Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2011, Article ID 831892, 13 pages
doi:10.1155/2011/831892
Research Article
On the Drazin Inverse of the Sum of Two Matrices
Xiaoji Liu,1, 2 Shuxia Wu,1 and Yaoming Yu3
1
College of Mathematics and Computer Science, Guangxi University for Nationalities,
Nanning 530006, China
2
Guangxi Key Laboratory of Hybrid Computational and IC Design Analysis, Nanning 530006, China
3
School of Mathematical Sciences, Monash University, Caulfield East, VIC 3800, Australia
Correspondence should be addressed to Xiaoji Liu, liuxiaoji.2003@yahoo.com.cn
Received 31 March 2011; Revised 25 August 2011; Accepted 21 October 2011
Academic Editor: Michela Redivo-Zaglia
Copyright q 2011 Xiaoji Liu et al. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
We deduce the explicit expressions for P QD and P QD of two matrices P and Q under the
conditions P 2 Q P QP and Q2 P QP Q. Also, we give the upper bound of ||P QD − P D ||2 .
1. Introduction
The symbol Cm×n stands for the set of m × n complex matrices, and In for short I stands for
the n × n identity matrix. For A ∈ Cn×n , its Drazin inverse, denoted by AD , is defined as the
unique matrix satisfying
Ak1 AD Ak ,
AD AAD AD ,
AAD AD A,
1.1
where k IndA is the index of A. In particular, if k 0, A is invertible and AD A−1
see, e.g., 1–3 for details. Recall that for A ∈ Cn×n with IndA k, there exists an n × n
nonsingular matrix X such that
AX
C 0
0 N
X −1 ,
1.2
2
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where C is a nonsingular matrix and N is nilpotent of index k, and
D
A X
C−1 0
0
X −1
0
1.3
see 1, 3. It is well known that if A is nilpotent, then AD 0. We always write Aπ : I−AAD .
Drazin 2 proved that in associative ring A BD AD BD when A, B are Drazin
invertible and AB BA 0. In 4, Hartwig et al. relaxed the condition to AB 0 and put
forward the expression for A BD where A, B ∈ Cn×n . In recent years, the Drazin inverse
of the sum of two matrices or operators has been extensively investigated under different
conditions see, 5–15. For example, in 7, the conditions are P Q λQP and P Q P QP ,
in 9 they are P 3 Q QP and Q3 P P Q, and in 15, they are P QP 0 and P Q2 0. These
results motivate us to investigate how to explicitly express the Drazin inverse of the sum
P Q under the conditions P 2 Q P QP and Q2 P QP Q, which are implied by the condition
P Q QP .
The paper is organized as follows. In Section 2, we will deduce some lemmas. In
Section 3, we will present the explicit expressions for P QD and P QD of two matrices P
and Q under the conditions P 2 Q P QP and Q2 P QP Q. We also give the upper bound of
P QD − P D 2 .
2. Some Lemmas
In this section, we will make preparations for discussing the Drazin inverse of the sum of two
matrices in next section. To this end, we will introduce some lemmas.
The first lemma is a trivial consequence of 16, Theorem 3.2.
Lemma 2.1. Let A ∈ Cn×n , B ∈ Cm×m , and C ∈ Cm×n with BC 0, and define
M
A 0
C B
.
2.1
Then,
⎛
MD ⎝
AD
CAD 0
2
BD
⎞
⎠.
2.2
Lemma 2.2. Let P, Q ∈ Cn×n . If P 2 Q P QP , then, for any positive integers i, j,
i P i1 Q P i QP P QP i , P 2i Q P i QP i ,
ii P i Qi P Qi .
Moreover, if Q2 P QP Q, then
P Qj P i P i1 Qj .
2.3
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Proof.
3
i By induction, we can easily get the results.
ii For i 1, it is evident. Assume that, for i k, the equation holds, that is, P k Qk P Qk . When i k 1, by i, we have
P i1 Qi1 P QP i Qi P QP Qi P Qi1 .
2.4
Hence, by induction, we have P i Qi P Qi for any i.
Assume Q2 P QP Q. By induction on j for 2.3. Obviously, when j 1, it holds by
statement i. Assume that it holds for j k, that is, P Qk P i P i1 Qk . When j k 1,
P Qk1 P i P Qk−1 Q2 P P i−1 P Qk P QP i−1 P Qk P i Q P i1 Qk Q P i1 Qk1 .
2.5
Hence 2.3 holds for any j.
