 Open Access
 Total Downloads : 233
 Authors : S. Krishnamoorthy, G.Bhuvaneswari
 Paper ID : IJERTV2IS90555
 Volume & Issue : Volume 02, Issue 09 (September 2013)
 Published (First Online): 19092013
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Secondary kGeneralized Inverse of a skNormal Matrices
S. Krishnamoorthy1 And G.Bhuvaneswari 2
1 Head & Professor of Mathematics, Ramanujan Research Center,
Govt. Arts College (Autonomous), Kumbakonam, Tamilnadu612001 , India.
2 Research Scholar, Lecturer in Mathematics, Ramanujan Research Center,
Govt. Arts College (Autonomous) , Kumbakonam, Tamilnadu 612001, India.
Abstract:
Secondary kgeneralized inverse of a given square matrix is defined and its characterizations are given. Secondary k generalized inverses of sk normal matrices are discussed.
Ams Classification : 15A09,15A57.
Keywords: sk normal, sk unitary, nilpotent, sk hermitian matrices.
1.Introduction:
Ann Lee initiated the study of secondary symmetric matrices in[1]. The concept of secondary k – normal matrices was introduced in [3]. Some equivalent conditions on secondary k normal matrices are given in [4]. In this paper we describe secondary k generalized inverse of a square matrix, as the unique solution of a certain set of equation . This secondary kgeneralized inverse exists for particular
kind of square matrices. Let Cnxn denote the space of nxn complex matrices. We deal with secondary kgeneralized inverse of sk normal matrices. Throught this paper, if ACnxn , then we assume that if A 0 then A(KVA*VK) 0
i.e., A(KVA*VK) = 0 A = 0 (1)
It is clear that the conjugate secondary k transpose satisfies the following properties.
KV(A+ B)*VK = (KVA*VK)+(KVB*VK)
KV(A)*VK = (KVA*VK) KV(BA)*VK = (KVA*VK)(KVB*VK)
Now if BA(KVA*VK) = CA(KVA*VK) then by (1)
BA(KVA*VK) CA(KVA*VK) = 0
(BA(KVA*VK) CA(KVA*VK))(KV(B C)*VK) = 0
(BA CA)(KV(BA CA)*VK) = 0
(BACA) = 0
BA = CA
Therefore BA(KVA*VK) = CA(KVA*VK) BA = CA (2)
Similarly,
B(KVA*VK)A = C(KVA*VK)A
B(KVA*VK) = C(KVA*VK) (3)
Definition 1.1: [3]
A Matrix ACnxn is said to be secondary knormal ( sk normal) if
A(KVA*VK) = (KVA*VK)A
Example 1.2:
i 2 3 4
A = 4 i 2 3
is a sk normal matrix for k=(1,3),(2,4) the permutation matrix be
3 4 i 2
2 3 4 i
0 0 1 0 0 0 0 1
0 0 0 1 0 0 1 0
K =
and
V =
1 0 0 0 0 1 0 0
0 1 0 0 1 0 0 0
Definition 1.3:
A matrix ACnxn is said to be secondary kunitary (sk unitary) if
A(KVA*VK) = (KVA*VK)A I
Example 1.4:
i 1 1 0
1 i 0 1
A=
is a sk unitary matrix
1 0 i 1
0 1 1 i
Section 2: Secondary k – Generalized inverses of a matrix Theorem 2.1:
For any ACnxn , the four equations
AXA = A (4)
XAX = X (5)
KV(AX)*VK = AX (6)
KV(XA)*VK = XA (7)
have a unique solution for any ACnxn .
Proof: First, we shall show that equations (5) & (6) are equivalent to the single equation
XKV(AX)*VK = X (8)
From equations (5) and (6), (7) follows, since it is merely (6) substituted in (5) Conversely, equation
(8) implies
AXKV(AX)*VK = AX
Since the left hand side is sk hermitian, (6) follows. By substituting (6) in (8), we get XAX = Xwhich is actually (5). Therefore (5) and (7) are equivalent to (8) Similarly, (4) & (7) are equivalent to the equation
XA(KVA*VK) = KVA*VK (9)
Thus to find a solution for the given set of equations, it is enough to find an X satisfying (8) & (9). Now the expressions ((KVA*VK)A),((KVA*VK)A)2, ((KVA*VK)A)3 cannot all be linearly independent ( i.e) there exists a relation
1((KVA*VK)A) + 2((KVA*VK)A)2 ++ k((KVA*VK)A)k = 0 (10)
Where 1, 2,, k are not all zero. Let r be the first non zero . (i.e) 1 = 2 r1 0 .
Therefore (10) implies that
r((KVA*VK)A)r = r+1((KVA*VK)A)r+1 ++ m((KVA*VK)A)m
r
r
If we take B = –1r+1I + r+2((KVA*VK)A)++ m((KVA*VK)A)mr1
Then
B((KVA*VK)A)r+1 = –1
B((KVA*VK)A)r+1 = –1
((KVA*VK)A)r+1 ++ ((KVA*VK)A)m
((KVA*VK)A)r+1 ++ ((KVA*VK)A)m
r r+1 m
B((KVA*VK)A)r+1 = ((KVA*VK)A)r . By using (2) & (3) repeatedly, we get
B(KVA*VK)A(KVA*VK) = KVA*VK (11)
Now if we take X = B(KVA*VK) then (11) implies that this X satisfies (9) implies (7), we have (KV(XA)*VK)(KVA*VK) = KVA*VK
B(KV(XA)*VK)(KVA*VK) = B(KVA*VK)
Therefore X = B(KVA*VK) satisfies (8). Thus X = B(KVA*VK) is a solution for the given set of equations.
Now let us prove that this X is unique. Suppose that X and Y satisfy (8) and (9). Then by substituting (7) in (5) and (6) in (4), we obtain
(KV(XA)*VK)X = X and (KV(AX)*VK)A = A
Also,
Y = (KV(YA)*VK)Y and KVA*VK = (KVA*VK)AY
Now X = X(KVX*VK)(KVA*VK)
= X(KVX*VK)(KVA*VK)AY
= X(KV(AX)*VK)AY
= XAY
= XA(KV(YA)*VK)Y
= XA(KVA*VK)(KVY*VK)Y
= (KVA*VK)(KVY*VK)Y
= (KV(YA)*VK)Y
X = Y
Therefore X is unique.
Definition 2.2: Let ACnxn . The unique solution of (4), (5), (6) and (7) is called secondary k
generalized inverse of A and is written as A sk .
Example 2.3:
1 1 1
sk
1/9 1/9 1/9
If A = 1 1 1
1 1 1
then A = 1/9 1/9 1/9
1/9 1/9 1/9
Note 2.4: By using (7) in (5), (6) in (4) and from (8) and (9) we obtain
sk
sk
sk *
sk *
sk
A (KV ( A ) VK )(KVA VK ) A (KVA VK )(KV ( A ) VK ) A
(12)
A sk A(KVA*VK ) (KVA*VK ) (KVA*VK )AA sk
If is a scalar, then sk means 1 when 0 and zero when = 0 .
Section3: Secondary kgeneralized inverse of sk normal matrices.
In this paper, characterizations of secondary kgeneralized inverse (skg) inverse of a matrix are obtained skg inverse of sknormal matrices are discussed .sk herimitian matrices are defined and the condition for sk normal matrices to be diagonal is investigated.
Theorem 3.1: For ACnxn .
sk
sk
*

