Intersections of various spectra of operator matrices

Nikola Sarajlija111e-mail address<EMAIL_ADDRESS>
University of Novi Sad, Faculty of Sciences, Novi Sad, 21000, Serbia

September 3, 2024

###### Abstract

This paper is concerned with general $n\times n$ upper triangular operator
matrices with given diagonal entries. We characterize perturbations of the
left (right) essential spectrum, the essential spectrum, as well as the left
(right) the Weyl spectrum of such operators, thus generalizing and improving
some results of [3], [4], [6], [16], [17], [25] from $n=2$ to case of $n\geq
2$. We show that related results hold in arbitrary Banach spaces without
assuming separability, thus extending and improving recent results from [22],
[23]. For the lack of adjointness and orthogonality in Banach spaces, we use
an adequate alternative: the concept of inner generalized inverse.

Keywords: essential spectrum, Weyl spectrum, perturbation theory, $n\times n$,
regular operators

##  1 Introduction

In the last few decades considerable attention has been devoted to the study
of spectral properties of operator matrices, having in mind their governing
importance in various areas of mathematics (see for example [9, Chapter
VIII]). One has soon realized that one way for successful work on problems
arising in spectral theory, is to see operator matrices as entries of simpler
blocks. Block operator matrices, and especially upper triangular operator
matrices, have been extensively studied by numerous authors (see [3], [4],
[6], [8], [11], [14], [15], [22], [23] and many others…). The reason for this
lies in the fact that if an operator $T$ is acting on a direct sum of Banach
spaces, it takes the upper triangular form under condition that certain number
of those spaces is invariant for $T$.

Development of this topic began in the last century, and is of great
importance ever since. In the beginning, authors have only considered the case
of $2\times 2$ operator matrices. Pioneering work in that direction was the
article of Du and Jin from 1994 ([8]) treating the usual spectrum. Han et al.
have generalized their result to Banach spaces ([11]), and Lee has proved some
facts concerning the Weyl spectrum ([14]). Afterwards, Djordjević in 2002 gave
some characterizations for $2\times 2$ upper triangular operators to be
Fredholm, Weyl, and Browder ([6]). After that, many authors have explored
various properties of $2\times 2$ block operators in a connection with
intersection of spectra, Weyl and Browder type theorems, etc. (see for example
[15], [3]).

Investigation of spectral properties of general $n\times n$ operators began no
sooner than 2013, when Zguitti published the article dealing with some
properties of the Drazin spectrum ([24]). Huang et al. continued his work in
2016 by investigating properties of the point, residual, and continuous
spectrum of $n\times n$ matrix operators ([12]). Fredholm and the Weyl
spectrum of such operators have been studied by Wu and Huang in [22], [23]
only a few years ago, and this article is concerned with generalizing their
results from separable Hilbert to arbitrary Banach spaces.

Partial operator matrix is an operator matrix with some entries specified and
the others unknown. Most commonly, those entries are linear and bounded
operators with appropriate domains. We mention that there has been some
interest in block operators with unbounded entries lately, see [2], [18],
[20], but we shall not pursue this point any further. Usually, it is required
that unknown entries satisfy some special properties, and finding those
entries is the task known as the completion of the matrix. It is sometimes a
rather difficult task, but there are some ways to make this task easier.
Perturbation of an upper triangular operator matrix is an intersection of its
certain spectrum, some of the entries being fixed, for example the diagonal
ones, and the others chosen arbitrarly with appropriate domains. Finding
perturbation is a very useful pre-step to the completion of a matrix. This
article is ought to be just a such pre-step.

Article is organized as follows. In Section 2 we give some basic concepts and
terminology that we use in the rest of this work. We introduce notion of
embedded spaces that serves us as a surrogate to a lack of separability, and
we present some of its variants. Perturbation results related to Fredholmness
and Weylness of upper triangular block operators belong to Section 3 and
Section 4, respectively. We supply each of those sections with an additional
subsection that contains an answer to the Question 3 raised in Section 2.

##  2 Notation and preliminaries

Let $X,Y,X_{1},...,X_{n}$ be arbitrary Banach spaces. We use the notation
$\mathcal{B}(X_{i},X_{j})$ for the collection of all linear and bounded
operators from $X_{i}$ to $X_{j}$. Particularly,
$\mathcal{B}(X_{i})=\mathcal{B}(X_{i},X_{i})$, and
$X^{\prime}=\mathcal{B}(X,\mathds{C})$ is the dual space of $X$. If
$T\in\mathcal{B}(X_{i},X_{j})$, then by $\mathcal{N}(T)$ and $\mathcal{R}(T)$
we denote the kernel and range of $T$. Those sets are subspaces of $X_{i}$ and
$X_{j}$, respectively, and $\mathcal{N}(T)$ is closed.

If $T\in\mathcal{B}(X,Y)$, then its dual operator is
$T^{\prime}\in\mathcal{B}(Y^{\prime},X^{\prime})$ defined by
$T^{\prime}f(x):=f(Tx),\ f\in Y^{\prime}$. We shall use some properties of
dual operators on several occasions. For example, it is known that
$\|T\|=\|T^{\prime}\|$ and $T\mapsto T^{\prime}$ is an isometric isomorphism
of $\mathcal{B}(X,Y)$ into $\mathcal{B}(X^{\prime},Y^{\prime})$. For other
basic features of dual operators we recommend [9].

For $U\subseteq X$ we define set $U^{\circ}\subseteq X^{\prime}$, and for
$V\subseteq X^{\prime}$ we define set ${}^{\circ}V\subseteq X$ as

$U^{\circ}:=\\{f\in X^{\prime}:\ f\upharpoonleft_{U}=0\\}.$
${}^{\circ}V:=\\{x\in X:\ V(x)=0\\}.$

$U^{\circ}$ and ${}^{\circ}V$ are called the left and right annihilator of $U$
and $V$, respectively. Above all interesting features that hold for
annihilators, we point out only one needed in the proof of Lemma 2.10, see
[21]:

$\overline{\mathcal{R}(A)}=^{\circ}[\mathcal{N}(A^{\prime})],$ (2.1)

where $A\in\mathcal{B}(X,Y)$.

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ be given. We denote
by $T_{n}^{d}(A)$ an $n\times n$ partial upper triangular operator matrix of
the form

$T_{n}^{d}(A)=\begin{bmatrix}D_{1}&A_{12}&A_{13}&...&A_{1,n-1}&A_{1n}\\\
0&D_{2}&A_{23}&...&A_{2,n-1}&A_{2n}\\\ 0&0&D_{3}&...&A_{3,n-1}&A_{3n}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\ 0&0&0&...&D_{n-1}&A_{n-1,n}\\\
0&0&0&...&0&D_{n}\end{bmatrix}\in\mathcal{B}(X_{1}\oplus
X_{2}\oplus\cdots\oplus X_{n}),$ (2.2)

where $A:=(A_{12},\ A_{13},...,\ A_{ij},...,\ A_{n-1,n})$ is an operator tuple
consisting of unknown variables $A_{ij}\in\mathcal{B}(X_{j},X_{i})$, $1\leq
i<j\leq n,\ n\geq 2$. For convenience, we denote by $\mathcal{B}_{n}$ the
collection of all such tuples $A$. One easily verifies that for $T_{n}^{d}(A)$
of the form (2.2), its dual operator matrix $T_{n}^{d}(A)^{\prime}$ is given
by

$T_{n}^{d}(A)^{\prime}=\scalebox{0.9}{\mbox{$\displaystyle\begin{bmatrix}D_{1}^{\prime}&0&0&...&0&0\\\
A_{12}^{\prime}&D_{2}^{\prime}&0&...&0&0\\\
A_{13}^{\prime}&A_{23}^{\prime}&D_{3}^{\prime}&...&0&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
A_{1,n-1}^{\prime}&A_{2,n-1}^{\prime}&A_{3,n-1}^{\prime}&...&D_{n-1}^{\prime}&0\\\
A_{1n}^{\prime}&A_{2n}^{\prime}&A_{3n}^{\prime}&...&A_{n-1,n}^{\prime}&D_{n}^{\prime}\end{bmatrix}$}}\in\mathcal{B}(X_{1}^{\prime}\oplus
X_{2}^{\prime}\oplus\cdots\oplus X_{n}^{\prime}).$ (2.3)

There are several questions that we are interested in:

Question 1. Can we find an appropriate characterization for Fredholmness,
Weylness, etc. for $T_{n}^{d}(A)$, in terms of Fredholmness, Weylness, etc. of
its diagonal entries $D_{i}$?

Question 2. What can we say about
$\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{*}(T_{n}^{d}(A))$, when
$\sigma_{*}$ runs through set
$\\{\sigma_{le},\sigma_{re},\sigma_{e},\sigma_{lw},\sigma_{rw}\\}$?

Question 3. Under what conditions the equality
$\sigma_{*}(T_{n}^{d}(A))=\bigcup\limits_{i=1}^{n}\sigma_{*}(D_{i})$ holds,
$\sigma_{*}\in\\{\sigma_{le},\sigma_{re},\sigma_{e},\sigma_{lw},\sigma_{rw}\\}$?

Here we give complete answers to Questions 2 i 3, while Question 1 remains
opened.

We use $\mathcal{G}_{l}(X)$ and $\mathcal{G}_{r}(X)$ to denote the sets of all
left and right invertible operators on $X$, respectively. It is well known
that if $T\in\mathcal{B}(X)$, than $T\in\mathcal{G}_{l}(X)$ if and only if
$\mathcal{N}(T)=\\{0\\}$ and $\mathcal{R}(T)$ is closed and complemented.
Similarly, $T\in\mathcal{G}_{r}(X)$ if and only if $\mathcal{R}(T)=X$ and
$\mathcal{N}(T)$ is complemented. The set of all invertible operators on $X$
is denoted by $\mathcal{G}(X)=\mathcal{G}_{l}(X)\cap\mathcal{G}_{r}(X)$.

Now we list some elementary notions from Fredholm theory (see [26]). Let
$T\in\mathcal{B}(X)$, $\alpha(T)=\dim\mathcal{N}(T)$ and $\beta(T)=\dim
X/\mathcal{R}(T)$. Quantities $\alpha$ and $\beta$ are called the nullity and
deficiency of $T$, respectively, and in the case where at least one of them is
finite we define $\mathrm{ind}(T)=\alpha(T)-\beta(T)$ to be the index of $T$.
Notice that $\mathrm{ind}(T)$ may be $\pm\infty$ or integer. Families of left
and right Fredholm operators, respectively, are defined as

$\displaystyle\Phi_{l}(X)=\\{T\in\mathcal{B}(X):\ \alpha(T)<\infty\ and\
\mathcal{R}(T)\ is\ closed\\}$

and

$\displaystyle\Phi_{r}(X)=\\{T\in\mathcal{B}(X):\beta(T)<\infty\\}.$

The set of Fredholm operators is

$\Phi(X)=\Phi_{l}(X)\cap\Phi_{r}(X)=\\{T\in\mathcal{B}(X):\alpha(T)<\infty\
and\ \beta(T)<\infty\\}.$

Families of left and right Weyl operators, respectively, are defined as

$\Phi_{l}^{-}(X)=\\{T\in\Phi_{l}(X):\ \mathrm{ind}(T)\leq 0\\}$

and

$\Phi_{r}^{+}(X)=\\{T\in\Phi_{r}(X):\ \mathrm{ind}(T)\geq 0\\}.$

The set of Weyl operators is

$\Phi_{0}(X)=\Phi_{l}^{-}(X)\cap\Phi_{r}^{+}(X)=\\{T\in\Phi(X):\
\mathrm{ind}(T)=0\\}.$

Corresponding spectra of an operator $T\in\mathcal{B}(X)$ are defined as
follows:
-the left spectrum: $\sigma_{l}(T)=\\{\lambda\in\mathds{C}:\lambda-T\not\in\mathcal{G}_{l}(X)\\}$;
-the right spectrum: $\sigma_{r}(T)=\\{\lambda\in\mathds{C}:\lambda-T\not\in\mathcal{G}_{r}(X)\\}$;
-the spectrum: $\sigma(T)=\\{\lambda\in\mathds{C}:\lambda-T\not\in\mathcal{G}(X)\\}$;
-the left essential spectrum: $\sigma_{le}(T)=\\{\lambda\in\mathds{C}:\lambda-T\not\in\Phi_{l}(X)\\}$;
-the right essential spectrum: $\sigma_{re}(T)=\\{\lambda\in\mathds{C}:\lambda-T\not\in\Phi_{r}(X)\\}$;
-the essential spectrum: $\sigma_{e}(T)=\\{\lambda\in\mathds{C}:\lambda-T\not\in\Phi(X)\\}$;
-the left the Weyl spectrum: $\sigma_{lw}(T)=\\{\lambda\in\mathds{C}:\lambda-T\not\in\Phi_{l}^{-}(X)\\}$;
-the right the Weyl spectrum: $\sigma_{rw}(T)=\\{\lambda\in\mathds{C}:\lambda-T\not\in\Phi_{r}^{+}(X)\\}$;
-the Weyl spectrum: $\sigma_{w}(T)=\\{\lambda\in\mathds{C}:\lambda-T\not\in\Phi_{0}(X)\\}$.
All of these spectra are compact nonempty subsets of the complex plane.
We write
$\rho_{l}(T),\rho_{r}(T),\rho(T),\rho_{le}(T),\rho_{re}(T),\rho_{e}(T),\rho_{lw}(T),\rho_{rw}(T),\rho_{w}(T)$
for the corresponding complements of the sets above, respectively.

The following lemma imposes a connection between $T$ and its dual operator
$T^{\prime}$ in terms of nullity and deficiency of $T$. This claim will be
crucial at some points.

