 Open Access
 Total Downloads : 156
 Authors : S. Krishnamoorthy, G. Manikandan
 Paper ID : IJERTV2IS120951
 Volume & Issue : Volume 02, Issue 12 (December 2013)
 Published (First Online): 25122013
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
On Products of Polynomial Conjugate EPr Matrices
On Products of Polynomial Conjugate EPr Matrices
S. Krishnamoorthy
Head of the Department of Mathematics, Ramanujam Research Centre, Department of Mathematics,
Government Arts College(Autonomous), Kumbakonam612 001, Tamil nadu(India).
G. Manikandan, Lecturer in Mathematics,
Ramanujam Research Centre, Department of Mathematics, Government Arts College(Autonomous),
Kumbakonam612 001, Tamil nadu(India).
Abstract
In this paper we disscuss the product of polynomial
inverse of A satisfying the following four equations:(1) AXA A, (2) XAX X ,
conjugate EPr (con EPr ) matrices is polynomial
(3) AX* AX , (4) XA* XA
[2].A* is the
con EPr .
Keywords: EP matrix, polynomial matrix,
conjugate transpose of A . In general product of two polynomial con EPr matrices need not be
i 0
Generalized inverse.
AMS classification:15A09, 15A15, 15A57.
polynomial con EPr . For instance,
0 0
0 0
and
0 0
0 i
are polynomial con EP1
matrices, but the

Introduction
Throughout this paper we deal with complex polynomial square matrices. An nn square matrix A() which is a polynomial in the
scalar variable from a field C represented by A() Amm Am1m1 ……. A1 A0 where the leading coefficient Am 0 , Ai, s are square
product is not polynomial con EP1 matrix. The
purpose of this paper is to answer the question of when the product of polynomial con EPr matrices is
polynomial con EPr , analogous to that of EPr matrices studied by [1]. We shall make use of the following results on range space, rank and generalized inverse of a matrix.
matrices in
Vnn
is defined a polynomial matrix.
(1) R(A) R(B) AA BB
Any matrix A is said to be polynomial con EPr if
(2)
R(A ) R(A*)
R(A) R(AT ) or equivalently N(A) N(AT ) or
(3)
(A) (A )
equivalently AA AA and is said to be
(AT ) (A)
polynomial con EPr if A is polynomial con EPr and
(4)
(A ) A .
(A) r , where R(A), N(A), A,AT and (A) denote
the range space, null space, conjugate, transpose and
rank of A respectively. A denotes the Moore Penrose

ON PRODUCTS OF POLYNOMIAL CONJUGATE EPr MATRICES
In this section, Explain the product of polynomial
conjugate EPr (con EPr ) matrices is polynomial con EPr
Theorem 2.1
A is polynomial con EPr . Hence the theorem.
Corollary 2.2
Let A and B be polynomial con EPr matrices.
Then AB is a polynomial con EPr matrices
(AB) r and R(A) R(B) .
Let
A1 and
An (n 1)
be polynomial con EPr
Proof:
matrices and let A A1A2A3……An . Then the following statements are equivalent.

A is polynomial con EPr .

R(A1) R(An ) and (A) r
1 n
1 n

R(A*) R(A* ) and (A) r
Proof follows from Theorem 1 for the product of two matrices A, B .
Remark 2.3
In the above corollary both the conditions that
(AB) r and R(A) R(B) are essential for a

A polynomial con EPr .

Proof:
n
n

(ii): Since R(A) R(A1) and rk(A) rk(A1) we get R(A) R(A1) . Similarly R(AT ) R(AT ) .
product of two polynomial con EPr matrices to be
polynomial con EPr . This can be seen in the following:
Example 2.4
Now, A is polynomial con EPr R(A) R(AT )
Let
A 1 i ,
B i 1
be polynomial
i 1 1 i
and (A) r (by definition of polynomial con EPr )
n
n
R(A1) R(AT ) and (A) r
R(A1) R(An ) and (A) r
(since An is polynomial con EPr )

(iii)
con EPr matrices. Here R(A) R(B) , (AB) 1
and AB is not polynomial con EP1 .
Example 2.5
R(A ) R(A ) A A A A
(by result (1))
Let A i 0 and
B i i by
1 n 1 1 n n
0 0
i i
A A A A
1 1 n n
polynomial con EPr matrices. Here R(A) R(B) ,
AA1 A An (since A1 , An
1 n
are polynomial con EPr )
(AB) 1 and AB is not polynomial con EP1 .
R(A ) R(A ) (by results
1 n
1 n
1 n
(1)and (4))
Remark 2.6
In particular for A B, corollary 1 reduces to the
Therefore,
R(A*) R(A* )
(by results (2))
following.
1 n
1 n
R(A1) R(An ) and (A) r R(A*) R(A* )
and (A) r .

