IJEDR1802035
International Journal of Engineering Development and Research (
www.ijedr.org
)
192
Quaternion Quasi-Normal Matrices And Their
Eigenvalues
1J. Rajeswari , 2Dr. K. Gunasekaran
1Ph.D Research Scholar, 2HOD and Assistant Professor,
1,2Ramanujan Research Centre, PG and Research Department of Mathematics,
Government Arts College(Autonomous), Kumbakonam. Tamil Nadu, India.
_____________________________________________________________________________________________________
Abstract - For square quaternion matrices
A
andB
of the same size, commutativity-like relations such asAB
=
BA
,
,
CT
AB
=
BA
AB
=
BA
T,
AB
=
B A
T,
etc., often cause a special structure ofA
to be reflected in some special structure forB
. We study eigenvalue pairing theorems forB
whenA
is quaternion quasi normal (QQN), a class of quaternion matrices that is a natural generalization of the real normal and complex normal matrices. A new canonical form for QQN matrices is an important tool for our development.AMS Classification: 15A99, 15A04, 15A15, 15A116, 15A48
Key words - Quaternion matrices, Spectrum, Canonical form, Direct sum, Skew- Hermitian, Orthonormal, Jordan block.
__________________________________________________________________________________________________
1. Introduction
Any eigenvalues (quaternion eigenvalues) of a real square matrix
A
come in conjugate pairs, and corresponding eigenvectors can be chosen in conjugate pairs (Ax
=
x
if and only ifC C C
Ax
=
x
); real eigenvectors ofA
can be associated with its real eigenvalues. IfA
is diagonalizable, it can therefore be diagonalized in a special way:A
=
SAS
−1,
C
A
=
L
L
R
is diagonal, the diagonal entries ofL
(if any) are in the open upper half partition of four dimension structure, the diagonal entries ofR
(if any) are real,S
=
[
Y
Y
CZ
]
is non-singular,Y
has the same number of columns asL
andZ
is real.If
A
is quaternion normal, it can be quaternion unitary diagonalized in the same way:A UAU
=
CT,
CA
=
L
L
R
is diagonal, the diagonal entries ofL
(if any) are in the open upper half partition of four dimension structure, the diagonal entries ofR
(if any) are real,U
=
[
Y
Y
CZ
]
is unitary,Y
has the same number of columns asL
andZ
is real. This canonical form is different from, but equivalent to, the classical real normal form [5, Theorem 2.5.8] and suggests a wide class of generalizations that play a role in the study of eigenvalue pairing theorems that motivated our investigations. We use standard terminology and notation, as in [5,6]. We let be the set ofm n
matrices with entries in F=¡
or
and write( )
n n nM
M
. The set of eigenvalues (spectrum) ofA
M
n( )
is denoted by
( )
A
.Two characterizing properties of a quaternion normal matrix
A
play an essential role in our discussion: (a)A
can be quaternion unitarily diagonalized and (b) a nonzero vectorx
is a right
−
eigenvector ofA
(Ax
=
x
for some scalar
) if and only if it is a left eigenvector, necessarily with the same eigenvalue (x
CTA
=
x
CT). Eigenvectors of a normal matrix associated with distinct eigenvalues are necessarily orthogonal. IfA
is quaternion normal, then the quaternion orthogonal complement of the span of any collection of eigenvectors is an invariant subspace ofA
.2. Quaternion Quasi unitary matrices Definition 2.1
A matrix
U
M
n( )
is said to be r-quaternion quasi unitary (r-QQU) ifU
is quaternion unitary,[
C]
U
=
Y
Y
Z
,Y
M
n r( )
, andZ
M
n n −2r( )
¡
Whenr
=
0
, thenU
=
Z
is real orthogonal; when2
r
=
n
thenU
=
Y
Y
C
. When the value of the parameterr
is not relevant, we say thatU
is QQU. For a givenY
M
n r( )
IJEDR1802035
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193
If
Y
Y
C
has quaternion orthonomal columns andn
2
r
, then there is always a quaternionX
M
n n −2r( )
such that[
Y
Y
CX
]
(and hence also[
Y
CY
X
C]
) is quaternion unitary. However, any suchX
has an important property: the column spaces ofX
andX
Care the identical namely, the quaternion orthogonal complement of the column space of
Y
Y
C
. We say that a subspace spanned by the columns ofX
M
n m( )
is self-conjugate if it is the same as the column space ofX
C.Lemma 2.2
Let 𝑋̂
M
n m( )
have rankm
1
and suppose that the column space of 𝑋̂ is self-conjugate. Then there is a real( )
n mZ
M
with quaternion orthonormal columns and the same column space as 𝑋̂. In particular, ifn
m
2
r
0
,( )
n rY
M
, 𝑋̂
M
n m −2r( )
, 𝑍̃
M
n n m −( )
¡
, and
Y
Y
C 𝑋̂𝑍̃]
M
n( )
