# Perturbation Bound for the Drazin Inverse of the Matrix-Value Function

Authors: Yonghui Qin; Zhenshu Xie; Xiaoji Liu
Journal: Advances in Mathematical Physics
 (2019)

---

## Abstract

The perturbation analysis of the differential for the Drazin inverse of the matrix-value functionA(t)∈Cn×n is investigated. An upper bound of the Drazin inverse and its differential is also considered. Applications to the perturbation bound for the solution of the matrix-value function coefficients some matrix equations are given.


---

## Body

## 1. Introduction

The Drazin inverse has been widely applied in the fields, like singular differential equations [1], linear operators theory analysis [2, 3], Markov chains [4], and so on. For the systems with a nonconstant coefficient, one required knowledge of how to differentiate the Drazin inverse as given in [5]. The necessary information for the derivative of the Drazin inverse is developed in [6]. Many results on perturbation analysis for the Drazin inverse of a matrix A have been investigated in [7–10] and reference therein.LetCm×n be the set of all m×n complex matrix. For any A∈Cn×n, there exists a unique n×n matrix X such that [1] (1)XAX=X,AX=XA,AkXA=A,X is called the Drazin inverse of A, denoted by X=AD, where k is the index of A, and denoted by ind(A)=k. The core-nilpotent decomposition of is A=C+N, where C is invertible and N is a nilpotent matrix and AD=CD as in [1]. Let Rn×n be a set of the matrix-value function. For A(t)∈Rm×n, the definition of the Moore-Penrose inverse for a matrix-valued function is given in [11]. For A(t)∈Rn×n, the equations in (1) are extended to the Drazin inverse of a matrix-value function in [6]. For A(t)∈Rn×n, then A(t)=C(t)+N(t) for all t∈[a,b]. The Drazin inverse of A(t) is defined by AD(t)=[A(t)]D with t∈[a,b] as in [6] and the core-nilpotent decomposition of A(t)∈Rn×n with t∈[a,b] is given as follows:(2)At=PtA1t00A2tP-1t,where P(t) is a nonsingular matrix-value function, A1(t) is nonsingular, and A2(t) is nilpotent for all t∈[a,b] in (2), respectively. From (2), we easy have the Drazin inverse of the matrix-value function A(t) as follows:(3)ADt=PtA1-1t000P-1t.As [6], we have(4)Ct=PtA1t000P-1t,Nt=Pt000A2tP-1t,ADt=CDt=PtA1-1t000P-1t,NDt=0.From (3), we have(5)AtADt=PtI000P-1t,Aπt≔I-AtADt=Pt000IP-1t.The basic continuity properties of AD were developed in [12]. For a differentiable matrix-valued function A, we denote dA(t)/dt by dA and (BD)m by ADm, respectively.The perturbation bound theory of the Drazin inverse on Banach algebra is considered in [13]. For a n×n matrix-value differentiable function, the derivative of the Drazin inverse is defined in [6]. In [12], Campbell et al. investigate the continuity properties of the Drazln inverse. They proved that a sequence of the Drazin inverse of matrix {AjD} converges to AD if and only if there exists a j0 such that core-rank Aj = core-rank A for j≥j0. In [5], Campbell et al. considered the applications of the Drazin inverse to the differential equation Ax+B=f. In [14], Kelly gave the results on the multivariate, real-valued functions on Rd induced matrix-valued functions on the space of d-tuples of n×n pairwise-commuting self-adjoint matrices. Kelly also examined the geometry of this space of matrices and also concluded that the best notion of differentiation of these matrix functions is differentiation along curves in [14]. There are not some results on the perturbation analysis for the differential of the Drazin inverse of the matrix-value function A(t). In this paper, we present some results of perturbation analysis for the differentiable of the Drazin inverse of the matrix-value function A(t) with respect to an interval t∈[a,b] for any complex numbers a,b. We will also consider the perturbation bound theory for the Drazin inverse of a matrix-value function A(t).


## 2. Preliminaries

Some notations and lemma will be presented in the section.Lemma 1 (see [6]).
Suppose thatA(t) and AD(t) are differentiable n×n matrix-valued functions. Let l be any integer such that l≥ind(A(t)) for all t. In particular, l can be taken to be n. Then(6)dADt=-ADtdAtADt+∑m=0l-1ADt2+mdAtAmtAπt+∑m=0l-1AπtAmtdAtADt2+m.From Lemma1, we derive the following lemma from the results in [1, 15].Lemma 2.
Let(7)M=AtCt0Bt∈Rn+m×n+m,N=At0DtBt∈Rn+m×n+m, where A(t)∈Rn×n,B(t)∈Rm×m, C(t)∈Rn×m, and D(t)∈Rm×n. If A(t) and B(t) are Drazin invertible, then M and N are Drazin invertible,(8)MD=ADtC0BDt,ND=ADt0DBDt,where(9)C=∑n=0∞ADtn+2CtBntBπt+Aπt∑n=0∞AntCtBDtn+2-ADtCtBDt,D=∑n=0∞BDtn+2DtAntAπt+Bπt∑n=0∞BntDtADtn+2-BDtDtADt.


