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 
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
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
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
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
, 
,···,
such that
Define an operator 
as follows:
Let
and
,
and
,
and
.
Since 
be and 
be are linear independent, 
is linear independent. Let
and
.
Then 
and 
is an operator from 
onto
. Define 
by
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
,
such that
Define an operator 
as follows:
Let
and
,
and 
and
.
Since 
is linear independent, 
is linear independent. Denote
and
.
Then
,
and 
is an operator from 
onto
. Define 
by
Let
. It is clear that 
is well defined and bounded with finite rank. By directly computation, we can get
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
, 
and 
is closed. Thus there is a vector 
such that
. Because
, there exist 
and 
such that
. Hence there exist 
and 
such that 
and
. The last equality is possible, because 
is onto
. Therefore,
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).