is quaternion unitary, then there exists aZ
M
n m −2r( )
¡
such that [𝑌 𝑌𝐶 𝑍 𝑍 ̃ ] is quaternion unitary.Proof:
Since 𝑋̂ has full rank, there is a matrix
X
M
n m( )
with quaternion orthonormal columns with the same column space as that of 𝑋̂. Since the column space of 𝑋̂, and hence ofX
, is self-conjugate, there exists a nonsingularW
M
r( )
such that
X
C=
XW
. Then,I
=
(
X
C CT)
X
C=
W
CTX
CTXW
=
W W
CT , soW
is quaternion unitary. Moreover,C C C
X
=
X W
=
XWW
, soX I WW
(
−
C)
=
0
. SinceX
has full column rank, we must haveWW
C=
I
, that is,W
is quaternion unitary and coninvolutory and hence it is also symmetric [6, Section 6.4].Let
p t
( )
be a polynomial such thatV
p W
(
)
is a square root ofW
. ThenV
is quaternion unitary and symmetric, and hence it is also coninvolutory. Moreover,X
C=
XV
2soX V
C −1=
X V
C C=
XV
Z
is real. Since it is obtained fromX
by a right quaternion unitary transformation,Z
has quaternion orthonormal columns and the same column space asX
.Note 2.3
If the assumption that 𝑋̂ has full rank is omitted in Lemma 2.2, one may still show that its column space has a real quaternion orthonormal basis [6, Theorem 6.4.24]. The following three assertions are easily verified.
Proposition 2.4
Let
U V
,
M
n( )
be r-QQU matrices and letQ
M
n( )
be realquaternion orthogonal. Thena.
0
20
rT CT C
n r r
I
U U
U U
I
I
−
=
=
is quaternion unitary, symmetric, and coninvolutory,b.
UV
CT is realquaternion orthogonal, and c.QU
is r-QQU.Proof
b. Suppose
U
=
Y
1Y
1CZ
1
andV
=
Y
2Y
2CZ
2
withY
1,Y
2
M
n r( )
. Then CTUV
is a product of quaternion unitary matrices and hence is quaternion unitary. However,UV
CT =YY
1 2CT+
Y Y
1C 2T+
Z Z
1 2T =1 2 1 2
2 Re(
Y Y
CT)
+
Z Z
T is real, so it is real quaternion orthogonal.3. Quaternion Quasi-Normal matrices Definition 3.1
A matrix
A
M
n( )
is said to be quaternion quasi-normal (QQN) if (i)A
is quaternion normal. (ii)x
C is an eigenvector ofA
wheneverx
is, and (iii) the nullspace ofA
is self-conjugate, that isAx
=
0
if and only ifAx
C=
0
.Every realquaternion normal matrix is QQN, but so are several other familiar symmetry classes of quaternion normal matrices. If
A
is QQN and Q is real quaternion orthogonal, it follows immediately from the definition thatA
CandQAQ
T are both QQN [1]. The basic structure of the eigenspaces of a QQN matrix is described in the following lemma, which leads directly to a pleasant canonical form.Lemma 3.2
Suppose
is a nonzero eigenvalue of a QQN matrixA
M
n( )
, and let the columns ofY
be an quaternion orthonormal basis of the
-eigenspace ofA
, so thatAY
=
Y
. Then there is a nonzero scalar
such thatIJEDR1802035
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)
194
C C
AY
=
Y
; if
, thenAY
C=
Y
C, the columns ofY
C are an quaternion orthonormal basis for the
C-eigenspace ofA
C and
Y
Y
C
has quaternion orthonormal columns.Proof
Let
x
be a unit
-eigenvector ofA
, so there is some scalar
such thatAx
C=
x
Cthat is,A x
C=
Cx
. Since the conjugate ofx
C is not in the nullspace ofA
, it follows that
0
. We claim that the
-eigenspace ofA
and the
C -eigenspace ofA
C have the same dimension.Since
A
is QQN if and only ifA
C is QQN [10], for purposes of obtaining a contradiction it suffices to suppose that the
-eigenspace ofA
has dimension greater than that of the
C-eigenspace ofA
C. Supposeu
is a unit vector in the
-eigenspace ofA
that is quaternion orthogonal tox
, soAx
=
x
,
Au
=
u
,
Ax
C=
x
C,
and there is some scalarv
such thatAu
C=
vu
C. Butx u
+
0
andA x u
(
+
)
=
(
x u
+
)
, soAx
C+
Au
C =A x u
(
+
)
C =
(
x u
+
)
C =C C
x
u
+
for some scalar
. It follows that
= =
v
. Thus, if the columns ofY
are a quaternion orthonormal basis of the
-eigenspace ofA
(soAY
=
Y
), thenAY
C=
Y
Cand the column space ofY
C is contained in the
C-eigenspace ofA
C. This shows that the dimension of the
C-eigenspace ofA
Ccannot be less than that of the
-eigenspace ofA
, so these two eigenspaces must have the same dimension. Moreover, this argument shows that each eigenspace is the conjugate of the other. If
=
, the eigenspace is self-conjugate; if
normality ofA
ensures that the two eigenspaces are quaternion orthogonal.Theorem 3.3
A matrix
A
M
n( )
is QQN if and only if there is a nonnegative integerr
, ar
-quaternion quasi-unitary matrix CU
=
Y
Y
Z
, and a diagonal matrixA
=L
1
L
2L
3such thatA
=UAU
CT,L L
1,
2
M
r( )
are non-singular, and there are nonnegative integersf
andg
, positive integersn
1...