###### Lemma 2.1

([5, p. 7-8]) For $T\in\mathcal{B}(X)$ with closed range the following holds:
$(a)$ $\alpha(T)=\beta(T^{\prime}),\beta(T)=\alpha(T^{\prime})$;
$(b)$ $T\in\Phi_{l}(X)$ if and only if $T^{\prime}\in\Phi_{r}(X^{\prime})$;
$(c)$ $T\in\Phi_{r}(X)$ if and only if $T^{\prime}\in\Phi_{l}(X^{\prime})$;
$(d)$ $\mathrm{ind}(T^{\prime})=-\mathrm{ind}(T).$

We emphasize the fact that our results are ought to hold in arbitrary Banach
spaces, not just the separable ones. In order to prove main theorems which
concern perturbation of various spectra of $T_{n}^{d}(A)$, we introduce a
concept that will compensate loss of separability: the notion of embedded
spaces. To our knowledge, this condition was first used in this context by D.
S. Djordjević in 2002 (see [6, Theorem 4.4]).

###### Definition 2.2

Let $X$ and $Y$ be Banach spaces. We say that $X$ can be embedded in $Y$ and
write $X\preceq Y$ if and only if there exists a left invertible operator
$J:X\rightarrow Y.$

###### Remark 2.3

Obviously, $X\preceq Y$ if and only if there exists a right invertible
operator $J_{1}:Y\rightarrow X$.
If $H$ and $K$ are Hilbert spaces, than $H\preceq K$ if and only if $\dim
H\leq\dim K$. Here, $\dim H$ stands for the orthogonal dimension of $H$.

For the left (right) the Weyl spectrum we shall need another variant of the
definition above (see [6, Definition 4.2]).

###### Definition 2.4

Let $X$ and $Y$ be Banach spaces. We say that $X$ can be essentially embedded
in $Y$ and write $X\prec Y$ if and only if:
(a) $X\preceq Y$;
(b) for every $T\in\mathcal{B}(X,Y)$, $Y/\mathcal{R}(T)$ is infinite
dimensional.

###### Remark 2.5

If $H$ and $K$ are Hilbert spaces, than $H\prec K$ if and only if $\dim H<\dim
K$ and $K$ is infinite dimensional, where $\dim H$ is the orthogonal dimension
of $H$.

For the essential spectrum we need another form of Definition 2.2, that we
introduce right now.

###### Definition 2.6

Let $X$ and $Y$ be Banach spaces. We say that $X$ can be strongly embedded in
$Y$ and write $X\preceq_{s}Y$ if and only if:
(a) $X\preceq Y$;
(b) there exists $J\in\mathcal{B}(X,Y)$ such that $Y/\mathcal{R}(J)$ is of
finite dimension.

###### Remark 2.7

Notice the similarity with the notion of isomorphism up to a finite
dimensional subspace introduced in [6]. Correct link between these two is the
following: if Banach spaces $X$ and $Y$ are isomorphic up to a finite
dimensional subspace, then at least one of them can be strongly embedded into
another.

If $X$ and $Y$ are isomorphic up to a finite dimensional subspace, we write

$X\cong Y\ (u.f.d.s)$

for brevity.

One important difference between Hilbert and Banach spaces is that closed
subspace of a Hilbert space is always complemented $(\mathcal{H}=M\oplus
M^{\bot})$. This is not true for the case of Banach spaces. Since we would
like to prove our results by decomposing Banach spaces in an appropriate way,
we introduce the following condition.

###### Definition 2.8

We say that $D\in\mathcal{B}(X)$ satisfy the complements condition if and only
if:
$(a)$ $\mathcal{N}(D)$ is complemented;
$(b)$ $\mathcal{R}(D)$ is closed and complemented.

We immediately give an example of one important class of operators which
satisfies the complements condition.

We say that an operator $T\in\mathcal{B}(X_{i},X_{j})$ is (relatively) regular
if and only if there is $T^{\prime}\in\mathcal{B}(X_{j},X_{i})$ such that
$TT^{\prime}T=T$ holds. In that case we say $T^{\prime}$ is inner generalized
inverse of $T$. Notice that existence of such $T^{\prime}$ does not imply its
uniqueness. One can prove the following characterization (see [7, Corollary
1.1.5]):

###### Theorem 2.9

$T\in\mathcal{B}(X_{i},X_{j})$ is regular if and only if $\mathcal{N}(T)$ and
$\mathcal{R}(T)$ are closed and complemented subspaces of $X_{i}$ and $X_{j}$,
respectively.

Thus, it follows that regular operators satisfy the complements condition
introduced in Definition 2.8.
It is important to highlight that operators in sets $\mathcal{G}_{l}(X),\
\mathcal{G}_{r}(X),\ \Phi_{l}(X),\ \Phi_{r}(X)$ are regular operators.
Moreover, it is easily proved that, following upper terminology, $TT^{\prime}$
and $T^{\prime}T$ are both projections, and so we have decompositions (see [7,
Theorem 1.1.3])

$X_{i}=\mathcal{N}(T)\oplus\mathcal{R}(T^{\prime}T),\quad
X_{j}=\mathcal{N}(TT^{\prime})\oplus\mathcal{R}(T).$ (2.4)

We continue with some introductory lemmas.

###### Lemma 2.10

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$. Then

$\overline{\mathcal{R}(T_{n}^{d}(A)-\lambda)}=X_{1}\oplus
X_{2}\oplus\cdots\oplus X_{n}\quad for\ every\ A\in\mathcal{B}_{n}$

if and only if

$\overline{\mathcal{R}(D_{1}-\lambda)}=X_{1},\
\overline{\mathcal{R}(D_{2}-\lambda)}=X_{2},...,\overline{\mathcal{R}(D_{n}-\lambda)}=X_{n}.$

Proof. Since necessity holds obviously, we prove only sufficiency. Hence,
assume that $\overline{\mathcal{R}(D_{1}-\lambda)}=X_{1},\
\overline{\mathcal{R}(D_{2}-\lambda)}=X_{2},...,\overline{\mathcal{R}(D_{n}-\lambda)}=X_{n}.$
Regarding the relation (2.1) we get
$\mathcal{N}(D_{1}^{\prime}-\lambda)=\mathbf{0},\
\mathcal{N}(D_{2}^{\prime}-\lambda)=\mathbf{0},...,\
\mathcal{N}(D_{n}^{\prime}-\lambda)=\mathbf{0}.$ But then
$\mathcal{N}(T_{n}^{d}(A)^{\prime}-\lambda)\subseteq\bigcup\limits_{i=1}^{n}\mathcal{N}(D_{i}^{\prime}-\lambda)$
implying $\mathcal{N}(T_{n}^{d}(A)^{\prime}-\lambda)=\mathbf{0}$, and again
with regards to (2.1) we have
$\overline{\mathcal{R}(T_{n}^{d}(A)-\lambda)}=X_{1}\oplus
X_{2}\oplus\cdots\oplus X_{n}$. $\square$

###### Lemma 2.11

Let $T_{n}^{d}(A)\in\mathcal{B}(X_{1}\oplus\cdots\oplus X_{n}).$ Then:

  * (i)

$\sigma_{le}(D_{1})\subseteq\sigma_{le}(T_{n}^{d}(A))\subseteq\bigcup\limits_{k=1}^{n}\sigma_{le}(D_{k})$;

  * (ii)

$\sigma_{re}(D_{n})\subseteq\sigma_{re}(T_{n}^{d}(A))\subseteq\bigcup\limits_{k=1}^{n}\sigma_{re}(D_{k});$

  * (iii)

$\sigma_{lw}(T_{n}^{d}(A))\subseteq\bigcup\limits_{k=1}^{n}\sigma_{lw}(D_{k})$;

  * (iv)

$\sigma_{rw}(T_{n}^{d}(A))\subseteq\bigcup\limits_{k=1}^{n}\sigma_{rw}(D_{k})$.

We continue with two more lemmas that will be essential in proving theorems to
come.

###### Lemma 2.12

([19, section 16, Lemma 2]) Let $T,F\in\mathcal{B}(X_{1},X_{2})$. If $F$ is a
finite rank operator, then $\mathcal{R}(T+F)$ is closed if and only if
$\mathcal{R}(T)$ is closed.

###### Lemma 2.13

([1, Lemma 1.3] and [13, Theorem 5.2]) Let $D_{i}\in\mathcal{B}(X_{i})$,
$i=1,2$ and $A_{12}\in\mathcal{B}(X_{2},X_{1})$ with
$\overline{\mathcal{R}(D_{1})}=X_{1}$. If $\mathcal{R}(T_{2}^{d}(A))$ is
closed, then $\mathcal{R}(D_{2})$ is closed.

##  3 Perturbations of left (right) essential spectrum and essential spectrum
of $T_{n}^{d}(A)$

Hereafter, if $D\in\mathcal{B}(X_{i},X_{j})$ satisfies the complements
condition, complements of $\mathcal{N}(T)$ and $\mathcal{R}(T)$ will be
denoted by ′ sign, that is $X_{i}=\mathcal{N}(T)\oplus\mathcal{N}(T)^{\prime}$
and $X_{j}=\mathcal{R}(T)\oplus\mathcal{R}(T)^{\prime}$. As already announced,
our intention is to exploit the property (2.4) of regular operators as a
substitution for orthogonality in Hilbert spaces. However, we will formulate
our main theorems in a wider context of Definition 2.8, and then obtain
corresponding results expressed in the language of (2.4) as consequences.

We start with a result which characterizes perturbations of left essential
spectrum of $T_{n}^{d}(A)$.

###### Theorem 3.1

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ satisfying the complements condition. Additionally, suppose that we
have

$\mathcal{N}(D_{k})\preceq X_{j}/\mathcal{R}(D_{j})\quad for\ all\ n\geq
k>j\geq 1.$ (3.1)

Then

$\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{le}(T_{n}^{d}(A))=\sigma_{le}(D_{1})\cup\Big{(}\bigcup\limits_{k=2}^{n}\Delta_{k}\Big{)},$

where

$\Delta_{k}:=\Big{\\{}\lambda\in\sigma_{le}(D_{k}):\
\sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty\Big{\\}},\ 2\leq k\leq n.$

Proof. Inclusion $(\supseteq)$ is already proved in [22]. For convenience of a
reader we provide the full proof, with a slight difference in that now we have
different space decompositions.

Write $\Delta_{1}:=\sigma_{le}(D_{1})$,
$\Delta_{k1}=\\{\lambda\in\mathds{C}:\mathcal{R}(D_{k}-\lambda)\ is\ not\
closed\ and\\\ \ \sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty\\}$ and
$\Delta_{k2}=\\{\lambda\in\mathds{C}:\mathcal{R}(D_{k}-\lambda)\ is\ closed\
and\\\ \sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty,\
\alpha(D_{k}-\lambda)=\infty\\}$ for $2\leq k\leq n$.
Lemma 2.11 implies
$\Delta_{1}\subseteq\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{le}(T_{n}^{d}(A))$
. Let $\lambda\in\bigcup\limits_{k=2}^{n}\Delta_{k}$.
Let $\lambda\in\Delta_{k1}$, that is $\mathcal{R}(D_{k}-\lambda)$ is not
closed and $\sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty$,
$k\in\\{2,...,n\\}$. Then for each $A\in\mathcal{B}_{k}$, operator matrix
$T_{k}^{d}(A)$ as an operator from
$X_{1}\oplus\mathcal{N}(D_{2})^{\prime}\oplus\mathcal{N}(D_{2})\oplus\mathcal{N}(D_{3})^{\prime}\oplus\mathcal{N}(D_{3})\oplus\cdots\oplus\mathcal{N}(D_{k})^{\prime}\oplus\mathcal{N}(D_{k})$
into
$\mathcal{R}(D_{1})\oplus\mathcal{R}(D_{1})^{\prime}\oplus\mathcal{R}(D_{2})\oplus\mathcal{R}(D_{2})^{\prime}\oplus\cdots\oplus\mathcal{R}(D_{k-1})\oplus\mathcal{R}(D_{k-1}^{\prime})\oplus\mathcal{R}(D_{k})\oplus\mathcal{R}(D_{k})^{\prime}$
admits the following block representation

$T_{k}^{d}(A)=\scalebox{0.85}{\mbox{$\displaystyle\begin{bmatrix}D_{1}^{(1)}&A_{12}^{(1)}&A_{12}^{(2)}&A_{13}^{(1)}&A_{13}^{(2)}&...&A_{1,k-1}^{(1)}&A_{1,k-1}^{(2)}&A_{1k}^{(1)}&A_{1k}^{(2)}\\\
0&A_{12}^{(3)}&A_{12}^{(4)}&A_{13}^{(3)}&A_{13}^{(4)}&...&A_{1,k-1}^{(3)}&A_{1,k-1}^{(4)}&A_{1k}^{(3)}&A_{1k}^{(4)}\\\
0&D_{2}^{(1)}&0&A_{23}^{(1)}&A_{23}^{(2)}&...&A_{2,k-1}^{(1)}&A_{2,k-1}^{(2)}&A_{2k}^{(1)}&A_{2k}^{(2)}\\\
0&0&0&A_{23}^{(3)}&A_{23}^{(4)}&...&A_{2,k-1}^{(3)}&A_{2,k-1}^{(4)}&A_{2k}^{(3)}&A_{2k}^{(4)}\\\
0&0&0&D_{3}^{(1)}&0&...&A_{3,k-1}^{(1)}&A_{3,k-1}^{(2)}&A_{3k}^{(1)}&A_{3k}^{(2)}\\\
0&0&0&0&0&...&A_{3,k-1}^{(3)}&A_{3,k-1}^{(4)}&A_{3k}^{(3)}&A_{3k}^{(4)}\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots\\\
0&0&0&0&0&...&D_{k-1}^{(1)}&0&A_{k-1,k}^{(1)}&A_{k-1,k}^{(2)}\\\
0&0&0&0&0&...&0&0&A_{k-1,k}^{(3)}&A_{k-1,k}^{(4)}\\\
0&0&0&0&0&...&0&0&D_{k}^{(1)}&0\\\ 0&0&0&0&0&...&0&0&0&0\\\ \end{bmatrix}$}}$
(3.2)

