Ann. Funct. Anal. 3 (2012), no. 2, 115–127
A nnals of F unctional A nalysis
ISSN: 2008-8752 (electronic)
URL: www.emis.de/journals/AFA/
RANK EQUALITIES FOR MOORE-PENROSE INVERSE AND
DRAZIN INVERSE OVER QUATERNION
HUASHENG ZHANG
Communicated by Q.-W. Wang
Abstract. In this paper, we consider the ranks of four real matrices Gi (i =
0, 1, 2, 3) in M † , where M = M0 + M1 i + M2 j + M3 k is an arbitrary quaternion
matrix, and M † = G0 + G1 i + G2 j + G3 k is the Moore-Penrose inverse of M .
Similarly, the ranks of four real matrices in Drazin inverse of a quaternion matrix are also presented. As applications, the necessary and sufficient conditions
for M † is pure real or pure imaginary Moore-Penrose inverse and N D is pure
real or pure imaginary Drazin inverse are presented, respectively.
1. Introduction
Throughout this paper, we denote the real number field by R, the set of all
m × n matrices over the quaternion algebra
H = {a0 + a1 i + a2 j + a3 k | i2 = j 2 = k 2 = ijk = −1, a0 , a1 , a2 , a3 ∈ R}
by Hm×n , the identity matrix with the appropriate size by I, the conjugate transpose of a matrix A by A∗ , the column right space, the row left space of a matrix A
over H by R (A) , N (A) , respectively. The Moore-penrose inverse of A ∈ Hm×n ,
denoted by A† , is defined to be the unique solution X to the four matrix equations
(i) AXA = A, (ii) XAX = X, (iii) (AX)∗ = AX, (iv) (XA)∗ = XA.
Let A ∈ Hm×mbe given with IndA = k, the smallest positive integer such that
r Ak+1 = r Ak . The Drazin inverse of matrix A, denoted by AD , is defined to
be the unique solution X of the following three matrix equations
(i) Ak XA = Ak , (ii) XAX = X, (iii) XA = AX.
Date: Received: 30 December 2012; Accepted: 9 April 2012.
2010 Mathematics Subject Classification. Primary 15A03; Secondary 15A09, 15A24, 15A33.
Key words and phrases. Moore-penrose inverse, rank, quaternion matrix, Drazin inverse.
115
116
H. ZHANG
Suppose
M = M0 + M1 i + M2 j + M3 k, N = N0 + N1 i + N2 j + N3 k
(1.1)
be a quaternion matrix, where Mi ∈ Rm×n , Ni ∈ Rm×m , i = 0, 1, 2, 3, and let



M0 −M1 −M2 −M3
N0 −N1 −N2 −N3
 M1 M0

M3 −M2 
N3 −N2 
 , N =  N1 N0
,
M =
 M2 −M3 M0
 N2 −N3 N0
M1 
N1 
M3 M2 −M1 M0
N3 N2 −N1 N0

(1.2)
and the Moore-Penrose inverse of M, the Drazin inverse of N are denoted by
M † = G0 + G1 i + G2 j + G3 k, N D = D0 + D1 i + D2 j + D3 k,
(1.3)
respectively, where Gi ∈ Rn×m , Di ∈ Rm×m , i = 0, 1, 2, 3.
Moore-Penrose inverse of matrix is an attractive topic in matrix theory and
have extensively been investigated by many authors (see, e.g., [1]-[11]). Drazin
inverse is also one of the important types of generalized inverses of matrices,
and have well been examined in the literatures, (see, e.g., [1]-[2], [13]-[16]). For
example, Campbell and Meyer gave a basic identity on Drazin inverse of a matrix
in [1]
†
AD = Ak A2k+1 Ak .
(1.4)
L. Zhang presented a characterization of the Drazin inverse of any n×n singular
matrix and proposed a method for solving the Drazin inverse and an algorithm
with detailed steps to compute the Drazin inverse in [13].
As well known, the expressions of Gi , Di (i = 0, 1, 2, 3) in M † , N D are quite
complicated if there are no restrictions (see, e.g., [3], [5]). In that case, it is
difficult to find properties of Gi , Di (i = 0, 1, 2, 3) in M † , N D . In this paper, we
derived the ranks of Gi , Di (i = 0, 1, 2, 3) in M † , N D through a simpler method,
and then give some interesting consequences.
As a continuation of the above works, we in this paper investigate the ranks
of real matrices Gi , Di (i = 0, 1, 2, 3) in M † and N D . In Section 2, we derive
the formulas of rank equalities of four real matrices G0 , G1 , G2 and G3 in M † =
G0 + G1 i + G2 j + G3 k. Moreover, we established the necessary and sufficient
conditions for M † is pure real or pure imaginary Moore-Penrose inverse. In
Section 3, the formulas of rank equalities of four real matrices D0 , D1 , D2 and D3
in N D = D0 + D1 i + D2 j + D3 k are established, and the necessary and sufficient
conditions for N D is pure real or pure imaginary Drazin inverse are presented.
Some further research topics related to this paper are also given.
2. Rank equality for Gi (i = 0, 1, 2, 3) in M †
We begin with the following lemmas which can be generalized to H.
Lemma 2.1. (see [6]) Let A1 , A2 , · · · , Ak ∈ Hm×n . Then the Moore-Penrose inverse of their sum satisfies
RANK FOR M-P INVERSE AND DRAZIN INVERSE
117
† 

