ESTIMATION OF A CONDITION NUMBER RELATED TO THE WEIGHTED DRAZIN INVERSE

Vol. 39, No. 1, 2009, 1-9

ESTIMATION OF A CONDITION NUMBER RELATED TO THE WEIGHTED DRAZIN INVERSE

Dijana Mosi´c¹

Abstract. In this paper we get the formula for the condition number of theW-weighted Drazin inverse solution of a linear system W AW x=b, whereA is a bounded linear operator between Hilbert spacesX and Y, W is a bounded linear operator between Hilbert spacesY andX,xis an unknown vector in the range of (AW)^D andbis a vector in the range of (W A)^D.

AMS Mathematics Subject Classification (2000): 47A05, 15A09

Key words and phrases: NW–weighted Drazin inverse, condition number of a linear system

1. Introduction

In this paper X and Y denote arbitrary Hilbert spaces. We use B(X, Y) to denote the set of all linear bounded operators from X to Y. Set B(X) = B(X, X).

LetA ∈ B(X, Y), W ∈ B(Y, X) be nonzero operators. If there exists S ∈ B(X, Y) satisfying

(AW)^k+1SW = (AW)^k, SW AW S=S, AW S=SW A,

for some nonnegative integer k, thenS is called the W-weighted Drazin inverse of A and denoted by S =Ad,W [5]. If there exists Ad,W, then we say that A is W-Drazin invertible and Ad,W must be unique [5]. If X = Y, A ∈ B(X) and W = I, then S = A^D, the ordinary Drazin inverse of A [1]. We use i(S) to denote the Drazin index of S ∈ B(X). IfS has a Drazin inverse, then i(S) = inf{k∈N :S^k =S^k+1S^D}.

Let us recall that if A ∈ B(X, Y) and W ∈ B(Y, X) then the following conditions are equivalent [4]:

(1) AisW-Drazin invertible, (2) AW is Drazin invertible, (3) W Ais Drazin invertible.

1Department of Mathematics and Informatics, Faculty of Sciences and Mathematics, Uni- versity of Niˇs, P.O. Box 224, Viˇsegradska 33, 18000 Niˇs, Serbia, e-mail: [email protected]

Let A ∈ B(X, Y), W ∈ B(Y, X) and let A be W-Drazin invertible. Then AW andW A are Drazin invertible and

X =N((W A)^D)⊕R((W A)^D), Y =N((AW)^D)⊕R((AW)^D).

Let X and Y be equipped with norms k · kX and k · kY. The Q-norm for a vectorx∈X, theP-norm for a vectory∈Y and theP Q-norm for an operator A∈ B(X, Y) are defined by

kxkQ = q

kx1k²_X+kx2k²_X, kykP =

q

ky1k²_Y +ky2k²_Y, kAkP Q= sup

kxkQ≤1

kAxkP

where

x=x1+x2, x1∈R((W A)^D), x2∈N((W A)^D), y=y1+y2, y1∈R((AW)^D), y2∈N((AW)^D).

Notice that we can also change the inner product inX in the following way:

hx, yiP =hx1, y1iX+hx2, y2iX

where

x=x1+x2, y=y1+y2, x1, y1∈R((W A)^D), x2, y2∈N((W A)^D).

Now,k · kP is induced byh·,·iP. Similarly, forh·,·iQ andk · kQ inY. From [3] we can writeA, W in the form

A=A₁⊕A₂, W =W₁⊕W₂,

with A1, W1 invertible and W2A2 and A2W2 quasinilpotent. Hence, the W–

weighted Drazin inverse ofAhas the form

Ad,W = (W1A1W1)⁻¹⊕0.

Let us consider the equation

W AW x=b, b∈R((W A)^D).

Then there exists a uniquex∈R((AW)^D) such that x=Ad,Wb.

We say thatB ∈ B(X, Y) obeys the condition (W) atAif

B−A=AW Ad,WW(B−A)W AW Ad,W and kAd,WW(B−A)kkWk<1.

SetE=B−A.

IfF is a continuously differentiable function F :B(X, Y)×X −→Y

(A, x)7−→F(A, x),

the absolute condition number of F at x is the scalar kF⁰(x)k. The relative condition number ofF atxis

kF⁰(x)kkxkX

kykY . Following [2] we introduce the operator

F :B(X, Y)×X −→Y (A, b)7−→F(A, b) =Ad,Wb=x.

We know that the operatorF is a differentiable function, if the perturbationE ofA fulfils the following condition:

Ad,W(W AW)EW =EW, W E(W AW)Ad,W =W E.

(1)

We need the following important theorem.

Theorem 1.1. ([4]) Let A, B ∈ B(X, Y), W ∈ B(Y, X), let A be W–Drazin invertible and let B obey condition (W) atA. Then B is W–Drazin invertible, (BW)(Bd,WW) = (AW)(Ad,WW), i(BW) =i(AW),

B_d,W = (I+A_d,WW EW)⁻¹A_d,W =A_d,W(I+W EW A_d,W)⁻¹, R(Bd,W) =R(Ad,W) and N(Bd,W) =N(Ad,W).

