Calcolo (2013) 50:305–311DOI 10.1007/s10092-012-0069-x

Some results on the Drazin inverse of a modified matrix

Dijana Mosic

Received: 15 May 2012 / Accepted: 22 August 2012 / Published online: 6 September 2012© Springer-Verlag 2012

Abstract We give new conditions under which the Drazin inverse of a modifiedmatrix A − C DD B can be expressed in terms of the Drazin inverse of A and itsgeneralized Schur complement Z = D − B ADC , generalizing some recent results inthe literature.

Keywords Drazin inverse · Sherman–Morrison–Woodbury formula ·Modified matrix

Mathematics Subject Classification (2010) 15A09 · 65F20

1 Introduction

The Drazin inverse is very useful, and the applications in singular differential ordifference equations, Markov chains, cryptography, iterative method and numericalanalysis can be found in [1,2].

Let Cn×n denote the set of n×n complex matrices. The Drazin inverse of A ∈ C

n×n

is the unique matrix AD satisfying the following three equations

AD AAD = AD, AAD = AD A, Ak+1 AD = Ak,

The author is supported by the Ministry of Science, Republic of Serbia, Grant No. 174007.

D. Mosic (B)Faculty of Sciences and Mathematics, University of Niš,P.O. Box 224, 18000 Niš, Serbiae-mail: dijana@pmf.ni.ac.rs

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306 D. Mosic

where k = ind(A) is the index of A. If ind(A) = 1, then the Drazin inverse of Ais reduced to the group inverse, denoted by A#. If ind(A) = 0, then AD = A−1. Inaddition, we denote Aπ = I − AAD , where I is the identity matrix with proper sizes.

The Sherman–Morrison–Woodbury formula (or SMW formula) [5,7] related tothe inverse of matrix, i.e. the formula (A − C D−1 B)−1 = A−1 + A−1C(D −B A−1C)−1 B A−1, is an useful computational tool in applications to statistics, net-works, structural analysis, asymptotic analysis, optimization and partial differentialequations.

Wei [6] studied the expressions of the Drazin inverse of a modified square matrixA − C B. Chen and Xu [3] discussed some representations for the weighted Drazininverse of a modified rectangular matrix A−C B under some conditions. These resultscan be applied to update finite Markov chains.

In [4] Dopazo and Martínez-Serrano presented some new formulae for the Drazininverse of the modified matrix A − C DD B in terms of the Drazin inverse of A andthe generalized Schur complement Z = D − B ADC of A under the cases:

(i) AπC = 0, C Dπ = 0, Dπ B = 0, Zπ B = 0 and C Zπ = 0;(ii) B Aπ = 0, Dπ B = 0, C Dπ = 0, Zπ B = 0 and C Zπ = 0.

So, they extended the Sherman–Morrison–Woodbury formula to the case where A,A − C DD B and Z = D − B ADC are not necessarily invertible.

In this paper, we consider the Drazin inverse of a modified matrix A − C DD Bin terms of the Drazin inverse of A and the Drazin inverse of its generalized Schurcomplement. Also we develop conditions which are weaker than conditions (i) and(ii) and, as a consequence, we obtain formulae for the Drazin inverse of a modifiedmatrix of Dopazo and Martínez-Serrano in [4].

2 The Drazin inverse of a modified matrix

We study the Drazin inverse of a modified matrix A−C DD B under certain conditionsdifferent of the conditions presented in [4]. Thus, the Drazin inverse of the matrixA − C DD B is expressed in terms of generalized SMW formula.

Theorem 1 Let A, B, C, D ∈ Cn×n, and let ind(A) = k. If Z = D − B ADC,

AπC = C Dπ , Dπ B = 0 and DZπ = 0,

then

(A − C DD B)D = AD + ADC Z D B AD

−k−1∑

i=0

(AD + ADC Z D B AD

)i+1ADC Z D B Ai Aπ (1)

and ind(A − C DD B) ≤ ind(A).

Proof Suppose that S = AD + ADC Z D B AD . The assumptions AπC = C Dπ andDZπ = 0 imply AADC = C DDD and DD Zπ = 0. Now we can obtain that

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Some results on the Drazin inverse of a modified matrix 307

(A − C DD B)S = AAD + AADC Z D B AD − C DD B AD − C DD(D − Z)Z D B AD

= AAD + (AADC − C DDD)Z D B AD − C DD Zπ B AD

= AAD. (2)

Since DDD B = B, ADC = ADC DD , and DZπ = 0, then

S(A − C DD B) = AD A − ADC Z D B Aπ − ADC Zπ DD B

= AD A − ADC Z D B Aπ − ADC DDD Zπ DD B

= AD A − ADC Z D B Aπ . (3)

Denote by X the right side of (1). Notice that, by AπC DD =C Dπ DD =0, (2) and (3),