· · · Ak
Im
· · · Ak−1   Im 
 . .
.. 
..
.
.   .. 
· · · A1
Im
A1 A2

A
1
k A1
(A1 + A2 + · · · + Ak )† = [In , In , · · · In ] 
.
..
 ..
k
.
A2 A3

Lemma 2.2. (see [6]) Let A1 , A2 , · · · , Ak ∈ Hm×n . Then the Drazin inverse of
their sum satisfies
D 
· · · Ak
Im
· · · Ak−1   Im
 .
.. 
..

 ..
.
.
· · · A1
Im
A1 A2
 Ak A1
1
(A1 + A2 + · · · + Ak )D = [In , In , · · · In ] 
..
 ...
k
.
A2 A3



.

Lemma 2.3. (see [7]) Let A ∈ Hm×n , B ∈ Hm×k , C ∈ Hl×n and D ∈ Hl×k be
given. Then the rank of the Schur complement S = D−CA† B satisfies the equality
†
r D − CA B = r
A∗ AA∗ A∗ B
CA∗
D
−r (A) .
(2.1)
Lemma 2.4. (see [8]) Let A ∈ Hm×n , B ∈ Hm×k and C ∈ Hl×n be given, and
suppose that
R (AQ) = R (A) , R [(P A)∗ ] = R (A∗ ) .
Then
r [AQ, B] = r [A, B] , r
PA
C
=r
A
C
.
where P and Q are arbitrary matrices over H.
Now we establish the main result about Moore-Penrose inverse.
Theorem 2.5. Let M, M and M + be
ranks of Gi (i = 0, 1, 2, 3) in (1.3) can
"
#
c0 M
f0
M
r (G0 ) = r
−r M ,
f 0
M
"
#
c
f
M2 M2
r (G2 ) = r
−r M ,
f 0
M
given by (1.1), (1.2) and (1.3). Then the
be determined by the following formulas
"
#
c1 M
f1
M
r (G1 ) = r
−r M ,
(2.2)
f 0
M
"
#
c
f
M3 M3
r (G3 ) = r
−r M ,
(2.3)
f 0
M
where
c0
M


∗ 

−M1 −M2 −M3 
∗
∗
∗
M
−M
M
M
M
M
−M
1
0
3
2
0
3
2
 M0
M3 −M2 
∗
∗  
  M3∗
M
−M
M2 −M3 M0
M1  ,
=
0
1
 −M3 M0
M1 
∗
∗
∗
−M2 M1
M0
M3 M2 −M1 M0
M2 −M1 M0
118
H. ZHANG


