On the Drazin Inverse of the Sum of Two Matrices

Abstract

We deduce the explicit expressions for (𝑃+𝑄)^𝐷 and (𝑃𝑄)^𝐷 of two matrices 𝑃 and 𝑄 under the conditions 𝑃²𝑄=𝑃𝑄𝑃 and 𝑄²𝑃=𝑄𝑃𝑄. Also, we give the upper bound of ||(𝑃+𝑄)^𝐷−𝑃^𝐷||₂.

1. Introduction

The symbol ℂ^𝑚×𝑛 stands for the set of 𝑚×𝑛 complex matrices, and 𝐼_𝑛 (for short 𝐼) stands for the 𝑛×𝑛 identity matrix. For 𝐴∈ℂ^𝑛×𝑛, its Drazin inverse, denoted by 𝐴^𝐷, is defined as the unique matrix satisfying𝐴^𝑘+1𝐴^𝐷=𝐴^𝑘,𝐴^𝐷𝐴𝐴^𝐷=𝐴^𝐷,𝐴𝐴^𝐷=𝐴^𝐷𝐴, (1.1) where 𝑘=Ind(𝐴) is the index of 𝐴. In particular, if 𝑘=0, 𝐴 is invertible and 𝐴^𝐷=𝐴⁻¹ (see, e.g., [1–3] for details). Recall that for 𝐴∈ℂ^𝑛×𝑛 withInd(𝐴)=𝑘, there exists an 𝑛×𝑛 nonsingular matrix 𝑋 such that𝐴=𝑋𝑋𝐶00𝑁⁻¹, (1.2) where 𝐶 is a nonsingular matrix and 𝑁 is nilpotent of index 𝑘, and𝐴^𝐷𝐶=𝑋⁻¹0𝑋00⁻¹ (1.3) (see [1, 3]). It is well known that if 𝐴 is nilpotent, then 𝐴^𝐷=0. We always write 𝐴^𝜋∶=𝐼−𝐴𝐴^𝐷.

Drazin [2] proved that in associative ring (𝐴+𝐵)^𝐷=𝐴^𝐷+𝐵^𝐷 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+𝐵)^𝐷 where 𝐴,𝐵∈ℂ^𝑛×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 𝑃𝑄=𝜆𝑄𝑃 and 𝑃𝑄=𝑃𝑄𝑃, in [9] they are 𝑃³𝑄=𝑄𝑃 and 𝑄³𝑃=𝑃𝑄, and in [15], they are 𝑃𝑄𝑃=0 and 𝑃𝑄²=0. These results motivate us to investigate how to explicitly express the Drazin inverse of the sum 𝑃+𝑄 under the conditions 𝑃²𝑄=𝑃𝑄𝑃 and 𝑄²𝑃=𝑄𝑃𝑄, which are implied by the condition 𝑃𝑄=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+𝑄)^𝐷 and (𝑃𝑄)^𝐷 of two matrices 𝑃 and 𝑄 under the conditions 𝑃²𝑄=𝑃𝑄𝑃 and 𝑄²𝑃=𝑄𝑃𝑄. We also give the upper bound of ‖(𝑃+𝑄)^𝐷−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∈ℂ^𝑛×𝑛,𝐵∈ℂ^𝑚×𝑚, and 𝐶∈ℂ^𝑚×𝑛 with 𝐵𝐶=0, and define 𝑀=𝐴0𝐶𝐵. (2.1) Then, 𝑀^𝐷=𝐴^𝐷0𝐶(𝐴^𝐷)²𝐵^𝐷. (2.2)

Lemma 2.2. Let 𝑃,𝑄∈ℂ^𝑛×𝑛. If 𝑃²𝑄=𝑃𝑄𝑃, then, for any positive integers 𝑖,𝑗,(i)𝑃^𝑖+1𝑄=𝑃^𝑖𝑄𝑃=𝑃𝑄𝑃^𝑖,𝑃^2𝑖𝑄=𝑃^𝑖𝑄𝑃^𝑖,(ii)𝑃^𝑖𝑄^𝑖=(𝑃𝑄)^𝑖.
Moreover, if 𝑄²𝑃=𝑄𝑃𝑄, then 𝑃𝑄^𝑗𝑃^𝑖=𝑃^𝑖+1𝑄^𝑗. (2.3)