X (A − C DD B) = S(A − C DD B) −k−1∑

i=0

Si+1 ADC Z D B Ai+1 Aπ

= AD A −k−1∑

i=0

Si ADC Z D B Ai Aπ (4)

and

(A − C DD B)X = (A − C DD B)S − (A − C DD B)Sk−1∑

i=0

Si ADC Z D B Ai Aπ

= AD A −k−1∑

i=0

Si ADC Z D B Ai Aπ . (5)

Thus, by the equalities (4) and (5), X (A − C DD B) = (A − C DD B)X .From (4) and Aπ X = 0, we have

X (A − C DD B)X =(

AD A −k−1∑

i=0

Si ADC Z D B Ai Aπ

)X = AD AX = X.

Using again (5) and AπC DD = 0,

(A − C DD B) − (A − C DD B)2 X = (I − (A − C DD B)X)(A − C DD B)

= AAπ − AπC DD B+k−1∑

i=0

Si ADC Z D B Ai+1 Aπ

= AAπ +k−1∑

i=0

Si ADC Z D B Ai+1 Aπ . (6)

By induction on integer n ≥ 1, we will get

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308 D. Mosic

((A − C DD B) − (A − C DD B)2 X

)n = An Aπ +k−1∑

i=0

Si ADC Z D B Ai+n Aπ . (7)

Indeed, for n = 1 the equality (7) is true, by (6). Now, we will assume that it holds forn and let us prove that it holds for n + 1. Applying (6), (7), Aπ S = 0 and Aπ AD = 0,we have

((A − C DD B) − (A − C DD B)2 X

)n+1 =(

An Aπ +k−1∑

i=0

Si ADC Z D B Ai+n Aπ

)

×(

AAπ +k−1∑

i=0

Si ADC Z D B Ai+1 Aπ

)

= An+1 Aπ +k−1∑

i=0

Si ADC Z D B Ai+n+1 Aπ .

Consequently, for k = ind(A),((A − C DD B) − (A − C DD B)2 X

)k = 0 whichgives (A − C DD B)k+1 X = (A − C DD B)k and ind(A − C DD B) ≤ ind(A). Hence,(A − C DD B)D = X .

We can check that AπC = C Dπ ⇔ AπC DD = 0 and ADC Dπ = 0 ⇔ AπC Dl =0 and AkC Dπ = 0, where k = ind(A) and l = ind(D). So, we can replace thecondition AπC = C Dπ in Theorem 1 with some of these equivalent conditions.

In the next result, we state the hypothesis AπC D = 0 and AC Dπ = 0 instead ofAπC = C Dπ in Theorem 1 and we obtain the same formula for the Drazin inverse(A − C DD B)D .

Corollary 1 Let A, B, C, D ∈ Cn×n, and let ind(A) = k. If Z = D − B ADC,

AπC D = 0, AC Dπ = 0, Dπ B = 0 and DZπ = 0,

then (A − C DD B)D is defined as in (1) and ind(A − C DD B) ≤ ind(A).

As we can see, the assumptions AπC D = 0 and AC Dπ = 0 imply AπC = C Dπ ,but the converse is not true obviously.

In the same way as in the proof of Theorem 1, we can show the following theorem.Observe that the conditions of the next theorem are weaker than conditions (i).

Theorem 2 Let A, B, C, D ∈ Cn×n, and let ind(A) = k. If Z = D − B ADC,

AπC = C Dπ , Dπ B = 0, Zπ B = 0 and C Zπ = 0,

then (A − C DD B)D is defined as in (1) and ind(A − C DD B) ≤ ind(A).

Proof Let X be the right side of (1), and let S = AD + ADC Z D B AD . Using AADC =C DDD and Zπ B = 0, we get

(A−C DD B)S = AAD +(AADC − C DDD)Z D B AD −C DD Zπ B AD = AAD . (8)

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Some results on the Drazin inverse of a modified matrix 309

By the hypothesis C Zπ = 0, observe that

S(A − C DD B) = AD A − ADC Z D B Aπ − ADC Zπ DD B

= AD A − ADC Z D B Aπ . (9)

The equalities AπC DD = C Dπ DD = 0, (8) and (9) imply

X (A − C DD B) = (A − C DD B)X = AD A −k−1∑

i=0

Si ADC Z D B Ai Aπ .

The rest of proof follows in the same way as in the proof of Theorem 1.

If we add that AπC = C Dπ = 0 in Theorem 2, we obtain [4, Theorem 2.1] as aconsequence.

Observe that the conditions AπC = 0, C Dπ = 0, Dπ B = 0 and DZπ = 0, implyC Zπ = 0 and Zπ B = 0.

The following theorem concerning the another formula for (A − C DD B)D can beproved in the similar way as in Theorem 1.

