1. Introduction
In the last decades considerable attention has been paid to upper triangular operator matrices, particularly to spectra of operator matrices, see [1-8]. H. Du and J. Pan firstly researched the intersection of the spectra of 2 × 2 upper triangular operator matrices, and also proposed some open problems. In this note, we mainly study these problems.
For the context, we give some notations. Let
and
be Hilbert spaces,
,
and
denote the sets of all linear bounded operators on
,
and from
into
, respectively. For
,
,
, define an operator
by
.
Let
,
,
,
and
denote the nullspace, the range, the spectrum, the point spectrum, the approximation point spectrum of the resolvent set, the nullity and the deficiency of an operator
, respectively, where
and ![](https://www.scirp.org/html/4-7401689\e0b140f2-4de3-452c-80f4-1e4a7440a1ec.jpg)
use
,
and
to denote the sets of left Fredholm operators, right Fredhlom operators and semi-Fredholm operators in
, respectively. If T is a semi-Fredholm operator, define the index of T,
, by
. Note that
and it is necessary for either
or
to be finite dimensional in order for (1) to make sense ([3]).
For
,
, denote
![](https://www.scirp.org/html/4-7401689\3e748951-7e17-4abf-b60f-16f550a1373e.jpg)
![](https://www.scirp.org/html/4-7401689\cfc992cb-57fe-48e5-9d6d-36f623466ed9.jpg)
Under the situation that do not cause confusion, we simplify
as
.
In [2], H. Du and J. Pan have proved that,
(1)
for given
and
, the author asked a question that whether there exists an operator
such that
?
In this note, when
(n is a natural number), an affirmative answer of the question has been obtained.
2. Main Results and Proofs
To prove the main result, we begin with some lemmas.
Lemma 1. ([2]). Given
,
, then
.
Lemma 2. ([9]). Let
be an open connected subset of
and suppose
such that
, then there is a finite-rank operator
such that
is invertible, and also
is invertible for every
.
For any
, it is clear that
.
If there exists a
such that
then
.
But how to construct the operator such that
?
In the next theorem, we give a necessary condition of the answer of the question.
Theorem 3. For a given pair
of operators, where
,
, if
(n is a natural number) and each
has finite simple connected open sets, then there exists an operator
such that
.
Proof. For convenience, we divide the proof into two cases.
Case 1. If n = 0, that is,
, let
.
It is easy to see that
from lemma 1. Thus
so the result is obtained.
Case 2. If
, that is,
. Then
has finite simple connected open sets, now reordering and denoting by
. Thus there exists a natural number
such that
![](https://www.scirp.org/html/4-7401689\dc723a31-5541-4cac-afca-fe7030d654e5.jpg)
For each
choose a
, then
is a finite subset of
and
.
Next, the rest of proof is divided into two steps.
Step 1. We construct
as follows:
Let
and
are orthonormal basis for
and
, respectively and denote
,
.
First define an operator
from
onto
by
,
. Then define
by
![](https://www.scirp.org/html/4-7401689\226e12ad-9d52-4cd1-ae0b-d8f8fb4221dd.jpg)
It is clear that
is well defined and
.
If
, then let
.
If
, let
and
be orthonormal basis for
and
, respectively.
It is clear that
and
are linear independent. then there must be unit vectors
,
,···,
![](https://www.scirp.org/html/4-7401689\d1a4bd9b-63d7-470f-83bf-cb778ac90aab.jpg)
such that
![](https://www.scirp.org/html/4-7401689\da05fdfe-cd29-4a70-af57-145a05c0eb07.jpg)
Define an operator
as follows:
Let
and
,
and
,![](https://www.scirp.org/html/4-7401689\6e33a189-9b97-44ef-8e42-e177733a5ca6.jpg)
and
.
Since
be and
be are linear independent,
is linear independent. Let
![](https://www.scirp.org/html/4-7401689\2e5db310-fd6a-4724-9ae0-4e0d3471f3cb.jpg)
and
.
Then
and
is an operator from
onto
. Define
by
![](https://www.scirp.org/html/4-7401689\8d96a891-bad9-4bfc-9cb6-8798820ba5dd.jpg)
The process can be similarly done continuously.
Let
and
be orthonormal basis for
and
, respectively. It is clear that
is linear independent. Then there must be unit vectors
,
![](https://www.scirp.org/html/4-7401689\22b8203c-0d29-4615-b052-640266a77ea5.jpg)
![](https://www.scirp.org/html/4-7401689\84ea0d6c-2940-4382-8b02-48f5a69f7e7e.jpg)
such that
![](https://www.scirp.org/html/4-7401689\fa33825f-dccd-456c-941e-27f1ecc75a3f.jpg)
![](https://www.scirp.org/html/4-7401689\3068a129-4f39-4c5f-b594-c30cf8cb72e6.jpg)
Define an operator
as follows:
Let
and
,
and ![](https://www.scirp.org/html/4-7401689\90602e1b-182e-46e1-ab1e-26953ead670d.jpg)
and
.
Since
is linear independent,
is linear independent. Denote
and
.
Then
,
and
is an operator from
onto
. Define
by
![](https://www.scirp.org/html/4-7401689\cb524fca-321e-4983-a540-2387b62c6393.jpg)
Let
. It is clear that
is well defined and bounded with finite rank. By directly computation, we can get
![](https://www.scirp.org/html/4-7401689\2dc79503-7ee2-4c37-87f4-4283677d7676.jpg)
Step 2. We prove that
defined as above such that
.
It is sufficient to prove that for any
,
is invertible. From Lemma 2, it is only to prove for any
,
is invertible. To finish it, it is to prove that
is injective and surjective.
If there exists a vector
with
where
and
, then
and
. By definition of
then
, thus
. On the other hand, since
is injective on
, then
and so,
. By assumption that
hence
. Therefore
is injective.
For any vector
, where
and
.
Since
and
, ![](https://www.scirp.org/html/4-7401689\b8562fe7-9f59-4b1a-bc2c-0e54fa7564d4.jpg)
and
is closed. Thus there is a vector ![](https://www.scirp.org/html/4-7401689\6fbc2113-1d08-4060-b82e-7608927e905d.jpg)
such that
. Because
, there exist
and
such that
. Hence there exist ![](https://www.scirp.org/html/4-7401689\1a2d7e50-ced0-47a4-aa5f-c313125085b5.jpg)
and
such that
and
. The last equality is possible, because
is onto
. Therefore,
![](https://www.scirp.org/html/4-7401689\71bce967-cbaa-469e-88a1-381ad82a5ac2.jpg)
As
is arbitrary,
is surjective.
Hence, for any
,
is invertible, i.e.,
. So
.
The proof is completed.
Example 4. If
and
,
is the shift operator on
, let
then
is invertible. From directly computation,
and
, where
is the interior of unit disk. For any
,
is invertible. Thus
.
3. Acknowledgements
This subject is supported by NSF of China (No. 11171197) and the Natural Science Basic Research Plan of Henan Province (No. 122300410420, 122300410427).