∗ 

M0 −M2 −M3 
∗
∗
∗
M
M
M
−M
M
M
M
1
0
3
2
1
2
3


c1 =  M1 M3 −M2   M3∗ M0∗ −M1∗   M2 −M3 M0
M1  ,
M
 M2 M0
M1 
∗
∗
∗
−M2 M1 M0
M3 M2 −M1 M0
M3 −M1 M0


∗ 

M0 −M1 −M3 
∗
∗
∗
M
M
M
−M
M
M
M
1
0
3
2
3
2
1


c2 =  M1 M0 −M2   M0∗ −M3∗ M2∗   M2 −M3 M0
M1  ,
M
 M2 −M3 M1 
∗
∗
∗
−M2 M1 M0
M3 M2 −M1 M0
M3 M2
M0

∗ 

M0 −M1 −M2  ∗
M1 M0
M3 −M2
M3∗
M1 M2∗
 M1 M0

M3   ∗
∗
∗  
c3 = 
M2 −M3 M0
M1  ,
M
 M2 −M3 M0  M0∗ −M∗3 M2 ∗
M3 M2 −M1 M0
M3 M0 −M1
M3 M2 −M1




M0
−M1
 M1 
 M0
f0 = 
 f

M
 M2  , M1 =  −M3
M3
M2




−M2
−M3

 M3 
 −M2
f2 = 
,M
 f


 M0  , M3 =  M1
−M1
M0


,

and
f = [M0 , −M1 , −M2 , −M3 ] .
M
Proof. According to Lemma 1, we have
(M0 + M1 i + M2 j + M3 k)†

† 

Im
M0 M1 i M2 j M3 k
 M1 i M0 M3 k M2 j   Im 
1

 
= [In , In , In , In ] 
 M2 j M3 k M0 M1 i   Im 
4
Im
M3 k M2 j M1 i M0


† 
M0 −M1 −M2 −M3
Im


 M1 M0
1
M3 −M2 
  −iIm 
= [Im , iIm , jIm , kIm ] 
 M2 −M3 M0
M1   −jIm 
4
M3 M2 −M1 M0
−kIm



G0 −G1 −G2 −G3
Im
 G1 G0


1
G3 −G2 
  −iIm  .
= [Im , iIm , jIm , kIm ] 
 G2 −G3 G0
G1   −jIm 
4
G3 G2 −G1 G0
−kIm
Obviously, G0 can be written as


Im
† 0 
†

G0 = [In , 0, 0, 0] M 
 0  = P M Q,
0
(2.4)
RANK FOR M-P INVERSE AND DRAZIN INVERSE
119
where


Im
 0 

P = [Im , 0, 0, 0] , Q = 
 0 .
0
Then it follows by Lemma 2, Lemma 3, (1.4) and (2.4) we get
∗
∗
∗
M MM M Q
r (G0 ) =
−r M
∗
PM
0
∗
MM M MP ∗
−r M
=
∗
QM
0


 
M0 −M1 −M2 −M3
M0 −M1 −M2 −M3


  M1 M0
∗
M
M0
M3 −M2
M
−M
1
3
2 
 
M 




M2 −M3 M0
M1
M1
=  M2 −M3 M0
 M3 M2 −M1 M0
M
M
−M
M0
3
2
1
M0 −M1 −M2 −M3
−r M
M0
M1
M2
M3
0











0
0
0
0
0 −M1 −M2 −M3


  0 M0
M3 −M2
M3 −M2  ∗  M1 M0
 
M 

 0 −M3 M0
M
−M
M
M1
M
=
2
3
0
1

 0 M2 −M1 M0
M3 M2 −M1 M0
M0 −M1 −M2 −M3
−r M

 


−M1 −M2 −M3
M1 M0
M3 −M2
  M0

M3 −M2  ∗ 
 
M2 −M3 M0
M1 
M
 −M3 M0
M1 
=

M3 M2 −M1 M0

M2 −M1 M0
M0 −M1 −M2 −M3
 





M0
M1
M2
M3
0
M0
M1
M2
M3
0









−r M


which is the first equality in (2.2). The other equalities (2.2) and (2.3) can also
be derived by the similar approach.
If M0 = 0, then the result in (2.2) and (2.3) can be simplified to the following.
Corollary 2.6. Let M = M1 i + M2 j + M3 k, and denote the Moore-Penrose
inverse of M as M † = G0 + G1 i + G2 j + G3 k,


0 −M1 −M2 −M3
0
M3 −M2 
f=
 M1
,
M
 M2 −M3
0
M1 
M3 M2 −M1
0
120
H. ZHANG
Then
A
C
r (G0 ) = r
f
+ r [A, B] − r M ,
r (G1 ) = r (C) , r (G2 ) = r (B) ,