The norm on the data is the norm inB(X, Y)×X defined as (A, b)7−→ k[αW AW, βb]k=

q

α²kW AWk²_QP+β²kbk²_Q.

In [2], Cui and Diao investigated the condition number of theW-weighted Drazin inverse solution of a linear system W AW x =b, whereA is an m×n rank deficient matrix, the index ofAW isk1, the index ofW Aisk2, b is a real vector of the sizenin the range of (W A)^k1,xis a real vector of the sizemin the range of (AW)^k2. For two positive real numbersαand β, they considered the weighted Frobenius normk[αW AW, βb]k^(F)_Q,P˜and gave the explicit formula of the condition number of the W-weighted Drazin inverse solution of a rectangular linear system. In this paper we extend the result obtained in [2] to linear bounded operators between Hilbert spaces.

2. Results

Now, we prove the following result.

Theorem 2.1. If the perturbation E in A fulfills the condition (1), then the absolute condition number of the W–weighted Drazin inverse solution of linear system, with the norm

k[αW AW, βb]k= q

α²kW AWk²_QP +β²kbk²_Q on the data(A, b)and the normkxkP on the solution, satisfies

C≤ kAd,WkP Q

s 1

β² +kxk²_P α² .

Let(En)n be a sequence of perturbations inAfulfilling the condition(1)and let (fn)n be a sequence of perturbations in b. If Cn is the corresponding absolute condition number, then

Cn→ kAd,WkP Q

s 1

β² +kxk²_P

α² , n→ ∞.

Hence,kAd,WkP Q

q

1

β² +^kxk_α2^2P is a sharp bound.

Proof. We know that F(A, b) = Ad,Wb. Under the condition (1), F is a differentiable function andF⁰ is defined as follows

F⁰(A, b)|_(E,f)= lim

²→0

(A+²E)d,W(b+²f)−Ad,Wb

² ,

whereE is the perturbation of Aandf is the perturbation ofb.

SinceE satisfies the condition (1), we have ([4])

(A+²E)d,W =Ad,W−²Ad,WW EW Ad,W +O(²²), and then we can easily get that

F⁰(A, b)|(E,f)=−Ad,WW EW x+Ad,Wf.

Then

kF⁰(A, b)|_(E,f)kP = kAd,W(W EW x−f)kP

≤ kAd,WkP Q(kW EWkQPkxkP +kfkQ).

The norm of a linear mapF⁰(A, b) is the supermum ofkF⁰(A, b)|_(E,f)kP on the unit ball ofB(X, Y)×X.Since

k[αW EW, βf]k²=α²kW EWk²_QP+β²kfk²_Q

we get

kF⁰(A, b)k

= sup

α²kW EWk²_QP+β²kfk²_Q≤1

kAd,W(W EW x−f)kP

≤ sup

α²kW EWk²_QP+β²kfk²_Q≤1

kAd,WkP Q(kW EWkQPkxkP+kfkQ)

= sup

α²kW EWk²_QP+β²kfk²_Q≤1

kAd,WkP Q

µ

αkW EWkQP

kxkP

α +βkfkQ

1 β

¶

=kA_d,Wk_{P Q} sup

α²kW EWk²_QP+β²kfk²_Q≤1

(αkW EWk_QP, βkfk_Q)· µkxkP

α , 1 β

¶

where (αkW EWkQP, βkfkQ) and

³kxkP

α ,¹_β

´

can be considered as vectors in R².

Therefore, from the Cauchy–Schwarz inequality we get:

kF⁰(A, b)k ≤ kAd,WkP Q

s kxk²_P

α² + 1 β².

Next, we show the other part of the theorem. For a sequence (un)ninR((W A)^D), kunk= 1, there exists a sequence (vn)ninR((AW)^D) ,kvnk ≤1 and lim

n→∞kvnk= 1, such that, for all n∈N,

(W1A1W1)⁻¹un=k(W1A1W1)⁻¹kvn =kAd,WkP Qvn. Taking, for alln∈N,

ˆ un=

· un

0

¸

, vˆn =

· vn

0

¸ ,

we obtain

Ad,Wuˆn =

· (W₁A₁W₁)⁻¹ 0

0 0

¸ · u_n 0

¸

=

· (W1A1W1)⁻¹un

0

¸

=

· k(W1A1W1)⁻¹kvn

0

¸

= k(W1A1W1)⁻¹k

· vn

0

¸

= kAd,WkP Qvˆn.

It is easy to check that kuˆnkQ= 1 and kˆvnkP ≤1, for alln∈N.

Letu∈R((W A)^D) andv∈R((AW)^D). DefineSu,v∈ B(R((AW)^D), R((W A)^D)) as follows: ifx∈R((AW)^D), then

Su,v(x)^def= hx, viu.

For allT ∈ B(R((W A)^D), R((AW)^D)) we have T Su,v(x) =T(u)hx, vi.

