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,

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.

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 ofthen, thus. On the other hand, since is injective on, thenand so,. By assumption thathence. 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.

This subject is supported by NSF of China (No. 11171197) and the Natural Science Basic Research Plan of Henan Province (No. 122300410420, 122300410427).