The Explicit Expression of the Drazin Inverse and Its Application

Abstract

We give explicit expressions of (𝑃±𝑄)_𝑑 of two matrices 𝑃 and 𝑄, in terms of 𝑃, 𝑄, 𝑃_𝑑, and 𝑄_𝑑, (𝑃±𝐼)_𝑑, under the condition that 𝑃𝑄=𝑃, and apply the result to finding an explicit representation for the Drazin inverse of some 2×2 block matrix.

1. Introduction

In recent years, the representations and perturbation analysis of the Drazin inverse for matrices or operators have been investigated (see [1–6]). In [7], the author presented the presentations of the Drazin inverse of sum and product of two operators over Banach spaces on the condition of the commutativity up to a factor. And, in [8], the same author discussed explicit representations of Drazin inverses of sums and differences of two idempotents over Hilbert spaces.

These investigations motivate us to deal with an explicit expression of the Drazin inverse of differences and sums of two matrices. The paper is organized as follows. In this section, we will introduce some notions and lemmas. In Section 2, we will present these explicit expressions of differences and sums of two matrices 𝑃 and 𝑄 under the condition 𝑃𝑄=𝑃. In Section 3, we will deduce an explicit representation for the Drazin inverse of the 2×2 block matrix 𝐴𝐵𝐶𝐷 with 𝐴=𝐵𝐶 and 𝐵=𝐵𝐷 in terms of its subblocks and their Drazin inverses and (𝐴+𝐼)_𝑑. In Section 4, we will present a numerical example to demonstrate the main result in Section 2.

Throughout this paper the symbol ℂ^𝑚×𝑛 stands for the set of 𝑚×𝑛 complex matrices, and 𝐼∈ℂ^𝑛×𝑛 stands for the unit matrix. Let 𝐴∈ℂ^𝑛×𝑛; the Drazin inverse, denoted by 𝐴_𝑑, of matrix 𝐴 is defined as the unique matrix satisfying 𝐴^𝑘+1𝐴_𝑑=𝐴^𝑘,𝐴_𝑑𝐴𝐴_𝑑=𝐴_𝑑,𝐴𝐴_𝑑=𝐴_𝑑𝐴, (1) where 𝑘=Ind(𝐴) is the index of 𝐴. In particular, if Ind(𝐴)=1, then 𝐴_𝑑 is called the group inverse, denoted by 𝐴_𝑔, of 𝐴. Apparently, if 𝐴 is nonsingular, then Ind(𝐴)=0; otherwise, Ind(𝐴)≥1, especially Ind(0)=1. If 𝐴 is nilpotent, then 𝐴_𝑑=0. If 𝑋 is nonsingular and 𝐵=𝑋𝐴𝑋⁻¹, then 𝐵_𝑑=𝑋𝐴_𝑑𝑋⁻¹ (see [6, 9, 10]). For convenience, we write 𝐴^𝜋=𝐼−𝐴𝐴_𝑑 and use the convention ∑^𝑚_𝑖=𝑛=0, if 𝑚<𝑛.

Before we start the discussion, we need some preparations.

Lemma 1 (see [11, Theorem 3.2]). Let 𝑀=^𝐴𝐵_0𝐷 with 𝑡=Ind(𝐴) and 𝑙=Ind(𝐷). Then, 𝑀_𝑑=𝐴_𝑑𝑆0𝐷_𝑑, (2) where 𝑆=𝐴^𝜋𝑡−1_𝑛=0𝐴^𝑛𝐵𝐷_𝑑^𝑛+2+^𝑙−1_𝑛=0𝐴_𝑑^𝑛+2𝐵𝐷^𝑛𝐷^𝜋−𝐴_𝑑𝐵𝐷_𝑑. (3)

Lemma 2. Let 𝑃,𝑄∈ℂ^𝑚×𝑚 with Ind(𝑄)=𝑠. If 𝑃𝑄=𝑃, then 𝑃𝑄_𝑑𝑄=𝑃,_𝑑𝑄𝑃^𝑘=𝑄_𝑑𝑄𝑃^𝑘,𝑄𝑃^𝜋𝑘=𝑄^𝑘𝑃^𝜋,𝑄_𝑑𝑃^𝜋𝑘=𝑄^𝑘_𝑑𝑃^𝜋,𝑘≥1. (4)

Proof. Since 𝑃𝑄=𝑃 and Ind(𝑄)=𝑠, 𝑃=𝑃𝑄^𝑠=𝑃𝑄^𝑠+1𝑄_𝑑=𝑃𝑄_𝑑, (5) and then 𝑃𝑄𝑄_𝑑=𝑃. So from this, by induction, it follows that (𝑄_𝑑𝑄𝑃)^𝑘=𝑄_𝑑𝑄𝑃^𝑘 for 𝑘≥1.
Now, we will show inductively that (𝑄𝑃^𝜋)^𝑘=𝑄^𝑘𝑃^𝜋. It holds for 𝑘=1. Assume it holds for 𝑘=𝑛; that is, (𝑄𝑃^𝜋)^𝑛=𝑄^𝑛𝑃^𝜋. Then, 𝑄𝑃^𝜋𝑛+1=𝑄𝑃^𝜋𝑄𝑃^𝜋𝑛=𝑄𝐼−𝑃_𝑑𝑃𝑄^𝑛𝑃^𝜋=𝑄^𝑛+1𝑃^𝜋. (6) So it holds for any 𝑘≥1.
From 𝑃𝑄_𝑑=𝑃, we can similarly show that (𝑄_𝑑𝑃^𝜋)^𝑘=𝑄^𝑘_𝑑𝑃^𝜋.