(i):
A is polynomial con EPr R(A ) R(A )T and
(A ) r (by definition of polynomial con EPr )
R(A) R(A) and (A ) r
R(AT ) R(A) and (A) r
Corollary 2.7
Let A be polynomial con EPr . Then Ak is polynomial con EPr (Ak ) r .
Theorem 2.8
Let (AB) (B) r1 and (BA) (A) r2 . If
AB , B are con EPr1 and A is con EPr2 , then
(by results (2)and (3))
BA is con EPr2 .
Proof
Since
(BA) (A) r2 ,it is enough to show that
Remark 2.11
For any two polynomial con EPr matrices
N(BA) N(BA)T . N(A) N(BA) and (BA) (A)
A and B , since AB , AB ,
A B ,
A B , A B ,
implies N(BA) N(A) . Similarly N(AB) N(B) .
BA
all have the same rank, the property of a
Now,
N(BA) N(A)
N(AT )
(Since A is polynomial
matrix being polynomial con EPr is preserved for its conjugate and MoorePenrose inverse, by applying Corollary 1 for a pair of polynomial con EPr matrices
r
r
con EP )
2
among A , B , A , B , A , B , A , B and using
r
r
con EP )
1
N(BTAT )
N((AB)T )
N(AB)
(Since AB is polynomial
the result 2, we can deduce the following.
Corollary 2.12
Let A , B be polynomial con EPr matrices.
Then the following statements are equivalent.

AB is polynomial con EPr matrices.
N(B) (Since N(AB) N(B) )
con EPr )
N(BT ) (Since B is polynomial

AB is polynomial con EPr matrices.

A B is polynomial con EPr matrices.
N(ATBT ) N(BAT ) .

A B
is polynomial con EPr matrices
Further, (BA) (BA)T implies N(BA) N(BA)T . Hence the Theorem.

