Abstract and Applied Analysis

Volume 2013 (2013), Article ID 373147, 8 pages

http://dx.doi.org/10.1155/2013/373147

Research Article

The Intersection of Upper and Lower Semi-Browder Spectrum of Upper-Triangular Operator Matrices

Shifang Zhang,¹ Huaijie Zhong,¹ and Long Long²

¹School of Mathematics and Computer Science, Fujian Normal University, Fuzhou 350007, China

²School of Mathematics, Central South University, Changsha 410075, China

Received 22 January 2013; Accepted 3 August 2013

Academic Editor: Alberto Fiorenza

Copyright © 2013 Shifang Zhang et al. 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.

Abstract.

When and are given, we denote by the operator acting on the infinite-dimensional separable Hilbert space of the form . In this paper, it is proved that there exists some operator such that is upper semi-Browder if and only if there exists some left invertible operator such that is upper semi-Browder. Moreover, a necessary and sufficient condition for to be upper semi-Browder for some is given, where denotes the subset of all of the invertible operators of .

1. Introduction

It is well known that if is a Hilbert space, is a bounded linear operator defined on , and is an invariant closed subspace of , then can be represented in the following form:

which motivated the interest in upper-triangular operator matrices. For recent investigations on this subject, see references [1–23].

Throughout this paper, let and be separable infinite-dimensional complex Hilbert spaces, and let be the set of all bounded linear operators from into ; when , we write as . For , , and , we have . For , let and denote the range and the kernel of , respectively, and denote that and . If , the ascent of is defined to be the smallest nonnegative integer which satisfies and . If such does not exist, then the ascent of is defined as infinity. Similarly, the descent of is defined as the smallest nonnegative integer for which holds. If such does not exist, then is defined as infinity, too. If the ascent and the descent of are finite, then they are equal (see [6]). For , if is closed and , then is said to be an upper semi-Fredholm operator; if , which implies that is closed, then is said to be a lower semi-Fredholm operator. If is either upper or lower semi-Fredholm operator, then is said to be a semi-Fredholm operator. If both and , then is said to be a Fredholm operator. For a semi-Fredholm operator , its index is defined by .

For a semi-Fredholm operator , its shift Samuel multiplicity and backward shift Samuel multiplicity are defined, respectively, by the following (see [24]):

Moreover, it has been proved that , and that . These two invariants refine the Fredholm index and can be regarded as the stabilized dimensions of the kernel and the cokernel (see [24]).

In this paper, the sets of invertible operators and left invertible operators from into are denoted by and , respectively; the sets of all Fredholm operators, upper semi-Fredholm operators, and lower semi-Fredholm operators from into are denoted by , , and , respectively; the sets of all Browder operators, upper semi-Browder operators, and lower semi-Browder operators, on are defined, respectively, by the following:

Moreover, for , we introduce its corresponding spectra as follows.

 The spectrum is given as . The left spectrum is given as . The essential spectrum is defined as . The upper semi-Fredholm spectrum is defined as . The lower semi-Fredholm spectrum is presented as . The Browder spectrum is presented as . The upper semi-Browder spectrum is defined as . The lower semi-Browder spectrum is presented as .

Using the Samuel multiplicities, Zhang and Wu (see [20]) gave a necessary and sufficient condition for which for some and characterized the set of . In this paper, our main goal is to characterize the intersection of and . This paper is organized as follows. In Section 2, we give a necessary and sufficient condition for which for some and get

In Section 3, we give a necessary and sufficient condition for which for some and get

For the sake of convenience, we now present some lemmas which will be used in the sequel.

Lemma 1 (see [20, 24]). An operator is semi-Fredholm if and only if   can be decomposed into the following form with respect to some orthogonal decomposition

where , is a right invertible operator, is a finite nilpotent operator, is a left invertible operator, and . Moreover, , , and .

Lemma 2 (see [18]). Let , , and .

(1)If , then if for some .(2)If for some , then .(3)If and , then for any .(4)If , then if for some ; if for some .(5)If for some , then and .(6)If two of , , and are Browder, then so is the third.

Lemma 3 (see [20]). Let . Then, is upper semi-Browder if can be decomposed into the following form with respect to some orthogonal decomposition

where , is nilpotent, is left invertible, and .

Lemma 4 (see [20]). Let . Then, is lower semi-Browder if can be decomposed into the following form with respect to some orthogonal decomposition :

where , is right invertible, is nilpotent, and .

Lemma 5 (see [20]). For any given and , for some if and

Lemma 6 (see [9]). For any given and , is left invertible for some if is left invertible and

Lemma 7 (see [25]). Let be a linear subspace of . Then, the following statements are equivalent.

(1)Any bounded operator with is compact.(2) contains no closed infinite-dimensional subspace.

2. and

In [1, 20], the authors have proved that

They, moreover, proved that

Comparing the above two kinds of spectra with the upper semi-Weyl spectrum and Weyl spectrum, one may expect that the following equality holds:

However, it is not that case, as the following example shows.

Example 8. Let be the unilateral shift on , that is,

and let the operators and be defined by

Then, we have , while . Moreover, , while and  . Thus, (13) does not hold.

In spite of the above counter example, we have the following.

Proposition 9. For any given and , one has

Proof. From the proof of Theorem  2.3 in [20], we know that when , if and only if . Combining this fact with Corollary 2.5 of [20], it is easy to see that

Noting that implies that is closed, it follows from corollary 2.5 of [2] that

where is not uppersemi-Fredholm operator with index less than or equal to .

Now, we are ready to present the main result of this section.