n
f ,m
1...
m
g , and2
f
+
g
distinct scalars
1...
f ,
1...
f,v
1...
v
g such thatn
1+
...
+
n
f =r
,m
1+
...
m
g =n
−
2
r
,L
1 =1
1
I
n...
fI
nf
,L
2 =1
1
I
n...
fI
nf
andL
3 =1
1 m
...
g mgv I
v I
.Proof
Suppose
A
is QQN. Since the nullspace of a QQN matrix is self-conjugate, Lemma 2.2 ensures that ifA
is singular then there is a real matrixZ
with quaternion orthonormal columns that span the nullspace ofA
. If the column space ofZ
is all of
n thenA
=Z Z
0
T and we are done. If not, let
be any eigenvalue ofA
acting on the quaternion orthogonal complement of the column space ofZ
and let the columns ofY
be a quaternion orthonormal basis for the
–eigenspace ofA
. Lemma 3.2 ensures that either the column space ofY
is self-conjugate or there is a nonzero scalar
such that the column space ofY
C is the
C–eigenspace ofA
C. In the first case, replaceY
with a real matrix with quaternion orthonormal columns and the same column space and append it toZ
, which then is a real matrix with quaternion orthonormal columns; in the second case, the matrix
Y
Y
CZ
has quaternion orthonormal columns.If the column space of
Y
Y
CZ
is all of
n, we are done. If not, proceed in the same way to consider any eigenvalue ofA
acting on the quaternion orthogonal complement of the column space of
Y
Y
CZ
. Augment eitherZ
or
Y
andY
C as before and continue until this process exhausts the finitely many distinct eigenvalues ofA
. At each stage, the construction ensures that any new eigenvalue considered is distinct from any eigenvalue ofA
previously encountered, so we obtain a QQN matrix that diagonalizesA
and gives a representation of the asserted form.Conversely, suppose that
A
has a representation of the asserted form. Any eigenvectorx
ofA
is in one and only one eigenspace ofA
, which is spanned by a set of contiguous columns ofU
corresponding to a unique diagonal block inA
. But the span of each such set of contiguous columns is either self-conjugate (the nullspace ofA
is of this type), or is the conjugate of an eigenspace ofA
corresponding to a different eigenvalue. In either event, the conjugate ofx
is an eigenvector ofA
.Note 3.4
QQN matrices have polar-type decompositions of all three classical types in which the factors commute.
Theorem 3.5
Let
A
M
n( )
be QQN. ThenIJEDR1802035
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195
b. A commutes with
A
TandA
=
QS
=
SQ
withQ
quaternion orthogonal andS
symmetric.c. A commutes with
A
C (that is,AA
C is real ) andA
=
RE
=
ER
withR
real andE
coninvolutory.Proof
Let
A UAU
=
CT be QQN, withA
=
L
1
L
2L
3 and a conformal QQU matrixU
. For any given nonzero complex numberz
, we writez
=
re
i for a uniquer
0
and a unique
[0, 2
𝜋)
; we representz
=
0
withr
=
0
and
=
0
and write0
=
0
e
i0. For any given diagonal matrixD
=diag d
( ...