AA∗ A AA∗ B B
f ,
r (G3 ) = r  CA∗ A CA∗ B 0  − r M
C
0
0
where



−M3
0 −M1 −M2
0
M3  , B =  −M2  , C = [M3 , M2 , −M1 ] .
A =  M1
M1
M2 −M3
0

c = M0 + M1 i. As a special case
Let M2 = M3 = 0, we get a complex matrix M
of Theorem 2.1, we have the following corollary.
c = M0 +M1 i and M
c† = G0 +G1 i. Then the ranks
Corollary 2.7. Suppose that M
of G0 , G1 can be determined by the following formulas
Vb0 V0
r (G0 ) = r
−r
W 0
Vb1 V1
r (G1 ) = r
−r
W 0
M0 −M1
M1 M0
M0 −M1
M1 M0
,
,
where
V0 =
V1 =
M0
M1
−M1
M0
, Vb0 =
, Vb1 =
M0
M1
−M1
M0
M0∗ [M1 , M0 ] ,
M1∗ [M1 , M0 ] , W = [M0 , −M1 ] .
Now we give a group of rank inequalities derived from (2.2) and (2.3).
RANK FOR M-P INVERSE AND DRAZIN INVERSE
121
Corollary 2.8. Let M, M and M † be given by (1.1), (1.2) and (1.3). Then the
ranks of G0 in M † satisfies the rank inequalities