Lemma 3. Let 𝑃 be nilpotent of index 𝑡>1 and 𝑆=∑^𝑡−1_𝑖=0𝑎_𝑖^[1]𝑃^𝑖. If 𝑎_𝑖^[1]=1, then 𝑆^𝑛=^𝑡−1_𝑖=0𝑎_𝑖^[𝑛]𝑃^𝑖,𝑛≥2, (7) where 𝑎_𝑖^[𝑛]=∑^𝑖_𝑢=0𝑎_𝑢^[𝑛−1], 𝑖=0,…,𝑡−1.

Proof. Since 𝑃^𝑡=0, we can easily write 𝑆^𝑛 as 𝑆^𝑛=^𝑡−1_𝑖=0𝑎_𝑖^[𝑛]𝑃^𝑖,𝑛≥2. (8) We will prove the relationship 𝑎_𝑖^[𝑛]=^𝑖_ℎ=0𝑎_ℎ^[𝑛−1],𝑖=0,…,𝑡−1 (9) by induction on 𝑛. Obviously, (9) holds for 𝑛=2. Assume inductively that (9) holds for 𝑛=𝑘.
Since 𝑎_𝑖^[1]=1 and 𝑃^𝑡=0, 𝑆^𝑘+1=𝑆^𝑘𝑆=^𝑡−1_𝑖=0𝑎_𝑖^[𝑘]𝑃^𝑖𝑡−1_𝑗=0𝑃^𝑗=^𝑡−1_{ℎ=0𝑖+𝑗=ℎ}𝑎_𝑖^[𝑘]𝑃^𝑖+𝑗=^𝑡−1ℎ_ℎ=0𝑖=0𝑎_𝑖^[𝑘]𝑃^ℎ. (10) On the other hand, 𝑆^𝑘+1=^𝑡−1_𝑖=0𝑎_𝑖^[𝑘+1]𝑃^𝑖. (11) So, (9) holds for 𝑛=𝑘+1. Hence, (9) holds for 𝑛≥2.

Lemma 4 (see [9, Lemma 7.7.2]). Let 𝑁=𝐴𝐵00𝑜𝑟𝑁=00𝐵𝐴. (12) Then, Ind(𝐴)≤Ind(𝑁)≤Ind(𝐴)+1.

2. The Drazin Inverse of Differences and Sums of Two Matrices

In this section, we will investigate how to express 𝑃±𝑄 as a function of 𝑃, 𝑄, 𝑃_𝑑, and 𝑄_𝑑, (𝑃±𝐼)_𝑑, under the condition 𝑃𝑄=𝑃. We begin with the following theorem, in which 𝑃 is assumed to be nilpotent.

Theorem 5. Let 𝑃,𝑄∈ℂ^𝑛×𝑛 with Ind(𝑄)=𝑠. If 𝑃𝑄=𝑃 and 𝑃 is nilpotent of index 𝑡, then (𝑃−𝑄)_𝑑=^𝑠−1_𝑛=0^𝑡−2_𝑖=0𝑎_𝑖^[𝑛+2]𝐼−𝑎_𝑖^[𝑛+1]𝑄_𝑑𝑄^𝑛𝑃^𝑖+1−𝑄_𝑑𝑄^𝑠𝑡−2_𝑖=0𝑎_𝑖^[𝑠+1]𝑃^𝑖+1−𝑄_𝑑, (13) where 𝑎_𝑖^[𝑛]=∑^𝑖_𝑘=0𝑎_𝑘^[𝑛−1] and 𝑎_𝑖^[1]=1, 𝑖=0,…,𝑡−1.