1d
p)
= 11
(
i...
i p)
p
diag r e
r e
we defineD
1/ 2
/2 1/2
1
(
i...
i p)
p
diag
+
r e
+
r e
,D
diag r
( ... )
1r
p , and
( )
D
1
(
i...
i p)
diag e
e
. The following factors give the asserted decompositions ofA
:(a)
P
=(
1 2 3)
CTU L
L
L U
andV
=U
( ( )
L
1
(
L
2)
(
L U
3))
CT(b)
Q
=U L
(
2−1/2 1/2L
1
L
−11/2 1/2L
2
I U
)
CT andS
=U L L
(
1/2 1/22 1
L L
1/2 1/22 1
L U
3)
CT, and(c)
R
=U L
((
C2)
1/2L
1/21
L
1/22(
L
1C)
1/2
L U
3)
CT andE
=U L
((
C2)
−1/2L
1/21
L
1/22(
L
1C)
−1/2
(
L U
3))
CTNote 3.6
Finally, we observe that quaternion normal matrices in all of the familiar symmetry classes are QQN.
Theorem 3.7
Let
A
M
n( )
be quaternion normal. In each of the following cases,A
is QQN,U
is r-QQU,A UAU
=
CT,1 2 3
A
=
L
L
L
, and the direct summandsL
s can be chosen to have the indicated pattern of eigenvalues:i)
A
is real(
A
C=
A
)
: the diagonal entries ofL
1lie in the open upper half plane,L
2=
L
1C and the diagonal entries of3
L
are real.ii)
A
is skew-symmetric(
A
T= −
A
)
: the diagonal entries ofL
1 are either positive or lie in the open upper half plane,L
2= −
L
1 andL
3=
0
.iii)
A
is coninvolutory(
A
C=
A
−1)
: the diagonal entries ofL
1 lie in the open exterior of the unit disc,L
2=
(
L
1C)
−1 and thediagonal entries of
L
3 have modulus one.iv)
A
is quaternion orthogonal(
A
T=
A
−1)
: the diagonal entries ofL
1 lie in the open exterior of the unit disc together withthe open circular arc
e
i: 0
,L
2=
L
1−1 and1 2
3 m m
L
=
I
−
I
.v)
A
is skew- quaternion orthogonal(
A
T= −
A
−1)
: the diagonal entries ofL
1lie in the open exterior of the unit disc togetherwith the open circular arc
e
i:
/ 2
3 / 2
,L
2= −
L
1−1 and1 2
3 m m
L
=
iI
−
iI
.vi)
A
is pure imaginary(
A
C= −
A
)
: the diagonal entries ofL
1 lie in the open left half plane,L
2= −
L
1C and the diagonalIJEDR1802035
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vii)
A
is skew-coninvolutory(
A
C= −
A
−1)
: the diagonal entries ofL
1 lie in the open exterior of the unit disc,L
2= −
(
L
1C)
−1 and the diagonal entries ofL
3 have modulus one.viii)
A
is symmetric(
A
T=
A
)
:r
=
0
,L
3 is a diagonal matrix with no restrictions on its entries, andU
=
Z
is real quaternion orthogonal.Proof
In each of the following cases, let
x
be a unit vector such thatAx
=
x
. In order to show thatA
is QQN, it suffices to show thatx
C is an eigenvector ofA
.i)
Ax
C =(
Ax
)
C =(
x
)
C =
Cx
C, sox
C is an eigenvector corresponding to the eigenvalue
C. ii)x A
T =(
−
Ax
)
T =(
−
x
)
T =−
x
T, sox
C is an eigenvector corresponding to the eigenvalue−
.iii)
x
=A Ax
C =
A x
C , soAx
C =(
C)
−1x
C;x
C is an eigenvector corresponding to the eigenvalue(
C)
−1.iv)
x
=A Ax
T =
A x
T , sox A
T =
−1x
T and henceAx
C =
−1x
C ;x
C is an eigenvector corresponding to the eigenvalue
−1.v)
−
x
=A Ax
T =
A x
T , sox A
T =−
−1x
T and henceAx
C =−
−1x
C;x
C is an eigenvector corresponding to the eigenvalue−
−1.vi)
iA
is real.vii)
iA
is coninvolutory.viii)
x A
T =(
Ax
)
T =(
x
)
T=
x
T, sox
C is an eigenvector corresponding to the eigenvalue
.Note 3.8
In the canonical form described in Theorem 3.3, the nonnegative integer
2
r
is the sum of the dimensions of the isotropic eigenspaces ofA
, or, equivalently,n
−
2
r
is the sum of the dimensions of the self-conjugate eigenspaces ofA
. Thus,r
is uniquely determined byA
. We say that a QQN matrixA
is degenerate ifr
=
0
, which Theorem 3.7 (viii) ensures is the case if and only ifA
is symmetric; otherwise, we say thatA
is nondegenerate.4. Eigenvalue pairing theorems
Throughout this section,