A
sk A

KVA*
VK = KVA
sk
VK

If A is non singular, then
sk = A1

(A sk =
sk
sk
A ) A

((KVA*VK)A sk = sk (KV sk VK)*
) A A
Proof: Let ACnxn .

By the definition of skg inverse, we have
sk A
sk =
sk
and
sk (
sk
sk
sk =
sk
A A A A A ) A A
These two equations imply that
sksk
A = A

From the definition of A sk , we have AA sk A = A
(KVA*VK)(KV( sk )*VK)(KVA*VK) = KVA*VK
A
)
)
Also (KVA*VK)(KV(A* sk VK)(KVA*VK) = KVA*VK
From these two equations, we have
(KV( sk )*VK) = KV(A* sk VK
A )

Since A is non singular, A1 exists
Now A sk A = A (By definition of )
A sk
Pre multiplying & post multiplying by A1 we have
A
A
sk
= A1

The equations, A sk A = A and
A
)
)
(A)(A sk (A) = (A) imply that
)
)
A
A
(A sk =
sk
(A sk =
sk
sk
where
sk = 1
) A

from (12) we have,

sk
sk * *
sk
A (KV(A ) VK)(KVA VK) = A
Also A sk A = A .
A
Therefore A sk (KV( sk )*VK)(KVA*VK)A = A
A A
Substitute this in the right hand side of the defining relation, we get
((KVA*VK)A sk = sk (KV(A* sk VK)
) A )
Theorem 3.2: A necessary and sufficient condition for the equation AXB= D to have a solution is
A sk D
sk B = D , in which case the general solution is
A B
X = sk D sk + Y sk AYB sk , where Y is arbitrary.
A B A B
Proof: Let us assume that X satisfies the equation AXB= D, then D = AXB
= A sk AXB
sk B = A
sk D
sk B (By the definition of )
A B
Conversely if
D = A
A
sk D
B
sk B ,then X =
sk D
sk
sk , then it is a particular solution
A
of AXB= D. since AXB = A
B
sk D
A B
sk B = D.
A B
A
A
If Y Cnxn , then any expression of the form X =
sk D
sk + Y
sk AYB
sk is
B
B
A
A
B
B
A
A
B
B
a solution of AXB= D.and conversely,if X is a solution AXB= D, then
A
A
B
B
X = sk D
sk + X
sk AX
sk B satisfies AXB= D. Hence the theorem.
Theorem 3.3: The matrix equations AX = B and XD = E have a common solution if and only if each equation has a solution and AE = BD.
Proof: It is easy to see that the conditions is necessary, conversely A sk B and E D sk
are solutions
of AX=B and XD=E and hence
AAsk B B
and
EDsk D E . Also AE=BD. By using these facts it
can be prove that
X Ask B EDsk Ask AEDsk
is a common solution of the given equations.
Definition 3.4: A matrix ECnxn
is said to be secondaryk hermitian idempotent matrix (sk. h.i) if
E(KVE*VK) = E (i.e) E = KVE*VK and E2 = E .
Theorem 3.5: (i)
A sk A ,
AAsk , 1
sk A, 1 A sk
are all the sk hermitian idempotent.
A A
)
)
(ii) J is idempotent there exist sk hermitian idempotents E and F such that J = (FE sk
in which case J = EJF.
Proof: Proof of (i) is obvious. If J is idempotent then J2=J. By (i) of theorem (3.1),
sk
J = (J
sk J)(JJ
sk )
. Now if we take
E = JJ sk
and
F = J
sk J they will satisfy our
)
)
requirements conversely if J = (FE sk
then J=EFPEF where
P = (KV((FE sk )*VK)(FE sk (KV(FE sk )*VK) . Therefore J=EJF and hence
) ) )
J2 = E(FE sk FE(FE sk F = E(FE sk F = J . Hence J is idempotent.
) ) )
Note 3.6:(i) sk hermitian idempotent matrices are sk normal matrices.
(ii) The skg inverse of an sk hermitian idempotent matrix is also sk hermitian idempotent
matrix.
Definition 3.7: For any square matrix A there exists a unique set of matrices J defined for each complex number such that