Proof. If 𝑡=1, then, from the convention in Section 1, (13) clearly holds. Now assume 𝑡>1. From 𝑠=Ind(𝑄), there exists a nonsingular matrix 𝑊₁ such that 𝑄=𝑊₁𝑄₁00𝑄₂𝑊₁⁻¹,𝑄_𝑑=𝑊₁𝑄₁⁻¹0𝑊00¹⁻¹, (14) where 𝑄₁ is nonsingular and 𝑄₂ is nilpotent of index 𝑠. Partitioning 𝑊₁⁻¹𝑃𝑊₁ conformably with 𝑊₁⁻¹𝑄𝑊₁, we have 𝑃=𝑊₁𝑃₁𝑃₄𝑃₃𝑃₂𝑊₁⁻¹. (15) Since 𝑃𝑄=𝑃, 𝑃𝑄_𝑑=𝑃 by Lemma 2, and then 𝑃₂=0, 𝑃₄=0, and 𝑃_𝑖𝑄₁=𝑃_𝑖, 𝑖=1,3. Let 𝑡 be the index of nilpotent matrix 𝑃. Then, 𝑃^𝑡=𝑊₁𝑃^𝑡₁0𝑃₃𝑃₁^𝑡−10𝑊₁⁻¹=0, (16) and therefore 𝑃₁ is also nilpotent, and (𝑃₁−𝐼)⁻¹=−∑^𝑡−1_𝑖=0𝑃^𝑖₁. Thus, 𝑃−𝑄=𝑊₁𝑃₁−𝑄₁0𝑃₃−𝑄₂𝑊₁⁻¹=𝑊₁𝑃₁−𝐼𝑄₁0𝑃₃−𝑄₂𝑊₁⁻¹. (17) By Lemma 1, we have (𝑃−𝑄)_𝑑=𝑊₁𝑄₁⁻¹𝑃₁−𝐼⁻¹0𝑊𝑋0¹⁻¹=𝑊₁−𝑄∑₁^−1𝑡−1_𝑖=0𝑃^𝑖₁0𝑊𝑋0¹⁻¹, (18) where 𝑋=^𝑠−1_𝑛=0(−1)^𝑛𝑄^𝑛₂𝑃₃𝑃₁−𝐼𝑄₁^−(𝑛+2). (19) By Lemma 3, (−1)^𝑛𝑃₃𝑃₁−𝐼𝑄₁^−(𝑛+2)=𝑃₃𝑄₁^−1𝑡−1_𝑖=0𝑃^𝑖₁^𝑛+2=𝑃₃^𝑡−1_𝑖=0𝑃^𝑖₁^𝑛+2=𝑃₃^𝑡−1_𝑖=0𝑎_𝑖^[𝑛+2]𝑃^𝑖₁=^𝑡−2_𝑖=0𝑎_𝑖^[𝑛+2]𝑃₃𝑃^𝑖₁. (20) Thus, 𝑊₁𝑊00𝑋0¹⁻¹=^𝑠−1_𝑛=0^𝑡−2_𝑖=0𝑊₁𝑎00_𝑖[𝑛+2]𝑄^𝑛₂𝑃₃𝑃^𝑖₁0𝑊₁⁻¹=^𝑠−1_𝑛=0^𝑡−2_𝑖=0𝑎_𝑖^[𝑛+2]𝑄^𝜋𝑄^𝑛𝑃^𝑖+1. (21) So, by (18), (𝑃−𝑄)_𝑑=−^𝑡−1_𝑖=0𝑄_𝑑𝑃^𝑖+𝑄^𝜋𝑠−1_𝑛=0^𝑡−2_𝑖=0𝑎_𝑖^[𝑛+2]𝑄^𝑛𝑃^𝑖+1=−^𝑡−2_𝑖=0𝑄_𝑑𝑃^𝑖+1−𝑄_𝑑+^𝑠−1_𝑛=0^𝑡−2_𝑖=0𝑎_𝑖^[𝑛+2]𝑄^𝑛𝑃^𝑖+1−𝑄_𝑑^𝑠_𝑛=1^𝑡−2_𝑖=0𝑎_𝑖^[𝑛+1]𝑄^𝑛𝑃^𝑖+1=−𝑄_𝑑+^𝑠−1_𝑛=0^𝑡−2_𝑖=0𝑎_𝑖^[𝑛+2]𝑄^𝑛𝑃^𝑖+1−𝑄_𝑑^𝑠_𝑛=0^𝑡−2_𝑖=0𝑎_𝑖^[𝑛+1]𝑄^𝑛𝑃^𝑖+1=^𝑠−1_𝑛=0^𝑡−2_𝑖=0𝑎_𝑖^[𝑛+2]𝐼−𝑎_𝑖^[𝑛+1]𝑄_𝑑𝑄^𝑛𝑃^𝑖+1−𝑄_𝑑𝑄^𝑠𝑡−2_𝑖=0𝑎_𝑖^[𝑠+1]𝑃^𝑖+1−𝑄_𝑑. (22)

If 𝑃𝑄=𝑃, then (−𝑃)𝑄=−𝑃. So, it immediately follows from the above theorem.

Corollary 6. Let 𝑃,𝑄∈ℂ^𝑛×𝑛 with Ind(𝑄)=𝑠. If 𝑃𝑄=𝑃 and 𝑃 is nilpotent of index 𝑘, then (𝑃+𝑄)_𝑑=^𝑠−1_𝑛=0^𝑡−2_𝑖=0(−1)^𝑖𝑎_𝑖^[𝑛+2]𝐼−𝑎_𝑖^[𝑛+1]𝑄_𝑑𝑄^𝑛𝑃^𝑖+1−𝑄_𝑑𝑄^𝑠𝑡−2_𝑖=0(−1)^𝑖𝑎_𝑖^[𝑠+1]𝑃^𝑖+1+𝑄_𝑑, (23) where 𝑎_𝑖^[𝑛]=∑^𝑖_𝑘=0𝑎_𝑘^[𝑛−1], 𝑎_𝑖^[1]=1, 𝑖=0,…,𝑡−1.