## 3. Perturbation Analysis for the Drazin Inverse of Matrix-Valued Function

Leta,b be constants and a matrix-value function A(t)∈Rn×n with all t∈[a,b], the perturbation matrix is given as A(t+ϵ)=A(t)+A(ϵ), and the index of A(t) is denoted by ind(A(t)). For simplicity, let P=I-AD(t+ϵ)A(t+ϵ) in this paper. We will analyze the perturbation bound for the Drazin inverse of matrix-value function with the condition A(t)A(ϵ)=A(ϵ)A(t) and A(t)A(ϵ)=0, respectively.For simplicity, we first consider that elements ofA(t+ϵ) are linear function, which is given as A(t+ϵ)=A(t)+A(ϵ). Let l=max⁡{ind(A(t)),ind(A(t+ϵ))} and Cmi be Binomial coefficients.Theorem 3.
LetA(t)∈Rn×n be matrix-value function with a small perturbation A(ϵ) and A(ϵ)=A(t+ϵ)-A(t). Assume that AD(t) exits and it is differentiable for all t∈[a,b]. If A(ϵ)A(t)=A(t)A(ϵ), then(10)iADt+ϵ=Ct+CϵDI+Ct+CϵDNt+Nϵ-1,(11)iidADt+ϵ=-L2dAtL2+∑m=0l-1L22+mdAtL1I-L2At+Aϵ+∑m=0l-1I-L2At+AϵL1dAtL22+m, where C(t)+N(t), C(ϵ)+N(ϵ) are both core-nilpotent decomposition (see [12]), and(12)L1=∑i=0mCmiAitAm-iϵ,L2=Ct+CϵDI+Ct+CϵDNt+Nϵ-1.Proof.
(i) SinceA(t) is Drazin invertible, A(t) and AD(t) are presented by (2) and (3), respectively. By A(ϵ)A(t)=A(t)A(ϵ) and from (2), we obtain A3(ϵ)=0 and A4(ϵ)=0. By (2), we obtain(13)At+ϵ=PtA1t+A1ϵ00A2t+A2ϵP-1t.From (13), we prove that AD(t+ϵ)⇔(Ai(t)+Ai(ϵ))D,i=1,2. Now, we will consider the Drazin inverse properties of the matrix-value function A(t+ϵ). Similar to (2), then A1(ϵ) and A2(ϵ) have the block matrix form as follows:(14)A1ϵ=Q1ϵA11ϵ00A22ϵQ1-1ϵ,A2ϵ=Q2ϵA33ϵ00A44ϵQ2-1ϵ,where A11(ϵ), A33(ϵ) are invertible and A22(ϵ), A44(ϵ) are nilpotent. From Ai(t)Ai(ϵ)=Ai(ϵ)Ai(t), i=1,2, we obtain(15)A1t=Q1ϵA11t00A22tQ1-1ϵ,A2t=Q2ϵA33t00A44tQ2-1ϵ.By (14) and (15), we get(16)At+ϵ=A11t+A11ϵ0000A22t+A22ϵ0000A33t+A33ϵ0000(A44t+A44ϵ.From (16), we obtain AD(t+ϵ) exists ⇔{(Ajj(t)+Ajj(ϵ))D}j=14 exist.
Next, we present the perturbation analysis for the Drazin inverse of{Ajj(t)+Ajj(ϵ)}j=14, respectively. If A1(t) is invertible, from A1(t)A1(ϵ)=A1(ϵ)A1(t), it implies that A1(t)+A1(ϵ) is invertible. Similarly, A22(t)+A22(ϵ) is invertible. Since A22(t) is invertible, A22(ϵ) is nilpotent, and A22(t)A22(ϵ)=A22(ϵ)A22(t), which imply that A22-1(t)A22(ϵ) is a nilpotent matrix and(17)A22t+A22ϵD=A22-1tI+A22ϵA22-1t-1.Similarly, I+A33(t)A33-1(ϵ) is invertible and [A33(t)+A33(ϵ)]D=(I+A33(ϵ)-1A33(t))-1A33-1(ϵ). Since A44(t),A44(ϵ) are nilpotent and A44(t)A44(ϵ)=A44(ϵ)A44(t), we can conclude that A44(t)+A44(ϵ) is nilpotent. Thus [A44(t)+A44(ϵ)]D=0. Since A11(t)+A11(ϵ) is Drazin invertible and by (16)-(17), we obtain(18)ADt+ϵ=A11t+A11ϵD0000A22-1tI+A22ϵA22-1t-10000A33-1ϵA33tA33-1ϵ+I-100000.From (18), we get(19)ADt+ϵ=Ct+CϵDI+Ct+CϵDNt+Nϵ-1,where(20)(Ct+CϵD=PtA11t+A11ϵD⊕A22-1t⊕A33-1ϵ⊕0P-1t,I+Ct+CϵDNt+Nϵ-1=PtI⊕I+A22ϵA22-1t-1⊕I+A33tA33-1ϵ-1⊕IP-1t.
Next, we consider the derivative of the polynomial matrix with respect tot∈[-1,1]. As [12], let A(t)=C(t)+N(t) be the core-nilpotent decomposition of the matrix-valued function A(t), where C(t) is invertible and N(t) is nilpotent, respectively. Since the elements of the matrix are polynomials, we have A(t) is differentiable and C(t)+N(t) and N(t)+N(ϵ) are also. To this end, firstly we consider the expression of the derivative AD(t+ϵ). Note that dA(t+ϵ)=d[A(t)+A(ϵ)]=dA(t). By Lemma 1, we have(21)dADt+ϵ=-ADt+ϵdAtADt+ϵ+∑m=0l-1ADt+ϵ2+mdAtAmtP)+∑m=0l-1PAmt+ϵdAtADt+ϵ2+m.Since A(t)A(ϵ)=A(ϵ)A(t), we have [A(t)+A(ϵ)]m=∑i=0mCmiAi(t)Am-i(ϵ)=L1, where Cmi is Binomial coefficients.