Lemma 2.3. Let P, Q ∈ Cn×n . Suppose that P 2 Q P QP and Q2 P QP Q. Then, for any positive
integer m,
P Qm i
P m−i Qi Qm−i P i ,
Cm−1
m−1
i0
2.6
where the binomial coefficient Cji j!/i!j − i!, j ≥ i.
Moreover, if P, Q are nilpotent with P s 0 and Qt 0, then P Q is nilpotent and its index
is less than s t.
Proof. We will show by induction that 2.6 holds. Trivially, 2.6 holds for m 1. Assume
that 2.6 holds for m k, that is,
P Qk k−1
i0
i
P k−i Qi Qk−i P i .
Ck−1
Then, for m k 1, we have, by Lemma 2.2,
P Qk1 k−1
i
P k−i Qi Qk−i P i P Q
Ck−1
i0
k−1
i
P k1−i Qi P k−i Qi1 Qk−i P i1 Qk1−i P i
Ck−1
i0
P k1 k−1 i1
Qk1 i
i−1
Ck−1
P k1−i Qi P Qk
Ck−1
k−1 i1
i
i−1
Ck−1
Qk1−i P i QP k
Ck−1
2.7
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P k1 k−1
k−1
Cki P k1−i Qi P Qk Qk1 Cki Qk1−i P i QP k
i1
i1
k
k
Cki P k1−i Qi Cki Qk1−i P i .
i0
i0
2.8
Hence 2.6 holds for any m ≥ 1.
If P, Q are nilpotent with P s 0 and Qt 0, then taking m s t − 1 in 2.6 yields
P Qst−1 0, that is, P Q is nilpotent of index less than s t.
Lemma 2.4 see 1, Theorem 7.8.4. Let P, Q ∈ Cn×n . If P Q QP , then P QD QD P D P D QD and P D Q QP D .
Lemma 2.5. Let P, Q ∈ Cn×n and P be invertible. If P Q QP , then
D
D
P QD I P −1 Q P −1 P −1 I P −1 Q .
2.9
Moreover, if Q is nilpotent of index t, then P Q is invertible and
P Q−1 t−1
t−1
P −i−1 −Qi .
−Qi P −i−1 i0
2.10
i0
Proof. Since P Q P I P −1 Q I P −1 QP , by Lemma 2.4,
D
D
P QD I P −1 Q P −1 P −1 I P −1 Q .
2.11
Note that the nilpotency of Q with commuting with P implies that P −1 Q is nilpotent
of index t. Thus, I P −1 Q is invertible and so is P Q, and
t−1
t−1
−1
P −i−1 −Qi .
−Qi P −i−1 P Q−1 I P −1 Q P −1 i0
2.12
i0
Lemma 2.6. Let P, Q ∈ Cn×n with Q Q1 ⊕ Q2 , where Q1 is invertible and Q2 is nilpotent of index
t, and let
P
P1 P3
P4 P2
2.13
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5
be partitioned conformably with Q. Suppose that Q2 P QP Q and P 2 Q P QP . Then, P3 0 and
Q1 P1 P1 Q1 ,
2.14
Q2 P4 P2 P4 0,
2.15
Q22 P2 Q2 P2 Q2 ,
2.16
Pi2 Qi Pi Qi Pi ,
2.17
i 1, 2.
Moreover, if P is nilpotent of index s, then P4 P1s−1 0.
Proof. Since Q2 P QP Q, Q2t P Qt P Qt by Lemma 2.2, that is,
Q12t 0
0
P1 P 3
t t P1 P 3
Q1 0
Q1 0
,
0 0
0 0
P4 P2
P4 P2
0
2.18
namely,
Q12t P1 Q12t P3
0
0
t
Q1 P1 Q1t 0
0
0
.
2.19
Thus, P3 0 because the invertibility of Q1 . So from Q2 P QP Q and P 2 Q P QP , it follows,
respectively, that
P1 Q1 Q1 P1 ,
Q22 P4 Q2 P4 Q1 ,
Q22 P2 Q2 P2 Q2 ,
2.20
and that
P2 P4 Q1 P2 Q2 P4 ,
Pi2 Qi Pi Qi Pi ,
i 1, 2.
2.21
Since Q2t 0,
Q2 P4 Q22 P4 Q1−1 Q2t P4 Q1−t1 0,
2.22
and then P2 P4 P2 Q2 P4 Q1−1 0. From this, we can easily verify
s
P Therefore, if P s 0, then P4 P1s−1 0.
P1s
0
P4 P1s−1 P2s
.