If the nilpotency of 𝑃 is taken out in Theorem 5, then we can obtain our main result, a more general result.

Theorem 7. Let 𝑃,𝑄∈ℂ^𝑛×𝑛 with Ind(𝑄𝑃^𝜋)=𝑠, Ind(𝑃)=𝑡, Ind[(𝑃−𝐼)𝑃𝑃_𝑑]=𝑙, and Ind[(𝑃−𝑄)𝑃^𝜋]=𝑘. If 𝑃𝑄=𝑃, then (𝑃−𝑄)_𝑑=^𝑠−1_𝑛=0^𝑡−2_𝑖=0𝑎_𝑖^[𝑛+2]𝐼−𝑎_𝑖^[𝑛+1]𝑄_𝑑𝑄^𝑛𝑃^𝑖+1𝑃^𝜋−𝑄_𝑑𝑃^𝜋−^𝑘−1_𝑛=0(−1)^𝑛𝑄^𝑛𝑄^𝜋(𝑄−𝐼)𝑃𝑃_𝑑(𝑃−𝐼)_𝑑^𝑛+2+𝑄^𝜋+𝑄_𝑑𝑃𝑃_𝑑(𝑃−𝐼)_𝑑+^𝑙−1_𝑛=0(−1)^𝑛+1𝑄^𝑛_𝑑𝑄_𝑑−𝑄²_𝑑𝑃𝑃_𝑑(𝑃−𝐼)^𝑛(𝑃−𝐼)^𝜋−𝑄_𝑑𝑄^𝑠𝑡−2_𝑖=0𝑎_𝑖^[𝑠+1]𝑃^𝜋𝑃^𝑖+1, (24) where 𝑎_𝑖^[𝑛]=∑^𝑖_𝑘=0𝑎_𝑘^[𝑛−1], 𝑎_𝑖^[1]=1, 𝑖=0,…,𝑡−1.