By (19) and (21), we obtain(22)dADt+ϵ=-L2dAt×L2+∑m=0l-1L22+mdAtL1I-L2At+Aϵ+∑m=0l-1I-L2At+AϵL1dAtL22+m,where l=max⁡{ind(A(t)),ind(A(t+ϵ))} and L2=[(C(t)+C(ϵ))D[I+(C(t)+C(ϵ))D(N(t)+N(ϵ))]-1.Theorem 4.
LetA(t) be an matrix-valued function with a small scalar ϵ and A(ϵ)=A(t+ϵ)-A(t). If A(t)A(ϵ)=0, then A(t+ϵ) is Drazin invertible and(23)ADt+ϵ=Aπϵ∑n=0l-1AnϵADtn+1+ADϵ∑n=0l-1AnϵDAntAπt.Furthermore, if AD(t+ϵ) differentiable, then(24)dADt+ϵ=Ω1dAtΩ1+Ω2+Ω2′+∑m=3l-1Ω12+mdAtAmtPP∑m=3l-1AmtdAtΩ12+ml≥3,Ω2+Ω2′0≤l≤2.where(25)Ω1=Aπϵ∑n=0l-1AnϵADtn+∑n=0l-1ADϵn+1AntAπt,(26)Ω2=Ω12dAt+Ω1dAt+Ω1dAtAt+AϵAt+AϵP,(27)Ω2′=PdAt+At+AϵdAt+At+AϵdAtΩ1Ω1Ω12.Proof.
Firstly, as the proof of Theorem3, we derive (13). From A(t)A(ϵ)=0, we have(28)AtAϵ=PtA1tA1ϵA1tA3ϵA2tA4ϵA2tA2ϵP-1t=0,where A1(t) is invertible and A2(t) is nilpotent. Thus, we prove that(29)A1ϵ=0,A3ϵ=0,A2tA4ϵ=0,A2tA2ϵ=0. Therefore, we get that(30)At+ϵ=PtA1t0A4ϵA2t+A2ϵP-1t. From Lemma 2, it implies that A1(t) and A2(t)+A2(ϵ) are both Drazin invertible. Note that A1(t) is invertible. Here, we need to consider the Drazin invertible properties of A2(t)+A2(ϵ) and its expression. Since A2(t)A2(ϵ)=0 and by the Theorem 5.1.3 of [[15], Chapter V], we prove that A2(t)+A2(ϵ) is Drazin invertible. As (2), A2(ϵ) and A2(t) can be rewritten as(31)A2ϵ=P1tA11ϵ00A22ϵP1-1t,A2t=P1tA11tA12tA21tA22tP1-1t,where A11(ϵ) is invertible and A22(t) is nilpotent. By A2(t)A2(ϵ)=0, we obtain(32)A12tA22ϵ=0,A22tA22ϵ=0,A11t=0,A21t=0.
Therefore, we get(33)A2t+A2ϵ=P1tA11ϵA12t0A22t+A22ϵP1-1t.Since A2(t)A2(ϵ)=0, A22(t)A22(ϵ)=0. By the Lemma 5.1.1 of [[15], Chapter V], A22(t)+A22(ϵ) is nilpotent. Thus, [A22(t)+A22(ϵ)]D=0. By Lemma 2, we get(34)A2t+A2ϵD=P1tA11-1ϵA12t00P1-1t,where A12(t)=∑n=0l[A11-1(ϵ)]n+2A12(t)[A22(t)]n. Note that(35)A2t+A2ϵD=A2Dϵ+A2D2ϵ∑n=0l-1A2DϵnA2ntI-A2DϵA2ϵ.where l≥ind(A2(t)). Since A2(t)A2D(ϵ)=0, it implies that A2(t)A2(ϵ)=0. According to (35), we have(36)A2t+A2ϵπ=A2πϵ-A2Dϵ∑n=1lA2Dϵn-1A2ntA2πϵ.Since A1(t) is invertible, by (8) of Lemma 2 and (35),(37)At+ϵD=PtA1-1t0A4ϵ∑n=0lA2Dϵn+2A2n+1tP-1t,where l=max⁡{ind(A(t)),ind(A(ϵ)),ind(A(t)+A(ϵ))}, PB=I-[A2t+A2ϵ][A2(t)+A2(ϵ)]D, and(38)A4ϵ=PB∑n=0lA2t+A2ϵnA4ϵA1-1tnA1-1t2-A2Dϵ∑n=0lA2DϵnA2ntA4ϵA1-1t=PB∑n=0lA2ϵnA4ϵA1-1tnA1-1t2-A2Dϵ∑n=0lA2DϵnA2ntA4ϵA1-1t=A2πϵ∑n=0∞A2ϵnA4ϵA1-1tnA1-1t2-A2Dϵ∑n=0lA2DϵnA2ntA4ϵA1-1t.By (37), AD(t+ϵ) exists and its expression is given as in (25).
According to Lemma1 and (ii), we obtain(39)dAt+ϵD=-ADt+ϵdAt+ϵADt+ϵ+∑m=0l-1ADt+ϵ2+mdAt+ϵAmt+ϵP+P∑m=0l-1Amt+ϵdAt+ϵADt+ϵ2+m.Since A(t+ϵ)=A(t)+A(ϵ), we derive(40)ADt+ϵdAt+ϵADt+ϵ=Ω1dAtΩ1,where(41)Ω1=Aπϵ∑n=0l-1AnϵADtn+∑n=0l-1ADϵn+1AntAπt. If A(t)A(ϵ)=0, by (10) and (40), we have(42)∑m=0l-1ADt+ϵ2+mdAt+ϵAmt+ϵP=Ω2+∑m=3l-1Ω12+mdAtAmtP,l≥3,Ω2,0≤l≤2,where(43)Ω2=Ω12dAt+Ω1dAt+Ω1dAtAt+AϵAt+AϵP. Similarly, we obtain(44)P∑m=0l-1Amt+ϵdAt+ϵADt+ϵ2+m=Ω2′+P∑m=3l-1AmtdAtΩ12+m,l≥3,Ω2′,0≤l≤2,where Ω2′=P{dA(t)+(A(t)+A(ϵ))[dA(t)+(A(t)+A(ϵ))dA(t)Ω1]Ω1}Ω12.
According to the result (10) of (i) and (39), (40), (42), and (44), we have the following simplification: (45)dAt+ϵD=Ω1dAtΩ1+Ω2+Ω2′+∑m=3l-1Ω12+mdAtAmtP+P∑m=3l-1AmtdAtΩ12+m,l≥3,Ω2+Ω2′,0≤l≤2.