Assume that $\mathcal{R}(T_{k}^{d}(A)-\lambda)$ is closed. Then, according to
Lemma 2.12, range of the following operator matrix

$\displaystyle\begin{bmatrix}D_{1}^{(1)}-\lambda&A_{12}^{(1)}&A_{12}^{(2)}&A_{13}^{(1)}&A_{13}^{(2)}&...&A_{1,k-1}^{(1)}&A_{1,k-1}^{(2)}&A_{1k}^{(1)}&A_{1k}^{(2)}\\\
0&0&0&0&0&...&0&0&0&0\\\
0&D_{2}^{(1)}-\lambda&0&A_{23}^{(1)}&A_{23}^{(2)}&...&A_{2,k-1}^{(1)}&A_{2,k-1}^{(2)}&A_{2k}^{(1)}&A_{2k}^{(2)}\\\
0&0&0&0&0&...&0&0&0&0\\\
0&0&0&D_{3}^{(1)}-\lambda&0&...&A_{3,k-1}^{(1)}&A_{3,k-1}^{(2)}&A_{3k}^{(1)}&A_{3k}^{(2)}\\\
0&0&0&0&0&...&0&0&0&0\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots\\\
0&0&0&0&0&...&D_{k-1}^{(1)}-\lambda&0&A_{k-1,k}^{(1)}&A_{k-1,k}^{(2)}\\\
0&0&0&0&0&...&0&0&0&0\\\ 0&0&0&0&0&...&0&0&D_{k}^{(1)}-\lambda&0\\\
0&0&0&0&0&...&0&0&0&0\\\ \end{bmatrix}$

is closed due to $\sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty$, and
hence range of the following operator matrix

$\displaystyle\begin{bmatrix}D_{1}^{(1)}-\lambda&A_{12}^{(1)}&A_{12}^{(2)}&A_{13}^{(1)}&A_{13}^{(2)}&...&A_{1,k-1}^{(1)}&A_{1,k-1}^{(2)}&A_{1k}^{(1)}&A_{1k}^{(2)}\\\
0&D_{2}^{(1)}-\lambda&0&A_{23}^{(1)}&A_{23}^{(2)}&...&A_{2,k-1}^{(1)}&A_{2,k-1}^{(2)}&A_{2k}^{(1)}&A_{2k}^{(2)}\\\
0&0&0&D_{3}^{(1)}-\lambda&0&...&A_{3,k-1}^{(1)}&A_{3,k-1}^{(2)}&A_{3k}^{(1)}&A_{3k}^{(2)}\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots\\\
0&0&0&0&0&...&D_{k-1}^{(1)}-\lambda&0&A_{k-1,k}^{(1)}&A_{k-1,k}^{(2)}\\\
0&0&0&0&0&...&0&0&D_{k}^{(1)}-\lambda&0\\\ \end{bmatrix}$

is closed. Since
$\overline{\mathcal{R}(D_{s}^{(1)}-\lambda)}=\overline{\mathcal{R}(D_{s}-\lambda)}$,
$s=1,2,...,k-1$ , then

$\displaystyle\mathcal{R}\left(\begin{bmatrix}D_{1}^{(1)}-\lambda&A_{12}^{(1)}&A_{12}^{(2)}&A_{13}^{(1)}&A_{13}^{(2)}&...&A_{1,k-1}^{(1)}&A_{1,k-1}^{(2)}\\\
0&D_{2}^{(1)}-\lambda&0&A_{23}^{(1)}&A_{23}^{(2)}&...&A_{2,k-1}^{(1)}&A_{2,k-1}^{(2)}\\\
0&0&0&D_{3}^{(1)}-\lambda&0&...&A_{3,k-1}^{(1)}&A_{3,k-1}^{(2)}\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&0&0&0&...&D_{k-1}^{(1)}-\lambda&0\\\ \end{bmatrix}\right)$

is dense in
$\overline{\mathcal{R}(D_{1}-\lambda)}\oplus\overline{\mathcal{R}(D_{2}-\lambda)}\oplus\cdots\oplus\overline{\mathcal{R}(D_{k-1}-\lambda)}$
from Lemma 2.10. By Lemma 2.13, it follows that
$\mathcal{R}(D_{k}^{(1)}-\lambda)$ is closed, which is contradiction to the
fact that $\mathcal{R}(D_{k}-\lambda)$ is not closed. This implies that
$\mathcal{R}(T_{k}^{d}(A)-\lambda)$ is not closed for every
$A\in\mathcal{B}_{k}$, and hence
$\lambda\in\bigcap\limits_{A\in\mathcal{B}_{k}}\sigma_{le}(T_{k}^{d}(A))$.
This proves
$\bigcup\limits_{k=2}^{n}\Delta_{k1}\subseteq\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{le}(T_{n}^{d}(A))$.

Let $\lambda\in\Delta_{k2}$, $k\in\\{2,...,n\\}.$ Without loss of generality,
put $\lambda=0$. We only consider
$0\in\Delta_{k2}\setminus(\bigcup\limits_{s=2}^{k}\Delta_{s1}\bigcup\limits\Delta_{1})$.
Therefore, $0\in\rho_{le}(D_{1})$, $\mathcal{R}(D_{s})$ is closed for
$s=2,...,k$, $\sum\limits_{s=1}^{k-1}\beta(D_{s})<\infty$ and
$\alpha(D_{k})=\infty$. Then for each $A\in\mathcal{B}_{k}$, operator matrix
$T_{k}^{d}(A)$ admits block representation as in (3.2). Evidently,
$D_{1}^{(1)},\ D_{2}^{(1)},...,D_{k}^{(1)}$ as in (3.2) are all invertible.
Thus, there exist invertible operator matrices $U$ and $V$ such that

$UT_{k}^{d}(A)V=\scalebox{0.85}{\mbox{$\displaystyle\begin{bmatrix}D_{1}^{(1)}&0&0&0&0&...&0&0&0&0\\\
0&0&A_{12}^{(4)}&0&A_{13}^{(4)}&...&0&A_{1,k-1}^{(4)}&0&A_{1k}^{(4)}\\\
0&D_{2}^{(1)}&0&0&0&...&0&0&0&0\\\
0&0&0&0&A_{23}^{(4)}&...&0&A_{2,k-1}^{(4)}&0&A_{2k}^{(4)}\\\
0&0&0&D_{3}^{(1)}&0&...&0&0&0&0\\\
0&0&0&0&0&...&0&A_{3,k-1}^{(4)}&0&A_{3k}^{(4)}\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots\\\
0&0&0&0&0&...&D_{k-1}^{(4)}&0&0&0\\\ 0&0&0&0&0&...&0&0&0&A_{k-1,k}^{(4)}\\\
0&0&0&0&0&...&0&0&D_{k}^{(1)}&0\\\ 0&0&0&0&0&...&0&0&0&0\\\ \end{bmatrix}$}}$
(3.3)

Note that $A_{ij}^{(4)}$ in (3.3) are not the original ones in (3.2) in
general, but we still use them for the convenience. Combine by (3.3),
$T_{k}^{d}(A)$ is left Fredholm if and only if

$\begin{bmatrix}A_{12}^{(4)}&A_{13}^{(4)}&A_{14}^{(4)}&...&A_{1k}^{(4)}\\\
0&A_{23}^{(4)}&A_{24}^{(4)}&...&A_{2k}^{(4)}\\\
0&0&A_{34}^{(4)}&...&A_{3k}^{(4)}\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\
0&0&0&...&A_{k-1,k}^{(4)}\end{bmatrix}:\begin{bmatrix}\mathcal{N}(D_{2})\\\
\mathcal{N}(D_{3})\\\ \mathcal{N}(D_{4})\\\ \vdots\\\ \mathcal{N}(D_{k})\\\
\end{bmatrix}\rightarrow\begin{bmatrix}\mathcal{R}(D_{1})^{\prime}\\\
\mathcal{R}(D_{2})^{\prime}\\\ \mathcal{R}(D_{3})^{\prime}\\\ \vdots\\\
\mathcal{R}(D_{k-1})^{\prime}\\\ \end{bmatrix}$ (3.4)

is left Fredholm. Since $\sum\limits_{s=1}^{k-1}\beta(D_{s})<\infty$ and
$\alpha(D_{k})=\infty$, it follows that

$\alpha\left(\begin{bmatrix}A_{1k}^{(4)}\\\ A_{2k}^{(4)}\\\ A_{3k}^{(4)}\\\
\vdots\\\ A_{k-1,k}^{(4)}\end{bmatrix}\right)=\infty,$

and hence operator defined in (3.4) is not left Fredholm for every
$A\in\mathcal{B}_{k}$. This proves
$\bigcup\limits_{k=2}^{n}\Delta_{k2}\subseteq\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{le}(T_{n}^{d}(A))$.

For the opposite inclusion, it suffices to prove that
$0\not\in\bigcup\limits_{k=1}^{n}\Delta_{k}$ implies
$0\not\in\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{le}(T_{n}^{d}(A))$, i.e.
there exists an operator tuple $A\in\mathcal{B}_{n}$ such that $T_{n}^{d}(A)$
is left Fredholm. Let $0\not\in\bigcup\limits_{k=1}^{n}\Delta_{k}$. Then,
direct calculation show that $0\in\bigcup\limits_{k=1}^{n}\Omega_{k}$, where
$\Omega_{1}=\\{\lambda\in\rho_{le}(D_{1}):\ \beta(D_{1}-\lambda)=\infty\\}\\\
\Omega_{k}=\\{\lambda\in\bigcap\limits_{s=1}^{k}\rho_{le}(D_{s}):\
\beta(D_{k}-\lambda)=\infty\\},\ 2\leq k\leq n-1,\\\
\Omega_{n}=\bigcap\limits_{s=1}^{n}\rho_{le}(D_{s}).\\\ $There are three cases
to consider.
Case 1: $0\in\Omega_{1}$

This means that $\alpha(D_{1})<\infty$ ($D_{1}$ is left Fredholm), and
$\beta(D_{1})=\infty$. We find $A\in\mathcal{B}_{n}$ such that
$\alpha(T_{n}^{d}(A))<\infty$ and $\mathcal{R}(T_{n}^{d}(A))$ is closed. We
choose $A=(A_{ij})_{1\leq i<j\leq n}$ so that $A_{ij}=0$ if $j-i\neq 1$, that
is we place all nonzero operators of tuple $A$ on the superdiagonal. It
remains to define $A_{ij}$ for $j-i=1$, $1\leq i<j\leq n$. First notice that
$A_{i,i+1}:X_{i+1}\rightarrow X_{i}$. Since all of diagonal entries satisfy
the complements condition, we know that
$X_{i+1}=\mathcal{N}(D_{i+1})\oplus\mathcal{N}(D_{i+1})^{\prime}$,
$X_{i}=\mathcal{R}(D_{i})^{\prime}\oplus\mathcal{R}(D_{i})$, and from
assumption (3.1) we get
$\mathcal{N}(D_{i+1})\preceq\mathcal{R}(D_{i})^{\prime}$. It follows that
there is left invertible operator
$J_{i}:\mathcal{N}(D_{i+1})\rightarrow\mathcal{R}(D_{i})^{\prime}$. We put
$A_{i,i+1}=\begin{bmatrix}J_{i}&0\\\
0&0\end{bmatrix}:\begin{bmatrix}\mathcal{N}(D_{i+1})\\\
\mathcal{N}(D_{i+1})^{\prime}\end{bmatrix}\rightarrow\begin{bmatrix}\mathcal{R}(D_{i})^{\prime}\\\
\mathcal{R}(D_{i})\end{bmatrix}$, and we implement this procedure for all
$1\leq i\leq n-1$. Notice that $\mathcal{R}(D_{i})$ is complemented to
$\mathcal{R}(A_{i,i+1})$ for each $i=1,...,n-1.$

Now we have chosen our $A$, we show that
$\mathcal{N}(T_{n}^{d}(A))\cong\mathcal{N}(D_{1})$, implying
$\alpha(T_{n}^{d}(A))=\alpha(D_{1})<\infty$. Let us put $T_{n}^{d}(A)x=0$,
where $x=x_{1}+\cdots+x_{n}\in X_{1}\oplus\cdots\oplus X_{n}$. The previous
equality is then equivalent to the following system of equations

$\begin{bmatrix}D_{1}x_{1}+A_{12}x_{2}\\\ D_{2}x_{2}+A_{23}x_{3}\\\ \vdots\\\
D_{n-1}x_{n-1}+A_{n-1,n}x_{n}\\\ D_{n}x_{n}\end{bmatrix}=\begin{bmatrix}0\\\
0\\\ \vdots\\\ 0\\\ 0\end{bmatrix}.$

The last equation gives $x_{n}\in\mathcal{N}(D_{n})$. Since
$\mathcal{R}(D_{s})$ is complemented to $\mathcal{R}(A_{s,s+1})$ for all
$1\leq s\leq n-1$, we have $D_{s}x_{s}=A_{s,s+1}x_{x+1}=0$ for all $1\leq
s\leq n-1$. That is, $x_{i}\in\mathcal{N}(D_{i})$ for every $i=1,...,n$, and
$J_{s}x_{s+1}=0$ for every $1\leq s\leq n-1.$ Due to left invertibility of
$J_{s}$ we get $x_{s}=0$ for $s=2,...,n$, which proves the claim. Therefore,
$\alpha(T_{n}^{d}(A))=\alpha(D_{1})<\infty$.