Proof. There exists a nonsingular matrix 𝑊₁ such that 𝑃=𝑊₁𝑃₁00𝑃₂𝑊₁⁻¹,𝑄=𝑊₁𝑄₁𝑄₃𝑄₄𝑄₂𝑊₁⁻¹, (25) where 𝑃₁ is nonsingular and 𝑃₂ is nilpotent of index 𝑡. From 𝑃𝑄=𝑃, we get 𝑄=𝑊₁𝑄𝐼0₄𝑄₂𝑊₁⁻¹, (26) where 𝑃₂𝑄₄=0 and 𝑃₂𝑄₂=𝑃₂. So 𝑃−𝑄=𝑊₁𝑃₁−𝐼0−𝑄₄𝑃₂−𝑄₂𝑊₁⁻¹. (27) By Lemma 1, we have (𝑃−𝑄)_𝑑=𝑊₁𝑃₁−𝐼_𝑑0𝑋𝑃₂−𝑄_2𝑑𝑊₁⁻¹, (28) where 𝑋=^𝑙−1_𝑛=0𝑃₂−𝑄_2𝑑^𝑛+2−𝑄₄𝑃₁−𝐼^𝑛𝑃₁−𝐼^𝜋+𝑃₂−𝑄₂^𝜋𝑘−1_𝑛=0𝑃₂−𝑄₂^𝑛−𝑄₄𝑃₁−𝐼_𝑑^𝑛+2−𝑃₂−𝑄_2𝑑−𝑄₄𝑃₁−𝐼_𝑑, (29) and 𝑘=Ind(𝑃₂−𝑄₂)=Ind[(𝑃−𝑄)𝑃^𝜋] and 𝑙=Ind(𝑃₁−𝐼)=Ind[(𝑃−𝐼)𝑃𝑃_𝑑].
Since 𝑃₂𝑄₄=0 and 𝑃₂𝑄₂=𝑃₂, by Theorem 5 and Lemma 2, 𝑃₂−𝑄_2𝑑𝑄₂^𝑗_𝑑𝑄₄=−^𝑡−1_𝑛=0𝑄_2𝑑𝑃^𝑛₂𝑄₂^𝑗_𝑑𝑄₄+^𝑠−1_𝑛=0𝑄^𝜋₂𝑄^𝑛₂𝑃₂^𝑡−2_𝑖=0𝑃^𝑖₂^𝑛+2𝑄₂^𝑗_𝑑𝑄₄=−𝑄_2𝑑^𝑗+1𝑄₄𝑃,𝑗≥0,₂−𝑄₂^𝜋𝑄₄=𝑄₄+𝑃₂−𝑄₂𝑄_2𝑑𝑄₄=𝑄₄−𝑄₂𝑄_2𝑑𝑄₄=𝑄^𝜋₂𝑄₄,𝑃₂𝑄^𝜋₂=0, (30) and then 𝑃₂−𝑄₂^𝜋𝑘−1_𝑛=0𝑃₂−𝑄₂^𝑛−𝑄₄=−^𝑘−1_𝑛=0𝑃₂−𝑄₂^𝑛𝑄^𝜋₂𝑄₄=−^𝑘−1_𝑛=0−𝑄₂^𝑛𝑄^𝜋₂𝑄₄. (31) Thus, 𝑋=^𝑙−1_𝑛=0(−1)^𝑛+1𝑄_2𝑑^𝑛+2𝑄₄𝑃₁−𝐼^𝑛𝑃₁−𝐼^𝜋−^𝑘−1_𝑛=0(−1)^𝑛𝑄^𝑛₂𝑄^𝜋₂𝑄₄𝑃₁−𝐼_𝑑^𝑛+2−𝑄_2𝑑𝑄₄𝑃₁−𝐼_𝑑. (32)
Since 𝑃−𝐼=𝑊₁𝑃₁−𝐼00𝑃₂𝑊−𝐼¹⁻¹, (33) we have that, for 𝑛≥0, 𝑃𝑃_𝑑(𝑃−𝐼)^𝑛_𝑑=𝑊₁𝑃₁−𝐼^𝑛_𝑑0𝑊00¹⁻¹,(𝑃−𝐼)^𝑛𝑃𝑃_𝑑(𝑃−𝐼)^𝜋=𝑊₁𝑃₁−𝐼^𝑛𝑃₁−𝐼^𝜋0𝑊00¹⁻¹. (34) Obviously, 𝑄𝑃^𝜋_𝑑=𝑊₁0𝑄00_2𝑑𝑊₁⁻¹=𝑄_𝑑𝑃^𝜋,(𝑄−𝐼)𝑃𝑃_𝑑=𝑊₁𝑄00₄0𝑊₁⁻¹,𝑌^𝑛𝑃^𝜋=𝑊₁000𝑌^𝑛₂𝑊₁⁻¹,𝑛≥0, (35) where the symbol 𝑌 denotes 𝑄 or 𝑃. Also, 𝑃^𝜋𝑄𝑃^𝜋𝜋=𝑃^𝜋−𝑃^𝜋𝑄𝑃^𝜋𝑄_𝑑𝑃^𝜋=𝑃^𝜋−𝑄𝑄_𝑑𝑃^𝜋=𝑄^𝜋𝑃^𝜋. (36) Note that 𝑃^𝜋(𝑄−𝐼)=(𝑄−𝐼). Then, by (32), 𝑊₁𝑊00𝑋0¹⁻¹=^𝑙−1_𝑛=0(−1)^𝑛+1𝑄_𝑑^𝑛+2𝑃^𝜋(𝑄−𝐼)𝑃𝑃_𝑑(𝑃−𝐼)^𝑛𝑃𝑃_𝑑(𝑃−𝐼)^𝜋−^𝑘−1_𝑛=0(−1)^𝑛𝑄^𝑛𝑃^𝜋𝑄𝑃^𝜋𝜋(𝑄−𝐼)𝑃𝑃_𝑑(𝑃−𝐼)_𝑑^𝑛+2−𝑄_𝑑𝑃^𝜋(𝑄−𝐼)𝑃𝑃_𝑑(𝑃−𝐼)_𝑑=^𝑙−1_𝑛=0(−1)^𝑛+1𝑄_𝑑^𝑛+2(𝑄−𝐼)𝑃𝑃_𝑑(𝑃−𝐼)^𝑛(𝑃−𝐼)^𝜋−^𝑘−1_𝑛=0(−1)^𝑛𝑄^𝑛𝑄^𝜋(𝑄−𝐼)𝑃𝑃_𝑑(𝑃−𝐼)_𝑑^𝑛+2−𝑄_𝑑(𝑄−𝐼)𝑃𝑃_𝑑(𝑃−𝐼)_𝑑. (37)
Note that 𝑃^𝜋𝑄^𝑛𝑃^𝜋=𝑄^𝑛𝑃^𝜋. So, by Theorem 5, 𝑊₁0𝑃00₂−𝑄_2𝑑𝑊₁⁻¹=^𝑠−1_𝑛=0^𝑡−2_𝑖=0𝑎_𝑖^[𝑛+2]𝐼−𝑎_𝑖^[𝑛+1]𝑄_𝑑𝑃^𝜋𝑄^𝑛𝑃^𝜋𝑃^𝑖+1𝑃^𝜋−𝑄_𝑑𝑃^𝜋−𝑄_𝑑𝑃^𝜋𝑄^𝑠𝑃^𝜋𝑡−2_𝑖=0𝑎_𝑖^[𝑠+1]𝑃^𝑖+1𝑃^𝜋=^𝑠−1_𝑛=0^𝑡−2_𝑖=0𝑎_𝑖^[𝑛+2]𝐼−𝑎_𝑖^[𝑛+1]𝑄_𝑑𝑄^𝑛𝑃^𝑖+1𝑃^𝜋−𝑄_𝑑𝑃^𝜋−𝑄_𝑑𝑄^𝑠𝑡−2_𝑖=0𝑎_𝑖^[𝑠+1]𝑃^𝜋𝑃^𝑖+1. (38) Hence, putting (34), (37), and (38) into (28) yields (24).