A UAU
=
CT
M
n( )
is a nondegenerate QQN, factored as in Theorem 3.3 withA
=1 2 3
L
L
L
andU
=
Y
Y
CZ
; letB
M
n( )
be given. IfA
is in one of the seven nondegenerate classes enumerated in Theorem 3.7 (i)-(viii), we assume without loss of generality that its eigenvalues have been ordered to achieve the locations stated there for the diagonal entries ofL L
1,
2 andL
3.To illustrate the realm of results we wish to study, consider the prototype case of ordinary commutativity:
AB
=
BA
. ThenUAU
CTB
=
BUAU
CT, so
A U
(
CTBU
)
=(
U
CTBU A
)
...(1)Let
CT
U
BU
=CT CT C CT
T T C T
T T C T
Y
BY
Y
BY
Y
BZ
Y BY
Y BY
Y BZ
Z BY
Z BY
Z BZ
11 12 13 21 22 23 31 32 33
B
B
B
B
B
B
B
B
B
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V
C 2( )
n r
Y
Y
M
,C
V
CTBV
=CT CT C
T T C
Y
BY
Y
BY
Y BY
Y BY
= 11 12 2 21 22( )
rB
B
M
B
B
...(3)and
L
1
L
2M
2r( )
. Writing out Eq.1 in block form gives the identity3 3 33
(
)
(
)
CT
T
C
V
BZ
L Z BV
L B
= 3 33 3(
)
(
)
CT TC
V
BZ L
Z BV
B L
...(4)Because
L L
1,
2 andL
3 (and hence also
andL
3) have pairwise disjoint spectra, Sylvester’s Theorem[6, Theorem 4.4.6] and equality of the (1,2) blocks in Eq.4, as well as the (2,1) blocks, implies thatV
CTBZ
=
0
andZ BV
T=
0
. Thus,
U
CTBU
=CT
T T
C
V
BZ
Z BV
Z BZ
= 330
0
C
B
,is block diagonal and unitarily similar to
B
; the column spaces ofV
=
Y
Y
C
andZ
are each invariant underB
. We are interested in the eigenstructure ofC
, which is the restriction ofB
to the column space of
Y
Y
C
.Writing out the equation
=
C
C
from the (1,1) blocks of Eq.4 gives the identity1 11 1 12
2 21 2 22
L B
L B
L B
L B
=11 1 12 2 21 1 22 2
B L
B L
B L
B L
...(5)which tells us that
B
12=
B
21=
0
since
( )
L
1I
(
L
2)
=
. Thus, the column spaces ofY
andY
Care each invariant under,
B C
=Y
CTBY
Y BY
T C, andU
CTBU
=Y
CTBY
Y BY
T C
Z BZ
T . Although there is nothing special about the eigenstructure of a quaternion symmetric matrix, in this case we get something interesting if we assume thatB
is symmetric:CT
Y
BY
=(
Y BY
T C T)
is similar toY BY
T C, so every block in the Jordan canonical form ofC
appears an even number of times. In particular, every eigenvalue ofC
has even multiplicity.Similar calculations, some with the help of Proposition 2.4, show that other commutativity-related assumptions about
AB
have useful consequences for
C
: IfAB
=
BA
, then =
C
C
; ifAB
=
BA
CT, then =
C
C
C; and if TAB
=
BA
, then =
C
C L
(
2
L
1)
. Under natural conditions on the spectra ofL
1 andL
2, these relations imply thatC
is block diagonal. Moreover, certain conditions onB
ensure various pairings of the eigenvalues ofC
.Other authors have considered implications of the condition
AB
=
BA
T for various classes of matrices (see [2,4,8,9]). The following results have their origin in a study of the spectral properties of quaternion unitary operators induced by ergodic measure preserving transformations[1,3].Theorem 4.1
Let
A UAU
=
CT
M
n( )
be a nondegenerate QQN, factored as in Theorem 3.3 withA
=
L
1L
2L
3 and CU
=
Y
Y
Z
. LetB
M
n( )
be given, and setC
=
V
CTBV
withV
=
Y
Y
C
.Suppose that any one of the following conditions is satisfied: i)
AB
=BA
,IJEDR1802035
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iii)
AB
=BA
CTand
( )
L
1I
(
L
C2)
=
iv)
AB
=−
BA
CTand
( )
L
1I
(
−
L
C2)
=
, orv)
AB
=−
BA
T,
( )
L
1I
(
−
L
1)
=
, and
(
L
2)
I
(
−
L
2)
=
.Then
C
=Y
CTBY
Y BY
T C is block diagonal. Moreover,a) If
B
T=
B
, then each block in the Jordan canonical form ofC
occurs an even number of times, so every eigenvalue ofC