Secondly, we show that $\mathcal{R}(T_{n}^{d}(A))$ is closed. It is not hard
to see that
$\mathcal{R}(T_{n}^{d}(A))=\mathcal{R}(D_{1})\oplus\mathcal{R}(J_{1})\oplus\mathcal{R}(D_{2})\oplus\mathcal{R}(J_{2})\oplus\cdots\oplus\mathcal{R}(D_{n-1})\oplus\mathcal{R}(J_{n-1})\oplus\mathcal{R}(D_{n})$.
Furthermore, due to left invertibility of $J_{i}$’s, there exist closed
subspaces $U_{i}$ of $\mathcal{R}(D_{i})^{\prime}$ such that
$\mathcal{R}(D_{i})^{\prime}=\mathcal{R}(J_{i})\oplus U_{i}$, $i=1,...,n-1$.
It means that $X_{1}\oplus X_{2}\oplus\cdots\oplus
X_{n}=\mathcal{R}(D_{1})\oplus\mathcal{R}(J_{1})\oplus
U_{1}\oplus\mathcal{R}(D_{2})\oplus\mathcal{R}(J_{2})\oplus
U_{2}\oplus\cdots\oplus\mathcal{R}(D_{n-1})\oplus\mathcal{R}(J_{n-1})\oplus
U_{n-1}\oplus\mathcal{R}(D_{n})$. Comparing these equalities, we conclude that
$\mathcal{R}(T_{n}^{d}(A))$ is closed.

Case 2: $0\in\Omega_{k}$, $2\leq k\leq n-1$

This means that $\alpha(D_{1}),...,\alpha(D_{k})<\infty$ ($D_{1},...,D_{k}$
are left Fredholm), and $\beta(D_{k})=\infty$. We find $A\in\mathcal{B}_{n}$
such that $\alpha(T_{n}^{d}(A))<\infty$ and $\mathcal{R}(T_{n}^{d}(A))$ is
closed. We choose $A=(A_{ij})_{1\leq i<j\leq n}$ analogously as in Case 1,
with the only exception of choosing first $k-1$ operators on the superdiagonal
to be zero. That is, $A_{i,i+1}=\begin{bmatrix}J_{i}&0\\\
0&0\end{bmatrix}:\begin{bmatrix}\mathcal{N}(D_{i+1})\\\
\mathcal{N}(D_{i+1})^{\prime}\end{bmatrix}\rightarrow\begin{bmatrix}\mathcal{R}(D_{i})^{\prime}\\\
\mathcal{R}(D_{i})\end{bmatrix}$, $k\leq i\leq n$, and $A_{ij}=0$ otherwise,
$1\leq i<j\leq n$, following notation of Case 1.

Now we have chosen our $A$, we show that
$\mathcal{N}(T_{n}^{d}(A))\cong\mathcal{N}(D_{1})\oplus\cdots\oplus\mathcal{N}(D_{k})$.
Let us put $T_{n}^{d}(A)x=0$, where $x=x_{1}+\cdots+x_{n}\in
X_{1}\oplus\cdots\oplus X_{n}$. The previous equality is then equivalent to

$\begin{bmatrix}D_{1}x_{1}\\\ \vdots\\\ D_{k-1}x_{k-1}\\\
D_{k}x_{k}+A_{k,k+1}x_{k+1}\\\ \vdots\\\ D_{n-1}x_{n-1}+A_{n-1,n}x_{n}\\\
D_{n}x_{n}\end{bmatrix}=\begin{bmatrix}0\\\ \vdots\\\ 0\\\ 0\\\ \vdots\\\ 0\\\
0\end{bmatrix}.$

The last $n-k$ equalities give that $x_{s}=0$ for $s=k+1,...,n$ by the same
reasoning seen in Case 1, which proves the claim. Therefore,
$\alpha(T_{n}^{d}(A))=\sum\limits_{s=1}^{k}\alpha(D_{s})<\infty$.

Secondly, we show that $\mathcal{R}(T_{n}^{d}(A))$ is closed. It is not hard
to see that
$\mathcal{R}(T_{n}^{d}(A))=\mathcal{R}(D_{1})\oplus\cdots\oplus\mathcal{R}(D_{k-1})\oplus\mathcal{R}(D_{k})\oplus\mathcal{R}(J_{k})\oplus\cdots\oplus\mathcal{R}(D_{n-1})\oplus\mathcal{R}(J_{n-1})\oplus\mathcal{R}(D_{n})$.
Furthermore, due to left invertibility of $J_{i}$’s, there exist closed
subspaces $U_{i}$ of $\mathcal{R}(D_{i})^{\prime}$ such that
$\mathcal{R}(D_{i})^{\prime}=\mathcal{R}(J_{i})\oplus U_{i}$, $k\leq i\leq
n-1$. It means that $X_{1}\oplus X_{2}\oplus\cdots\oplus
X_{n}=X_{1}\oplus\cdots\oplus
X_{k-1}\oplus\mathcal{R}(D_{k})\oplus\mathcal{R}(J_{k})\oplus
U_{k}\oplus\cdots\oplus\mathcal{R}(D_{n-1})\oplus\mathcal{R}(J_{n-1})\oplus
U_{n-1}\oplus X_{n}$. Comparing these equalities, we conclude that
$\mathcal{R}(T_{n}^{d}(A))$ is closed.

Case 3: $0\in\Omega_{n}$

In this case $0\in\bigcap\limits_{s=1}^{n}\rho_{le}(D_{s})$, which implies
$\alpha(D_{1}),...,\alpha(D_{n})<\infty$ ($D_{1},...,D_{n}$ are left
Fredholm). Choose trivially $A_{ij}=0$ for every $1\leq i<j\leq n$. Then
$\mathcal{N}(T_{n}^{d}(A))\cong\mathcal{N}(D_{1})\oplus\cdots\oplus\mathcal{N}(D_{n})$
and
$\mathcal{R}(T_{n}^{d}(A))=\mathcal{R}(D_{1})\oplus\cdots\oplus\mathcal{R}(D_{n})$,
and since $X_{1}\oplus X_{2}\oplus\cdots\oplus
X_{n}=\mathcal{R}(D_{1})\oplus\mathcal{R}(D_{1})^{\prime}\oplus\cdots\oplus\mathcal{R}(D_{n})\oplus\mathcal{R}(D_{n}^{\prime})$,
it follows that $\mathcal{R}(T_{n}^{d}(A))$ is closed and
$\alpha(T_{n}^{d}(A))=\sum\limits_{s=1}^{n}\alpha(D_{s})<\infty$. $\square$

###### Corollary 3.2

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ being regular. Additionally, suppose that we have

$\mathcal{N}(D_{k})\preceq X_{j}/\mathcal{R}(D_{j})\quad for\ all\ n\geq
k>j\geq 1.$

Then

$\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{le}(T_{n}^{d}(A))=\sigma_{le}(D_{1})\cup\Big{(}\bigcup\limits_{k=2}^{n}\Delta_{k}\Big{)},$

where

$\Delta_{k}:=\Big{\\{}\lambda\in\sigma_{le}(D_{k}):\
\sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty\Big{\\}},\ 2\leq k\leq n.$

The following is a dual result of Theorem 3.1.

###### Theorem 3.3

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ satisfying the complements condition. Additionally, suppose that we
have

$X_{k}/\mathcal{R}(D_{k})\preceq\mathcal{N}(D_{j})\quad for\ all\ 1\leq
k<j\leq n.$ (3.5)

Then

$\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{re}(T_{n}^{d}(A))=\sigma_{re}(D_{n})\cup\Big{(}\bigcup\limits_{k=1}^{n-1}\Delta_{k}\Big{)},$

where

$\Delta_{k}:=\Big{\\{}\lambda\in\sigma_{re}(D_{k}):\
\sum\limits_{s=k+1}^{n}\alpha(D_{s}-\lambda)<\infty\Big{\\}},\ 1\leq k\leq
n-1.$

Proof: From Lemma 2.1 it follows that
$\sigma_{re}(T_{n}^{d}(A))=\sigma_{le}(T_{n}^{d}(A)^{\prime})$. Now, the proof
immediately follows from Theorem 3.1, and with regards to (2.3). $\square$

###### Corollary 3.4

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ being regular. Additionally, suppose that we have

$X_{k}/\mathcal{R}(D_{k})\preceq\mathcal{N}(D_{j})\quad for\ all\ 1\leq
k<j\leq n.$

Then

$\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{re}(T_{n}^{d}(A))=\sigma_{re}(D_{n})\cup\Big{(}\bigcup\limits_{k=1}^{n-1}\Delta_{k}\Big{)},$

where

$\Delta_{k}:=\Big{\\{}\lambda\in\sigma_{re}(D_{k}):\
\sum\limits_{s=k+1}^{n}\alpha(D_{s}-\lambda)<\infty\Big{\\}},\ 1\leq k\leq
n-1.$

Now, we conclude with perturbation of essential spectrum. We put into use
Definition 2.6.

###### Theorem 3.5

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ satisfying the complements condition. Additionally, suppose that we
have

$\mathcal{N}(D_{k})\preceq_{s}X_{j}/\mathcal{R}(D_{j})\quad for\ all\ n\geq
k>j\geq 1.$ (3.6)

Then

$\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{e}(T_{n}^{d}(A))=\sigma_{le}(D_{1})\bigcup\limits\sigma_{re}(D_{n})\bigcup\limits\Big{(}\bigcup\limits_{k=2}^{n-1}\Delta_{k}\Big{)}\bigcup\limits\\{\lambda\in\mathds{C}:\
\beta(D_{1}-\lambda)=\infty,\
\sum\limits_{s=2}^{n}\alpha(D_{s}-\lambda)<\infty\\}\bigcup\limits\\{\lambda\in\mathds{C}:\
\alpha(D_{n}-\lambda)=\infty,\
\sum\limits_{s=1}^{n-1}\beta(D_{s}-\lambda)<\infty\\},$

where

$\Delta_{k}:=\Big{\\{}\lambda\in\sigma_{le}(D_{k}):\
\sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty\Big{\\}}\bigcup\limits\Big{\\{}\lambda\in\sigma_{re}(D_{k}):\
\newline \sum\limits_{s=k+1}^{n}\alpha(D_{s}-\lambda)<\infty\Big{\\}},\ 2\leq
k\leq n-1.$

Proof: Write $\Delta_{1}:=\sigma_{le}(D_{1})\bigcup\limits\sigma_{re}(D_{n})$
and $\Delta_{n}:=\\{\lambda\in\mathds{C}:\ \beta(D_{1}-\lambda)=\infty,\
\sum\limits_{s=2}^{n}\alpha(D_{s}-\lambda)<\infty\\}\bigcup\limits\\{\lambda\in\mathds{C}:\
\alpha(D_{n}-\lambda)=\infty,\
\sum\limits_{s=1}^{n-1}\beta(D_{s}-\lambda)<\infty\\}.$ By Theorems 3.1 and
3.3 we clearly have
$\bigcup\limits_{k=1}^{n}\Delta_{k}\subseteq\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{e}(T_{n}^{d}(A))$.
Without loss of generality, we only prove the case when $\lambda=0$ in what
follows.
For the opposite inclusion, it suffices to prove that
$0\not\in\bigcup\limits_{k=1}^{n}\Delta_{k}$ implies
$0\not\in\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{e}(T_{n}^{d}(A))$, i. e.
there exists an operator tuple $A\in\mathcal{B}_{n}$ such that $T_{n}^{d}(A)$
is Fredholm.
Let $0\not\in\bigcup\limits_{k=1}^{n}\Delta_{k}$. Then, direct calculations
show that $0\in\bigcup\limits_{l=1}^{n-1}\bigcup\limits_{k=2}^{n}\Omega_{lk}\
(l<k)$ or $0\in\bigcap\limits_{s=1}^{n}\rho_{e}(D_{s})$, where
$\Omega_{1k}=\Big{\\{}\lambda\in\bigcap\limits_{s=k+1}^{n}\rho_{e}(D_{s})\bigcap\limits\rho_{le}(D_{1})\bigcap\limits\rho_{re}(D_{k}):\\\
\beta(D_{1}-\lambda)=\alpha(D_{k}-\lambda)=\infty\Big{\\}},\ 2\leq k\leq
n-1;\\\
\Omega_{1n}=\\{\lambda\in\rho_{le}(D_{1})\bigcap\limits\rho_{re}(D_{n}):\
\beta(D_{1}-\lambda)=\alpha(D_{n}-\lambda)=\infty\\},\\\
\Omega_{lk}=\Big{\\{}\lambda\in\bigcap\limits_{s=1}^{l-1}\rho_{e}(D_{s})\bigcap\limits\rho_{le}(D_{l})\bigcap\limits\rho_{re}(D_{k})\bigcap\limits\bigcap\limits_{s=k+1}^{n}\rho_{e}(D_{s}):\\\
\beta(D_{l}-\lambda)=\alpha(D_{k}-\lambda)=\infty\Big{\\}},\ 2\leq l<k\leq
n-1;\\\
\Omega_{ln}=\Big{\\{}\lambda\in\bigcap\limits_{s=1}^{l-1}\rho_{e}(D_{s})\bigcap\limits\rho_{le}(D_{l})\bigcap\limits\rho_{re}(D_{n}):\\\
\beta(D_{l}-\lambda)=\alpha(D_{n}-\lambda)=\infty\Big{\\}},\ 2\leq l\leq n-1.$
There are three cases to consider:
Case 1: $0\in\Omega_{1k},\ 2\leq k\leq n-1$. In this case
$\beta(D_{1})=\alpha(D_{k})=\infty$, $\alpha(D_{1}),\beta(D_{k})<\infty$ and
$D_{k+1},...,D_{n}$ are Fredholm. We find $A\in\mathcal{B}_{n}$ such that
$\alpha(T_{n}^{d}(A))<\infty$ and $\beta(T_{n}^{d}(A))<\infty$. We choose
$A=(A_{ij})_{1\leq i<j\leq n}$ so that $A_{ij}=0$ if $j-i\neq 1$ or $j\neq k$,
that is we place all nonzero operators of tuple $A$ on the superdiagonal or in
the k-th column. It remains to define $A_{ij}$ for $j-i=1$, $1\leq i<j\leq n$,
and for $j=k$. First notice that $A_{i,i+1}:X_{i+1}\rightarrow X_{i}$. Since
all of diagonal entries satisfy the complements condition, we know that
$X_{i+1}=\mathcal{N}(D_{i+1})\oplus\mathcal{N}(D_{i+1})^{\prime}$,
$X_{i}=\mathcal{R}(D_{i})^{\prime}\oplus\mathcal{R}(D_{i})$, and due to (3.6)
we get $\mathcal{N}(D_{i+1})\preceq_{s}\mathcal{R}(D_{i})^{\prime}$, and so
there is left invertible operator
$J_{i}:\mathcal{N}(D_{i+1})\rightarrow\mathcal{R}(D_{i})^{\prime}$ with
$\dim(\mathcal{R}(D_{i})^{\prime}/\mathcal{R}(J_{i}))<\infty$. We put
$A_{i,i+1}=\begin{bmatrix}J_{i}&0\\\
0&0\end{bmatrix}:\begin{bmatrix}\mathcal{N}(D_{i+1})\\\
\mathcal{N}(D_{i+1})^{\prime}\end{bmatrix}\rightarrow\begin{bmatrix}\mathcal{R}(D_{i})^{\prime}\\\
\mathcal{R}(D_{i})\end{bmatrix}$, and we implement this procedure for all
$1\leq i\leq k-2$. If $j=k$, then the reasoning analogous to the previous one
gives us left invertible
$J_{i}^{\prime}:\mathcal{N}(D_{k})\rightarrow\mathcal{R}(D_{i})^{\prime}$ with
$\dim(\mathcal{R}(D_{i})^{\prime}/\mathcal{R}(J_{i}^{\prime}))<\infty$, and so
we put $A_{ik}=\begin{bmatrix}J_{i}^{\prime}&0\\\ 0&0\\\
\end{bmatrix}:\begin{bmatrix}\mathcal{N}(D_{k})\\\
\mathcal{N}(D_{k})^{\prime}\end{bmatrix}\rightarrow\begin{bmatrix}\mathcal{R}(D_{i})^{\prime}\\\
\mathcal{R}(D_{i})\end{bmatrix}$ for all $1\leq i\leq k-1$. Finally, we place
$A_{ij}=0$ if $j>k.$