Proof. (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^𝑘𝑄^𝑘=(𝑃𝑄)^𝑘. When 𝑖=𝑘+1, by (i), we have𝑃^𝑖+1𝑄^𝑖+1=𝑃𝑄𝑃^𝑖𝑄^𝑖=𝑃𝑄(𝑃𝑄)^𝑖=(𝑃𝑄)^𝑖+1. (2.4) Hence, by induction, we have 𝑃^𝑖𝑄^𝑖=(𝑃𝑄)^𝑖 for any 𝑖.
Assume 𝑄²𝑃=𝑄𝑃𝑄. 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, 𝑃𝑄^𝑘𝑃^𝑖=𝑃^𝑖+1𝑄^𝑘. When 𝑗=𝑘+1, 𝑃𝑄^𝑘+1𝑃^𝑖=𝑃𝑄^𝑘−1𝑄²𝑃𝑃^𝑖−1=𝑃𝑄^𝑘𝑃𝑄𝑃^𝑖−1=𝑃𝑄^𝑘𝑃^𝑖𝑄=𝑃^𝑖+1𝑄^𝑘𝑄=𝑃^𝑖+1𝑄^𝑘+1. (2.5) Hence (2.3) holds for any 𝑗.

Lemma 2.3. Let 𝑃,𝑄∈ℂ^𝑛×𝑛. Suppose that 𝑃²𝑄=𝑃𝑄𝑃 and 𝑄²𝑃=𝑄𝑃𝑄. Then, for any positive integer 𝑚, (𝑃+𝑄)^𝑚=^𝑚−1_𝑖=0𝐶^𝑖_𝑚−1𝑃^𝑚−𝑖𝑄^𝑖+𝑄^𝑚−𝑖𝑃^𝑖, (2.6) where the binomial coefficient 𝐶^𝑖_𝑗=𝑗!/𝑖!(𝑗−𝑖)!,𝑗≥𝑖.
Moreover, if 𝑃,𝑄 are nilpotent with 𝑃^𝑠=0 and 𝑄^𝑡=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+𝑄)^𝑘=^𝑘−1_𝑖=0𝐶^𝑖_𝑘−1𝑃^𝑘−𝑖𝑄^𝑖+𝑄^𝑘−𝑖𝑃^𝑖. (2.7) Then, for 𝑚=𝑘+1, we have, by Lemma 2.2, (𝑃+𝑄)^𝑘+1=^𝑘−1_𝑖=0𝐶^𝑖_𝑘−1𝑃^𝑘−𝑖𝑄^𝑖+𝑄^𝑘−𝑖𝑃^𝑖()=𝑃+𝑄_{𝑘−1𝑖=0}𝐶^𝑖_𝑘−1𝑃^{𝑘+1−𝑖}𝑄^𝑖+𝑃^𝑘−𝑖𝑄^𝑖+1+𝑄^𝑘−𝑖𝑃^𝑖+1+𝑄^{𝑘+1−𝑖}𝑃^𝑖=𝑃^𝑘+1+^𝑘−1_𝑖=1𝐶^𝑖_𝑘−1+𝐶^𝑖−1_𝑘−1𝑃^{𝑘+1−𝑖}𝑄^𝑖+𝑃𝑄^𝑘+𝑄^𝑘+1+^𝑘−1_𝑖=1𝐶^𝑖_𝑘−1+𝐶^𝑖−1_𝑘−1𝑄^{𝑘+1−𝑖}𝑃^𝑖+𝑄𝑃^𝑘=𝑃^𝑘+1+^𝑘−1_𝑖=1𝐶^𝑖_𝑘𝑃^{𝑘+1−𝑖}𝑄^𝑖+𝑃𝑄^𝑘+𝑄^𝑘+1+^𝑘−1_𝑖=1𝐶^𝑖_𝑘𝑄^{𝑘+1−𝑖}𝑃^𝑖+𝑄𝑃^𝑘=^𝑘_𝑖=0𝐶^𝑖_𝑘𝑃^{𝑘+1−𝑖}𝑄^𝑖+^𝑘_𝑖=0𝐶^𝑖_𝑘𝑄^{𝑘+1−𝑖}𝑃^𝑖. (2.8) Hence (2.6) holds for any 𝑚≥1.
If 𝑃,𝑄 are nilpotent with 𝑃^𝑠=0 and 𝑄^𝑡=0, then taking 𝑚=𝑠+𝑡−1 in (2.6) yields (𝑃+𝑄)^{𝑠+𝑡−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∈ℂ^𝑛×𝑛. If 𝑃𝑄=𝑄𝑃, then (𝑃𝑄)^𝐷=𝑄^𝐷𝑃^𝐷=𝑃^𝐷𝑄^𝐷 and 𝑃^𝐷𝑄=𝑄𝑃^𝐷.