## 4. Perturbation Bound for the Drazin Inverse of the Matrix-Value Function

In this subsection, we consider the perturbation bound for the Drazin inverse and its differential of the matrix-value functionA(t) with any t in Ω.Theorem 5.
LetA(t)∈Rn×n, A(ϵ)∈Cn×n be both Drazin invertible matrix-value function and A(ϵ)A(t)=A(t)A(ϵ). Then(46)iADt+ϵ-AtD≤KPt1+2β1+λminAtI+AϵADtDλmin2At+KPtΩ02ADϵI+AtADϵDλminAϵ. Furthermore, we obtain(47)iiAD2t+ϵ-AD2t≤KPt1+2β1+λminAtI+AϵADtDλmin2At+KPtΩ02ADϵI+AtADϵDλminAϵ×KPt1+2β1+λminAtI+AϵADtDλmin2At+KPtΩ02ADϵI+AtADϵDλminAϵ+ADt,(48)iiidAt+ϵD-dAtD≤2ADt2dADt1+I+ADtAϵDI+AϵADtD+2dCtCt+CϵD2I+Nt+NϵCt+CϵD-13×I-Ct+NtCt+CϵD+CD2tI-CDtCt, where K(P(t))=P(t)P-1(t), 0<β≤1, and(49)Ω0=I-ADtAtAϵDI-ADtAtAϵ.Proof.
(i) As Theorem3, let A(t+ϵ)=A(t)+A(ϵ). If A(ϵ)A(t)=A(t)A(ϵ), then(50)AtAϵ=AϵAt⇒A1tA1ϵ=A1ϵA1t⇒A11tA11ϵ=A11ϵA11t.Thus, we have A11(t)(Ir1+A11-1(t)A11(ϵ))=Ir1+A11-1tA11ϵA11t and(51)A11tI+A11-1tA11ϵD-A11-1t=Ir1+A11-1tA11ϵD-Ir1A11-1t.As (51), we also have(52)Ir1+A11ϵA11-1tA11tD-A11-1t=A11-1tIr1-Ir1+A11ϵA11-1tD. Similarly, we derive(53)A22-1tIr2+A22ϵA22-1t-1-A22-1t=-A22-2tA22ϵIr2+A22ϵA22-1t-1,(54)Ir2+A22-1tA22ϵ-1A22-1t-A22-1t=-Ir2+A22ϵA22-1t-1A22ϵA22-2t.From (10), (51), and (53), we have(55)ADt+ϵ-ADt=Pt[A11-1tIr1-Ir1+A11ϵA11-1tD0000-A22-2tA22ϵIr2+A22ϵA22-1t-10000A33-1ϵIr3+A33tA33-1ϵ-100000P-1t. Therefore, we get(56)ADt+ϵ-ADt≤KPtA11-1tIr1-Ir1+A11ϵA11-1tD+A22-1t2A22ϵIr2+A22ϵA22-1t-1+A33-1ϵIr3+A33tA33-1ϵ-1≤KPtλmin2AtIr1-Ir1+A11ϵA11-1tD+λminAtA22ϵIr2+A22ϵA22-1t-1+KPtλminAϵIr3+A33tA33-1ϵ-1,where K(P(t))=P(t)P-1(t), λmin(A(t)), and λmin(A(ϵ)) denote the minimum absolute value and nonzero eigenvalues for the matrix-value function A(t) and the constant matrix A(ϵ), respectively.
If0<β≤1, then(57)Ir1+A11ϵA11-1tD+Ir2+A22ϵA22-1t-1≤2βI+AϵADtD.Similarly, we have(58)Ir3+A33tA33-1ϵ-1=AπtAϵDAπtAϵI+AtADϵD.By (57) and (58), we have(59)Φ1≤2Ψ1+Ψ21/2≤2β1+λminAtI+AϵADtD1/2, where (60)Φ1=λminAtIr1+A11ϵA11-1tD+A22ϵIr2+A22ϵA22-1t-1,Ψ1=Ir1+A11ϵA11-1tD2,Ψ2=A22ϵ2Ir2+A22ϵA22-1t-12. From (56)-(58), we have(61)ADt+ϵ-ADt≤KPt1+2β1+λminAtI+AϵADtDλmin2At+KPtΩ02ADϵI+AtADϵDλminAϵ,where Ω0=[Aπ(t)A(ϵ)]D[Aπ(t)A(ϵ)]. From (19), we derive(62)ADt+ϵ≤KPtλmin2AtIr1+A11ϵA11-1tD+λminAtA22ϵIr2+A22ϵA22-1t-1+KPtλminAϵIr3+A33tA33-1ϵ-1≤KPt2β1+λminAtI+AϵADtDλmin2At+KPtΩ02ADϵI+AtADϵDλminAϵ.By (62), we get(63)AD2t+ϵ-AD2t≤ADt+ϵ-ADtADt+ϵ+ADt≤KPt1+2β1+λminAtI+AϵADtDλmin2At+KPtΩ02ADϵI+AtADϵDλminAϵ×KPt1+2β1+λminAtI+AϵADtDλmin2At+KPtΩ02ADϵI+AtADϵDλminAϵ+ADt.As (51), we obtain(64)A11t+A11ϵDdA11tA11t+A11ϵD-dA11-1t≤A11-1t2dA11t1+Ir1+A11-1tA11ϵDIr1+A11ϵA11-1tD≤ADt2dADt1+I+ADtAϵDI+AϵADtD,A22t+A22ϵ-1dA22tA22t+A22ϵ-1-dA22-1t≤A22-1t2dA22t1+Ir2+A22ϵA22-1t-1Ir2+A22-1tA22ϵ-1≤ADt2dADt1+I+ADtAϵDI+AϵADtD.Since