Now we choos, forn= 1,2,3, . . ., η=

s kxk²_P

α² + 1

β², fn= 1 β²ηuˆn, En =− 1

α²η

· W₁⁻¹ 0

0 0

¸ · Sun,x 0

0 0

¸ · W₁⁻¹ 0

0 0

¸ . Then we have, for a fixedn,

EnW = − 1 α²η

· W₁⁻¹ 0

0 0

¸ · S_un,x 0

0 0

¸ · W₁⁻¹ 0

0 0

¸ · W₁ 0 0 W2

¸

= − 1

α²η

· W₁⁻¹Sun,x 0

0 0

¸ · I 0 0 0

¸

= − 1

α²η

· W₁⁻¹Sun,x 0

0 0

¸ . Since

Ad,W(W AW) =I⊕0,

we can verify thatEn fulfills the first equation of condition (1):

Ad,W(W AW)EnW = − 1 α²η

· I 0 0 0

¸ · W₁⁻¹Sun,x 0

0 0

¸

= − 1

α²η

· W₁⁻¹Sun,x 0

0 0

¸

= EnW.

In the same way we have W En = − 1

α²η

· W1 0 0 W2

¸ · W₁⁻¹ 0

0 0

¸ · Sun,x 0

0 0

¸ · W₁⁻¹ 0

0 0

¸

= − 1

α²η

· I 0 0 0

¸ · Sun,xW₁⁻¹ 0

0 0

¸

= − 1

α²η

· Sun,xW₁⁻¹ 0

0 0

¸ . Since

(W AW)Ad,W =I⊕0,

we know

W En(W AW)Ad,W = − 1 α²η

· Sun,xW₁⁻¹ 0

0 0

¸ · I 0 0 0

¸

= − 1

α²η

· Sun,xW₁⁻¹ 0

0 0

¸

= W En.

Thus En fulfills the condition (1), for alln∈ N. Now we want to verify that the perturbation (En, fn) satisfiesα²kW EnWk²_QP +β²kfnk²_Q ≤1.

α²kW EnWk²_QP +β²kfnk²_Q

= 1

α²η²

°°

°°

· W₁ 0 0 W2

¸ · W₁⁻¹ 0

0 0

¸ · S_un,x 0

0 0

¸ · W₁⁻¹ 0

0 0

¸

·

· W1 0 0 W2

¸°°

°°

2 QP

+ 1

β²η²kˆunk²_Q

= 1

α²η²

°°

°°

· I 0 0 0

¸ · Sun,x 0

0 0

¸ · I 0 0 0

¸°°

°°

2 QP

+ 1

β²η²kˆu_nk²_Q

= 1

α²η²

°°

°°

· Sun,x 0

0 0

¸ · I 0 0 0

¸°°

°°

2 QP

+ 1

β²η²

= 1

α²η²

°°

°°

· Sun,x 0

0 0

¸°°

°°

2 QP

+ 1

β²η²

= 1

α²η²kSun,xk²+ 1 β²η²

≤ 1

α²η²kunk²kxk²_P+ 1 β²η²

= 1 η²

µkxk²_P α² + 1

β²

¶

= 1.

Thus, we have

F⁰(A, b)|(E_n,f_n) = −Ad,WW EnW x+Ad,Wfn

= 1

α²η((W1A1W1)⁻¹⊕0)(Sun,x⊕0)x+ 1

β²ηAd,Wuˆn

= 1

α²η((W1A1W1)⁻¹Sun,x⊕0)x+ 1

β²ηAd,Wuˆn

= 1

α²η

· (W1A1W1)⁻¹hx, xiun

0

¸ + 1

β²ηAd,Wuˆn

= 1

α²η

· kxk²_P(W1A1W1)⁻¹un

0

¸ + 1

β²ηAd,Wuˆn

= 1

α²ηkxk²_P

· k(W1A1W1)⁻¹kvn

0

¸ + 1

β²ηAd,Wuˆn

= 1

α²ηkxk²_Pk(W1A1W1)⁻¹k

· vn

0

¸ + 1

β²ηkAd,WkP Qˆvn

= 1

α²ηkxk²_PkA_d,Wk_{P Q}vˆ_n+ 1

β²ηkA_d,Wk_{P Q}vˆ_n

= kAd,WkP Q

η

µkxk²_P α² + 1

β²

¶ ˆ vn

= kAd,WkP Qηˆvn. So

kF⁰(A, b)|(En,fn)kP → kAd,WkP Q

s kxk²_P

α² + 1

β² (n→ ∞), withα²kW EnWk²_QP +β²kfnk²_Q≤1, we get

kF⁰(A, b)k → kAd,WkP Q

s kxk²_P

α² + 1

β², (n→ ∞) and we complete the proof. 2

References

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Second Ed., Springer 2003.

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Comput. 20 (2006), 35–59.

[3] Daji´c, A., Koliha, J. J., The weighted g–Drazin inverse for operators. J. Aus- tralian Math. Soc. 82 (2007), 163–181.

[4] Rakoˇcevi´c, V., Wei, Y., A weighted Drazin inverse and applications. Linear Algebra Appl. 350 (2002), 25–39.

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Received by the editors July 14, 2006