Now we have chosen our $A$, we show that
$\mathcal{N}(T_{n}^{d}(A))\cong\mathcal{N}(D_{1})\oplus\mathcal{N}(D_{k+1})\oplus\cdots\oplus\mathcal{N}(D_{n})$.
Let us put $T_{n}^{d}(A)x=0$, where $x=x_{1}+\cdots+x_{n}\in
X_{1}\oplus\cdots\oplus X_{n}$. The previous equality is then equivalent to

$\begin{bmatrix}D_{1}x_{1}+A_{12}x_{2}+A_{1k}x_{k}\\\
D_{2}x_{2}+A_{23}x_{3}+A_{2k}x_{k}\\\ \vdots\\\
D_{k-2}x_{k-2}+A_{k-2,k-1}x_{k-1}+A_{k-2,k}x_{k}\\\
D_{k-1}x_{k-1}+A_{k-1,k}x_{k}\\\ D_{k}x_{k}\\\ D_{k+1}x_{k+1}\\\ \vdots\\\
D_{n}x_{n}\end{bmatrix}=\begin{bmatrix}0\\\ 0\\\ \vdots\\\ 0\\\ 0\\\ 0\\\ 0\\\
\vdots\\\ 0\end{bmatrix}.$

The k-th equality yields $x_{k}\in\mathcal{N}(D_{k})$. Therefore, by the
procedure already seen, the (k-1)-th equality gives $x_{k-1}=x_{k}=0$. Now,
the last system of equations implies

$\begin{bmatrix}D_{1}x_{1}+A_{12}x_{2}\\\ D_{2}x_{2}+A_{23}x_{3}\\\ \vdots\\\
D_{k-2}x_{k-2}\\\ 0\end{bmatrix}=\begin{bmatrix}0\\\ 0\\\ \vdots\\\ 0\\\
0\end{bmatrix}.$

It is not hard to see that continuing this procedure iteratively, we get
$x_{k}=x_{k-1}=\cdots=x_{2}=0$, and this proves our statement. This means that
$\alpha(T_{n}^{d}(A))<\infty.$

Now, we prove that $\beta(T_{n}^{d}(A))<\infty$. We know that
$\mathcal{R}(T_{n}^{d}(A))=\mathcal{R}(D_{1})\oplus\mathcal{R}(J_{1})\oplus\mathcal{R}(J_{1}^{\prime})\oplus\mathcal{R}(D_{2})\oplus\mathcal{R}(J_{2})\oplus\mathcal{R}(J_{2}^{\prime})\oplus\cdots\oplus\mathcal{R}(D_{k-2})\oplus\mathcal{R}(J_{k-2})\oplus\mathcal{R}(J_{k-2}^{\prime})\oplus\mathcal{R}(D_{k-1})\oplus\mathcal{R}(J_{k-1}^{\prime})\oplus\mathcal{R}(D_{k})\oplus\mathcal{R}(D_{k+1})\oplus\cdots\oplus\mathcal{R}(D_{n})$.
Furthermore, we know that there exist closed finite dimensional subspaces
$U_{i},\ U_{i}^{\prime}$ of $\mathcal{R}(D_{i})^{\prime}$ such that
$\mathcal{R}(D_{i})^{\prime}=\mathcal{R}(J_{i})\oplus
U_{i}=\mathcal{R}(J_{i}^{\prime})\oplus U_{i}^{\prime}$, $1\leq i\leq k-1$. It
means that $X_{1}\oplus X_{2}\oplus\cdots\oplus
X_{n}=\mathcal{R}(D_{1})\oplus\mathcal{R}(J_{1})\oplus
U_{1}\oplus\mathcal{R}(J_{1}^{\prime})\oplus
U_{1}^{\prime}\oplus\mathcal{R}(D_{2})\oplus\mathcal{R}(J_{2})\oplus
U_{2}\oplus\mathcal{R}(J_{2}^{\prime})\oplus
U_{2}^{\prime}\oplus\cdots\oplus\mathcal{R}(D_{k-2})\oplus\mathcal{R}(J_{k-2})\oplus
U_{k-2}\oplus\mathcal{R}(J_{k-2}^{\prime})\oplus
U_{k-2}^{\prime}\oplus\mathcal{R}(D_{k-1})\oplus\mathcal{R}(J_{k-1}^{\prime})\oplus
U_{k-1}^{\prime}\oplus X_{k}\oplus X_{k+1}\oplus\cdots\oplus X_{n}$. Comparing
these equalities, we conclude that $\dim((X_{1}\oplus\cdots\oplus
X_{n})/\mathcal{R}(T_{n}^{d}(A)))=\dim(U_{1})+\dim(U_{1}^{\prime})+\dim(U_{2})+\dim(U_{2}^{\prime})+\cdots+\dim(U_{k-2})+\dim(U_{k-2}^{\prime})+\dim(U_{k-1}^{\prime})+\beta(D_{k})+\beta(D_{k+1})+\cdots+\beta(D_{n})<\infty$.

Case 2: $0\in\Omega_{1n}$

In this case $\alpha(D_{1})<\infty,\ \beta(D_{n})<\infty$, and
$\beta(D_{1})=\alpha(D_{n})=\infty$. We choose $A\in\mathcal{B}_{n}$
analogously as in Case 1. We shall get
$\mathcal{N}(T_{n}^{d}(A))\cong\mathcal{N}(D_{1})$, and therefore
$\alpha(T_{n}^{d}(A))<\infty$. We also get $\dim((X_{1}\oplus\cdots\oplus
X_{n})/\mathcal{R}(T_{n}^{d}(A)))=\dim(U_{1})+\dim(U_{1}^{\prime})+\dim(U_{2})+\dim(U_{2}^{\prime})+\cdots+\dim(U_{n-2})+\dim(U_{n-2}^{\prime})+\dim(U_{n-1}^{\prime})+\beta(D_{n})<\infty$,
with subspaces $U_{i},\ U_{i}^{\prime}$ as in Case 1.

Case 3: $0\in\Omega_{lk}$, $2\leq l<k\leq n-1.$

In this case $\beta(D_{l})=\alpha(D_{k})=\infty$,
$\alpha(D_{l}),\beta(D_{k})<\infty$ and $D_{k+1},...,D_{n}$ are Fredholm, as
well as $D_{1},...,D_{l-1}$. We find $A\in\mathcal{B}_{n}$ such that
$\alpha(T_{n}^{d}(A))<\infty$ and $\beta(T_{n}^{d}(A))<\infty$. We choose
$A=(A_{ij})_{1\leq i<j\leq n}$ so that $A_{ij}=0$ if $j-i\neq 1$ or $j\neq k$,
that is we place all nonzero operators of tuple $A$ on the superdiagonal or in
the k-th column. It remains to define $A_{ij}$ for $j-i=1$, $1\leq i<j\leq n$,
and for $j=k$. First notice that $A_{i,i+1}:X_{i+1}\rightarrow X_{i}$. Since
all of diagonal entries satisfy the complements condition, we know that
$X_{i+1}=\mathcal{N}(D_{i+1})\oplus\mathcal{N}(D_{i+1})^{\prime}$,
$X_{i}=\mathcal{R}(D_{i})^{\prime}\oplus\mathcal{R}(D_{i})$, and due to (3.6)
we get $\mathcal{N}(D_{i+1})\preceq_{s}\mathcal{R}(D_{i})^{\prime}$, and so
there is left invertible operator
$J_{i}:\mathcal{N}(D_{i+1})\rightarrow\mathcal{R}(D_{i})^{\prime}$ with
$\dim(\mathcal{R}(D_{i})^{\prime}/\mathcal{R}(J_{i}))<\infty$. We put
$A_{i,i+1}=\begin{bmatrix}J_{i}&0\\\
0&0\end{bmatrix}:\begin{bmatrix}\mathcal{N}(D_{i+1})\\\
\mathcal{N}(D_{i+1})^{\prime}\end{bmatrix}\rightarrow\begin{bmatrix}\mathcal{R}(D_{i})^{\prime}\\\
\mathcal{R}(D_{i})\end{bmatrix}$, and we implement this procedure for all
$l\leq i\leq k-2$. If $j=k$, then the reasoning analogous to the previous one
gives us left invertible
$J_{i}^{\prime}:\mathcal{N}(D_{k})\rightarrow\mathcal{R}(D_{i})^{\prime}$ with
$\dim(\mathcal{R}(D_{i})^{\prime}/\mathcal{R}(J_{i}^{\prime}))<\infty$, and so
we put $A_{ik}=\begin{bmatrix}J_{i}^{\prime}&0\\\ 0&0\\\
\end{bmatrix}:\begin{bmatrix}\mathcal{N}(D_{k})\\\
\mathcal{N}(D_{k})^{\prime}\end{bmatrix}\rightarrow\begin{bmatrix}\mathcal{R}(D_{i})^{\prime}\\\
\mathcal{R}(D_{i})\end{bmatrix}$ for all $l\leq i\leq k-1$. Finally, we place
$A_{ij}=0$ if $j>k.$

Now we have chosen our $A$, we show that
$\mathcal{N}(T_{n}^{d}(A))\cong\mathcal{N}(D_{1})\oplus\cdots\oplus\mathcal{N}(D_{l})\oplus\mathcal{N}(D_{k+1})\oplus\cdots\oplus\mathcal{N}(D_{n})$.
Let us put $T_{n}^{d}(A)x=0$, where $x=x_{1}+\cdots+x_{n}\in
X_{1}\oplus\cdots\oplus X_{n}$. The previous equality is then equivalent to

$\begin{bmatrix}D_{1}x_{1}+A_{1k}x_{k}\\\ D_{2}x_{2}+A_{2k}x_{k}\\\ \vdots\\\
D_{l-1}x_{l-1}+A_{l-1,k}x_{k}\\\ D_{l}x_{l}+A_{l,l+1}x_{l+1}+A_{l,k}x_{k}\\\
\vdots\\\ D_{k-2}x_{k-2}+A_{k-2,k-1}x_{k-1}+A_{k-1,k}x_{k}\\\
D_{k-1}x_{k-1}+A_{k-1,k}x_{k}\\\ D_{k}x_{k}\\\ D_{k+1}x_{k+1}\\\ \vdots\\\
D_{n}x_{n}\end{bmatrix}=\begin{bmatrix}0\\\ 0\\\ \vdots\\\ 0\\\ 0\\\ \vdots\\\
0\\\ 0\\\ 0\\\ 0\\\ \vdots\\\ 0\end{bmatrix}.$

By the same reasoning as seen in Case 1 we get $x_{k}=0$. Now, the last system
of equations is equivalent to

$\begin{bmatrix}D_{1}x_{1}\\\ D_{2}x_{2}\\\ \vdots\\\ D_{l-1}x_{l-1}\\\
D_{l}x_{l}+A_{l,l+1}x_{l+1}\\\ \vdots\\\ D_{k-2}x_{k-2}+A_{k-2,k-1}x_{k-1}\\\
D_{k-1}x_{k-1}\\\ 0\\\ D_{k+1}x_{k+1}\\\ \vdots\\\
D_{n}x_{n}\end{bmatrix}=\begin{bmatrix}0\\\ 0\\\ \vdots\\\ 0\\\ 0\\\ \vdots\\\
0\\\ 0\\\ 0\\\ 0\\\ \vdots\\\ 0\end{bmatrix}.$

It is not hard to see that continuing this reasoning iteratively, we get
$x_{k}=x_{k-1}=\cdots=x_{l+1}=0$, and this proves our statement. This means
that $\alpha(T_{n}^{d}(A))<\infty.$