Corollary 8. Let 𝑃,𝑄∈ℂ^𝑛×𝑛 with Ind(𝑄𝑃^𝜋)=𝑠, Ind(𝑃)=𝑡, Ind[(𝑃+𝐼)𝑃𝑃_𝑑]=𝑙, and Ind[(𝑃+𝑄)𝑃^𝜋]=𝑘. If 𝑃𝑄=𝑃, then (𝑃+𝑄)_𝑑=^𝑠−1_𝑛=0^𝑡−2_𝑖=0(−1)^𝑖𝑎_𝑖^[𝑛+2]𝐼−𝑎_𝑖^[𝑛+1]𝑄_𝑑𝑄^𝑛𝑃^𝑖+1𝑃^𝜋+𝑄_𝑑𝑃^𝜋+^𝑘−1_𝑛=0𝑄^𝑛𝑄^𝜋(𝑄−𝐼)𝑃𝑃_𝑑(𝑃+𝐼)_𝑑^𝑛+2+𝑄^𝜋+𝑄_𝑑𝑃𝑃_𝑑(𝑃+𝐼)_𝑑+^𝑙−1_𝑛=0𝑄^𝑛_𝑑𝑄_𝑑−𝑄²_𝑑𝑃𝑃_𝑑(𝑃+𝐼)^𝑛(𝑃+𝐼)^𝜋−𝑄_𝑑𝑄^𝑠𝑡−2_𝑖=0(−1)^𝑖𝑎_𝑖^[𝑠+1]𝑃^𝜋𝑃^𝑖+1, (39) where 𝑎_𝑖^[𝑛]=∑^𝑖_𝑘=0𝑎_𝑘^[𝑛−1], 𝑎_𝑖^[1]=1, 𝑖=0,…,𝑡−1.

3. The Drazin Inverse of Some 2 × 2 Block Matrix

In this section, we will apply the results in Section 2 to studying the representation for the Drazin inverse of a 2×2 block matrix, in terms of its subblocks.

Theorem 9. Let 𝑀=^𝐴𝐵_𝐶𝐷∈ℂ^𝑛×𝑛 with Ind[(𝐶𝐵)^𝜋𝐷]=𝑠−1, Ind(𝐴)=𝑡−1, Ind[(𝐴+𝐼)𝐴𝐴_𝑑]=𝑙−1, and Ind^𝐴𝜋0_{0(𝐶𝐵)𝜋}𝑀=𝑘. If 𝐴=𝐵𝐶 and 𝐵=𝐵𝐷, then𝑀_𝑑=𝐼−(𝐴+𝐼)_𝑑(𝐴+𝐼)_𝑑𝐴_𝑑0𝐷_𝑑+𝐷^𝜋𝐶(𝐴+𝐼)²_𝑑𝐴_𝑑+𝐷²_𝑑−𝐷_𝑑𝐶(𝐴+𝐼)_𝑑𝐴_𝑑𝐷²_𝑑(𝐶𝐵)^𝜋𝑀+^𝑠−1_𝑛=1^𝑡−2_𝑖=0(−1)^𝑖𝑎00_𝑖[𝑛+2]𝐼−𝑎_𝑖^[𝑛+1]𝐷_𝑑𝐷^𝑛−1𝐶𝐴^𝑖𝐴^𝜋0𝑀+^𝑘−1_𝑛=1𝐷00_𝜋𝐷^𝑛−𝐷^𝑛−1𝐶(𝐴+𝐼)_𝑑^𝑛+2𝐴_𝑑0𝑀+^𝑙−1_𝑛=0𝐷00_𝑑𝑛+2−𝐷_𝑑^𝑛+3𝐶(𝐴+𝐼)^𝑛(𝐴+𝐼)^𝜋𝐴_𝑑0𝑀+^𝑡−2_𝑖=0(−1)^𝑖𝑎_𝑖^[2]𝐴^𝑖𝐴^𝜋0−𝑎_𝑖^[1]𝐷²_𝑑+𝑎_𝑖^[𝑠+1]𝐷_𝑑𝐷^𝑠−1𝐶𝐴^𝑖𝐴^𝜋0𝑀, (40)where 𝑎_𝑖^[𝑛]=∑^𝑖_𝑘=0𝑎_𝑘^[𝑛−1], 𝑎_𝑖^[1]=1, 𝑖=0,…,𝑡−1.