Theorem 10. For any given and , one has

Proof. Since is obvious, it is sufficient to prove that if , then there exists some left invertible operator such that .

Suppose that . It follows from Lemma 5 that and

There are two cases to consider.

Case  1. Assume that , , and . Then, it follows from Lemma 3 that can be decomposed into the following form:

where , is nilpotent, is a left invertible operator, and . So, we can let

where is unitary. Obviously, is left invertible. Now, can be rewritten as

Since is left invertible and is invertible, then there exist unique and such that and , and

This implies that is left invertible. And, hence, Lemma 2 leads to .

Case  2. Assume that , , and . Then, it follows from Lemma 3 that can be decomposed into the following form:

where , is nilpotent, is a left invertible operator, and . By the assumption that and Lemma 1, we know that can be decomposed into the following form with respect to some orthogonal decomposition

where , is a right invertible operator, is a left invertible operator, is a finite nilpotent operator, and the parts marked by can be any operators. Moreover, . Thus, , and then there exists some left invertible . Noting that , we can let . Consider

Obviously, is left invertible, and can be rewritten as

where and are invertible and and are left invertible. Similar to the proof of Case 1, through direct calculation we can show that is left invertible. Also since and , we have and . Thus, it follows from Lemma 2 that .

By duality, we have the following.

Theorem 11. For any given and , one has

3. and

In this section, we give the characterization of invertible and Fredholm perturbations of upper semi-Browder spectra of upper-triangular matrices. We begin with some lemmas.

Lemma 12 (see [19]). For a given pair , if either or is a compact operator, then, for each , is not a semi-Fredholm operator.

In particular, if is not compact, then is not semi-Browder for any invertible operator .

Lemma 13. The following statements are equivalent.

(i) is not compact. (ii)For each given , if , then there exists an operator such that is an upper semi-Browder operator.(iii)For each given , if , then there exists an operator such that is an upper semi-Browder operator.

Proof. Obviously, we only need to prove the implications (i) (ii) and (iii) (i).

 (iii) (i). If is compact, then it follows from Lemma 12 that is not a semi-Fredholm operator for each , which contradicts with (iii). Thus, is not compact. (i) (ii). Suppose that is not compact. Then, we consider the following two cases.

Case  1. Assume that is closed. It follows from Lemma 3 that can be decomposed into the following form with respect to some orthogonal decomposition

where , is nilpotent, and is a left invertible operator. Noting that , we have . Since the assumption that is not compact, we have that . Also since , let with and . Define an operator by

where and are invertible operators. Obviously, is invertible. Next, we claim that is an upper semi-Browder operator. To see this, can be rewritten as

where is invertible and is left invertible. By Lemma 2 and the fact that , it is sufficient to prove that

is semi-Browder. For this, we only need to show that is left invertible. In fact, since is invertible and and are left invertible, we can set , , and such that

Direct calculation shows that

which implies that is left invertible. Noting that , by Lemma 2 we have that is upper semi-Browder.

Case  2. Assume that is not closed. If is not compact, then by Lemma 7, contains a closed infinite-dimensional subspace. Without loss of generality, suppose that is a closed subspace of with and . Let . Thus, is a closed subspace of , and . Denote . Without loss of generality, we may assume that (otherwise, suppose that is an orthonormal basis of . Denote and , then and can be replaced by and , resp.). Since , let with and . Define an operator by

where , , and are unitary operators. Obviously, is invertible. can be rewritten as

where and are invertible and .

Next, we prove that . Noting that , then, by Lemma 2, it is sufficient to prove that

is left invertible. For this, let , , , and be operators satisfying

Direct calculation shows that

which implies that is left invertible.

Combining Case 1 with Case 2, the lemma is proved.

Similarly, we have the following.

Lemma 14. The following statements are equivalent:

(i) is not compact. (ii)For each given , if , then there exists an operator such that is a lower semi-Browder operator.(iii)For each given , if , then there exists an operator such that is a lower semi-Browder operator.

One is now ready to prove the main result of this section.

Theorem 15. For a given pair , one has

Proof. According to Lemma 12, it is clear that

For the conversion, without loss of generality, suppose that

Then, is not compact, and there exists some such that , and, hence, .

Case  1. . It follows from Lemma 13 that there exists some such that is an upper semi-Browder operator. This implies that . In this case, we have proved. Consider that

Case  2. Consider . This implies that , and, thus, since . It follows from Lemma 5 that . Moreover, using Lemmas 1 and 3, we have

where , is nilpotent, is a left invertible operator, , is a right invertible operator, is a left invertible operator, is a finite nilpotent operator, and the parts marked by can be any operators. Moreover, , and . Hence, . Now, put , where . Noting that , there exist unitaries and . Let . Obviously, .

Consider that operator

where

We claim that . In fact, since and are Browder operators, then, by Lemma 2, it is sufficient to show that

is upper semi-Browder. Observe that and are left invertible; and are invertible. Direct calculation shows that is injective. Since and , we have , and, hence, is an upper semi-Fredholm operator. Thus, is left invertible. Combining this with Lemma 2 yields , which means that . Thus,

Combining Case 1 with Case 2 leads to

This completes the proof.

By duality, we have

Theorem 16. For a given pair , one has

Acknowledgments

Long Long research is supported by the Freedom Explore Program of Central South University (Grant no. 2012QNZT040) and the Post Doctoral Program Foundation of Central South University. This work is also supported by the National Natural Science Foundation of China (11226113, 11301077, 11171066, and 10771191), the Foundation of the Education Department of Fujian Province (JA12074), and the Natural Science Foundation of Fujian Province (2012J05003).

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