2.23
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Journal of Applied Mathematics
3. Main Results
In this section, we will give the explicit expressions for P QD and P QD , under the
conditions P 2 Q P QP and Q2 P QP Q. Now, we begin with the following theorem.
Theorem 3.1. Let Q ∈ Cn×n with IndQ t and P ∈ Cn×n be nilpotent with P s 0. If P 2 Q P QP
and Q2 P QP Q, then
P QD s−1 QD
i1
−P i Qπ P
i0
s−2
i2
Pi
−1i i 1 QD
i0
s−1
s−2
i1
i2
QQD −P i QD
Qπ P QQD −1i i 1P i QD
.
i0
3.1
i0
Proof. If t 0, then Q is invertible, and therefore QP P Q. So, by Lemma 2.5, 3.1 holds.
Now assume that t > 0 and, without loss of generality, Q can be written as Q Q1 ⊕Q2 ,
where Q1 is invertible and Q2 is nilpotent of index t. So QD Q1−1 ⊕ 0. Since P 2 Q P QP and
Q2 P QP Q, we can write P , partitioned conformably with Q, by Lemma 2.6, as follows:
P
P1 0
P4 P 2
,
3.2
where P1 , P2 are nilpotent since P is nilpotent. We also write I I1 ⊕ I2 , partitioned
conformably with Q.
Since P1 is nilpotent and Q1 is invertible, by Lemma 2.5,
P1 Q1 −1 s−1
i0
Q1−i−1 −P1 i s−1
i0
−P1 i Q1−i−1 .
3.3
Also, the nilpotency of P2 , Q2 implies P2 Q2 D 0 by Lemma 2.3.
By 2.15, P2 Q2 P4 0. Hence, by Lemma 2.1, the argument above, and 2.14, we
have
P QD P1 Q1
0
P4
P2 Q2
D
⎛
−1
Q1−1 I1 Q1−1 P1 ⎝
−2
P4 Q1−2 I1 Q1−1 P1 ⎞
0
⎠.
0
3.4
By 3.3, it is easy to verify that
I1 Q1−1 P1
−2
s−1
s−1
−1i i 1Q1−i P1i −1i i 1P1i Q1−i .
i0
i0
3.5
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7
Since P s 0, we have P4 P1s−1 0 by Lemma 2.6, and, therefore, by 3.5,
−i2 i
s−2
s−2
i2
0 0
P1 0
P1 0
0
Q1
i
i
D
i
P Q P −1 i 1 Q
−1 i 1
0 I2
P4 P 2
∗ P2i
0
0
i0
i0
s−2
0
0
i
−1 i 1
−i2 i
P4 Q1
P1 0
i0
⎞
⎛
0
0
⎠.
⎝
−2
P4 Q1−2 I Q1−1 P1
0
3.6
π
Analogous to the argument above, we can see, by Lemma 2.5,
s−1 QD
i0
i1
−1 ⎞
−1
−1
Q
P
0
I
Q
1
1
1
⎠.
−P i ⎝ 1
0
0
⎛
3.7
Thus, putting 3.6 and 3.7 into 3.4 yields the first equation of 3.1.
Similar to the discussion of 3.6, we have
−1 ⎞
−1
−1
P
0
I
Q
Q
1
1
⎠,
QQD −P i QD
⎝ 1
0
0
i0
⎛
⎞
s−2
i2
0
0
Qπ P QQD −1i i 1P i QD
⎝
−2 ⎠,
−2
−1
P
Q
P
0
I
Q
i0
4 1
1
1
s−1
i1
⎛
3.8
and then putting them into 3.4 yields the second equation of 3.1.
The following theorem is our main result, and Theorem 3.1 and Lemma 2.5 can be
regarded as its special cases.
Theorem 3.2. Let P, Q ∈ Cn×n with IndP s ≥ 1 and IndQ t. If P 2 Q P QP and Q2 P QP Q. Then,
i
2
P QD P D QD P P D QD P D P QD P D ,
3.9
Q2 P D QP D Q,
3.10
PD
2
Q P D QP D .
3.11
8
Journal of Applied Mathematics
ii
s−1 i1
D
D 2 QD
−P i P π
P QD P D I P D Q P π Q P D I P D Q
i0
3.12
s−2
i2
Q P −1i i 1 QD
P iP π .