Proof. Let 𝑃=𝐴𝐵00,𝑄=00𝐶𝐷. (41) Then, 𝑀=𝑃+𝑄, and, for 𝑛≥1, 𝑄^𝑛=𝐷00_𝑛−1𝐶𝐷^𝑛,𝑄^𝑛_𝑑=𝐷00_𝑑𝑛+1𝐶𝐷^𝑛_𝑑,𝑄^𝜋=𝐼0−𝐷_𝑑𝐶𝐷^𝜋,𝑃^𝑛=𝐴^𝑛𝐴^𝑛−1𝐵,𝑃00_𝑑=𝐴_𝑑𝐴²_𝑑𝐵00,𝑃𝑃_𝑑=𝐴𝐴_𝑑𝐴_𝑑𝐵=𝐴00^𝑑0𝑃00𝑀,_𝜋=𝐴^𝜋−𝐴_𝑑𝐵,()0𝐼𝑃+𝐼^𝑛_𝛼=(𝐴+𝐼)^𝑛_𝛼∗,0𝐼 (42) where the subscript 𝛼 stands for 𝑑 or its absence.
Since 𝑊𝑊_𝑑(𝑊+𝐼)=(𝑊+𝐼)𝑊𝑊_𝑑, 𝑊𝑊_𝑑(𝑊+𝐼)^𝑛_𝑑=(𝑊+𝐼)^𝑛_𝑑𝑊𝑊_𝑑 where 𝑊 denotes 𝑃 or 𝐴. Thus, 𝑃𝑃_𝑑(𝑃+𝐼)^𝑛_𝛼=(𝑃+𝐼)^𝑛_𝛼𝑃𝑃_𝑑=(𝐴+𝐼)^𝑛_𝛼𝐴_𝑑000𝑀,𝑃𝑃_𝑑(𝑃+𝐼)^𝜋=(𝑃+𝐼)^𝜋𝑃𝑃_𝑑=(𝐴+𝐼)^𝜋∗0𝐼𝑃𝑃_𝑑=(𝐴+𝐼)^𝜋𝐴_𝑑0𝑎00𝑀,_𝑖[𝑛+2]𝐼−𝑎_𝑖^[𝑛+1]𝑄_𝑑=𝑎_𝑖^[𝑛+2]𝐼00𝑎_𝑖^[𝑛+2]𝐼−𝑎00_𝑖[𝑛+1]𝐷²_𝑑𝐶𝑎_𝑖^[𝑛+1]𝐷_𝑑=𝑎_𝑖^[𝑛+2]𝐼0−𝑎_𝑖^[𝑛+1]𝐷²_𝑑𝐶𝑎_𝑖^[𝑛+2]𝐼−𝑎_𝑖^[𝑛+1]𝐷_𝑑,𝑃^𝑖+1𝑃^𝜋=𝐴^𝑖+1𝐴^𝑖𝐵𝐴00^𝜋−𝐴_𝑑𝐵=𝐴0𝐼_𝑖+1𝐴^𝜋𝐴^𝑖𝐴^𝜋𝐵=𝐴00_𝑖𝐴^𝜋0𝑄00𝑀,_𝑛𝑃^𝑖+1𝑃^𝜋=𝐷00_𝑛−1𝐶𝐷^𝑛𝑃^𝑖+1𝑃^𝜋=𝐷00_𝑛−1𝐶𝐴^𝑖𝐴^𝜋0𝑀. (43)(44) So, for 𝑛≥1, 𝑎_𝑖^[𝑛+2]𝐼−𝑎_𝑖^[𝑛+1]𝑄_𝑑𝑄^𝑛𝑃^𝑖+1𝑃^𝜋=𝑎00_𝑖[𝑛+2]𝐼−𝑎_𝑖^[𝑛+1]𝐷_𝑑𝐷^𝑛−1𝐶𝐴^𝑖𝐴^𝜋0𝑎𝑀,_𝑖[2]𝐼−𝑎_𝑖^[1]𝑄_𝑑𝑃^𝑖+1𝑃^𝜋=𝑎_𝑖^[2]𝐴^𝑖𝐴^𝜋0−𝑎_𝑖^[1]𝐷²_𝑑𝐶𝐴^𝑖𝐴^𝜋0𝑎𝑀,_𝑖[𝑠+1]𝑄_𝑑𝑄^𝑠𝑃^𝜋𝑃^𝑖+1=𝑎_𝑖^[𝑠+1]𝐷00_2𝑑𝐶𝐷_𝑑𝐷00_𝑠−1𝐶𝐴^𝑖𝐴^𝜋0𝑀=𝑎00_𝑖[𝑠+1]𝐷_𝑑𝐷^𝑠−1𝐶𝐴^𝜋𝐴^𝑖0𝑀. (45)
Also, for 𝑛≥1, 𝑄^𝜋𝑄^𝑛+1−𝑄^𝑛𝑃𝑃_𝑑(𝑃+𝐼)_𝑑^𝑛+2=𝐼0−𝐷_𝑑𝐶𝐷^𝜋𝐷00_𝑛−𝐷^𝑛−1𝐶𝐷^𝑛+1−𝐷^𝑛×(𝐴+𝐼)_𝑑^𝑛+2𝐴_𝑑0𝑀=𝐷0000_𝜋𝐷^𝑛−𝐷^𝑛−1𝐶(𝐴+𝐼)_𝑑^𝑛+2𝐴_𝑑0𝑄𝑀,_𝜋(𝑄−𝐼)𝑃𝑃_𝑑(𝑃+𝐼)²_𝑑=𝐼0−𝐷_𝑑𝐶𝐷^𝜋−𝐼0()𝐶𝐷−𝐼𝐴+𝐼²_𝑑𝐴_𝑑0𝑀=−()00𝐴+𝐼²_𝑑𝐴_𝑑0𝐷_𝑑+𝐷^𝜋𝐶(𝐴+𝐼)²_𝑑𝐴_𝑑0𝑄𝑀,_𝜋+𝑄_𝑑𝑃𝑃_𝑑(𝑃+𝐼)_𝑑=𝐷𝐼0_2𝑑𝐶−𝐷_𝑑𝐶𝐷^𝜋+𝐷_𝑑(𝐴+𝐼)_𝑑𝐴_𝑑0𝑀=()00𝐴+𝐼_𝑑𝐴_𝑑0𝐷²_𝑑−𝐷_𝑑𝐶(𝐴+𝐼)_𝑑𝐴_𝑑0𝑀, (46) and, for 𝑛≥0, 𝑄_𝑑^𝑛+1−𝑄_𝑑^𝑛+2𝑃𝑃_𝑑(𝑃+𝐼)^𝑛(𝑃+𝐼)^𝜋=𝐷00_𝑑𝑛+2−𝐷_𝑑^𝑛+3𝐶𝐷^𝑛+1−𝐷^𝑛+2×(𝐴+𝐼)^𝑛𝐴_𝑑0𝑀()00𝐴+𝐼^𝜋𝐴_𝑑0𝑀=𝐷0000_𝑑𝑛+2−𝐷_𝑑^𝑛+3𝐶(𝐴+𝐼)^𝑛(𝐴+𝐼)^𝜋𝐴_𝑑0𝑀. (47)
Since 𝐷−𝐶𝐴_𝑑𝐵=𝐷−𝐶𝐵(𝐶𝐵)²_𝑑𝐶𝐵𝐷=(𝐶𝐵)^𝜋𝐷,𝐶𝐴^𝜋=𝐶−𝐶𝐵𝐶𝐵(𝐶𝐵)²_𝑑𝐶=(𝐶𝐵)^𝜋𝐶, (48) we have 𝑄_𝑑𝑃^𝜋=𝐷00_2𝑑𝐶𝐷_𝑑𝐴^𝜋−𝐴_𝑑𝐵=𝐷0𝐼00_2𝑑(𝐶𝐵)^𝜋𝐶𝐷²_𝑑(𝐶𝐵)^𝜋𝐷=000𝐷²_𝑑(𝐶𝐵)^𝜋𝑀,𝑄𝑃^𝜋=00𝐶𝐴^𝜋𝐷−𝐶𝐴_𝑑𝐵=()00𝐶𝐵^𝜋𝐶(𝐶𝐵)^𝜋𝐷,𝑀𝑃^𝜋=𝐴𝐴^𝜋𝐵−𝐴𝐴_𝑑𝐵𝐶𝐴^𝜋𝐷−𝐶𝐴_𝑑𝐵=𝐴𝐴^𝜋𝐵𝐴^𝜋(𝐶𝐵)^𝜋𝐶(𝐶𝐵)^𝜋𝐷=𝐴^𝜋00(𝐶𝐵)^𝜋𝑀, (49)(50) and then, by Lemma 4, Ind(𝑄𝑃^𝜋)≤Ind[(𝐶𝐵)^𝜋𝐷]+1=𝑠, and Ind[(𝑃+𝑄)𝑃^𝜋]≤Ind^𝐴𝜋0_{0(𝐶𝐵)𝜋}𝑀=𝑘.
By Lemma 4, Ind(𝑃)≤Ind(𝐴)+1=𝑡. Further, by (43), Ind[(𝑃+𝐼)𝑃𝑃_𝑑]≤Ind[(𝐴+𝐼)𝐴𝐴_𝑑]+1=𝑙.
Hence, putting (45)~(49) into (39) yields (40).