Lemma 2.5. Let 𝑃,𝑄∈ℂ^𝑛×𝑛 and 𝑃 be invertible. If 𝑃𝑄=𝑄𝑃, then (𝑃+𝑄)^𝐷=𝐼+𝑃⁻¹𝑄^𝐷𝑃⁻¹=𝑃⁻¹𝐼+𝑃⁻¹𝑄^𝐷. (2.9)
Moreover, if 𝑄 is nilpotent of index 𝑡, then 𝑃+𝑄 is invertible and (𝑃+𝑄)⁻¹=^𝑡−1_𝑖=0(−𝑄)^𝑖𝑃^−𝑖−1=^𝑡−1_𝑖=0𝑃^−𝑖−1(−𝑄)^𝑖. (2.10)

Proof. Since 𝑃+𝑄=𝑃(𝐼+𝑃⁻¹𝑄)=(𝐼+𝑃⁻¹𝑄)𝑃, by Lemma 2.4, (𝑃+𝑄)^𝐷=𝐼+𝑃⁻¹𝑄^𝐷𝑃⁻¹=𝑃⁻¹𝐼+𝑃⁻¹𝑄^𝐷. (2.11)
Note that the nilpotency of 𝑄 with commuting with 𝑃 implies that 𝑃⁻¹𝑄 is nilpotent of index 𝑡. Thus, 𝐼+𝑃⁻¹𝑄 is invertible and so is 𝑃+𝑄, and (𝑃+𝑄)⁻¹=𝐼+𝑃⁻¹𝑄⁻¹𝑃⁻¹=^𝑡−1_𝑖=0(−𝑄)^𝑖𝑃^−𝑖−1=^𝑡−1_𝑖=0𝑃^−𝑖−1(−𝑄)^𝑖. (2.12)

Lemma 2.6. Let 𝑃,𝑄∈ℂ^𝑛×𝑛 with 𝑄=𝑄₁⊕𝑄₂, where 𝑄₁ is invertible and 𝑄₂ is nilpotent of index 𝑡, and let 𝑃𝑃=¹𝑃₃𝑃₄𝑃₂ (2.13) be partitioned conformably with 𝑄. Suppose that 𝑄²𝑃=𝑄𝑃𝑄 and 𝑃²𝑄=𝑃𝑄𝑃. Then, 𝑃₃=0 and 𝑄₁𝑃₁=𝑃₁𝑄₁,𝑄₂𝑃₄=𝑃₂𝑃₄𝑄=0,₂₂𝑃₂=𝑄₂𝑃₂𝑄₂𝑃_2𝑖,𝑄_𝑖=𝑃_𝑖𝑄_𝑖𝑃_𝑖,𝑖=1,2. (2.14)(2.15)(2.16)(2.17)
Moreover, if 𝑃 is nilpotent of index 𝑠, then 𝑃₄𝑃₁^𝑠−1=0.