JJ = J (13)
J 1
(14)
AJ = JA
(15)
(A I)J is nilpotent (16)
Then the non zero J s are called the principal idempotent elements of A.
Theorem 3.8: If
E =1(A I)nsk (A I)n
and F =1(A I)n (A I)nsk ,
where n is sufficiently large, then the principal idempotent element of A are given by
sk
J FE and n can be taken as unity iff A is diagonable.
Proof: Assume that A is diagonable.
Let
E =1(A I)sk (A I) , and
F =1(A I)(A I)sk
Then by 3.5(i) E and F are sk hermitian idempotent matrices. If is not an eigen value of
A, then
A I 0
and hence F and E are zero by (iii) of theorem(3.1). Clearly,
(AI)E 0 and F(A – I) = 0 (17)
Therefore FE
Hence FE 0
FAE
if
= FE
(18)
Now if we take
J F E sk
then by (ii) of theorem (3.5),
J = E {F E
sk F
(19),
}
Now(18) implies JJ = J .Also by( 18), FJE = FE (20) If Z
is an eigen vector of A corresponding to the eigen value then EZ = Z. As A is diagonable, any column vector X conformable with A is expressible as a sum of eigen vectors (i.e) it is expressible in the form X E X . This is a finite sum over all complex .
Similarly, if Y* is conformable with A, it is expressible as Y* Y*F
Now by
equations (18) and (20) Y*( J )X (Y*F )( J )( E X )
Y*F E X
= Y*( J )X Y*X
Y*( J I)X 0
J I
Also(17) and (19) lead to AJ = J = JA (21)
This implies (A I)J is nilpotent and (15) and (16) are satisfied.
Moreover A= J
Conversely if J I
(22)
and A is not diagonable (n=1) then X = JX
gives X as a sum
of eigen vectors of A, since (21) was derived without assuming the diagonability of A. If A is not diagonable. It seems more convenient simply to prove that for any set of Js satisfying (19), (20), (21) &
(22) each J = (F E )sk
where F
and E
are defined as in the theorem.
If the Js satisfy (13), (14), (15) & (16) J = I and
(A I)n J = 0 = J(A I)n
(23)
Which comes by using the fact that (A I)J is nilpotent, where n is sufficiently large.
From (23) and the definition of E and F , we have
EF J JF
(24)
By using Euclids algorithm, there exist P and Q which are polynomials in A such that
I = (A I)n P+ Q(AI)n
if .
Now F(A – I)n 0 (AI)n E . Hence FE = 0 if
From (24) FJ = 0 = JE if . Since J = I , we get
FJ = F and JE = E (25) using (24) and (25 ) it is easy to see that
FE sk = J . Hence the theorem
Theorem 3.9: If A is sk normal, it is diagonable and its principal idempotent elements are sk hermitian.
Proof: If A is sk normal then (A I) is sk normal. By using (viii)of the theorem (3.1) in the
definition of E
and F
of theorem (3.8) we obtain E =1(A I)sk (A I)
and
F =1(A I)(A I)sk
Hence A is diagonable. since E F , J = E is sk hermitian.
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IsarelAdi Ben and Greville Thomas ME;Generalized inverses:Theory and Applications; A wiley interscience publications Newyork,(1974).

S. Krishnamoorthy, G. Bhuvaneswari , Secondary k normal Matrices, International Journal of Recent scientific Research Vol , 4,issue 5,pp 576578, May 2013.

S. Krishnamoorthy, G. Bhuvaneswari , Some Characteristics on Secondary k normal Matrices,Open Journal of Mathematical Modeling July 2013,1(1):8084.

Weddurburn, J.H.M.,Lectures on matrices, colloq.Publ.Amer. Math.Soc.No.17,1934.