l=max⁡{ind(A(t)),ind(A(t+ϵ))} (see [[6], Theorem 2] for detail), we have(65)AD2+ltdCtArtAπt=0,AD2+lt+ϵdCtArt+ϵP=0.According to (65) (see [6, (21),(22)] for detail), then(66)∑m=0l-1ADt+ϵm+2dAtAmt+ϵP=AD2t+ϵdCtP,∑m=0l-1ADtm+2dAtAmtAπt=AD2tdCtAπt.Similarly, we also have(67)∑m=0l-1PAmt+ϵdAtADt+ϵm+2=PdCt)AD2t+ϵ,∑m=0l-1tAπtAmtdAtADtm+2=AπtdCt)AD2t.From (66) and for AD(X)=CD(X), X=t or t+ϵ, we obtain(68)∑m=0l-1ADt+ϵm+2dAtAmt+ϵP-∑m=0l-1ADtm+2dAtAmtAπt≤dCt(Ct+CϵD2I+Nt+NϵCt+CϵD-13×I-Ct+NtCt+CϵD+CD2tCπt.Similarly, we have(69)∑m=0l-1PAmt+ϵdAtADt+ϵm+2-∑m=0l-1AπtAmtdAtADtm+2≤dCt(Ct+CϵD2I+Nt+NϵCt+CϵD-13×I-Ct+NtCt+CϵD+CD2tCπt.By (64),(68), and (69), we obtain(70)dAt+ϵD-dADt≤2ADt2dADt1+I+ADtAϵDI+AϵADtD+2dCt(Ct+CϵD2I+Nt+NϵCt+CϵD-13×I-Ct+NtCt+CϵD+CD2tCπt.Theorem 6.
LetA(t)∈Rn×n and A(ϵ)∈Cn×n be two Drazin invertible matrix-value functions. If A(t)A(ϵ)=0, then(71)iADt+ϵ-AtD≤ω3,(72)iiAD2t+ϵ-AD2t≤ω3Ω1+ADt.
Furthermore, we have(73)iiidAt+ϵD-dADt≤ω3dAtADt+ϵ+ADt+2×PADt2dCt+ω42dCt+AD2tdCtω3At+ADt+ϵAϵ,l≥3,2ω3′PdAt1+Ω1At+ϵ+Ω1AϵAt+A2ϵ+2ADt2×ω41+K1+Ω12A2t+AϵAt+A2ϵ+ADtAϵ+ω3At+ϵdAtP+AϵAt+A2ϵ+ω3′dAtAt2Aπt,0≤l≤2.Proof.
SinceA(t)A(ϵ)=0 and as the proof of Theorem 4, we have(74)At+ϵD-ADt≤AϵADϵADt+Aπϵ∑n=1lAϵnADtn+1+∑n=0lADϵn+1AtnAπt=ω3,where l=max⁡{ind(A(t)),ind(A(ϵ)),ind(A(t+ϵ))}. From (i), we have(75)Ω12-AD2t≤Ω1Ω1-ADt+Ω1-ADtADt≤Ω1Ω1-ADt+Ω1-ADtADt≤ω3Ω1+ADt.Note that(76)ADt+ϵdAtADt+ϵ-ADtdAtADt=ADt+ϵdAtADt+ϵ-ADt+ADt+ϵ-ADtdAtADt.As (66), we have(77)∑m=0l-1ADt+ϵm+2dAtAmt+ϵP=ADt+ϵ2dCtP,∑m=0l-1ADtm+2dAtAmtI-ADtAt=ADt2dCtAπt,∑m=0l-1PAmt+ϵdAtADt+ϵm+2=PdCt)ADt+ϵ2,∑m=0l-1AπtAmtdAtADtm+2=AπdCtAD2t. Thus, we obtain (78)∑m=0l-1ADt+ϵm+2dAtAmt+ϵP-∑m=0l-1ADtm+2dAtAmtAπt=ADt+ϵ2-ADt)2dCtP)+AD2tdCtADt-ADt+ϵAt-ADt+ϵAϵ,∑m=0l-1PAmt+ϵdAtADt+ϵm+2-∑m=0l-1tAπtAmtdAtADtm+2=PdCtAD2t+ϵ-AD2t+ADt-ADt+ϵAt-ADt+ϵAϵdCtAD2t. Note that(79)ADt+ϵ≤Aπϵ∑n=0lAϵnADtn+1+∑n=0lADϵn+1AtnAπt=ω4 and from (76), we have(80)ADt+ϵdAtADt+ϵ-ADtdAtADt≤ω3dAtADt+ϵ+ADt,∑m=0l-1ADt+ϵm+2dAtAmt+ϵP-∑m=0l-1ADtm+2dAtAmtAπt≤PADt2dCt+ω42dCt+ADt2dCtω3At+ADt+ϵAϵ,∑m=0l-1PAmt+ϵdAtADt+ϵm+2-∑m=0l-1AπtAmtdAtADtm+2≤PADt2dCt+ω72dCt+ADt2dCtω3At+ADt+ϵAϵ.For 0≤l≤2 and from Theorem 4 and Lemma 1, we obtain(81)Ω12dAt+Ω1dAt+Ω1dAtAt+ϵAt+ϵP-AD2tdAt+ADtdAt+ADtdAtAtAtAπt≤PΩ12-AD2tdAt+Ω1dAt+Ω1dAtAt+ϵAt+ϵ+ADt2dAtP-Aπt+Ω1-ADtdAtAt+ϵP+ADtdAtAt+ϵP-AtAπt+Ω12dAtAt+ϵP-AD2tdAtAtAπt≤ω3′PdAt1+Ω1At+ϵ+Ω1AϵAt+A2ϵ+ADt2×ω41+K1+Ω12A2t+AϵAt+A2ϵ+ADtAϵ+ω3At+ϵdAtP+AϵAt+A2ϵ+ω3′dAtA2tAπt,where K1=A(t)AD(t) and ω4=dA(t)(-AD(t)A(ϵ)+ADt-Ω1At+ϵ).It is proved by (iii) of Theorem 5.


