 of [1],
(
provided
.
(Note:
y > I sin a I
y>1
=
sin
i6
101)
2N
i
0
0.
_i).
0.
0
)
=(
-1 0
Thus
.
)
0
1
is contradefinite. If
5
1
0
2-y
1
S
is *-congruent to
Thus
)
i
is of type 1. If
S
is *-congruent to
Choose
I. ) Thus
1.
=
is
5
Without loss of generality, let S = (
)
then
e
Without loss of generality, assume
eiP = ±i.
First suppose
*-congruent to
Let
det S
- ±1. Without loss of generality, assume
I det SI
eiP = +i.
Suppose
S
is of type 2,
i(
1
0
2y
1
0
(
i sin a
e
a = Tr/2.
S
is of type 2, hence is of type 1.
q. e. d.
Now suppose
S
is singular. If
which is contradefinite and real then
S
S
is *-congruent to
S3,
is of type 2, hence is of
type 1. Therefore we assume S = 54 Here
S
is *-congruent to
S
69
S*
iff there exists
e
= cc*
iff
CE
,i3
ci(ce = e
C such that
e13 = ±,./7"Tv
iff
E
S
is over
iff
q. e. d.
.
The above interpretation cannot be applied to higher dimensional
matrices. Consider
T1 = diag(i,
i)
0
j1
T3 =
is the y-axis. Thus
F(T1)
F(S) C r(T)
F(S)
10
T1
63-
is bidefinite. If S = 2y
is the complex plane. Thus
definite. Det T3 = 1. But
T., j = 1,2,3
3
K. =
J
2
(T.-T. )
J
J
are not of signature zero.
o
(
F(T2)
1
),
then
is contra-
are not of type 1, because
70
BIBLIOGRAPHY
Ballantine, C.S. Products of Positive Definite Matrices III,
Linear Algebra and Its Applications 3 (1970) 79-114.
Carlson, D.H. A Note on Matrices Over Extention Fields, Duke
Mathematics Journal, Vol. 33 pp. 503-505, 1966.
Gantmacher, F. R. The Theory of Matrices, Vol. II, Chelsea
Publishing Company, New York. 1964.
Greub, W.H. Linear Algebra, 3rd Edition. New York,
Spring-Verlag, 1967.
S. Haynsworth, E. V. and Ostrowski, A. M. On the Inertia of Some
Classes of Partitioned Matrices. Linear Algebra and Its
Application. Vol. I. pp. 299-316, 1968.
Jacobson, N. Lectures in Abstract Algebra, Vol. II, New York.
Van-Nostrand, 1953.
Malicev, A. I. Foundations of Linear Algebra, San Francisco,
W.H. Freeman, 1963.
Ng, D.N. An Effective Criterion for Congruence of Real Symmetric Pairs, PhD. Thesis, Oregon State University, 1973.
Smiley, M. F. Algebra of Matrices, Boston, Allyn and Bacon,
1965.
Taussky, 0. and Zassenhaus , H. On the Similarity Transformation Between a Matrix and Its Transpose. Pacific Journal of
Mathematics. Vol. 9, pp. 893-896, 1959.
Turnbull, H.W. On the Equivalence of Pencils of Hermitian
Forms. London Math Soc. (2) 1935. Vol. 39, pp. 232-48.
Turnbull, H. W. and Aiken, A.C. An Introduction to the Theory
of Canonical Matrices, 4th Edition, New York, Dover, 1961.
APPENDICES
71
APPENDIX I
Pictures of Matrices
_
jB
jB
B
B
0
T1 =
jB
jB
_
4
B
B
.1=1.
n1
blocks
B
B
jB
jB
T=
oo
B
B
jB
jB
_
n1
blocks
72
TR
B
0
o
D
B
0
'FIB
:0
D
I
0
=
B
0
1
p.B
0
1
D
1.1.B
0
1
I
.
I
a
0
--I
0
- p.*D 1
i
1
1
0
-ii*Di
n1
Let
p. = (3+j ,
11* =
p.*=
blocks
0
-1.1.*D
I
0
- 1.1.*D
73
APPENDIX II
Miscellaneous Results
Here we observe that any non-singular
8,
has determinant over
-1
2x2
type 1 matrix
We also make observations on fields with
.
as a norm.
Proposition IV. 1.1.
Lemma 8. If
of type 1, then
and some
5
S
non-singular matrix of
is a 2 x 2
is over
det S
type 1, then
If
is a
S
non-singular matrix and
2x2
S
S*
-s+as
s
[
is *-congruent to
]
for some
is
S
SE
aE
It has been proved earlier that
Proof of Lemma 8.
is *-congruent to A* and A-1. Therefore the char-1,
acteristic polynomial of A is x2 - ax + 1, i.e., A = [01 a J
A = S-15*
-
Suppose
*
s
12
-s11+as12
S22
-s21+as22
-1S* = SA =
*
*
s12, s22 = s 12,
*
s21 = -s11 + as12. Let s = s11
Therefore
If
T
T*
s
11
11
s11 - s22 and
s*
s
We have S = (_s*.as
t a
so
=
is similar to
s
.