Proof. Since 𝑄²𝑃=𝑄𝑃𝑄,𝑄^2𝑡𝑃=𝑄^𝑡𝑃𝑄^𝑡 by Lemma 2.2, that is, 𝑄₁^2𝑡0𝑃00¹𝑃₃𝑃₄𝑃₂=𝑄^𝑡₁0𝑃00¹𝑃₃𝑃₄𝑃₂𝑄^𝑡₁000, (2.18) namely, 𝑄₁^2𝑡𝑃₁𝑄₁^2𝑡𝑃₃=𝑄00^𝑡1𝑃₁𝑄^𝑡₁000. (2.19) Thus, 𝑃₃=0 because the invertibility of 𝑄₁. So from 𝑄²𝑃=𝑄𝑃𝑄 and 𝑃²𝑄=𝑃𝑄𝑃, it follows, respectively, that 𝑃₁𝑄₁=𝑄₁𝑃₁,𝑄²₂𝑃₄=𝑄₂𝑃₄𝑄₁,𝑄²₂𝑃₂=𝑄₂𝑃₂𝑄₂, (2.20) and that 𝑃₂𝑃₄𝑄₁=𝑃₂𝑄₂𝑃₄,𝑃²_𝑖𝑄_𝑖=𝑃_𝑖𝑄_𝑖𝑃_𝑖,𝑖=1,2. (2.21) Since 𝑄^𝑡₂=0, 𝑄₂𝑃₄=𝑄²₂𝑃₄𝑄₁⁻¹=𝑄^𝑡₂𝑃₄𝑄₁^−𝑡+1=0, (2.22) and then 𝑃₂𝑃₄=𝑃₂𝑄₂𝑃₄𝑄₁⁻¹=0. From this, we can easily verify 𝑃^𝑠=𝑃^𝑠₁0𝑃₄𝑃₁^𝑠−1𝑃^𝑠₂. (2.23) Therefore, if 𝑃^𝑠=0, then 𝑃₄𝑃₁^𝑠−1=0.

3. Main Results

In this section, we will give the explicit expressions for (𝑃+𝑄)^𝐷 and (𝑃𝑄)^𝐷, under the conditions 𝑃²𝑄=𝑃𝑄𝑃 and 𝑄²𝑃=𝑄𝑃𝑄. Now, we begin with the following theorem.

Theorem 3.1. Let 𝑄∈ℂ^𝑛×𝑛 with Ind(𝑄)=𝑡 and 𝑃∈ℂ^𝑛×𝑛 be nilpotent with 𝑃^𝑠=0. If 𝑃²𝑄=𝑃𝑄𝑃 and 𝑄²𝑃=𝑄𝑃𝑄, then (𝑃+𝑄)^𝐷=^𝑠−1_𝑖=0𝑄^𝐷𝑖+1(−𝑃)^𝑖+𝑄^𝜋𝑃^𝑠−2_𝑖=0(−1)^𝑖(𝑖+1)𝑄^𝐷𝑖+2𝑃^𝑖=𝑄𝑄^𝐷𝑠−1_𝑖=0(−𝑃)^𝑖𝑄^𝐷𝑖+1+𝑄^𝜋𝑃𝑄𝑄^𝐷𝑠−2_𝑖=0(−1)^𝑖(𝑖+1)𝑃^𝑖𝑄^𝐷𝑖+2. (3.1)