Now we prove that $\beta(T_{n}^{d}(A))<\infty$. We know that
$\mathcal{R}(T_{n}^{d}(A))=\mathcal{R}(D_{1})\oplus\mathcal{R}(J_{1}^{\prime})\oplus\mathcal{R}(D_{2})\oplus\mathcal{R}(J_{2}^{\prime})\oplus\cdots\oplus\mathcal{R}(D_{l-1})\oplus\mathcal{R}(J_{l-1}^{\prime})\oplus\mathcal{R}(D_{l})\oplus\mathcal{R}(J_{l})\oplus\mathcal{R}(J_{l}^{\prime})\oplus\cdots\oplus\mathcal{R}(D_{k-2})\oplus\mathcal{R}(J_{k-2})\oplus\mathcal{R}(J_{k-2}^{\prime})\oplus\mathcal{R}(D_{k-1})\oplus\mathcal{R}(J_{k-1}^{\prime})\oplus\mathcal{R}(D_{k})\oplus\mathcal{R}(D_{k+1})\oplus\cdots\oplus\mathcal{R}(D_{n})$.
Furthermore, we know that there exist closed finite dimensional subspaces
$U_{i},\ U_{i}^{\prime}$ of $\mathcal{R}(D_{i})^{\prime}$ such that
$\mathcal{R}(D_{i})^{\prime}=\mathcal{R}(J_{i})\oplus
U_{i}=\mathcal{R}(J_{i}^{\prime})\oplus U_{i}^{\prime}$ with $\dim U_{i},\dim
U_{i}^{\prime}<\infty$, $l\leq i\leq k-1$. It means that $X_{1}\oplus
X_{2}\oplus\cdots\oplus
X_{n}=\mathcal{R}(D_{1})\oplus\mathcal{R}(J_{1}^{\prime})\oplus
U_{1}^{\prime}\oplus\mathcal{R}(D_{2})\oplus\mathcal{R}(J_{2}^{\prime})\oplus
U_{2}^{\prime}\oplus\cdots\oplus\mathcal{R}(D_{l-1})\oplus\mathcal{R}(J_{l-1}^{\prime})\oplus
U_{l-1}^{\prime}\oplus\mathcal{R}(D_{l})\oplus\mathcal{R}(J_{l})\oplus
U_{l}\oplus\mathcal{R}(J_{l}^{\prime})\oplus
U_{l^{\prime}}\oplus\cdots\oplus\mathcal{R}(D_{k-2})\oplus\mathcal{R}(J_{k-2})\oplus
U_{k-2}\oplus\mathcal{R}(J_{k-2}^{\prime})\oplus
U_{k-2}^{\prime}\oplus\mathcal{R}(D_{k-1})\oplus\mathcal{R}(J_{k-1}^{\prime})\oplus
U_{k-1}^{\prime}\oplus X_{k}\oplus X_{k+1}\oplus\cdots\oplus X_{n}$. Comparing
these equalities we conclude that $\dim((X_{1}\oplus\cdots\oplus
X_{n})/\mathcal{R}(T_{n}^{d}(A)))=\dim(U_{1}^{\prime})+\dim(U_{2}^{\prime})+\cdots+\dim(U_{l-1}^{\prime})+\dim(U_{l})+\dim(U_{l}^{\prime})+\cdots+\dim(U_{k-2})+\dim(U_{k-2}^{\prime})+\dim(U_{k-1}^{\prime})+\beta(D_{k})+\beta(D_{k+1})+\cdots+\beta(D_{n})<\infty$.

Case 4: $0\in\Omega_{ln}$

In this case $\beta(D_{l})=\alpha(D_{n})=\infty$,
$\alpha(D_{l}),\beta(D_{n})<\infty$ and $D_{1},...,D_{l-1}$ are Fredholm. We
choose $A\in\mathcal{B}_{n}$ analogously as in Case 3. Namely, we choose
$A=(A_{ij})_{1\leq i<j\leq n}$ so that $A_{ij}=0$ if $j-i\neq 1$ or $j\neq n$,
that is we place all nonzero operators of tuple $A$ on the superdiagonal or in
the n-th column. Following the same method as in the previous case, we shall
get
$\mathcal{N}(T_{n}^{d}(A))\cong\mathcal{N}(D_{1})\oplus\cdots\oplus\mathcal{N}(D_{l})$,
and therefore $\alpha(T_{n}^{d}(A))<\infty$. We also get
$\dim((X_{1}\oplus\cdots\oplus
X_{n})/\mathcal{R}(T_{n}^{d}(A)))=\dim(U_{1})+\dim(U_{1}^{\prime})+\dim(U_{2})+\dim(U_{2}^{\prime})+\cdots+\dim(U_{n-2})+\dim(U_{n-2}^{\prime})+\dim(U_{n-1}^{\prime})+\beta(D_{n})<\infty$,
with subspaces $U_{i},\ U_{i}^{\prime}$ as in Case 3.

Case 5: $0\in\bigcap\limits_{s=1}^{n}\rho_{e}(D_{s})$

Then we know $\alpha(D_{1}),...,\alpha(D_{n})<\infty$,
$\beta(D_{1}),...,\beta(D_{n})<\infty$. Choose trivially $A_{ij}=0$ for all
$1\leq i<j\leq n$ to get
$\alpha(T_{n}^{d}(A))=\sum\limits_{s=1}^{n}\alpha(D_{s})<\infty,\
\beta(T_{n}^{d}(A))=\sum\limits_{s=1}^{n}\beta(D_{s})<\infty$. The proof is
completed. $\square$

###### Corollary 3.6

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ being regular. Additionally, suppose that we have

$\mathcal{N}(D_{k})\preceq_{s}X_{j}/\mathcal{R}(D_{j})\quad for\ all\ n\geq
k>j\geq 1.$

Then

$\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{e}(T_{n}^{d}(A))=\sigma_{le}(D_{1})\bigcup\limits\sigma_{re}(D_{n})\bigcup\limits\Big{(}\bigcup\limits_{k=2}^{n-1}\Delta_{k}\Big{)}\bigcup\limits\\{\lambda\in\mathds{C}:\
\beta(D_{1}-\lambda)=\infty,\
\sum\limits_{s=2}^{n}\alpha(D_{s}-\lambda)<\infty\\}\bigcup\limits\\{\lambda\in\mathds{C}:\
\alpha(D_{n}-\lambda)=\infty,\
\sum\limits_{s=1}^{n-1}\beta(D_{s}-\lambda)<\infty\\},$

where

$\Delta_{k}:=\Big{\\{}\lambda\in\sigma_{le}(D_{k}):\
\sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty\Big{\\}}\bigcup\limits\Big{\\{}\lambda\in\sigma_{re}(D_{k}):\
\newline \sum\limits_{s=k+1}^{n}\alpha(D_{s}-\lambda)<\infty\Big{\\}},\ 2\leq
k\leq n-1.$

Having in mind the statement of Remark 2.7, we conclude that the following
Theorem also holds. Its proof is similar to the proof of Theorem 3.5, but
technically much more demanding.

###### Theorem 3.7

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ satisfying the complements condition. Additionally, suppose that we
have

$\mathcal{N}(D_{k})\cong X_{j}/\mathcal{R}(D_{j})\ (u.f.d.s)\quad for\ all\
n\geq k>j\geq 1.$

Then

$\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{e}(T_{n}^{d}(A))=\sigma_{le}(D_{1})\bigcup\limits\sigma_{re}(D_{n})\bigcup\limits\Big{(}\bigcup\limits_{k=2}^{n-1}\Delta_{k}\Big{)}\bigcup\limits\\{\lambda\in\mathds{C}:\
\beta(D_{1}-\lambda)=\infty,\
\sum\limits_{s=2}^{n}\alpha(D_{s}-\lambda)<\infty\\}\bigcup\limits\\{\lambda\in\mathds{C}:\
\alpha(D_{n}-\lambda)=\infty,\
\sum\limits_{s=1}^{n-1}\beta(D_{s}-\lambda)<\infty\\},$

where

$\Delta_{k}:=\Big{\\{}\lambda\in\sigma_{le}(D_{k}):\
\sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty\Big{\\}}\bigcup\limits\Big{\\{}\lambda\in\sigma_{re}(D_{k}):\
\sum\limits_{s=k+1}^{n}\alpha(D_{s}-\lambda)<\infty\Big{\\}},\ 2\leq k\leq
n-1.$

###### Corollary 3.8

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ being regular. Additionally, suppose that we have

$\mathcal{N}(D_{k})\cong X_{j}/\mathcal{R}(D_{j})\ (u.f.d.s)\quad for\ all\
n\geq k>j\geq 1.$

Then

$\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{e}(T_{n}^{d}(A))=\sigma_{le}(D_{1})\bigcup\limits\sigma_{re}(D_{n})\bigcup\limits\Big{(}\bigcup\limits_{k=2}^{n-1}\Delta_{k}\Big{)}\bigcup\limits\\{\lambda\in\mathds{C}:\
\beta(D_{1}-\lambda)=\infty,\
\sum\limits_{s=2}^{n}\alpha(D_{s}-\lambda)<\infty\\}\bigcup\limits\\{\lambda\in\mathds{C}:\
\alpha(D_{n}-\lambda)=\infty,\
\sum\limits_{s=1}^{n-1}\beta(D_{s}-\lambda)<\infty\\},$

where

$\Delta_{k}:=\Big{\\{}\lambda\in\sigma_{le}(D_{k}):\
\sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty\Big{\\}}\bigcup\limits\Big{\\{}\lambda\in\sigma_{re}(D_{k}):\
\sum\limits_{s=k+1}^{n}\alpha(D_{s}-\lambda)<\infty\Big{\\}},\ 2\leq k\leq
n-1.$

###  3.1 When does
$\sigma_{*}(T_{n}^{d}(A))=\bigcup\limits_{s=1}^{n}\sigma_{*}(D_{s})$ hold for
arbitrary $A\in\mathcal{B}_{n}$,
$\sigma_{*}\in\\{\sigma_{le},\sigma_{re},\sigma_{e}\\}$?

In this subsection we provide an answer to Question 3 stated in Section 2.
This question has been solved for the case of infinite dimensional separable
Hilbert spaces by Wu and Huang in [22]. It turns out that their results stated
in Section 3 of [22] hold, word by word, in arbitrary Banach spaces, with the
only distinction that in every claim we have to assume validity of the
complements condition for all $D_{i}$’s. We invite readers to copy the
statements of Theorems 4, 5, 6 from [22]t and their Corollaries, by including
the complements condition assumption and substituting Hilbert spaces
$\mathcal{H}_{1}$,…, $\mathcal{H}_{n}$ by Banach spaces $X_{1},...,X_{n}$. As
of special interest, we recommend studying Corollaries 2, 6 and 10 in a
connection with an interesting notion of he single valued extension property
of Finch ([10]). However, for the convenience of a reader, we decided to
provide a couple of mentioned statements for illustration, and so we give
analogues of Corollaries 4, 8 and 12 from [22]. Their proofs can be found in
[22], Section 3.

###### Corollary 3.9

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ satisfying the complements condition. Then

$\sigma_{le}(T_{n}^{d}(A))=\bigcup\limits_{k=1}^{n}\sigma_{le}(D_{k})$

holds for every $A\in\mathcal{B}_{n}$ if and only if

$\displaystyle\bigcup\limits_{k=2}^{n}\\{\lambda\in\sigma_{le}(D_{k}):\
\sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)=\infty\\}\subseteq\sigma_{le}(D_{1})\cup$
$\displaystyle\bigcup\limits_{k=2}^{n-1}\\{\lambda\in\sigma_{le}(D_{k}):\
\sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty\\}.$

###### Corollary 3.10

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ satisfying the complements condition. Then

$\sigma_{re}(T_{n}^{d}(A))=\bigcup\limits_{k=1}^{n}\sigma_{re}(D_{k})$

holds for every $A\in\mathcal{B}_{n}$ if and only if

$\displaystyle\bigcup\limits_{k=1}^{n-1}\\{\lambda\in\sigma_{re}(D_{k}):\
\sum\limits_{s=k+1}^{n}\alpha(D_{s}-\lambda)=\infty\\}\subseteq\sigma_{re}(D_{n})\cup$
$\displaystyle\bigcup\limits_{k=2}^{n-1}\\{\lambda\in\sigma_{re}(D_{k}):\
\sum\limits_{s=k+1}^{n}\alpha(D_{s}-\lambda)<\infty\\}.$

###### Corollary 3.11

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ satisfying the complements condition. Then

$\sigma_{e}(T_{n}^{d}(A))=\bigcup\limits_{k=1}^{n}\sigma_{e}(D_{k})$

holds for every $A\in\mathcal{B}_{n}$ if and only if

$\displaystyle\bigcup\limits_{k=1}^{n-1}\\{\lambda\in\mathds{C}:\
\beta(D_{k}-\lambda)=\sum\limits_{s=k+1}^{n}\alpha(D_{s}-\lambda)=\infty\\}\subseteq$
$\displaystyle\sigma_{le}(D_{1})\cup\sigma_{re}(D_{n})\cup\Big{(}\bigcup\limits_{k=2}^{n-1}\Delta_{k}\Big{)},$

where $\Delta_{k}$ is defined in Theorem 3.5.

##  4 Perturbations of left (right) the Weyl spectrum of $T_{n}^{d}(A)$

We start with a result which characterizes perturbations of left the Weyl
spectrum of $T_{n}^{d}(A)$. We need to use stronger form of Definition 2.2,
namely Definition 2.4.

###### Theorem 4.1

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ satisfying the complements condition. Additionally, suppose that we
have

$\mathcal{N}(D_{k})\prec X_{j}/\mathcal{R}(D_{j})\quad for\ all\ n\geq k>j\geq
1.$ (4.1)

Then

$\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{lw}(T_{n}^{d}(A))=\sigma_{le}(D_{1})\cup\Big{(}\bigcup\limits_{k=2}^{n+1}\Delta_{k}\Big{)},$

where

$\Delta_{k}:=\Big{\\{}\lambda\in\sigma_{le}(D_{k}):\
\sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty\Big{\\}},\ 2\leq k\leq n,$
$\Delta_{n+1}:=\\{\lambda\in\mathds{C}:\
\sum\limits_{s=1}^{n}\beta(D_{s}-\lambda)<\sum\limits_{s=1}^{n}\alpha(D_{s}-\lambda)\\}.$

Proof. Inclusion $(\supseteq)$ is already proved in [23]. For convenience of a
reader we provide the full proof, with a slight difference in that now we have
different space decompositions.

