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- Authors : S. Krishnamoorthy, G.Bhuvaneswari
- Paper ID : IJERTV2IS90555
- Volume & Issue : Volume 02, Issue 09 (September 2013)
- Published (First Online): 19-09-2013
- ISSN (Online) : 2278-0181
- Publisher Name : IJERT
- License: This work is licensed under a Creative Commons Attribution 4.0 International License
Secondary k-Generalized Inverse of a s-k-Normal 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 k-generalized inverse of a given square matrix is defined and its characterizations are given. Secondary k- generalized inverses of s-k normal matrices are discussed.
Ams Classification : 15A09,15A57.
Keywords: s-k normal, s-k unitary, nilpotent, s-k 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 k-generalized inverse exists for particular
kind of square matrices. Let Cnxn denote the space of nxn complex matrices. We deal with secondary k-generalized inverse of s-k 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
(BA-CA) = 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 k-normal ( s-k normal) if
A(KVA*VK) = (KVA*VK)A
Example 1.2:
i 2 3 4
A = 4 i 2 3
is a s-k 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 k-unitary (s-k unitary) if
A(KVA*VK) = (KVA*VK)A I
Example 1.4:
i 1 1 0
1 i 0 1
A=
is a s-k 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 s-k 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 r-1 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)m-r-1
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- k-generalized inverse of s-k normal matrices.
In this paper, characterizations of secondary k-generalized inverse (s-k-g) inverse of a matrix are obtained s-k-g inverse of s-k-normal matrices are discussed .s-k herimitian matrices are defined and the condition for s-k 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 = A-1
-
(A sk =
sk
sk
A ) A
-
((KVA*VK)A sk = sk (KV sk VK)*
) A A
Proof: Let ACnxn .
-
By the definition of s-k-g 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, A-1 exists
Now A sk A = A (By definition of )
A sk
Pre multiplying & post multiplying by A-1 we have
A
A
sk
= A-1
-
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 secondary-k hermitian idempotent matrix (s-k. 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 s-k hermitian idempotent.
A A
)
)
(ii) J is idempotent there exist s-k 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) s-k hermitian idempotent matrices are s-k normal matrices.
(ii) The s-k-g inverse of an s-k hermitian idempotent matrix is also s-k 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 s-k 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,
(A-I)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(A-I)n
if .
Now F(A – I)n 0 (A-I)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 s-k = J . Hence the theorem
Theorem 3.9: If A is s-k normal, it is diagonable and its principal idempotent elements are s-k hermitian.
Proof: If A is s-k normal then (A- I) is s-k 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 s-k hermitian.
References:
-
Ann Lee: Secondary symmetric , Secondary skew symmetric, Secondary orthogonal matrices:
Period math. Hungary 7(1976),63-76
-
Isarel-Adi Ben and Greville Thomas ME;Generalized inverses:Theory and Applications; A wiley interscience publications Newyork,(1974).
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S. Krishnamoorthy, G. Bhuvaneswari , Secondary k- normal Matrices, International Journal of Recent scientific Research Vol , 4,issue 5,pp 576-578, May 2013.
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S. Krishnamoorthy, G. Bhuvaneswari , Some Characteristics on Secondary k- normal Matrices,Open Journal of Mathematical Modeling July 2013,1(1):80-84.
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Weddurburn, J.H.M.,Lectures on matrices, colloq.Publ.Amer. Math.Soc.No.17,1934.