## 5. Applications

In the section, we state applications of our results given in Section4 to the perturbation bound for three matrix equations.


### 5.1. For CaseA(t)X=B(t)

We first consider the following equation [16]:(82)AtX=Bt,where A(t)∈Rn×n is a matrix-value function with ind(A)=k for all t∈[a,b] and B(t)∈Rn×n is given and X is unknown. If the range space R(B(t))⊂R(A(t)k), then(83)Xt=ADtBt.where R(X)⊂R(Ak(t)) [16]. Assume that there are perturbation constant matrices A(ϵ) and B(ϵ) such that A(t)+A(ϵ) and B(t)+B(ϵ) are both Drazin invertible. Thus, there exists a matrix-value function X^ with R(X^)⊂R(Ak(t+ϵ)), which satisfies(84)At+AϵX^=Bt+Bϵ.For simplicity, in the following results, we defineA(t+ϵ)≔A(t)+A(ϵ) and B(t+ϵ)≔B(t)+B(ϵ). We also assume that α,β,c are given in Theorem 5 and B(ϵ)/B(t)≪1, and define(85)G=I+Ct+CϵDNt+Nϵ,K1At=AtADt.From Theorem 3 and Theorem 5, we have the following result.Theorem 7.
LetA(t) and B(t) satisfy (82). Assume that A(t)A(ϵ)=A(ϵ)A(t). If R(B(t+ϵ))⊂R(A(t+ϵ)k) and R(X)⊂R(Ak(t+ϵ)), then(86)X^-XX≤4αβKPtK1At+cσmaxG-1Ct+CϵDAt+4αβKPtK1AtI+AϵADtD1+αADtA22ϵ+KPtΩ02AtADϵI+AtADϵD.Proof.
IfR(B(t+ϵ))⊂R(A(t+ϵ)k), then X^=AD(t+ϵ)B(t) satisfies (84). The following proof can be derived from Theorem 3 and Theorem 5. Note that easily(87)X^-X=ADt+ϵBt+ϵ-ADt+ϵBt+ADt+ϵBt-ADtBt=ADt+ϵBt+ϵ-Bϵ+ADt+ϵ-ADtBt.From (62), we obtain(88)ADt+ϵ≤(Ct+CϵDI+Ct+CϵDNt+Nϵ-12≤cσmaxG-1(Ct+CϵD,where G=I+[C(t)+C(ϵ)]D[N(t)+N(ϵ)]. Note that B=AX≤AX⇒1/X≤A/B⇒∥X≥B/A. According to B(ϵ)≪B(t), Theorem 3, and Theorem 5, we have(89)X^-XX≤ADt+ϵBϵAt+ADt+ϵ-ADtBtAtBt≤ADt+ϵBϵAtBt+ADt+ϵ-ADtAt≤At/λminAt+K1AtKPt2β1+λminAtI+AϵADtDλminAt+KPtΩ02AtADϵI+AtADϵDλminAϵ+KPtK1At+AtΩ02ADϵI+AtADϵDλminAϵ.
Similarly, by Theorem4 and Theorem 6, we obtain the following results.Theorem 8.
LetA(t) and B(t) satisfy (82) with A(t+ϵ)=A(t)+A(ϵ) and B(t+ϵ)=B(t)+B(ϵ). Assume that A(t)A(ϵ)=0. If R(B(t+ϵ))⊂R(A(t+ϵ)k) and R(X)⊂R(Ak(t+ϵ)), then X^-X/X≤K1(A(t))(ω4+ω3)/AD(t).