A B is polynomial con EPr matrices
BA is polynomial con EPr matrices
Lemma 2.9
If A , B are polynomial con EPr matrices and
AB has rank r , then BA has rank r .
Proof:
(AB) (B) dimN(A) N(B* ) .
Theorem 2.13
If A , B are polynomial con EPr matrices.
R(A) R(B) then (AB) BA .
Proof:
Since A is polynomial con EPr and R(A) R(B) ,
Since (AB) (B) r , N(A) N(B*) 0
we have
R(A ) R(B) . That is given
x Cn
(the
N(A) N(B*) 0 N(A) N(B) 0
(Since
set of all n1
complex matrices) there exists a
B is polynomial con EPr )
N(A) N(B) 0
N(A*) N(B) 0
(Since
yCn such that Bx A y . Now,
Bx A y BAABx BAAA y BA y
BBx
A is polynomial con EPr ).Now,
Since BB is hermitian, it follows that
BAAB is
*
hermitian. Similarly, R(A ) R(B) implies
(BA) (A) dimN(B) N(A ) r 0 r
ABBA is hermitian. Further by result (1),
Hence the Lemma.
Theorem 2.10
If A , B and AB AB are polynomial con EPr
matrices, then BA is polynomial con EPr .
Proof:
Since A , B are polynomial con EPr matrices and
(AB) r , by Lemma 1, (AB) r . Now the result follows from Theorem 2, for r1 r2 r .
AA BB . Hence,
AB(BA )AB ABB(BB) B
AB
(BA)AB(BA) B(BB) BBA
BA
Thus BA satisfies the defining equations of the MoorePenrose inverse, that is, (AB) BA . Hence the theorem.
Remark 2.14
In the above Theorem, the condition that
Now, set C Ir 0
K I
and consider
R(A) R(B) is essential.
nr
Ir 0 E EL I KT
CACT r
Example 2.15
K Inr KE KEL 0 Inr
Let
A i i and
B i 0
Here A and
E EL I
KT E EKT EL
i i
0 0
r
0 0
0 I 0 0
B are polynomial con EP1 matrices, (AB) 1 ,
nr
R(A) R(B) and (AB) BA
Remark 2.16
The converse of Theorem 4, need not be true in general. For,
CACT is polynomial con EPr From
N(A) N(CACT ) it follows that EL EKT 0 , and so L KT ,completing the proof.
Theorem 2.18
Let
A i 0
0 0
0 0
and
B 0 0 . A and B are
0 i
0 i
If A , B are polynomial con EPr matrices,
(AB) r and (AB) BA , then R(A) R(B) .
polynomial con EPr matrices, such that
(AB) BA , but R(A) R(B) .
Next to establish the validity of the converse of the Theorem 4, under certain condition, first let us prove a Lemma.
Proof:
Since A is polynomial con EPr , by Theorem 3 in [3], there is a unitary matrix U such that,
0 0
0 0
UTAU D 0 , where D is rr nonsingular
Lemma 2.17
E F
matrix.
Set U*BU B1
B3
B2
B4
Let
A G H be an nn polynomial con
UT ABU UTAUU*BU D 0 B1
B2
EPr matrix where E is an rr matrix and if E F
0 0 B B
has rank r then E is nonsingular. Moreover there is
3 4
DB1 DB2
E EKT
0 0
an (n r) r matrix K such that A .
1 2
1 2
KE KEKT
D 0 B B
Proof:
has
r
r
I 0
0 Inr 0 0
Since A is polynomial con EPr , is
rankr and thus.
0 0
B B D 0
polynomial con EPr and E Fhas rank r , the U*BAU U*BUUTAU 1 2
r
r
I 0 E F
E F
B3
B4 0 0
product
is a product of
B1D 0
0 0 G H 0 0
B D 0
polynomial con EPr matrices which has rank r .
3
Therefore by Lemma 1 the product
B1 0 D 0
has
B3 0 0 I
E F Ir 0 E 0
has rank r . Hence there
n r
G H 0 0 G 0
B1 B2 B1 0
rankr . It follows that 0 0 and B 0 have
is an (n r) r matrix K and an r(n r) matrix L such that G KE , F EL , and E is nonsingular.
KE KEL
KE KEL
Therefore, A E EL
3
rankr , so that B1 is nonsingular.
* B1
B1KT
Theorem 2.19
By Lemma 2,
U BU , with
KB KB KT
Let A , B are polynomial con EPr matrices,
1 1
(AB) r and (AB) BA , then AB is
(U*BU) (B1 ) r . By using Penrose representation for the generalized inverse [4], we get
polynomial con EPr .
Proof:
B*PB* B*PB*K* T
(U*BU) 1 1 1 1
where
R(B) R(B ) (Since B is polynomial con EPr )
KB*PB* KB*PB*K* *
1
1
1
1
1
1
1
1
1 1 1 1
P (B1 B* B1 KT KB*)1 B1(B*B1 B*K* KB1)1
R(B) R(B )
R(B*A*)
(Since R(B*A*) R(B*)
UTBU (U*BU)
Q QK*
KQ KQK*
where
and (AB)* (AB) r (B*)
R(AB)* R(AB)
(by result (2))
Q (I KT K)1B1(I K*K)1
R(AB )
(by hypothesis)
* T
1
1
D1 0
R(A ) R(A*) R(A) (by result (2)
U A U (U AU)
0 0 and A is polynomial con EPr ).
UTABU UTABU(UTABU) UTABU
UTABU(UT (AB) U)UTABU
is unitary)
UTABU(UTB A U)UTABU
(byhypothesis)
(since U
R(B) R(A) R(B) R(A)
Since (AB) r R(B) R(A) , by Corollary 1, AB
is polynomial con EPr . Hence the Theorem.
UTABU(UTB U)(U*A U)UTABU
(since U is unitary). On simplification, we get,
DB1QB1 DB2 KQB1 DB1
DB1(I B1 B2 K)QB1 DB1
1
1
Since B2 B1KT , QB1 (I KT K)1 . Hence (I KT K) (QB1)1 I . Thus KT K 0 which implies K*K 0 so that K 0 .
0 0
0 0
U*BU B1 0
UTAU D 0
0 0
U* A U D 0
0 0
0 0
Since D and B1 are rr nonsingular matrices we have
References:

T.S. Baskett and I.J.Katz, Theorems on Products of EPr matrices, Linear Algebra Appl., 2(1969), 87103.

A. Ben Israel and T.N.E.Greville, Generalized inverses Theory and Applications, Wiley, Interscience, New York, 1974.

AR. Meenakshi and R. Indira, On conjugate EP matrices periodica math., Hung.,

R. Penrose, On best approximate solutions of linear matrix equations, Proc.Cambridge Phil.
Soc., 52(1956), 1719

C.R. Rao and S.K. Mitra, Generalized inverse of matrices and its applications Wiley and Sons, New York, (1971).
D 0 B
1
1
R(D) R(B ) R R
0
0 0
0 0
0 0
0 0
1
R(U* A U) R(U*BU)
R(A) R(B).
Hence the Theorem.