π
i0
Proof. If s 0, then P is invertible and P Q QP . So, by Lemmas 2.4 and 2.5, 3.9 and 3.12
hold, respectively. Therefore, assume that s > 0, and, without loss of generality, let P P1 ⊕P2 ,
where P1 is invertible and P2 is nilpotent of index s. From hypotheses, by Lemma 2.6, we can
write
Q
Q1
0
,
Q4 Q2
3.13
partitioned conformably with P , and those equations in Lemma 2.6 hold. By Lemma 2.1,
therefore, we have
⎛
Q ⎝
Q1D
D
Q4 Q1D 0
2
Q2D
⎞
⎠,
Q 2
Q12
0
Q4 Q1 Q22
.
3.14
i By 2.14 and 2.15,
D
Q12 P1−1
0
Q1 P1−1 Q1 0
QP D Q,
Q4 Q1 P1−1 0
Q4 P1−1 Q1 0
−1
−2
2
P1 Q1 0
P1 Q1 P1−1 0
D
P D QP D ,
P
Q
0
0
0
0
−1 D 2
P1 Q1D P1−2 0
P1 Q1 0
D
D
P D QD
PQ P
0
0
0
0
D −1 Q1 P1 0
P P D QD P D ,
0
0
Q1 P1 0
P1 Q1 0
.
PQ 0 P2 Q2
0 P2 Q2
Q P
2
3.15
3.16
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9
By 2.17 and Lemma 2.2, P2 Q2 s P2s Q2s 0. By Lemma 2.4 and 3.16, we have
D
P Q −1 D P1 Q1 0
Q1 P1 D
0
0
P2 Q2 D
0
0
P D QD .
3.17
As a result, 3.9 holds.
ii By Lemma 2.6, P2 Q2 Q4 0 and then, by Lemma 2.1, we have
P QD D
P1 Q1
0
Q4
P2 Q2
⎞
0
P1 Q1 D
⎠.
2
⎝ Q4 P1 Q1 D P2 Q2 D
⎛
3.18
By Lemma 2.5, we have
P
D
D
I P Q
D
P1−1 0
0
⎛
I1 P1−1 Q1 0
0
P −1
⎝ 1
0
D
P1−1 Q1
I1 0
D
I2
⎞ P1 Q1 D 0
0
⎠
0
0
0
3.19
and, therefore,
⎛
⎝
0
0
⎞
2 ⎠ D
Q4 P1 Q1 0
0 0
2 ⎞
⎛
D
Q
0⎠
P
1
⎝ 1
Q4 Q2
0
0
Q1
0 I2
0
D 2
P π Q P D I P DQ
.
3.20
By 3.1, we have
0
0
0 P2 Q2 D
⎛
⎜
⎜
⎝
0
0
s−1
i0
s−1 i0
⎞
0
QD
i1
Q2D
−P2 i
i1
Q2π P2
−P i P π Qπ P
s−2
i
−1 i 1
i0
⎟
⎟
i2 i ⎠
P2
Q2D
s−2
i2
P iP π .
−1i i 1 QD
i0
Thus, substituting 3.19, 3.21, and 3.20 in 3.18 yields 3.12.
Note that P Q QP implies P 2 Q P QP and Q2 P QP Q.
3.21
10
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Corollary 3.3 see 14, Theorem 2. If P, Q ∈ Cn×n with P Q QP and IndP s, then
s−1 i1
D
QD
P QD I P D Q P D P π
−P i .
3.22
i0
Proof. From 3.19 and Lemma 2.5, we can obtain
I P DQ
D
D
P D P D I P DQ .
3.23
Since P k Q 0 for some k, P D Q 0. Thus, we have the following corollary.
Corollary 3.4. Let P, Q ∈ Cn×n with IndP s ≥ 1 and IndQ t. Suppose P 2 Q P QP and
Q2 P QP Q. If there exist two positive integers k and h such that P k Q 0 and Qh P 0, then
2
P QD P D QD Q P D .
3.24
If Q is a perturbation of P , then, we have the following result in which P QD − P D 2 has
an upper bound. Before the theorem, let us recall that if A2 < 1, then I A is invertible and
I A−1 ≤
2
1
,
1 − A2
I − I A−1 ≤
2
3.25
A2
.
1 − A2
Theorem 3.5. Let P, Q ∈ Cn×n with IndP s ≥ 1 and IndQ t. Suppose P 2 Q P QP and
Q2 P QP Q. If P D Q2 < 1, then
D D 2
P P Q
P π 2 Q2 P D 2
D
2
2
D
P Q − P ≤
2
2
1 − P D Q2
1 − P D Q2
D
Q P π 1 − QD s P s
QD 2 Qπ P π P 2
2
2
2
2
2
2
2
2
1 − QD 2 P 2
1 − QD 2 P 2
s
s−1
D
s
× 1 − sQD P s−1
−
1
s
P
Q
2 .