### 5.2. For CaseXA(t)=B(t)

In the following, we consider the solutions for the equation as follows:(90)XAt=Btwhere A(t) and B(t) are given in (90). If the null space N(B(t))⊃N(A(t)k), then(91)Xt=BtADt,where N(X)⊃N(Ak(t)) [16]. As (84), we have(92)At+AϵX^=Bt+Bϵ.The following result can derived from Theorem3 and Theorem 5.Theorem 9.
LetA(t) and B(t) satisfy (90). Assume that A(t)A(ϵ)=A(ϵ)A(t). If N(B(t+ϵ))⊃N(A(t+ϵ)k) and N(X)⊃N(Ak(t+ϵ)) satisfy (92), then(93)X^-XX≤4αβKPtK1At+cσmaxG-1Ct+CϵDAt+4αβKPtK1AtI+AϵADtD1+αADtAϵ+KPtΩ02AtADϵI+AtADϵD.The following result can derived from Theorem4 and Theorem 6.Theorem 10.
LetA(t) and B(t) satisfy (90). Assume that A(t)A(ϵ)=0. If R(B(t+ϵ))⊂R(A(t+ϵ)k) and R(X)⊂R(Ak(t+ϵ)), then X^-X/X≤K1(A(t))(ω4+ω3)/AD(t).


### 5.3. For CaseA(t)XB(t)=D(t)

Similarly, we consider the equation as follows:(94)AtXBt=Dtwhere A(t)∈Rn×n with ind(A)=k1, B(t)∈Rm×m with ind(B)=k2 and D(t)∈Rn×m are all given, k=max⁡{k1,k2}, and X is unknown. If the range space R(D(t))⊂R(A(t)k1) and the null space N(D(t))⊃N(B(t)k2), then(95)Xt=ADtDtBDt.where R(X)⊂R(Ak(t)) and N(X)⊂N(Bk(t)).For simplicity, letB(t)=C′(t)+N′(t) and B(ϵ)=C′(ϵ)+N′(ϵ) be the core-nilpotent decomposition of B(t) and B(ϵ) in the following result, respectively. We also assume that(96)G2=I+C′t+C′ϵDN′t+N′ϵ,K2Bt=BtBDt.Thus, we have the following result as Theorems 9 and 10.Theorem 11.
LetA(t) and B(t) satisfy the equation (94). Assume that(97)AtAϵ=AϵAt,BtBϵ=BϵBt,RBt+ϵ⊂RAt+ϵk1,NBt+ϵ⊃NAt+ϵk2. If R(X)⊂R(Ak(t+ϵ)) and N(X)⊃N(Ak(t+ϵ)), then(98)X^-XX≤c1σmaxG1-1Ct+CϵDAtBtc2σmaxG2-1C′t+C′ϵD+4αβKPtADtI+AϵADtD1+αADtAϵ+KPtΩ02ADϵI+AtADϵD+4αβKPtADt+4αβK2BtK1AtKPtI+AϵADtD1+αADtAϵ+K2BtAtKPtΩ02ADϵI+AtADϵD+4αβKPtK2BtK1At,X^-XX≤ω4K1AtK2BtADt×ω3+ω4′+ω3′BDt.


