** Advances in Linear Algebra & Matrix Theory** Vol.3 No.4(2013), Article ID:40927,4 pages DOI:10.4236/alamt.2013.34013

Commuting Outer Inverses

^{1}Department of Mathematics, Faculty of Science, Kanagawa University, Kanagawa, Japan

^{2}Department of Research and Graduate Studies, Linda Vista University, Pueblo Nuevo Solistahuacán, Mexico

Email: chiyom01@kanagawa-u.ac.jp, gkantun@ulv.edu.mx

Copyright © 2013 Muneo Chō, Gabriel Kantún-Montiel. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Received November 12, 2013; revised December 9, 2013; accepted December 16, 2013

**Keywords:** Generalized Inverse; Koliha-Drazin Inverse; Outer Inverse

ABSTRACT

The group, Drazin and Koliha-Drazin inverses are particular classes of commuting outer inverses. In this note, we use the inverse along an element to study some spectral conditions related to these inverses in the case of bounded linear operators on a Banach space.

1. Introduction

Several of the useful properties of the group, Drazin and Koliha-Drazin inverses can be related to their spectral characterizations. Some of these can be traced to the property of being commuting outer inverses.

Let be the set of bounded linear operators on a Banach space, and let. We denote the range by and the null space of by.

Let and be closed subspaces of. The outer inverse with prescribed range and null space, denoted is the unique operator which satisfies:

There is some advantage in prescribing the null space and range of an outer inverse by means of a third operator. In doing so, we will use the notion of invertibility along an element introduced by X. Mary ([1]). We say is invertible along if there exists such that

In this case, the inverse along is unique and we write.

From we have that BA and AB are projections such that and . Thus, we are effectively prescribing the range of the projection and the null space of the projection.

One of the useful properties of a generalized inverse is that, although the operator is not invertible, there is a subspace for which the reduction of the operator to that subspace is indeed invertible:

Theorem 1. ([2, Theorem 2]) Let be nonzero operators. The following statements are equivalent.

1. is invertible along.

2. is a closed and complemented subspace of, is closed such that and the reduction is invertible.

Recall an operator is said to be group invertible if there exists such that

In this case, such is unique and we write for the group inverse of.

Proposition 2. ([2, Theorem 3]) If is invertible along, then AT and TA are group invertible and.

Example 3. Let the space of square-summable sequences. Let be defined by and.

Then is invertible along with.

In the following section, we study an operator such that is invertible along with. Then, in Section 3 we study some projections related to the outer inverse with prescribed range and null space. Finally, in Section 4 we specialize to spectral projections, covering results from Dajić and Koliha ([3]).

2. Invertibility along a Commuting Operator

Proposition 4. Let be invertible along. If, then.

Proof. From Proposition 2 we have:

If is invertible along, then we have the following matrix form ([2]):

where is invertible and is a complement of, that is,.

When and commute, we can say a little more:

Theorem 5. Let be invertible along and. Then there exist an invertible operator on and an operator on such that

and

Proof. Suppose is invertible along and. Then by Proposition 4. Thus, since is a projection, we have that

.

Since

and

we also have

and hence we can consider the following matrix decomposition of:

In this case, , , is invertible. Indeed, to see that it is onto note that since and

, we have and hence . To see that it is also 1-1, let. Since, there exists such that. Then, .

Moreover, since, subspaces and are -invariant and maps onto, we get. Thus,

and clearly .

Example 6. Let the space of square-summable sequences. Let be defined by

Then it is easy to verify that is the operator such that

It is clear that, but we have

3. Projections

Commuting outer inverses are naturally linked to projections.

Proposition 7. Let. If is invertible along and, then there exists a bounded projection such that is invertible along.

Proof. From Theorem 5 we have that if is invertible along and, then

Thus, there exists a bounded projection such that and. Hence, is invertible along.

For a sort of converse, we give a necessary condition in Theorem 9.

Example 8. An operator and a projection such that is invertible along and.

Let be the set of two by two matrices with real entries. Let be the (rotation) matrix defined by

and let be the projection defined by

Then, it is easy to check that is group invertible, with group inverse.

We have

Thus, is invertible along but since

, we see. Notice that

.

Theorem 9. Let be a projection and suppose is invertible along. If , then.

Proof. Since is invertible along, has the following matrix form [2, Corollary 1]:

where is invertible and is the complement of, that is,.

Since is a complement for, we can take. Then, we have

Now, suppose. From Theorem 1 we know that is invertiblewhich implies. Thus,

It follows.

Note that the theorem above together with Proposition 4 implies that if, then. However, we can prove:

Proposition 10. Let be a projection, and suppose is invertible along. Then.

Proof. Using Proposition 4,

Example 11. An operator and a projection such that but is not invertible along.

Let and let be defined by

Then

However, the reduction can not be invertible, and by Theorem 1 is not invertible along.

Theorem 12. Let be a projection and be such that. Then, is invertible along if and only if and.

Proof. Suppose is invertible along. Since and from Proposition 4 and the definition of we have

Similarly, from

Proposition 4 and the definition of we get

Conversely, suppose and . We use Theorem 1. The reduction is clearly onto. From and we get that it is also 1-1.To see that is closed and we will show. It is clear that. For the other inclusion, let. Since, there exists such that. Then, from it follows that. Thus,. Finally, it is clear that is closed and complemented. Hence is invertible along.

4. Spectral Projections

Recall the spectrum of an operator is the set

Suppose is invertible along and. Then we have the matrix form

.

From and are invariant under, we have:

Since is invertible but is not, we know that.

Note that from and the matrix form of, there exists a projection such that, and. Thus, without loss of generality, we can suppose is invertible along a projection. A very important class of projections which commutes with is a class of spectral projections, which we now discuss.

The resolvent set of is and for the resolvent function is

A subset is said to be a spectral set of if and are both closed in. For a spectral set of, the spectral projection associated with is defined by

where is a Cauchy contour that separates from.

If is a point of the resolvent set or an isolated point of the spectrum, then the operator is called quasipolar. Let be quasipolar and let be the spectral projection associated with the spectral set, then ([4]):

Let. If, then we say that is a quasinilpotent operator. Recall that is nilpotent if for some, and nilpotent operators are quasinilpotent.

Quasipolar operators are generalized invertible in the sense of Koliha: an operator is Koliha-Drazin invertible if there exists such that is quasinilpotent,

In this case, by Lemma 2.4 of [5], Koliha-Drazin inverse is unique and we write.

An operator is Koliha-Drazin invertible if and only if 0 is an isolated point of. If is a pole of the resolvent of order, then is Drazin invertible with Drazin index. If 0 is a simple pole then it is group invertible.

As noted above, the Koliha-Drazin is a particular case when we consider the spectral set. For the general case when is a spectral set such that, Dajić and Koliha have defined a generalized inverse and studied its properties ([3]).

Theorem 13. Let and be a spectral set for. If then is invertible along.

Proof. Let. Then and are closed and. Now, since

is -invariant, and we have that is invertible. Thus, is closed, and is invertible. Therefore, by Theorem 1 is invertible along.

Corollary 14. Let and be a spectral set for. If then is invertible along.

Proof. If, then. From the theorem above, is invertible along .

5. Acknowledgements

This research is partially supported by Grant-in-Aid Scientific Research No.24540195.

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