2
2
3.26
2
Proof. Since P D Q2 < 1, I P D Q is invertible. Then by 3.12, we have
P QD − P D P D
s−1
i0
I P DQ
D i1
Q −1 2
− I P π Q P D I P DQ
−1
s−2
i2
P iP π .
−P P Q P −1i i 1 QD
i
π
π
i0
3.27
Journal of Applied Mathematics
11
In order to verity 3.26, we need to calculate the 2-norms of the right-hand side of the above
equation. By 3.25,
D D P P Q
−1
D
2
D
P
2,
−I I P Q
≤ 1−
D Q
P
2
2
2
−1 2 P π 2 Q2 P D 2
π
D
D
≤ P Q P I P Q
2 ,
1 − P D Q
2
2
s−1 s−1 i1
D i1
i π
D
Q
−P P ≤
Q P i2 P π 2
i0
2
i0
2
s D i
π
Q P i−1
2 P 2
i1
3.28
2
D
Q P π 1 − QD s P s
2
2
2
2
,
1 − QD P 2
2
s−2
s−2
i2
i2
π i
D
i π
π
π
PP ≤
i 1QD P i1
Q P −1 i 1 Q
2 Q 2 P 2
2
i0
i0
2
s−1 i1
iQD P i2 Qπ 2 P π 2 .
i1
Let q : QD 2 P 2 and S :
s−1
i1
2
iqi . Then,
s−1
q 1 − qs−1
q − sqs s − 1qs1
i
s
− s − 1qs .
1−q S
q − s − 1q 1−q
1−q
i1
3.29
Thus
s−1 i1
iQD P i2 Qπ 2 P π 2
i1
2
D 2 π
Q Q P π P 1 − sQD s−1 P s−1 s − 1QD s P s
2
2
2
2
2
2
2
2
.
2
1 − QD 2 P 2
By the above argument, we can get 3.26.
Finally, we give an example to illustrate our results.
3.30
12
Journal of Applied Mathematics
Example 3.6. Consider the matrices
⎛
1 0 0
⎜
⎜0 1 0
⎜
P ⎜
⎜0 0 0
⎝
0 0 0
⎞
0
⎟
0⎟
⎟
⎟,
1⎟
⎠
0
⎛1
⎞
0 0 0
⎜3
⎟
⎜
⎟
⎜0 0 0 0⎟
⎜
⎟
Q⎜
⎟.
⎜0 0 0 0⎟
⎜
⎟
⎝
⎠
1
0 0 0
3
3.31
We observe that P 2 Q P QP and Q2 P QP Q, but P Q /
QP . It is obvious that s IndP 2,
and
⎞
⎞
⎛
⎛
1 0 0 0
3 0 0 0
⎟
⎟
⎜
⎜
⎜0 1 0 0⎟
⎜0 0 0 0⎟
⎟
⎟
⎜
⎜
D
D
Q ⎜
3.32
P ⎜
⎟,
⎟.
⎜0 0 0 0⎟
⎜0 0 0 0⎟
⎠
⎠
⎝
⎝
0 0 0 0
0 0 0 3
Since P D Q2 1/3 < 1, I P D Q is invertible and
⎞
3
0
0
0
⎟
⎜4
⎟
⎜
⎜ 0 1 0 0⎟
⎟.
⎜
⎜
⎟
⎜ 0 0 1 0⎟
⎠
⎝
0 0 0 1
3.33
⎞
1
0
0
0
−
⎟
⎜ 4
⎟
⎜
⎜ 0 0 0 0⎟
⎟.
⎜
⎟
⎜
⎜ 0 0 0 9⎟
⎠
⎝
0 0 0 3
3.34
⎛
I P DQ
−1
By 3.12,
⎛
P QD − P D
√
We can compute P QD − P D 2 3 10. On the other hand, it is easy to get that P 2 P D 2 P π 2 Qπ 2 1, Q2 1/3, QD 2 3. By 3.26, we get the upper bound of
P QD − P D 2 is 161/4, it is bigger than and close to the exact norm.
Acknowledgments
The authors would like to thank the referees for their helpful comments and suggestions. This
work was supported by the National Natural Science Foundation of China 11061005, the
Ministry of Education Science and Technology Key Project under Grant 210164, and Grants
HCIC201103 of Guangxi Key Laboratory of Hybrid Computational and IC Design Analysis
Open Fund.
Journal of Applied Mathematics
13
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