DRAZIN INVERSE PDF

For a matrix A ∈ Cn×n with index 1, the group inverse A. # is the unique solution of the and its unique solution is the Drazin inverse, or. {1k,2,5}–inverse, of A. PDF | The main theme of this paper can be described as a study of the Drazin inverse for bounded linear operators in a Banach space X when 0 is an isolated. 1. Introduction. The main theme of this paper can be described as a study of the Drazin inverse. In , Drazin [4] introduced a different kind of generalized.

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To receive news and publication updates for Journal drazkn Applied Mathematics, enter your email address in the box below. This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We deduce the explicit expressions for and of two matrices and under the conditions and.

Also, we give the upper bound of. The symbol stands for the set of complex matrices, and for short stands for the identity matrix. Forits Drzin inverse, denoted byis defined as the unique matrix satisfying where is the index of.

In particular, ifis invertible and see, e. Recall that for withthere exists an nonsingular matrix such that where is a nonsingular matrix and is nilpotent of indexand see [ 13 ]. It is well known that if is nilpotent, then. Drazin [ 2 ] proved that in associative ring when are Drazin invertible and. In [ 4 ], Hartwig et al. 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 andin [ 9 ] they are andand in [ 15 ], they are and. These results motivate us to investigate how to explicitly express the Drazin inverse of the sum under the incerse andwhich are implied by the condition.

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King : A note on Drazin inverses.

The paper is organized as follows. In Section 2we will deduce some lemmas. In Section 3we will present the explicit expressions for and of two matrices and incerse the conditions and.

We also give the upper bound of. In this section, we will make preparations for discussing the Drazin inverse of the sum of two matrices in next niverse. To this end, we will introduce some lemmas. Letand withand define Then.

Ifthen, for any positive integersiii. Moreover, ifthen. Assume that, forthe equation holds, that is. Whenby iwe have Hence, by induction, we have for any. By inveese on for 2. Obviously, whenit holds by statement i. Assume that it holds forthat is. WhenHence 2. Then, for any positive integerwhere the binomial coefficient. Drazzin, if are nilpotent with andfrazin is nilpotent and its index is less than.

We will show by induction that 2. If are nilpotent with andthen taking in 2. Ifthen and. Let and be invertible. Ifthen Moreover, if is nilpotent of indexthen is invertible and. Sinceby Lemma 2. Thus, is invertible and so isand. Imverse withwhere is invertible and is nilpotent of indexand let be partitioned conformably with. Then, and Moreover, if is nilpotent of indexthen.

Since by Lemma 2. So from andit follows, respectively, that and that Sinceand then. From this, we can easily verify Therefore, ifthen.

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In this section, we will give the explicit expressions for andunder the conditions and. Now, we begin with the following theorem. Let with and be nilpotent with. If andthen.

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Ifthen is invertible, and therefore. So, by Lemma 2. Now assume that and, without loss of generality, can be written aswhere is invertible and is nilpotent of index.

Since andwe can writepartitioned conformably withby Lemma 2. We also writepartitioned conformably with. Since is nilpotent and is invertible, invere Lemma 2. Hence, by Lemma 2. Similar to the discussion of 3. The following theorem is our main result, and Theorem 3. Ifthen is invertible and. So, by Vrazin 2.

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Therefore, assume thatand, without loss of generality, letwhere is invertible and is nilpotent of index. From hypotheses, by Lemma 2. Note that implies and. If with andthen. Thus, we have the following corollary.

If there exist two positive integers and such that andthen If is a perturbation ofthen, we have invesre following result in which has an upper bound.

Before the theorem, let us drzzin that ifthen is invertible and. Sinceis invertible. Then, Thus By the above argument, we can get 3. Consider the matrices We observe that andbut. It is obvious thatand Sinceis invertible and By 3. On the other hand, it is easy to get that. The authors would like to thank the referees for their helpful comments invere suggestions. Journal of Applied Mathematics. Subscribe to Table of Contents Alerts.

Table of Contents Alerts. Abstract We deduce the explicit expressions for and of two matrices and under the conditions and. Introduction The symbol stands for the set of complex matrices, and for short stands for the identity matrix.

Some Lemmas In this rrazin, we will make preparations for discussing the Drazin inverse of the sum of two matrices in next section. Letand withand define Then, Lemma 2. Moreover, ifthen Proof.

Ifthen Moreover, if is nilpotent of indexthen is invertible and Proof. Thus, is invertible and so isand Lemma 2.

Main Results In this section, we will give the explicit expressions for andunder the conditions and. If andthen Proof. Then, i ii Proof. If with andthen Proof. Before the theorem, draizn us recall that ifthen is invertible and Theorem 3.

Ifthen Proof. Finally, drqzin give an example to illustrate our results. Acknowledgments The authors would like to thank the referees for their helpful comments and suggestions.