has even multiplicity.b) If
B
is real andJ
m( )
is a Jordan block ofC
, then so isJ
m(
C)
. Each Jordan block ofC
corresponding to a real eigenvalue occurs an even number of times, so each real eigenvalue ofC
(if any) has even multiplicity.c) If
B
T= −
B
andJ
m( )
is a Jordan block ofC
, then so isJ
m(
−
)
. Any nilpotent Jordan block ofC
occurs an even number of times, so ifC
is singular, zero is an eigenvalue with even multiplicity.d) If
B
C= −
B
and ifJ
m( )
is a Jordan block ofC
, then so is(
)
C mJ
−
. Each Jordan block ofC
corresponding to a pure imaginary eigenvalue occurs an even number of times, so each pure imaginary eigenvalue ofC
(if any) has even multiplicity.Proof
Under each of the assumptions (i)-(v), inspection of the analog of Eq.5 in each case shows that the off-diagonal blocks
12
B
andB
21 are zero.a) If
B
is symmetric, thenB
22=
B
11T, soC
=B
11
B
11T . The Jordan canonical form of a square quaternion matrix and its transpose are the same, so every block in the Jordan canonical form ofC
occurs an even number of times.b) If
B
is real, thenB
22=
(
Y
CTBY
)
C=
B
11C, so the Jordan blocks ofC
occur in conjugate pairs of the form( )
(
C)
m m
J
J
. If
is real, then its Jordan blocks occur in pairs of the formJ
m( )
J
m( )
, so each real eigenvalue has even multiplicity.c) If
B
is skew-symmetric, thenB
22= −
B
11T, so the Jordan blocks ofC
occur in pairs of the form( )
(
)
m m
J
J
−
. IfC
is singular, then its nilpotent Jordan blocks occur in pairs of the formJ
m(0)
J
m(0)
, so zero is an eigenvalue with even multiplicity.d) If
B
is pure imaginary, thenB iD
=
for some realD
M
n( )
and the assertions follow from (b).Note 4.2
Of course, the eigenvalues of
C
need not be eigenvalues ofB
. However, certain additional conditions ensure that CTV
BZ
andZ BV
T are both zero.B
13,B
23,B
31 andB
32 are all zero,U
CTBU
is block diagonal, and the eigenvalues ofC
are also eigenvalues ofB
.Theorem 4.3
Let
A UAU
=
CT
M
n( )
be a nondegenerate QQN, factored as in Theorem 3.3, withA
=
L
1L
2L
3 and CU
=
Y
Y
Z
. LetB
M
n( )
be given, and setC
=
V
CTBV
withV
=
Y
Y
C
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ii)
AB
=−
BA
,
( )
L
1I
(
−
L
2)
=
,
( )
L
1I
(
−
L
3)
=
, and
(
L
2)
I
(
−
L
3)
=
(these conditions are satisfied ifA
is coninvolutory or skew-coninvolutory),iii)
AB
=BA
CT,
( )
L
1I
(
L
C2)
=
,
( )
L
1I
(
L
C3)
=
, and
(
L
2)
I
(
L
C3)
=
(these conditions are satisfied ifA
is skew-quaternion orthogonal, coninvolutory, or skew-coninvolutory).iv)
AB
=−
BA
CT,
( )
L
1I
(
−
L
C2)
=
,
( )
L
1I
(
−
L
C3)
=
, and
(
L
2)
I
(
−
L
C3)
=
(these conditions are satisfied ifA
is quaternion orthogonal, coninvolutory, or skew-coninvolutory), orv)
AB
=−
BA
T,
( )
L
1I
(
−
L
1)
=
,
( )
L
1I
(
−
L
3)
=
,
(
L
2)
I
(
−
L
2)
=
, and
(
L
2)
I
(
−
L
3)
=
(these conditions are satisfied if
A
is real, pure imaginary, or skew-symmetric).Then
U
CTBU
=Y
CTBY
Y BY
T C
Z BZ
T =C
Z BZ
T , every eigenvalue ofC
is an eigenvalue ofB
, andC
satisfies each of the conclusions (a)-(d) of Theorem 4.1.Certain conditions force the diagonal blocks
B
11 andB
22 to be zero, and certain conditions onB
ensure that the eigenvalues ofC
are paired.Theorem 4.4
Let
A UAU
=
CT
M
n( )
be a nondegenerate QQN, factored as in Theorem 3.3, withA
=
L
1L
2L
3 and CU
=
Y
Y
Z
. LetB
M
n( )
be given, and setC
=
V
CTBV
withV
=
Y
Y
C
.Suppose that any one of the following conditions is satisfied:
i)
AB
=−
BA
,
( )
L
1I
(
−
L
1)
=
, and
(
L
2)
I
(
−
L
2)
=
, ii)AB
=BA
CT,
( )
L
1I
(
L
1C)