M0 −M3 M2
M0 −M1  + r [M0 , −M1 , −M2 , −M3 ]
r (G0 ) ≤ r  M3
−M2 M1
M0


−M3
 −M2 
−r M ,
+r
(2.5)
 M1 
M0


−M3
 −M2 

r (G0 ) ≥ r [M0 , −M1 , −M2 , −M3 ] + r 
(2.6)
 M1  − r M ,
M0




M0 −M3 M2
M1 M0
M3 −M2
M0 −M1  − r  M2 −M3 M0
M1 
r (G0 ) ≥ r  M3
−M2 M1
M0
M3 M2 −M1 M0


−M1 −M2 −M3
 M0
M3 −M2 
+r M .
−r
(2.7)
 −M3 M0

M1
M2 −M1 M0
Proof. It is clearly that
∗

"
#
∗
∗
∗
M
−M
M
0
3
2
c f
f +r M
f0 ,
f0 +r M
f ≤ r M0 M0 ≤ r  M3∗
M0∗ −M1∗  +r M
r M
f 0
M
M0∗
−M2∗ M1∗
f0 , M
f0 and M
f are defined as same as Theorem 2.1.
where M
Putting them in the first rank equality in (2.2), we obtain (2.5) and (2.6). To
show (2.7), we need the following rank equality
A B
A
†
r CA B ≥ r
−r
− r [A, B] + r (A) ,
C 0
C
†
Now applying above inequality to P M Q in (2.4), we have
†
M Q
M
r (G0 ) = r P M Q ≥ r
−r
− r M, Q + r M ,
P 0
P
which is (2.7).
Rank inequalities for the G1 , G2 and G3 in M † can also be derived in the similar
way shown above. We omit them here for simplicity.
Using the result of Theorem 2.1 and Corollary 2.2, we give a necessary and
sufficient condition for an arbitrary quaternion matrix M to have a pure real
or pure imaginary Moore-Penrose inverse. As a special case, a necessary and
sufficient condition for an arbitrary complex matrix to have a pure real or pure
imaginary Moore-Penrose inverse is also presented.
122
H. ZHANG
Theorem 2.9. Let M, M and M † be given by (1.1), (1.2) and (1.3). Then
(a) the Moore-Penrose inverse of M is a pure real matrix if and only if
c1 M1
c2 M2
c3 M3
M
M
M
r M =r
=r
=r
,
M
0
M
0
M
0
(b) the Moore-Penrose inverse of M is a pure imaginary matrix if and only if
c0 M0
M
r
=r M
M
0
ci and Mi (i = 0, 1, 2, 3) are defined as Theorem 2.1.
where M, M
Proof. From Theorem 2.1, the Moore-Penrose inverse of M is a pure real matrix
if and only if
r (G1 ) = r (G2 ) = r (G1 ) = 0.
That is
c1 M1
c2 M2
c3 M3
M
M
M
−r M = 0, r
r
−r M = 0, r
−r M = 0.
M
0
M
0
M
0
Thus we have part (a) . By the same manner, we can get part (b) .
c = M0 + M1 i and M
c† = G0 + G1 i. Then
Corollary 2.10. Suppose that M
c is a pure real matrix if and only if
(a) the Moore-Penrose inverse of M
M0 −M1
Vb0 V0
r
=r
M1 M0
W 0
c is a pure imaginary matrix if and only if
(b) the Moore-Penrose inverse of M
M0 −M1
Vb1 V1
r
=r
,
M1 M0
W 0
where
V0 =
−M1
M0
, Vb0 =
−M1
M0
M0∗ [M1 , M0 ] ,
and
V1 =
M0
M1
, Vb1 =
M0
M1
M1∗ [M1 , M0 ] , W = [M0 , −M1 ] .
3. Rank equality for Di (i = 0, 1, 2, 3) in N D
In this section, we derive the formulas of rank equalities of four real matrices
D0 , D1 , D2 and D3 in N D = D0 + D1 i + D2 j + D3 k. Moreover, we established
the necessary and sufficient conditions for N have a pure real or pure imaginary
Drazin inverse.
RANK FOR M-P INVERSE AND DRAZIN INVERSE
123
Theorem 3.1. Let N, N and N D be given by (1.1), (1.2) and (1.3) with IndM ≥
1. Then the ranks of in (1.3) can be determined by the following formulas
"
r (D0 ) = r
"
r (D1 ) = r
"
r (D2 ) = r
"
r (D1 ) = r
k
c0 N
N N
f0 N k
N
k
c1 N
N N
f1 N k
N
k
c2 N
N N
f2 N k
N
k
c3 N
N N
f3 N k
N
k
N
k−1
e
N
#
e
N
#
e
N
#
e
N
#
−r N
0
k
N
k−1
0
k
N
k−1
0
k
N
k−1
0
k
,
(3.1)
k
−r N ,
(3.2)
k
−r N ,
(3.3)
k
−r N ,
(3.4)
where



N0
0
0
0
N0



N0
N3 −N2 
f = [N0 , −N1 , −N2 , −N3 ] ,
c  0
e =  N1  , N
,N
N
 N2  0 =  0 −N3 N0
N1  0
0
N2 −N1 N0
N3



N0 −N1 −N2 −N3

N0
0
0 
c1 =  0
f1 = N1 N0 N3 −N2 ,
,N
N
 N2 −N3 N0
N1 
N3 N2 −N1 N0


N0 −N1 −N2 −N3

N3 −N2 
c2 =  N1 N0
f2 = N2 −N3 N0 N1 ,
,N
N
 0 −N3

0
0
N3 N2 −N1 N0
and


N0 −N1 −N2 −N3

N3 −N2 
c3 =  N1 N0
f3 = N3 N2 −N1 N0 .
,N
N
 0 −N3
0
0 
N3 N2 −N1 N0
124
H. ZHANG
Proof. According to Lemma 1, we have
(N0 + N1 i + N2 j + N3 k)D