Write $\Delta_{1}:=\sigma_{le}(D_{1})$,
$\Delta_{k1}=\\{\lambda\in\mathds{C}:\mathcal{R}(D_{k}-\lambda)\ is\ not\
closed\ and\\\ \ \sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty\\}$ and
$\Delta_{k2}=\\{\lambda\in\mathds{C}:\mathcal{R}(D_{k}-\lambda)\ is\ closed\
and\\\ \sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty,\
\alpha(D_{k}-\lambda)=\infty\\}$ for $2\leq k\leq n$.
Lemma 2.11 implies
$\Delta_{1}\subseteq\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{lw}(T_{n}^{d}(A))$
. Let $\lambda\in\bigcup\limits_{k=2}^{n}\Delta_{k}$.

Let $\lambda\in\Delta_{k1}$, that is $\mathcal{R}(D_{k}-\lambda)$ is not
closed and $\sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty$,
$k\in\\{2,...,n\\}$. Then for each $A\in\mathcal{B}_{k}$, operator matrix
$T_{k}^{d}(A)$ as an operator from
$\mathcal{N}(D_{1})^{\prime}\oplus\mathcal{N}(D_{1})\oplus\mathcal{N}(D_{2})^{\prime}\oplus\mathcal{N}(D_{2})\oplus\mathcal{N}(D_{3})^{\prime}\oplus\mathcal{N}(D_{3})\oplus\cdots\oplus\mathcal{N}(D_{k})^{\prime}\oplus\mathcal{N}(D_{k})$
into
$\mathcal{R}(D_{1})\oplus\mathcal{R}(D_{1})^{\prime}\oplus\mathcal{R}(D_{2})\oplus\mathcal{R}(D_{2})^{\prime}\oplus\cdots\oplus\mathcal{R}(D_{k-1})\oplus\mathcal{R}(D_{k-1}^{\prime})\oplus\mathcal{R}(D_{k})\oplus\mathcal{R}(D_{k})^{\prime}$
admits the following block representation

$T_{k}^{d}(A)=\scalebox{0.95}{\mbox{$\displaystyle\begin{bmatrix}D_{1}^{(1)}&0&A_{12}^{(1)}&A_{12}^{(2)}&...&A_{1,k-1}^{(1)}&A_{1,k-1}^{(2)}&A_{1k}^{(1)}&A_{1k}^{(2)}\\\
0&0&A_{12}^{(3)}&A_{12}^{(4)}&...&A_{1,k-1}^{(3)}&A_{1,k-1}^{(4)}&A_{1k}^{(3)}&A_{1k}^{(4)}\\\
0&0&D_{2}^{(1)}&0&...&A_{2,k-1}^{(1)}&A_{2,k-1}^{(2)}&A_{2k}^{(1)}&A_{2k}^{(2)}\\\
0&0&0&0&...&A_{2,k-1}^{(3)}&A_{2,k-1}^{(4)}&A_{2k}^{(3)}&A_{2k}^{(4)}\\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots\\\
0&0&0&0&...&D_{k-1}^{(1)}&0&A_{k-1,k}^{(1)}&A_{k-1,k}^{(2)}\\\
0&0&0&0&...&0&0&A_{k-1,k}^{(3)}&A_{k-1,k}^{(4)}\\\
0&0&0&0&...&0&0&D_{k}^{(1)}&0\\\ 0&0&0&0&...&0&0&0&0\\\ \end{bmatrix}$}}$
(4.2)

Assume that $\mathcal{R}(T_{k}^{d}(A)-\lambda)$ is closed. Then, according to
Lemma 2.12, range of the following operator matrix

$\begin{bmatrix}D_{1}^{(1)}-\lambda&0&A_{12}^{(1)}&A_{12}^{(2)}&...&A_{1,k-1}^{(1)}&A_{1,k-1}^{(2)}&A_{1k}^{(1)}&A_{1k}^{(2)}\\\
0&0&D_{2}^{(1)}-\lambda&0&...&A_{2,k-1}^{(1)}&A_{2,k-1}^{(2)}&A_{2k}^{(1)}&A_{2k}^{(2)}\\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots\\\
0&0&0&0&...&D_{k-1}^{(1)}-\lambda&0&A_{k-1,k}^{(1)}&A_{k-1,k}^{(2)}\\\
0&0&0&0&...&0&0&D_{k}^{(1)}-\lambda&0\\\ \end{bmatrix}$

is closed due to $\sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty$. Since
$\overline{\mathcal{R}(D_{s}^{(1)}-\lambda)}=\overline{\mathcal{R}(D_{s}-\lambda)}$,
$s=1,2,...,k-1$ , then

$\displaystyle\mathcal{R}\left(\begin{bmatrix}D_{1}^{(1)}-\lambda&A_{12}^{(1)}&A_{12}^{(2)}&...&A_{1,k-1}^{(1)}&A_{1,k-1}^{(2)}\\\
0&D_{2}^{(1)}-\lambda&0&...&A_{2,k-1}^{(1)}&A_{2,k-1}^{(2)}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&0&...&D_{k-1}^{(1)}-\lambda&0\\\ \end{bmatrix}\right)$

is dense in
$\overline{\mathcal{R}(D_{1}-\lambda)}\oplus\overline{\mathcal{R}(D_{2}-\lambda)}\oplus\cdots\oplus\overline{\mathcal{R}(D_{k-1}-\lambda)}$
from Lemma 2.10. By Lemma 2.13, it follows that
$\mathcal{R}(D_{k}^{(1)}-\lambda)$ is closed, which is contradiction to the
fact that $\mathcal{R}(D_{k}-\lambda)$ is not closed. This implies that
$\mathcal{R}(T_{k}^{d}(A)-\lambda)$ is not closed for every
$A\in\mathcal{B}_{k}$, and hence
$\lambda\in\bigcap\limits_{A\in\mathcal{B}_{k}}\sigma_{le}(T_{k}^{d}(A))$.
This proves
$\bigcup\limits_{k=2}^{n}\Delta_{k1}\subseteq\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{lw}(T_{n}^{d}(A))$.

Let $\lambda\in\Delta_{k2}$, $k\in\\{2,...,n\\}.$ Without loss of generality,
put $\lambda=0$. We only consider
$0\in\Delta_{k2}\setminus(\bigcup\limits_{s=2}^{k}\Delta_{s1}\bigcup\limits\Delta_{1})$.
Therefore, $0\in\rho_{le}(D_{1})$, $\mathcal{R}(D_{s})$ is closed for
$s=2,...,k$, $\sum\limits_{s=1}^{k-1}\beta(D_{s})<\infty$ and
$\alpha(D_{k})=\infty$. Then for each $A\in\mathcal{B}_{k}$, operator matrix
$T_{k}^{d}(A)$ admits block representation as in (4.2). Evidently,
$D_{1}^{(1)},\ D_{2}^{(1)},...,D_{k}^{(1)}$ as in (4.2) are all invertible.
Thus, there exist invertible operator matrices $U$ and $V$ such that

$UT_{k}^{d}(A)V=\scalebox{0.85}{\mbox{$\displaystyle\begin{bmatrix}D_{1}^{(1)}&0&0&0&...&0&0&0&0\\\
0&0&0&A_{12}^{(4)}&...&0&A_{1,k-1}^{(4)}&0&A_{1k}^{(4)}\\\
0&0&D_{2}^{(1)}&0&...&0&0&0&0\\\
0&0&0&0&...&0&A_{2,k-1}^{(4)}&0&A_{2k}^{(4)}\\\ 0&0&0&0&...&0&0&0&0\\\
0&0&0&0&...&0&A_{3,k-1}^{(4)}&0&A_{3k}^{(4)}\\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots\\\
0&0&0&0&...&D_{k-1}^{(4)}&0&0&0\\\ 0&0&0&0&...&0&0&0&A_{k-1,k}^{(4)}\\\
0&0&0&0&...&0&0&D_{k}^{(1)}&0\\\ 0&0&0&0&...&0&0&0&0\\\ \end{bmatrix}$}}$
(4.3)

Note that $A_{ij}^{(4)}$ in (4.3) are not the original ones in (4.2) in
general, but we still use them for the convenience. Combine by (4.3),
$T_{k}^{d}(A)$ is left Weyl if and only if

$\begin{bmatrix}0&A_{12}^{(4)}&A_{13}^{(4)}&A_{14}^{(4)}&...&A_{1k}^{(4)}\\\
0&0&A_{23}^{(4)}&A_{24}^{(4)}&...&A_{2k}^{(4)}\\\
0&0&0&A_{34}^{(4)}&...&A_{3k}^{(4)}\\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\\ 0&0&0&0&...&A_{k-1,k}^{(4)}\\\
0&0&0&0&...&0\\\ \end{bmatrix}:\begin{bmatrix}\mathcal{N}(D_{1})\\\
\mathcal{N}(D_{2})\\\ \mathcal{N}(D_{3})\\\ \vdots\\\ \mathcal{N}(D_{k-1})\\\
\mathcal{N}(D_{k})\\\
\end{bmatrix}\rightarrow\begin{bmatrix}\mathcal{R}(D_{1})^{\prime}\\\
\mathcal{R}(D_{2})^{\prime}\\\ \mathcal{R}(D_{3})^{\prime}\\\ \vdots\\\
\mathcal{R}(D_{k-1})^{\prime}\\\ \mathcal{R}(D_{k})^{\prime}\\\ \end{bmatrix}$
(4.4)

is left Weyl. Since $\sum\limits_{s=1}^{k-1}\beta(D_{s})<\infty$ and
$\alpha(D_{k})=\infty$, it follows that

$\alpha\left(\begin{bmatrix}A_{1k}^{(4)}\\\ A_{2k}^{(4)}\\\ A_{3k}^{(4)}\\\
\vdots\\\ A_{k-1,k}^{(4)}\end{bmatrix}\right)=\infty,$

and hence the operator defined in (4.4) is not left Weyl for every
$A\in\mathcal{B}_{k}$. This proves
$\bigcup\limits_{k=2}^{n}\Delta_{k2}\subseteq\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{lw}(T_{n}^{d}(A))$.

Let $0\in\Delta_{n+1}\setminus\bigcup\limits_{s=1}^{n}\Delta_{s}$. Then
$0\in\bigcap\limits_{s=1}^{n}\rho_{le}(D_{s})$ and
$\sum\limits_{s=1}^{n}\beta(D_{s})<\sum\limits_{s=1}^{n}\alpha(D_{s})$. Thus
for each $A\in\mathcal{B}_{n}$, operator matrix $T_{n}^{d}(A)$ admits block
representation as (4.2), and we still use (4.3) and (4.4), where $k=n$. Since
$0\in\bigcap\limits_{s=1}^{n}\rho_{le}(D_{s})$, it follows that $T_{n}^{d}(A)$
is left Weyl if and only if operator (4.4) is left Weyl. From
$\sum\limits_{s=1}^{n}\beta(D_{s})<\sum\limits_{s=1}^{n}\alpha(D_{s})$, we
know (4.4) is not left Weyl for every $A\in\mathcal{B}_{n}$. Therefore,
$\Delta_{n+1}\subseteq\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{lw}(T_{n}^{d}(A))$.

For the opposite inclusion, it suffices to prove that
$0\not\in\bigcup\limits_{k=1}^{n+1}\Delta_{k}$ implies
$0\not\in\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{lw}(T_{n}^{d}(A))$, i.e.
there exists an operator tuple $A\in\mathcal{B}_{n}$ such that $T_{n}^{d}(A)$
is left Weyl. Let $0\not\in\bigcup\limits_{k=1}^{n+1}\Delta_{k}$. Then, direct
calculation show that $0\in\bigcup\limits_{k=1}^{n}\Omega_{k}$, where
$\Omega_{1}=\\{\lambda\in\rho_{le}(D_{1}):\ \beta(D_{1}-\lambda)=\infty\\}\\\
\Omega_{k}=\\{\lambda\in\bigcap\limits_{s=1}^{k}\rho_{le}(D_{s}):\
\beta(D_{k}-\lambda)=\infty\\},\ 2\leq k\leq n-1,\\\
\Omega_{n}=\\{\lambda\in\bigcap\limits_{s=1}^{n}\rho_{le}(D_{s}):\
\sum\limits_{s=1}^{n}\alpha(D_{s}-\lambda)\leq\sum\limits_{s=1}^{n}\beta(D_{s}-\lambda)\\}.\\\
$There are three cases to consider.
Case 1: $0\in\Omega_{1}$

This means that $\alpha(D_{1})<\infty$ ($D_{1}$ is left Fredholm), and
$\beta(D_{1})=\infty$. We find $A\in\mathcal{B}_{n}$ such that
$\alpha(T_{n}^{d}(A))<\infty$, $\beta(T_{n}^{d}(A))=\infty$ and
$\mathcal{R}(T_{n}^{d}(A))$ is closed. We choose $A=(A_{ij})_{1\leq i<j\leq
n}$ completely the same as in Case 1 in the proof of Theorem 3.1. We get
$\mathcal{N}(T_{n}^{d}(A))\cong\mathcal{N}(D_{1})$ and
$\mathcal{R}(T_{n}^{d}(A))=\mathcal{R}(D_{1})\oplus\mathcal{R}(J_{1})\oplus\mathcal{R}(D_{2})\oplus\mathcal{R}(J_{2})\oplus\cdots\oplus\mathcal{R}(D_{n-1})\oplus\mathcal{R}(J_{n-1})\oplus\mathcal{R}(D_{n})$,
$X_{1}\oplus X_{2}\oplus\cdots\oplus
X_{n}=\mathcal{R}(D_{1})\oplus\mathcal{R}(J_{1})\oplus
U_{1}\oplus\mathcal{R}(D_{2})\oplus\mathcal{R}(J_{2})\oplus
U_{2}\oplus\cdots\oplus\mathcal{R}(D_{n-1})\oplus\mathcal{R}(J_{n-1})\oplus
U_{n-1}\oplus\mathcal{R}(D_{n})$. Therefore, $\alpha(T_{n}^{d}(A))<\infty$ and
$\beta(T_{n}^{d}(A))=\dim(U_{1})+\dim(U_{2})+\cdots+\dim(U_{n-1})+\beta(D_{n})=\infty$
due to (4.5). It follows that $T_{n}^{d}(A)$ is left Weyl.