Proof. If 𝑡=0, then 𝑄 is invertible, and therefore 𝑄𝑃=𝑃𝑄. 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 𝑄^𝐷=𝑄₁⁻¹⊕0. Since 𝑃²𝑄=𝑃𝑄𝑃 and 𝑄²𝑃=𝑄𝑃𝑄, we can write 𝑃, partitioned conformably with 𝑄, by Lemma 2.6, as follows: 𝑃𝑃=¹0𝑃₄𝑃₂, (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_𝑖=0𝑄₁^−𝑖−1−𝑃₁^𝑖=^𝑠−1_𝑖=0−𝑃₁^𝑖𝑄₁^−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+𝑄)^𝐷=𝑃₁+𝑄₁0𝑃₄𝑃₂+𝑄₂^𝐷=𝑄₁⁻¹(𝐼₁+𝑄₁⁻¹𝑃₁)⁻¹0𝑃₄𝑄₁⁻²(𝐼₁+𝑄₁⁻¹𝑃₁)⁻²0. (3.4) By (3.3), it is easy to verify that 𝐼₁+𝑄₁⁻¹𝑃₁⁻²=^𝑠−1_𝑖=0(−1)^𝑖(𝑖+1)𝑄₁^−𝑖𝑃^𝑖₁=^𝑠−1_𝑖=0(−1)^𝑖(𝑖+1)𝑃^𝑖₁𝑄₁^−𝑖. (3.5) Since 𝑃^𝑠=0, we have 𝑃₄𝑃₁^𝑠−1=0 by Lemma 2.6, and, therefore, by (3.5), 𝑄^𝜋𝑃^𝑠−2_𝑖=0(−1)^𝑖(𝑖+1)𝑄^𝐷𝑖+2𝑃^𝑖=^𝑠−2_𝑖=0(−1)^𝑖(𝑖+1)000𝐼₂𝑃₁0𝑃₄𝑃₂𝑄₁^−(𝑖+2)0𝑃00^𝑖10∗𝑃^𝑖₂=^𝑠−2_𝑖=0(−1)^𝑖()𝑃𝑖+100₄𝑄₁^−(𝑖+2)𝑃ⁱ₁0=𝑃00₄𝑄₁⁻²𝐼+𝑄₁⁻¹𝑃₁⁻²0. (3.6) Analogous to the argument above, we can see, by Lemma 2.5, ^𝑠−1_𝑖=0𝑄^𝐷𝑖+1(−𝑃)^𝑖=𝑄₁⁻¹𝐼₁+𝑄₁⁻¹𝑃₁⁻¹000. (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 𝑄𝑄^𝐷𝑠−1_𝑖=0(−𝑃)^𝑖𝑄^𝐷𝑖+1=𝑄₁⁻¹𝐼+𝑄₁⁻¹𝑃₁⁻¹0,𝑄00_𝜋𝑃𝑄𝑄^𝐷𝑠−2_𝑖=0(−1)^𝑖(𝑖+1)𝑃^𝑖𝑄^𝐷𝑖+2=𝑃00₄𝑄₁⁻²𝐼+𝑄₁⁻¹𝑃₁⁻²0, (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∈ℂ^𝑛×𝑛 with Ind(𝑃)=𝑠≥1 and Ind(𝑄)=𝑡. If 𝑃²𝑄=𝑃𝑄𝑃 and 𝑄²𝑃=𝑄𝑃𝑄. Then,(i)(𝑃𝑄)^𝐷=𝑃^𝐷𝑄^𝐷=𝑃𝑃^𝐷𝑄^𝐷𝑃^𝐷=𝑃𝑄^𝐷𝑃^𝐷2,𝑄²𝑃^𝐷=𝑄𝑃^𝐷𝑃𝑄,_𝐷2𝑄=𝑃^𝐷𝑄𝑃^𝐷. (3.9)(3.10)(3.11)(ii)(𝑃+𝑄)^𝐷=𝑃^𝐷𝐼+𝑃^𝐷𝑄^𝐷+𝑃^𝜋𝑄𝑃^𝐷𝐼+𝑃^𝐷𝑄^𝐷2+^𝑠−1_𝑖=0𝑄^𝐷𝑖+1(−𝑃)^𝑖𝑃^𝜋+𝑄^𝜋𝑃^𝑠−2_𝑖=0(−1)^𝑖(𝑖+1)𝑄^𝐷𝑖+2𝑃^𝑖𝑃^𝜋. (3.12)