D 

N0 N1 i N2 j N3 k
Im
 N1 i N0 N3 k N2 j   Im 
1
 

= [Im , Im , Im , Im ] 
 N2 j N3 k N0 N1 i   Im 
4
N3 k N2 j N1 i N0
Im

D 

N0 −N1 −N2 −N3
Im
 N1 N0


1
N3 −N2 
  −iIm 
= [Im , iIm , jIm , kIm ] 
 N2 −N3 N0
N1   −jIm 
4
N3 N2 −N1 N0
−kIm



D0 −D1 −D2 −D3
Im
 D1 D0


1
D3 −D2 
  −iIm  .
= [Im , iIm , jIm , kIm ] 
 D2 −D3 D0
D1   −jIm 
4
D3 D2 −D1 D0
−kIm
Obviously, G0 can be written as

Im
D 0
G0 = [Im , 0, 0, 0] N 
 0
0


D

 = P N D Q = P N k N 2k+1 N k Q,


Im
 0 

where P = [Im , 0, 0, 0] and Q = 
 0 .
0
Then it follows by Lemma 2, Lemma 3, (1.4) and (2.4) we get
 2k+1 ∗ 2k+1 2k+1 ∗ 2k+1 ∗ k 
N
N
N
N
N Q
 − r N 2k+1
r (D0 ) = 
∗
k
2k+1
PN N
0
" 2k+1
#
k
2k+1 N
N Q
=
−
r
N
k
PN
0
" 2k+1
#
k
k
k
k
k
k
N
− N QP N N − N N QP N N Q
=
−
r
N
k
PM
0
" k
#
k
k
k
N N − QP N − N QP N N Q
−
r
N
=
k
PN
0



 

N0
0
0
0
N0


 k
N0
N3 −N2 
k
 N k N k−1  N1  
 N  0




0
−N
N
N
N
=
,
3
0
1
2  −r N




0
N2 −N1 N0
N3
[N0 , −N1 , −N2 , −N3 ] N
k
0
RANK FOR M-P INVERSE AND DRAZIN INVERSE
125
which is the equality in (3.1). The equalities (3.2-3.4) can also be derived by the
similar approach.
b = N0 + N1 i. As a special case of
Let N2 = N3 = 0, we get a complex matrix N
Theorem 3.1, we have the following corollary.
b = N0 + N1 i and N
b + = D0 + D1 i. Then the ranks
Corollary 3.2. Suppose that N
of D0 , D1 can be determined by the following formulas
"
#
f k Vb W
fk W
f k−1 V1
W
fk ,
r (D0 ) = r
−r W
fk
Vb W
0
"
#
f k Ve W
fk W
f k−1 V1
W
fk ,
r (D1 ) = r
−
r
W
fk
Ve W
0
where
N0
N1
N0
N1
V1 =
f=
W
N0 0
0 −N1
b
e
,V =
,V =
,
0 N0
N1
0
−N1
, W1 = [N0 , −N1 ] , W2 = [N1 , N0 ] .
N0
Using the result of Theorem 3.1 and Corollary 3.2, we give a necessary and
sufficient condition for an arbitrary quaternion matrix N to have a pure real
or pure imaginary Drazin inverse. As a special case, a necessary and sufficient
condition for an arbitrary square complex matrix to have a pure real or pure
imaginary Drazin inverse is also presented.
Theorem 3.3. Let N, N and N D be given by (1.1), (1.2) and (1.3) with IndM ≥
1. Then
(a) the Drazin inverse of N is a pure real matrix if and only if
" k
#
" k
#
k
c1 N k N k−1 N
e
c2 N k N k−1 N
e
N N
N N
r N =r
=r
f1 M k
f2 N k
N
N
0
0
#
" k
c3 N k N k−1 N
e
N N
,
=r
f3 N k
N
0
(b) the Drazin inverse of N is a pure imaginary matrix if and only if
" k
#
k
c0 N k N k−1 N
e
N N
r
=
r
N ,
f0 N k
N
0
e, N
ci and N
f1 (i = 0, 1, 2, 3) are defined as Theorem 3.1.
where N
b = N0 + N1 i and N
b D = D0 + D1 i. Then
Corollary 3.4. Suppose that N
b is a pure real matrix if and only if
(a) the Drazin inverse of N
"
#
f k Ve W
fk W
f k−1 V1
W
fk ,
r
=
r
W
fk
Ve W
0
126
H. ZHANG
b is a pure imaginary matrix if and only if
(b) the Drazin inverse of N
"
#
k b f k f k−1
f
W V W W V1
fk ,
r
=
r
W
k
b
f
VW
0
where
N0
N1
N0
N1
V1 =
f=
W
N0 0
0 −N1
b
e
,V =
,V =
,
0 N0
N1
0
−N1
, W1 = [N0 , −N1 ] , W2 = [N1 , N0 ] .
N0
Acknowledgement. Supported by the youth teacher development plan of
Shandong province.
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Department of Mathematics, Liaocheng University, Shandong252059, P.R. China.
E-mail address: zhsh0510@163.com; zhsh0510@yahoo.com.cn