Case 2: $0\in\Omega_{k}$, $2\leq k\leq n-1$

This means that $\alpha(D_{1}),...,\alpha(D_{k})<\infty$ ($D_{1},...,D_{k}$
are left Fredholm), and $\beta(D_{k})=\infty$. We find $A\in\mathcal{B}_{n}$
such that $\alpha(T_{n}^{d}(A))<\infty$, $\beta(T_{n}^{d}(A))=\infty$ and
$\mathcal{R}(T_{n}^{d}(A))$ is closed. We choose $A=(A_{ij})_{1\leq i<j\leq
n}$ completely the same as in Case 2 in the proof of Theorem 3.1. We get
$\alpha(T_{n}^{d}(A))=\sum\limits_{s=1}^{k}\alpha(D_{s})<\infty$ and
$\beta(T_{n}^{d}(A))=\beta(D_{1})+\cdots+\beta(D_{k-1})+\dim(U_{k})+\cdots+\dim(U_{n-1})+\beta(D_{n})=\infty$
due to (4.5). Therefore $T_{n}^{d}(A)$ is left Weyl.

Case 3: $0\in\Omega_{n}$

In this case $0\in\bigcap\limits_{s=1}^{n}\rho_{le}(D_{s})$ and
$\sum\limits_{s=1}^{n}\alpha(D_{s})-\sum\limits_{s=1}^{n}\beta(D_{s})\leq 0$,
which implies $\alpha(D_{1}),...,\alpha(D_{n})<\infty$ ($D_{1},...,D_{n}$ are
left Fredholm). Choose trivially $A_{ij}=0$ for every $1\leq i<j\leq n$. Then
$\alpha(T_{n}^{d}(A))=\sum\limits_{s=1}^{n}\alpha(D_{s})<\infty$,
$\beta(T_{n}^{d}(A))=\sum\limits_{s=1}^{n}\beta(D_{s})$. Therefore,
$\alpha(T_{n}^{d}(A))-\beta(T_{n}^{d}(A))\leq 0$, hence $T_{n}^{d}(A)$ is left
Weyl. $\square$

###### Corollary 4.2

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ being regular. Additionally, suppose that we have

$\mathcal{N}(D_{k})\prec X_{j}/\mathcal{R}(D_{j})\quad for\ all\ n\geq k>j\geq
1.$ (4.5)

Then

$\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{lw}(T_{n}^{d}(A))=\sigma_{le}(D_{1})\cup\Big{(}\bigcup\limits_{k=2}^{n+1}\Delta_{k}\Big{)},$

where

$\Delta_{k}:=\Big{\\{}\lambda\in\sigma_{le}(D_{k}):\
\sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty\Big{\\}},\ 2\leq k\leq n,$
$\Delta_{n+1}:=\\{\lambda\in\mathds{C}:\
\sum\limits_{s=1}^{n}\beta(D_{s}-\lambda)<\sum\limits_{s=1}^{n}\alpha(D_{s}-\lambda)\\}.$

Inspecting the proof of Theorem 4.1, one easily spots that we can replace
condition (4.5) with what seems to be looser condition. And so, we state the
following.

###### Theorem 4.3

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ satisfying the complements condition. Additionally, suppose that we
have

$\mathcal{N}(D_{k})\preceq X_{j}/\mathcal{R}(D_{j})\quad for\ all\ n\geq
k>j\geq 1$,

$\beta(D_{n})=\infty.$

Then

$\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{lw}(T_{n}^{d}(A))=\sigma_{le}(D_{1})\cup\Big{(}\bigcup\limits_{k=2}^{n+1}\Delta_{k}\Big{)},$

where

$\Delta_{k}:=\Big{\\{}\lambda\in\sigma_{le}(D_{k}):\
\sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty\Big{\\}},\ 2\leq k\leq n,$
$\Delta_{n+1}:=\\{\lambda\in\mathds{C}:\
\sum\limits_{s=1}^{n}\beta(D_{s}-\lambda)<\sum\limits_{s=1}^{n}\alpha(D_{s}-\lambda)\\}.$

###### Corollary 4.4

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ being regular. Additionally, suppose that we have

$\mathcal{N}(D_{k})\preceq X_{j}/\mathcal{R}(D_{j})\quad for\ all\ n\geq
k>j\geq 1$,

$\beta(D_{n})=\infty.$

Then

$\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{lw}(T_{n}^{d}(A))=\sigma_{le}(D_{1})\cup\Big{(}\bigcup\limits_{k=2}^{n+1}\Delta_{k}\Big{)},$

where

$\Delta_{k}:=\Big{\\{}\lambda\in\sigma_{le}(D_{k}):\
\sum\limits_{s=1}^{k-1}\beta(D_{s}-\lambda)<\infty\Big{\\}},\ 2\leq k\leq n,$
$\Delta_{n+1}:=\\{\lambda\in\mathds{C}:\
\sum\limits_{s=1}^{n}\beta(D_{s}-\lambda)<\sum\limits_{s=1}^{n}\alpha(D_{s}-\lambda)\\}.$

The following is a dual result of Theorem 4.1.

###### Theorem 4.5

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ satisfying the complements condition. Additionally, suppose that we
have

$X_{k}/\mathcal{R}(D_{k})\prec\mathcal{N}(D_{j})\quad for\ all\ 1\leq k<j\leq
n.$ (4.6)

Then

$\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{rw}(T_{n}^{d}(A))=\sigma_{re}(D_{n})\cup\Big{(}\bigcup\limits_{k=1}^{n-1}\Delta_{k}\Big{)}\cup\Delta_{n+1},$

where

$\Delta_{k}:=\Big{\\{}\lambda\in\sigma_{re}(D_{k}):\
\sum\limits_{s=k+1}^{n}\alpha(D_{s}-\lambda)<\infty\Big{\\}},\ 1\leq k\leq
n-1.$ $\Delta_{n+1}=\\{\lambda\in\mathds{C}:\
\sum\limits_{s=1}^{n}\alpha(D_{s}-\lambda)<\sum\limits_{s=1}^{n}\beta(D_{s}-\lambda)\\}.$

Proof: From Lemma 2.1 it follows that
$\sigma_{rw}(T_{n}^{d}(A))=\sigma_{lw}(T_{n}^{d}(A)^{\prime})$. Now, proof
immediately follows from Theorem 4.1, and with regards to the form (2.3).
$\square$

###### Corollary 4.6

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ being regular. Additionally, suppose that we have

$X_{k}/\mathcal{R}(D_{k})\prec\mathcal{N}(D_{j})\quad for\ all\ 1\leq k<j\leq
n.$

Then

$\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{rw}(T_{n}^{d}(A))=\sigma_{re}(D_{n})\bigcup\limits\Big{(}\bigcup\limits_{k=1}^{n-1}\Delta_{k}\Big{)},$

where

$\Delta_{k}:=\Big{\\{}\lambda\in\sigma_{re}(D_{k}):\
\sum\limits_{s=k+1}^{n}\alpha(D_{s}-\lambda)<\infty\Big{\\}},\ 1\leq k\leq
n-1.$ $\Delta_{n+1}=\\{\lambda\in\mathds{C}:\
\sum\limits_{s=1}^{n}\alpha(D_{s}-\lambda)<\sum\limits_{s=1}^{n}\beta(D_{s}-\lambda)\\}.$

###### Theorem 4.7

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ satisfying the complements condition. Additionally, suppose that we
have

$X_{k}/\mathcal{R}(D_{k})\preceq\mathcal{N}(D_{j})\quad for\ all\ 1\leq
k<j\leq n$,

$\alpha(D_{1})=\infty.$

Then

$\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{rw}(T_{n}^{d}(A))=\sigma_{re}(D_{n})\cup\Big{(}\bigcup\limits_{k=1}^{n-1}\Delta_{k}\Big{)}\cup\Delta_{n+1},$

where

$\Delta_{k}:=\Big{\\{}\lambda\in\sigma_{re}(D_{k}):\
\sum\limits_{s=k+1}^{n}\alpha(D_{s}-\lambda)<\infty\Big{\\}},\ 1\leq k\leq
n-1.$ $\Delta_{n+1}=\\{\lambda\in\mathds{C}:\
\sum\limits_{s=1}^{n}\alpha(D_{s}-\lambda)<\sum\limits_{s=1}^{n}\beta(D_{s}-\lambda)\\}.$

###### Corollary 4.8

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ being regular. Additionally, suppose that we have

$X_{k}/\mathcal{R}(D_{k})\preceq\mathcal{N}(D_{j})\quad for\ all\ 1\leq
k<j\leq n$,

$\alpha(D_{1})=\infty.$

Then

$\bigcap\limits_{A\in\mathcal{B}_{n}}\sigma_{rw}(T_{n}^{d}(A))=\sigma_{re}(D_{n})\cup\Big{(}\bigcup\limits_{k=1}^{n-1}\Delta_{k}\Big{)}\cup\Delta_{n+1},$

where

$\Delta_{k}:=\Big{\\{}\lambda\in\sigma_{re}(D_{k}):\
\sum\limits_{s=k+1}^{n}\alpha(D_{s}-\lambda)<\infty\Big{\\}},\ 1\leq k\leq
n-1.$ $\Delta_{n+1}=\\{\lambda\in\mathds{C}:\
\sum\limits_{s=1}^{n}\alpha(D_{s}-\lambda)<\sum\limits_{s=1}^{n}\beta(D_{s}-\lambda)\\}.$

###  4.1 When does
$\sigma_{*}(T_{n}^{d}(A))=\bigcup\limits_{s=1}^{k}\sigma_{*}(D_{s})$ hold for
arbitrary $A\in\mathcal{B}_{n}$, $\sigma_{*}\in\\{\sigma_{lw},\sigma_{rw}\\}$?

In this subsection we provide an answer to Question 3 stated in Section 2.
This question has been solved for case of infinite dimensional separable
Hilbert spaces by Wu and Huang in [23]. Again, it turns out that their results
stated in Section 3 of [23] hold, word by word, in arbitrary Banach spaces,
with the only distinction that in every claim we have to assume validity of
the complements condition for all $D_{i}$’s. We invite readers to copy the
statements of Theorems 3.1, 3.6 from [23] and their Corollaries, by including
the complements condition assumption and substituting Hilbert spaces
$\mathcal{H}_{1}$,…, $\mathcal{H}_{n}$ by Banach spaces $X_{1},...,X_{n}$. As
of special interest, we recommend studying Remark 3.5 of [23] in a connection
with an interesting notion of single valued extension property of Finch
([10]). However, for the convenience of a reader, we decided to provide a
couple of mentioned statements for illustration, and so we give analogues of
Corollaries 3.3 and 3.8. Their proofs can be found in [23], Section 3.

###### Corollary 4.9

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ satisfying the complements condition. Then

$\sigma_{lw}(T_{n}^{d}(A))=\bigcup\limits_{k=1}^{n}\sigma_{lw}(D_{k})$

holds for every $A\in\mathcal{B}_{n}$ if and only if

$\Delta_{1}\cup\Delta_{2}=\emptyset$

where $\Delta_{k}$, $k=1,2$ are defined as follows

$\Delta_{1}:=\bigcup\limits_{k=2}^{n}\\{\lambda\in\bigcap\limits_{s=1}^{k-1}\rho_{le}(D_{s})\cap[\bigcup\limits_{s=1,\
s\neq k-1}^{n}\sigma_{lw}(D_{s})]:\ \beta(D_{k-1}-\lambda)=\infty\\},$
$\displaystyle\Delta_{2}:=\bigcup\limits_{k=1}^{n}\\{\lambda\in\bigcap\limits_{s=1}^{n}\rho_{le}(D_{s}):\
\sum\limits_{s=1}^{n}\alpha(D_{s}-\lambda)\leq\sum\limits_{s=1}^{n}\beta(D_{s}-\lambda),$
$\displaystyle\alpha(D_{k}-\lambda)>\beta(D_{k}-\lambda)\\}.$

###### Corollary 4.10

Let $D_{1}\in\mathcal{B}(X_{1}),\
D_{2}\in\mathcal{B}(X_{2}),...,D_{n}\in\mathcal{B}(X_{n})$ with all $D_{i},\
i=1,...,n$ satisfying the complements condition. Then

$\sigma_{rw}(T_{n}^{d}(A))=\bigcup\limits_{k=1}^{n}\sigma_{rw}(D_{k})$

holds for every $A\in\mathcal{B}_{n}$ if and only if

$\Delta_{1}\bigcup\limits\Delta_{2}=\emptyset$

where $\Delta_{k}$, $k=1,2$ are defined as follows

$\Delta_{1}:=\bigcup\limits_{k=1}^{n-1}\\{\lambda\in\bigcap\limits_{s=k+1}^{n}\rho_{re}(D_{s})\cap[\bigcup\limits_{s=1,\
s\neq k+1}^{n}\sigma_{rw}(D_{s})]:\ \alpha(D_{k+1}-\lambda)=\infty\\},$
$\displaystyle\Delta_{2}:=\bigcup\limits_{k=1}^{n}\\{\lambda\in\bigcap\limits_{s=1}^{n}\rho_{re}(D_{s}):\
\sum\limits_{s=1}^{n}\beta(D_{s}-\lambda)\leq\sum\limits_{s=1}^{n}\alpha(D_{s}-\lambda),$
$\displaystyle\beta(D_{k}-\lambda)>\alpha(D_{k}-\lambda)\\}.$

Acknowledgments
This work was supported by the Ministry of Education, Science and
Technological Development of the Republic of Serbia under Grant No.
451-03-9/2021-14/ 200125
I wish to express my gratitude to professor Dragan S. Djordjević for
introduction to this topic and for useful comments that greatly improved the
form of this paper.

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