## 5.1. For CaseA(t)X=B(t)

We first consider the following equation [16]:(82)AtX=Bt,where A(t)∈Rn×n is a matrix-value function with ind(A)=k for all t∈[a,b] and B(t)∈Rn×n is given and X is unknown. If the range space R(B(t))⊂R(A(t)k), then(83)Xt=ADtBt.where R(X)⊂R(Ak(t)) [16]. Assume that there are perturbation constant matrices A(ϵ) and B(ϵ) such that A(t)+A(ϵ) and B(t)+B(ϵ) are both Drazin invertible. Thus, there exists a matrix-value function X^ with R(X^)⊂R(Ak(t+ϵ)), which satisfies(84)At+AϵX^=Bt+Bϵ.For simplicity, in the following results, we defineA(t+ϵ)≔A(t)+A(ϵ) and B(t+ϵ)≔B(t)+B(ϵ). We also assume that α,β,c are given in Theorem 5 and B(ϵ)/B(t)≪1, and define(85)G=I+Ct+CϵDNt+Nϵ,K1At=AtADt.From Theorem 3 and Theorem 5, we have the following result.Theorem 7.
LetA(t) and B(t) satisfy (82). Assume that A(t)A(ϵ)=A(ϵ)A(t). If R(B(t+ϵ))⊂R(A(t+ϵ)k) and R(X)⊂R(Ak(t+ϵ)), then(86)X^-XX≤4αβKPtK1At+cσmaxG-1Ct+CϵDAt+4αβKPtK1AtI+AϵADtD1+αADtA22ϵ+KPtΩ02AtADϵI+AtADϵD.Proof.
IfR(B(t+ϵ))⊂R(A(t+ϵ)k), then X^=AD(t+ϵ)B(t) satisfies (84). The following proof can be derived from Theorem 3 and Theorem 5. Note that easily(87)X^-X=ADt+ϵBt+ϵ-ADt+ϵBt+ADt+ϵBt-ADtBt=ADt+ϵBt+ϵ-Bϵ+ADt+ϵ-ADtBt.From (62), we obtain(88)ADt+ϵ≤(Ct+CϵDI+Ct+CϵDNt+Nϵ-12≤cσmaxG-1(Ct+CϵD,where G=I+[C(t)+C(ϵ)]D[N(t)+N(ϵ)]. Note that B=AX≤AX⇒1/X≤A/B⇒∥X≥B/A. According to B(ϵ)≪B(t), Theorem 3, and Theorem 5, we have(89)X^-XX≤ADt+ϵBϵAt+ADt+ϵ-ADtBtAtBt≤ADt+ϵBϵAtBt+ADt+ϵ-ADtAt≤At/λminAt+K1AtKPt2β1+λminAtI+AϵADtDλminAt+KPtΩ02AtADϵI+AtADϵDλminAϵ+KPtK1At+AtΩ02ADϵI+AtADϵDλminAϵ.
Similarly, by Theorem4 and Theorem 6, we obtain the following results.Theorem 8.
LetA(t) and B(t) satisfy (82) with A(t+ϵ)=A(t)+A(ϵ) and B(t+ϵ)=B(t)+B(ϵ). Assume that A(t)A(ϵ)=0. If R(B(t+ϵ))⊂R(A(t+ϵ)k) and R(X)⊂R(Ak(t+ϵ)), then X^-X/X≤K1(A(t))(ω4+ω3)/AD(t).


## 5.2. For CaseXA(t)=B(t)

In the following, we consider the solutions for the equation as follows:(90)XAt=Btwhere A(t) and B(t) are given in (90). If the null space N(B(t))⊃N(A(t)k), then(91)Xt=BtADt,where N(X)⊃N(Ak(t)) [16]. As (84), we have(92)At+AϵX^=Bt+Bϵ.The following result can derived from Theorem3 and Theorem 5.Theorem 9.
LetA(t) and B(t) satisfy (90). Assume that A(t)A(ϵ)=A(ϵ)A(t). If N(B(t+ϵ))⊃N(A(t+ϵ)k) and N(X)⊃N(Ak(t+ϵ)) satisfy (92), then(93)X^-XX≤4αβKPtK1At+cσmaxG-1Ct+CϵDAt+4αβKPtK1AtI+AϵADtD1+αADtAϵ+KPtΩ02AtADϵI+AtADϵD.The following result can derived from Theorem4 and Theorem 6.Theorem 10.
LetA(t) and B(t) satisfy (90). Assume that A(t)A(ϵ)=0. If R(B(t+ϵ))⊂R(A(t+ϵ)k) and R(X)⊂R(Ak(t+ϵ)), then X^-X/X≤K1(A(t))(ω4+ω3)/AD(t).


## 5.3. For CaseA(t)XB(t)=D(t)

Similarly, we consider the equation as follows:(94)AtXBt=Dtwhere A(t)∈Rn×n with ind(A)=k1, B(t)∈Rm×m with ind(B)=k2 and D(t)∈Rn×m are all given, k=max⁡{k1,k2}, and X is unknown. If the range space R(D(t))⊂R(A(t)k1) and the null space N(D(t))⊃N(B(t)k2), then(95)Xt=ADtDtBDt.where R(X)⊂R(Ak(t)) and N(X)⊂N(Bk(t)).For simplicity, letB(t)=C′(t)+N′(t) and B(ϵ)=C′(ϵ)+N′(ϵ) be the core-nilpotent decomposition of B(t) and B(ϵ) in the following result, respectively. We also assume that(96)G2=I+C′t+C′ϵDN′t+N′ϵ,K2Bt=BtBDt.Thus, we have the following result as Theorems 9 and 10.Theorem 11.
LetA(t) and B(t) satisfy the equation (94). Assume that(97)AtAϵ=AϵAt,BtBϵ=BϵBt,RBt+ϵ⊂RAt+ϵk1,NBt+ϵ⊃NAt+ϵk2. If R(X)⊂R(Ak(t+ϵ)) and N(X)⊃N(Ak(t+ϵ)), then(98)X^-XX≤c1σmaxG1-1Ct+CϵDAtBtc2σmaxG2-1C′t+C′ϵD+4αβKPtADtI+AϵADtD1+αADtAϵ+KPtΩ02ADϵI+AtADϵD+4αβKPtADt+4αβK2BtK1AtKPtI+AϵADtD1+αADtAϵ+K2BtAtKPtΩ02ADϵI+AtADϵD+4αβKPtK2BtK1At,X^-XX≤ω4K1AtK2BtADt×ω3+ω4′+ω3′BDt.



---

*Source: 8136215-2019-01-02.xml*