=
, and
(
L
2)
I
(
L
C2)
=
, iii)AB
=−
BA
CT,
( )
L
1I
(
−
L
1C)
=
, and
(
L
2)
I
(
−
L
C2)
=
, iv)AB
=BA
T, orv)
AB
=−
BA
T, and
( )
L
1I
(
−
L
2)
=
.Then
C
=0
0
CT CT
Y
BY
Y BY
Moreover,
a) Every non-singular Jordan block of
C
2occurs an even number of times, and every eigenvalue ofC
2 has even multiplicity.b) If
B
is either real or pure imaginary, and ifJ
m( )
is a Jordan block ofC
, then so areJ
m(
−
)
andJ
m(
C)
. Thus, the eigenvalues ofC
occur in
conjugate quadruplets with the same multiplicities.c) If
B
CT =B
, then the eigenvalues ofC
are real and occur in
pairs with the same multiplicities. In fact they are singular values ofY
CTBY
Ctogether with their negatives.Proof
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a) We have
C
2 =B B
12 21
B B
21 12, and the non-singular Jordan blocks ofB B
12 21andB B
21 12are always the same; their nilpotent Jordan structures can be different, but zero is an eigenvalue of the same multiplicity for both. Hence,C
2 has an even number of zero eigenvalues.b) If
B
is real, thenC
= 1212
0
0
CB
B
. Since every square quaternion matrix is consimilar to a real matrix, there isa real
R
and a non-singularS
such thatB
12 =SR S
(
C)
−1. LetX
S
S
C and letK
1
2
I
I
I
I
−
, so1 T
K
−=
K
. A calculation reveals that(
XK
−1)
−1C XK
(
−1)
=−
R
R
. Thus, ifJ
m( )
is a Jordan block ofC
, then so areJ
m(
−
)
,J
m(
C)
, and(
)
C m
J
−
.c) If
B
=
B
CT, thenC
= 1212
0
0
CTB
B
. LetB
12=
U V
be a singular value decomposition ofB
12, and set CTX
U
V
. Then(
XK
−1)
−1C XK
(
−1)
=−
[5, Theorem 7.3.7]Note 4.5
Certain conditions on
L
1,L
2 andL
3 ensure thatU
CTBU
=C
Z BZ
T , so the eigenvalues ofC
are also eigenvalues ofB
.Theorem 4.6
Let
A UAU
=
CT
M
n( )
be a nondegenerate QQN, factored as in Theorem 3.3, withA
=
L
1L
2L
3 and CU
=
Y
Y
Z
. LetB
M
n( )
be given, and setC
=
V
CTBV
withV
=
Y
Y
C
.Suppose that any one of the following conditions is satisfied:
i)
AB
=−
BA
,
( )
L
1I
(
−
L
1)
=
,
( )
L
1I
(
−
L
3)
=
, and
(
L
2)
I
(
−
L
2)
=
, and2 3
(
L
)
(
L
)
I
−
=
(these conditions are satisfied ifA
is real, pure imaginary, or skew-symmetric).ii)
AB
=BA
CT,
( )
L
1I
(
L
C1)
=
,
(
L
2)
I
(
L
C2)
=
,
( )
L
1I
(
L
C3)
=
, and
(
L
2)
I
(
L
3C)
=
(these conditions are satisfied if
A
is real),iii)
AB
=−
BA
CT,
( )
L
1I
(
−
L
1C)
=
, and
(
L
2)
I
(
−
L
C2)
=
,
( )
L
1I
(
−
L
C3)
=
, and2 3
(
L
)
(
L
C)
I
−
=
(these conditions are satisfied ifA
is pure imaginary),iv)
AB
=BA
T, orv)
AB
=−
BA
T,
( )
L
1I
(
−
L
2)
=
,
( )
L
1I
(
−
L
3)
=
, and
(
L
2)
I
(
−
L
3)
=
(these conditions are satisfied ifA
is coninvolutory or skew-coninvolutory),Then
U
CTBU
=0
0
CT CT T
Y
BY
Z BZ
Y BY
Every eigenvalue of
C
is an eigenvalue ofB
, andC
satisfies the conclusions (a)-(c) of Theorem 4.4.Proof
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Note 4.7The relations
AB
=
B A
T andAB
T
BA
also lead to eigenvalue pairings, but via a somewhat different path.Lemma 4.8
Let
T
M
2r( )
be given.a) If
T
is skew-quaternion Hamiltonian, then every Jordan block ofT
occurs an even number of times.b) Suppose
T
is quaternion Hamiltonian. ThenT
2is skew-quaternion Hamiltonian. IfJ
m( )
is a nonsingular Jordan block ofT
, then so isJ
m(
−
)
. Every odd-sized singular Jordan block ofT
occurs an even number of times, and zero is an eigenvalue ofT
with even multiplicity.Proof
a) The asserted pairing follows from the fact that a skew-quaternion Hamiltonian matrix is similar (via a symplectic similarity) to the direct sum of a matrix and its transpose [7, Theorem 6].