A = (01
1
a
),
i. e.,
1
74
if there exists a non-singular
then
S = D*TD
such that D1 (T1 T*)D = A,
D
has the canonical form we have just shown. Thus it
is enough to consider
A
being a companion matrix.
Proof of Proposition IV. 1. 1. From Lemma 8, when
2x2
non-singular type
8*
=
SO
a
- s ,;(-1-as
).
matrix
-1
2
(det C)(det C*)(det S) = det S0,
Therefore
is *-congruent to
S
2
det S0 = s + (s*) + ass*
non-singular matrix, such that
C*SC =
is
S
S0'
If
C
is a
then
-
det S = (det So)[(det C)(det C*)]1
so
det S E
Proposition IV. 1.2.
1
H=
Suppose
(5+5*), K = 231(S-S*).
S
If
is an n x n matrix over
is a norm, and one of the
-1
following holds:
H-1K
is nilpotent,
(s) K-1H
is nilpotent,
(1)
then
S
is of type 1.
Proof: From Theorem 11.3.4, it is enough to show every
is *-congruent to its negative. Since
diagonal matrix over
is a norm,
over
e
,
-1 = act*,
for some
(aI)*B(aI) = -B.
Proposition IV. 1. 3.
aE
;
if
B
is any matrix
q. e. d.
Here we assume
E
and
to be
-1
75
finite fields, and S, H, and K are as defined in Proposition
IV. 1.2. If the following holds:
are not nilpotent.
H-1K
and
K-1H
is similar to
are in é3.,
then
K-1H
and all eigenvalues of K-1H
-K-1H
is of type 2, hence is of type 1.
S
In view of Theorems II. 3. 2, II. 3. 3, II. 3. 4, II. 3. 5, and
Proof:
is
11.3.7, it is enough to show that every diagonal matrix over
*-congruent to any diagonal matrix of the same order over
Let
diag(E , E2, .
B
. .
is a norm. Let
finite fields, every element in
that
a.a. = E.
1
1
1
Let
-1 -1
C = diag(al , az ,
Thus every diagonal matrix over
matrix.
and
Since
Ei E
, Ek),
E
-1
, ak
)
are
such
a. E
then C*BC = Ik.
is *-congruent to the identity
q. e. d.
Proposition IV. 1.4. Here es.
S,
,
H and K are as
defined in Proposition IV. 1.3. Suppose one of the following holds
H-1K
is nilpotent,
K-1H
is nilpotent.
In view of Theorems II. 3.3 and 11.3.6, we may, without loss of gen-
erality, assume S = Tz , z = 00, 0.
(Refer to pictures in Appendix
I.) Then the following holds:
(i) if
z = 0, (recall
diagonal matrix over
)
S=
then
0
= (En+jFn)
B,
where
B
is a
76
when
n
is odd,
when
n
is even,
(ii) if
z = 00
oo
n
is even,
when
n
is odd,
(b)
is even;
where
is
B
then
),
when
(i)
is of type 2 iff dim B
5
(recall S = Tn = (Fn+jEn) 0 B,
a diagonal matrix over
Proof:
is of type 2,
S
is of type 2
S
S
is of type 2 iff dim B
is even.
(a) Refer to Theorem II. 3. 4.
In view of Theorem 11.3.4, it is enough to show that if
is a diagonal matrix over
is even. Suppose
dim B
e
,
then
jB
is of type 2 iff dim B
is even. In the proof of Proposition
IV. 1.3, we have shown that every diagonal matrix over
is
*-congruent to any other diagonal matrix over Ex of the same order
without loss of generality, assume
and rank. If dim B=2m, m > 1,
B = (1,-1,1, -1,...,1,-1); let
then
C*BC = diag(Do, Do,
C = diag(C0),
, Do)
where
(c*) 2-c
0
DO =
Thus
jC*BC = C*jBC
c2-(c*) 2
is over
versely, suppose dim B
E
.
0
Thus
is odd and jB
jB
is of type 2.
Con-
is of type 2. I.e. , there
77
exists
over
non-singular matrix
a
8,
such that
C
C4<jBC = D, and D is
Thus
.
(det C*) det(jB)(det C) = jn(det B)(det C)*(det C)
= det W
L
which is over
But
.
n
.
is odd,
we have a contradiction; so jB
SO
n
J
Corollary 1.
If
E
.
Thus
is not of type Z.
q. e. d.
The proof is similar to that of (i).
(ii)
is not over
S, H, K
are finite fields,
and
are as Proposition IV. 1. 4, and if one of the following holds:
over
H-1K
is nilpotent
K-1H
is similar to -K-1H and has all its eigenvalues
then
,
Proof:
is of type 1.
S
Refer to Propositions IV. 1. 2 and IV. 1.4.
Corollary 2.
If
2 x 2 matrix over
the pencil
XH + I.LK
is similar to
-K-1H,
Z( and
such that
4"
are finite fields,
if1
H=
(s+s*), K =
is
S
1
(S=S*),
is non-singular. K is non-singular,
then
S
K-1H
is of type 2.
Proof: Since the characteristic polynomial of K-1H would be
of the form x2 - c,
where
cE
.
Since
is finite,
.4
is
78
the only quadratic field over
e
,
thus it contains all the roots of
all second degree polynomials. So we can apply Propositions IV. 1. 3
and IV. 1.4.