4. Example

In this section, we present a numerical example to demonstrate Theorem 7.

Example 1. Taking 𝑃,𝑄 as follows: ,,𝑃=0.13210.84590.2893−0.0597−0.28300.08020.09430.3066−0.33960.14621.11320.6179−0.62260.97640.20750.6745𝑄=0.54721.52830.1509−0.50940.00001.0000−0.000000.1132−0.38210.96230.1274−0.45281.52830.15090.4906 (51) we can get 𝑃_𝑑=,𝑃0.28300.00310.2390−0.2233−0.28300.08020.09430.30660.84910.00940.7170−0.6698−0.84910.24060.28300.9198_𝜋=,𝑄0.9057−0.0566−0.30190.01890.28300.9198−0.0943−0.3066−0.2830−0.16980.09430.05660.8491−0.2406−0.28300.0802_𝑑=,𝑄0.54721.52830.1509−0.50940.00001.0000−0.000000.1132−0.38210.96230.1274−0.45281.52830.15090.4906_𝜋=,0.4528−1.5283−0.15090.50940000−0.11320.38210.0377−0.12740.4528−1.5283−0.15090.5094 (52) and Ind(𝑄𝑃^𝜋)=1, Ind(𝑃)=2, Ind[(𝑃−𝐼)𝑃𝑃_𝑑]=2, and Ind[(𝑃−𝑄)𝑃^𝜋]=1. By Theorem 7, we have (𝑃−𝑄)_𝑑=−0.2264−0.73580.07550.2453−0.2830−0.91980.09430.30660.11320.3679−0.0377−0.1226−0.1698−0.55190.05660.1840. (53)