Proof. If 𝑠=0, then 𝑃 is invertible and 𝑃𝑄=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 𝑄𝑄=¹0𝑄₄𝑄₂, (3.13) partitioned conformably with 𝑃, and those equations in Lemma 2.6 hold. By Lemma 2.1, therefore, we have 𝑄^𝐷=𝑄^𝐷₁0𝑄₄(𝑄^𝐷₁)²𝑄^𝐷₂,𝑄²=𝑄²₁0𝑄₄𝑄₁𝑄²₂. (3.14)(i) By (2.14) and (2.15),𝑄²𝑃^𝐷=𝑄²₁𝑃₁⁻¹0𝑄₄𝑄₁𝑃₁⁻¹0=𝑄₁𝑃₁⁻¹𝑄₁0𝑄₄𝑃₁⁻¹𝑄₁0=𝑄𝑃^𝐷𝑃𝑄,_𝐷2𝑃𝑄=¹⁻²𝑄₁0=𝑃00¹⁻¹𝑄₁𝑃₁⁻¹000=𝑃^𝐷𝑄𝑃^𝐷,𝑃𝑄^𝐷𝑃^𝐷2=𝑃₁𝑄^𝐷₁𝑃₁⁻²0=𝑃00¹⁻¹𝑄^𝐷₁000=𝑃^𝐷𝑄^𝐷=𝑄^𝐷₁𝑃₁⁻¹000=𝑃𝑃^𝐷𝑄^𝐷𝑃^𝐷,𝑃𝑃𝑄=¹𝑄₁00𝑃₂𝑄₂=𝑄₁𝑃₁00𝑃₂𝑄₂. (3.15)(3.16) By (2.17) and Lemma 2.2, (𝑃₂𝑄₂)^𝑠=𝑃^𝑠₂𝑄^𝑠₂=0. By Lemma 2.4 and (3.16), we have (𝑃𝑄)^𝐷=(𝑄₁𝑃₁)^𝐷00𝑃₂𝑄₂^𝐷=𝑃₁⁻¹𝑄^𝐷₁000=𝑃^𝐷𝑄^𝐷. (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+𝑄)^𝐷=𝑃₁+𝑄₁0𝑄₄𝑃₂+𝑄₂^𝐷=𝑃₁+𝑄₁^𝐷0𝑄₄𝑃₁+𝑄₁^𝐷2𝑃₂+𝑄₂^𝐷. (3.18)By Lemma 2.5, we have 𝑃^𝐷𝐼+𝑃^𝐷𝑄^𝐷=𝑃₁⁻¹0𝐼00¹+𝑃₁⁻¹𝑄₁00𝐼₂^𝐷=𝑃₁⁻¹𝐼₁+𝑃₁⁻¹𝑄₁^𝐷0=𝑃00¹+𝑄₁^𝐷000 (3.19) and, therefore, 𝑄00₄𝑃₁+𝑄₁^𝐷20=000𝐼₂𝑄₁0𝑄₄𝑄₂𝑃₁+𝑄₁^𝐷2000=𝑃^𝜋𝑄𝑃^𝐷𝐼+𝑃^𝐷𝑄^𝐷2. (3.20) By (3.1), we have 0𝑃00₂+𝑄₂^𝐷=000_{𝑠−1𝑖=0}𝑄^𝐷₂^𝑖+1−𝑃₂^𝑖+𝑄^𝜋₂𝑃₂^𝑠−2_𝑖=0(−1)^𝑖(𝑖+1)𝑄^𝐷₂^𝑖+2𝑃^𝑖₂=^𝑠−1_𝑖=0𝑄^𝐷𝑖+1(−𝑃)^𝑖𝑃^𝜋+𝑄^𝜋𝑃^𝑠−2_𝑖=0(−1)^𝑖(𝑖+1)𝑄^𝐷𝑖+2𝑃^𝑖𝑃^𝜋. (3.21) Thus, substituting (3.19), (3.21), and (3.20) in (3.18) yields (3.12).

Note that 𝑃𝑄=𝑄𝑃 implies 𝑃²𝑄=𝑃𝑄𝑃 and 𝑄²𝑃=𝑄𝑃𝑄.