b) If
T
is quaternion Hamiltonian, block multiplication reveals thatT
2 is skew-quaternion Hamiltonian and that0
0
I
I
−
TE
F
G
E
−
0
0
I
I
−
=T
T
E
F
G
E
−
,so
−
T
is similar to the transpose ofT
, and hence toT
itself. This observation proves the asserted pairing of the nonsingular Jordan blocks ofT
andT
2. One checks that(
J
2m+1(0))
2is similar toJ
m(0)
J
m+1(0)
and that(
J
2m(0))
2 is similar to(0)
(0)
m m
J
J
. IfJ
2m+1(0)
is an odd-sized nilpotent Jordan block ofT
, the fact thatT
2 is Skew-Quaternion Hamiltonian means that its Jordan canonical form contains each of the blocksJ
m(0)
andJ
m+1(0)
an even number of times; their respective parities are unaffected by the presence or absence ofJ
2m(0)
in the Jordan form ofT
. Thus, the Jordan form ofT
must contain an even number of copies ofJ
2m+1(0)
.Theorem 4.9
Let
A
M
n( )
be quaternion normal and letB
M
n( )
be given.a) Suppose that
A
is not quaternion symmetric and thatA
andB
satisfy at least one of the four conditionsAB
=B A
T,
AB
=−
B A
T ,AB
T =BA
, orAB
T =−
BA
. Suppose either thatA
is coninvolutory and factored as in Theorem 3.7(iii), or thatA
is skew-quaternion coninvolutory and factored as in Theorem 3.7(vii). LetU
CTBU
=
B
st be defined as in eqn. (2). ThenU
CTBU
=B
11
B
22
B
33is block diagonal. IfAB
=B A
T orAB
T =BA
, then every Jordan block of11 22
B
B
occurs with even multiplicity. IfAB
=−
B A
T orAB
T =−
BA
, and ifJ
m( )
is a Jordan block ofB
11
B
22, then so isJ
m(
−
)
.b) Suppose that
A
M
n( )
is nonzero and skew-quaternion symmetric, factored as in Theorem 3.7(ii).b1) Suppose that
AB
=B A
T andBA
=AB
T. ThenU
CTBU
=C
B
33. Moreover,C
is Skew-quaternion Hamiltonian, so every block in the Jordan canonical form ofC
occurs an even number of times.IJEDR1802035
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202
C
occurs an even number of times, and zero is an eigenvalue ofC
with even multiplicity. Every Jordan block ofC
2 occurs an even number of times.Proof
a) The assumptions ensure that
A
is a nondegenerate QQN withL
2=
(
L
1C)
−1. Moreover,L
3is non-singular and the diagonal entries ofL
1lie outside the open unit disk. Using the notation and invoking its conclusions, it suffices to observe that1
s t s t
for alls t
,
=
1...
r
. Moreover,B
11is similar to
B
22, which ensures that the assertions about pairings of the Jordan blocks ofB
11
B
22 are correct.b1) Again,
A
is a nondegenerate QQN. The diagonal entries ofL
1 are either positive or in the open upper half plane, and2
L
=−
L
1. The key observation is that, under these conditions,L
1 is a polynomial inL
12, andL
12commutes withB
11, it follows thatL
1commutes withB
11. Thus,B
22T =( )
L
1 −1B L
11 1 =B
11. Also,B
12T =( )
L
1 −1B L
12 2 =−
( )
L
1 −1B L
12 1 =−
B
12 , soB
12is skew-quaternion symmetric. A similar computation shows thatB
21is also skew-quaternion symmetric, soC
is skew-Quaternion Hamiltonian.b2) One can argue as in (b1).
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