Theorem 3 Let A, B, C, D ∈ Cn×n, and let ind(A) = k. If Z = D − B ADC,

B Aπ = Dπ B, C Dπ = 0 and Zπ D = 0,

then

(A − C DD B)D = AD + ADC Z D B AD

−k−1∑

i=0

Aπ Ai C Z D B AD(

AD + ADC Z D B AD)i+1

(10)

and ind(A − C DD B) ≤ ind(A).

Remark (a) The assumption B Aπ = Dπ B in Theorem 3 can be replaced with equiv-alent conditions DD B Aπ = 0 = Dπ B AD or Dl B Aπ = 0 = Dπ B Ak , wherek = ind(A) and l = ind(D).

(b) We can state in Theorem 3 the conditions DB Aπ = 0 = Dπ B A instead ofB Aπ = Dπ B. It is clear that DB Aπ = 0 = Dπ B A imply B Aπ = Dπ B.

(c) If we assume that B Aπ = Dπ B, C Zπ = 0, Zπ B = 0 and C Dπ = 0, we can getthe representation for (A − C DD B)D as in (10), generalizing [4, Theorem 2.2].

3 Special cases

We can use the results proved in Sect. 2 to consider some interesting cases in thissection.

Applying Theorem 1 (or Corollary 1 or Theorem 2) with an additional condition,we get the generalized SMW formula for the Drazin inverse.

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310 D. Mosic

Corollary 2 Let A, B, C, D ∈ Cn×n be matrices such that the conditions of Theorem

1 (or Corollary 1 or Theorem 2) are satisfied. If ADC DD B Aπ = 0, then

(A − C DD B)D = AD + ADC Z D B AD. (11)

Proof From B = DDD B, ADC = ADC DDD and DD Z D Z = DD , we get

ADC Z D B Aπ = ADC Z D DDD B Aπ = ADC Z D(Z + B ADC)DD B Aπ

= ADC D(DD Z D Z)DD B Aπ + ADC Z D B ADC DD B Aπ

= ADC DD B Aπ + ADC Z D B ADC DD B Aπ = 0.

Using this equality, the formula (1) become (11).

Adding the condition AπC DD B AD = 0 in Theorem 3, we obtain the formula (11).If we suppose that D = I and Z = I − B ADC is nonsingular in Theorem 1 and

Theorem 3, we get the following corollary.

Corollary 3 Let A, B, C, D ∈ Cn×n, Z = I − B ADC is nonsingular and let

ind(A) = k.

(a) If AπC = 0, then (A − C B)D is equal to the right hand side of (1);(b) If B Aπ = 0, then (A − C B)D is equal to the right hand side of (10);(c) [6, Corollary 2.1] If AπC = 0 and B Aπ = 0, then (A − C B)D is equal to the

right hand side of (11);

where Z D = Z−1.

By theorems proved in Sect. 2, an extra condition ind(A) = 1 gives that the groupinverse of A − C DD B exists and the next formulae for (A − C DD B)# follow.

Corollary 4 Let A, B, C, D ∈ Cn×n, and let ind(A) = 1.

(a) If the conditions of Theorem 1 (or Corollary 1 or Theorem 2) are satisfied, then

(A − C DD B)# = (A# + A#C Z D B A#)(I − A#C Z D B Aπ ).

(b) If the conditions of Theorem 3 are satisfied, then

(A − C DD B)# = (I − AπC Z D B A#)(A# + A#C Z D B A#).

Note that in Corollary 2 and 4 we obtain the same formula as in [4, Corollary 2.3 andTheorem 4.1] but under different conditions.

In the end of this section, an example is given to illustrate our results. The followingexample describes a 2 × 2 matrices A, B, C and D which do not satisfy the conditionsof [4, Theorem 4.1], whereas the conditions of Theorem 1[or Corollary 4(a)] are met,which allows us to compute (A − C DD B)#.

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Some results on the Drazin inverse of a modified matrix 311

Example 3.1 Consider a 2 × 2 block matrices A =[

1 00 0

], B = D =

[1 10 0

]and

C =[

0 00 1

]. Notice that ind(A) = 1 and ind(D) = 1. Then Aπ =

[0 00 1

], Dπ =

[0 −10 1

]and Z = D. Since AπC = C Dπ = C �= 0, the mentioned results from [4]

fail to apply. It is evident that Dπ B = 0 and DZπ = 0, so we can apply Theorem 1[(or Corollary 4(a)] obtaining

(A − C DD B)# =[

1 00 0

].

Acknowledgments The author is grateful to the referee for constructive comments towards improvementof the original version of this paper.

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Comput. 203, 202–209 (2008)4. Dopazo, E., Martínez-Serrano, M.F.: On deriving the Drazin inverse of a modified matrix. Linear Algebra

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