Corollary 3.3 (see [14, Theorem 2]). If 𝑃,𝑄∈ℂ^𝑛×𝑛 with 𝑃𝑄=𝑄𝑃 and Ind(𝑃)=𝑠, then (𝑃+𝑄)^𝐷=𝐼+𝑃^𝐷𝑄^𝐷𝑃^𝐷+𝑃^𝜋𝑠−1_𝑖=0𝑄^𝐷𝑖+1(−𝑃)^𝑖. (3.22)

Proof. From (3.19) and Lemma 2.5, we can obtain 𝐼+𝑃^𝐷𝑄^𝐷𝑃^𝐷=𝑃^𝐷𝐼+𝑃^𝐷𝑄^𝐷. (3.23)

Since 𝑃^𝑘𝑄=0 for some 𝑘,𝑃^𝐷𝑄=0. Thus, we have the following corollary.

Corollary 3.4. Let 𝑃,𝑄∈ℂ^𝑛×𝑛 with Ind(𝑃)=𝑠≥1 and Ind(𝑄)=𝑡. Suppose 𝑃²𝑄=𝑃𝑄𝑃 and 𝑄²𝑃=𝑄𝑃𝑄. If there exist two positive integers 𝑘 and ℎ such that 𝑃^𝑘𝑄=0 and 𝑄^ℎ𝑃=0, then (𝑃+𝑄)^𝐷=𝑃^𝐷+𝑄^𝐷+𝑄𝑃^𝐷2. (3.24)
If 𝑄 is a perturbation of 𝑃, then, we have the following result in which ‖(𝑃+𝑄)^𝐷−𝑃^𝐷‖₂ has an upper bound. Before the theorem, let us recall that if ‖𝐴‖₂<1, then 𝐼+𝐴 is invertible and (𝐼+𝐴)⁻¹₂≤11−‖𝐴‖₂,𝐼−(𝐼+𝐴)⁻¹₂≤‖𝐴‖₂1−‖𝐴‖₂. (3.25)

Theorem 3.5. Let 𝑃,𝑄∈ℂ^𝑛×𝑛 with Ind(𝑃)=𝑠≥1 and Ind(𝑄)=𝑡. Suppose 𝑃²𝑄=𝑃𝑄𝑃 and 𝑄²𝑃=𝑄𝑃𝑄. If ‖𝑃^𝐷𝑄‖₂<1, then (𝑃+𝑄)^𝐷−𝑃^𝐷₂≤𝑃^𝐷₂𝑃^𝐷𝑄₂1−𝑃^𝐷𝑄₂+𝑃^𝜋₂‖𝑄‖₂𝑃^𝐷2₂1−𝑃^𝐷𝑄₂²+𝑄^𝐷₂𝑃^𝜋₂1−𝑄^𝐷𝑠₂‖𝑃‖^𝑠₂1−𝑄^𝐷₂‖𝑃‖₂+𝑄^𝐷2₂𝑄^𝜋₂𝑃^𝜋₂‖𝑃‖₂1−𝑄^𝐷₂‖𝑃‖₂²×1−𝑠𝑄^𝐷₂^𝑠−1‖𝑃‖₂^𝑠−1+(𝑠−1)𝑄^𝐷𝑠₂‖𝑃‖^𝑠₂. (3.26)

Proof. Since ‖𝑃^𝐷𝑄‖₂<1, 𝐼+𝑃^𝐷𝑄 is invertible. Then by (3.12), we have (𝑃+𝑄)^𝐷−𝑃^𝐷=𝑃^𝐷𝐼+𝑃^𝐷𝑄⁻¹−𝐼+𝑃^𝜋𝑄𝑃^𝐷𝐼+𝑃^𝐷𝑄⁻¹²+^𝑠−1_𝑖=0(𝑄^𝐷)^𝑖+1(−𝑃)^𝑖𝑃^𝜋+𝑄^𝜋𝑃^𝑠−2_𝑖=0(−1)^𝑖(𝑖+1)𝑄^𝐷𝑖+2𝑃^𝑖𝑃^𝜋. (3.27) 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), 𝑃^𝐷𝐼+𝑃^𝐷𝑄⁻¹−𝐼₂≤𝑃^𝐷₂𝑃^𝐷𝑄₂1−𝑃^𝐷𝑄₂,𝑃^𝜋𝑄𝑃^𝐷𝐼+𝑃^𝐷𝑄⁻¹²₂≤𝑃^𝜋₂‖𝑄‖₂𝑃^𝐷2₂1−𝑃^𝐷𝑄₂²_{𝑠−1𝑖=0},𝑄^𝐷𝑖+1(−𝑃)^𝑖𝑃^𝜋₂≤^𝑠−1_𝑖=0𝑄^𝐷₂^𝑖+1‖𝑃‖^𝑖₂𝑃^𝜋₂=^𝑠_𝑖=1𝑄^𝐷𝑖₂‖𝑃‖₂^𝑖−1𝑃^𝜋₂=𝑄^𝐷₂𝑃^𝜋₂1−𝑄^𝐷𝑠₂‖𝑃‖^𝑠₂1−𝑄^𝐷₂‖𝑃‖₂,𝑄^𝜋𝑃^𝑠−2_𝑖=0(−1)^𝑖(𝑖+1)𝑄^𝐷𝑖+2𝑃^𝑖𝑃^𝜋₂≤^𝑠−2_𝑖=0(𝑖+1)𝑄^𝐷₂^𝑖+2‖𝑃‖₂^𝑖+1𝑄^𝜋₂𝑃^𝜋₂=^𝑠−1_𝑖=1𝑖𝑄^𝐷₂^𝑖+1‖𝑃‖^𝑖₂𝑄^𝜋₂𝑃^𝜋₂. (3.28) Let 𝑞∶=‖𝑄^𝐷‖₂‖𝑃‖₂ and 𝑆∶=∑^𝑠−1_𝑖=1𝑖𝑞^𝑖. Then, 1−𝑞𝑆=^𝑠−1_𝑖=1𝑞^𝑖−(𝑠−1)𝑞^𝑠=𝑞1−𝑞^𝑠−1−()𝑞1−𝑞𝑠−1^𝑠=𝑞−𝑠𝑞^𝑠+(𝑠−1)𝑞^𝑠+1.1−𝑞 (3.29) Thus ^𝑠−1_𝑖=1𝑖𝑄^𝐷₂^𝑖+1‖𝑃‖^𝑖₂𝑄^𝜋₂𝑃^𝜋₂=𝑄^𝐷2₂𝑄^𝜋₂𝑃^𝜋₂‖𝑃‖₂1−𝑠𝑄^𝐷₂^𝑠−1‖𝑃‖₂^𝑠−1+(𝑠−1)𝑄^𝐷𝑠₂‖𝑃‖^𝑠₂1−𝑄^𝐷₂‖𝑃‖₂². (3.30) By the above argument, we can get (3.26).

Finally, we give an example to illustrate our results.

Example 3.6. Consider the matrices 1𝑃=1000010000010000,𝑄=31000000000000003. (3.31) We observe that 𝑃²𝑄=𝑃𝑄𝑃 and 𝑄²𝑃=𝑄𝑃𝑄, but 𝑃𝑄≠𝑄𝑃. It is obvious that 𝑠=Ind(𝑃)=2, and 𝑃^𝐷=1000010000000000,𝑄^𝐷=3000000000000003. (3.32) Since ‖𝑃^𝐷𝑄‖₂=(1/3)<1, 𝐼+𝑃^𝐷𝑄 is invertible and 𝐼+𝑃^𝐷𝑄⁻¹=34000010000100001. (3.33) By (3.12), (𝑃+𝑄)^𝐷−𝑃^𝐷=−14000000000090003. (3.34) We can compute ‖(𝑃+𝑄)^𝐷−𝑃^𝐷‖₂=3√10. On the other hand, it is easy to get that ‖𝑃‖₂=‖𝑃^𝐷‖₂=‖𝑃^𝜋‖₂=‖𝑄^𝜋‖₂=1, ‖𝑄‖₂=1/3, ‖𝑄^𝐷‖₂=3. By (3.26), we get the upper bound of ‖(𝑃+𝑄)^𝐷−𝑃^𝐷‖₂ is 16(1/4), 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.