Weyl-Type Theorems of Upper Triangular Relation Matrices

Abstract

The spectral theory of operator matrices has several applications in elasticity, quantum mechanics, fluid dynamics, and other fields of mathematical physics. The study of operator matrices is more challenging when the involved operators are not single-valued and should be studied in the context of the theory of relations. In this paper, we utilize the connection between linear relations and their induced operators and use space decomposition methods to characterize the distribution of the spectrum for upper triangular relation matrices. We undertake the same for the essential spectrum, Weyl spectrum, and Browder spectrum. Under certain conditions, we obtain a Browder-type theorem and a Weyl-type theorem for such relation matrices.

1. Introduction

A linear relation is also called a multi-valued linear operator, which is a generalization of the concept of an operator in the multi-valued case. We call $T:H\to K$ a linear relation if for all nonzero $\alpha ,\beta \in \mathbb{C}$ and $x_1,x_2\in D\left(T\right)\subseteq H$ (the domain of T),

$$T(\alpha x_1+\beta x_2)=\alpha T\left(x_1\right)+\beta T\left(x_2\right),$$

where H and K are infinite dimensional complex Hilbert spaces, and $T\left(x_1\right)$ and $T\left(x_2\right)$ are nonempty subsets of K. $\mathcal{L}\mathcal{R}(H,K)$ is the class of all linear relations with $D\left(T\right)=H$ into K. $\mathcal{B}(H,K)$ is the set of bounded linear operators from H into K, and we write $\mathcal{L}\mathcal{R}\left(H\right):=\mathcal{L}\mathcal{R}(H,H)$, $\mathcal{B}\left(H\right):=\mathcal{B}(H,H)$.

Let $T\in \mathcal{L}\mathcal{R}(H,K)$. The graph $G\left(T\right)$ of T is given by

$$G\left(T\right)=\left\{\right(u,v)\in H\oplus K:u\in D(T),v\in T(u\left)\right\}.$$

The closure of T, denoted by $\overline{T}$, is the linear relation defined by $G\left(\overline{T}\right):=\overline{G\left(T\right)}.$ The inverse relation $T^{-1}$ of T is given by $G\left(T^{-1}\right)=\{(v,u)\in K\oplus H:(u,v)\in G\left(T\right)\}.$ The relation T is called closed if $G\left(T\right)$ is a closed subspace of $H\oplus K$, continuous if the inverse image $T^{-1}\left(V\right)$ of any neighborhood $V\subseteq R\left(T\right)$ is again a neighborhood in H, and bounded if $T\in \mathcal{L}\mathcal{R}(H,K)$ and T is continuous. We denote by $\mathcal{C}\mathcal{R}(H,K)$ the set of all closed relations from H into K, and we denote by $\mathcal{B}\mathcal{R}(H,K)$ the class of bounded relations. $\mathcal{B}\mathcal{C}\mathcal{R}(H,K)$ is the class of all closed and bounded linear relations from H into K, and $\mathcal{BCR}\left(H\right):=\mathcal{BCR}(H,H)$. We denote the range and the kernel of T by $R\left(T\right):=T\left(D\right(T\left)\right)$ and $N\left(T\right):=\{x\in H:(x,0)\in G(T\left)\right\}$; we write $n\left(T\right):=\dim N\left(T\right)$, $d\left(T\right):=\dim {\left(R\left(T\right)\right)}^{\perp }$, $\beta \left(T\right):=\dim (K/R(T\left)\right)$, and $i\left(T\right):=n\left(T\right)-\beta \left(T\right)$. We write $Q_T:K\to K/\overline{T\left(0\right)}$ the quotient map. It is clear that $Q_{T}T$ is an operator so that we can define $\Vert Tx\Vert :=\Vert Q_{T}Tx\Vert$ for $x\in D\left(T\right)$ and $\Vert T\Vert :=\Vert Q_{T}T\Vert$. We can see that T is continuous if and only if $\Vert T\Vert <+\infty$.

Let $T\in \mathcal{B}\mathcal{C}\mathcal{R}(H,K)$, then $\beta \left(T\right)<\infty \iff$$R\left(T\right)$ is closed and $d\left(T\right)<\infty$ (see ([1], Theorem 2.13)).

Let $T\in \mathcal{L}\mathcal{R}(H,K)$, the adjoint relation $T^{*}\in \mathcal{L}\mathcal{R}(H,K)$ be defined by $G\left(T^{*}\right)=\{(v,v^{\prime })\in K\oplus H:\langle u^{\prime },v\rangle =\langle u,v^{\prime }\rangle forall(u,u^{\prime })\in G\left(T\right)\},$ and

$$N\left(T^{*}\right)={\left(R\left(T\right)\right)}^{\perp },T^{*}\left(0\right)={\left(D\left(T\right)\right)}^{\perp },N\left(\overline{T}\right)={\left(R\left(T^{*}\right)\right)}^{\perp },\overline{T}\left(0\right)={\left(D\left(T^{*}\right)\right)}^{\perp }.$$

Let $T\in \mathcal{C}\mathcal{R}(H,K)$. Then $R\left(T\right)$ is closed if and only if $R\left(T^{*}\right)$ is closed (see ([2], Theorem III.4.4)).

Let relation $T:H\to K$ be closed with $R\left(T\right)$ closed, then we say T is left Fredholm, denoted by $T\in {\mathsf{\Phi }}_{+}(H,K)$ if $n\left(T\right)<\infty$; it is right Fredholm, denoted by $T\in {\mathsf{\Phi }}_{-}(H,K)$, if $d\left(T\right)<\infty$; it is Fredholm, denoted by $T\in \mathsf{\Phi }(H,K)$, if T is both left and right Fredholm; it is Weyl, denoted by $T\in {\mathsf{\Phi }}_0(H,K)$, if T is a Fredholm relation and $i\left(T\right)=0$; it is left Weyl, denoted by $T\in {\mathsf{\Phi }}_{+}^{-}(H,K)$, if T is left Fredholm and $i\left(T\right)\le 0$; it is Browder, denoted by $T\in {\mathsf{\Phi }}_b(H,K)$, if T is Weyl with finite ascent and descent; and it is left Browder, denoted by $T\in {\mathsf{\Phi }}_0^{-}(H,K)$, if T is left Weyl of finite ascent, where the ascent $\alpha \left(T\right)$ and descent $\delta \left(T\right)$ of relation T are defined by

$$\begin{array}{c}\alpha \left(T\right)=min\{k\in \mathbb{N}:N\left(T^k\right)=N\left(T^{k+1}\right)\},\hfill \\ \delta \left(T\right)=min\{k\in \mathbb{N}:R\left(T^k\right)=R\left(T^{k+1}\right)\}.\hfill \end{array}$$

Let $T\in \mathcal{L}\mathcal{R}\left(H\right)$. The resolvent set of T is the set

$$\rho \left(T\right)=\{\lambda \in \mathbb{C}:T-\lambda I\mathrm{is}\mathrm{injective},\mathrm{open}\mathrm{with}\mathrm{dense}\mathrm{range}\}.$$

Moreover, if T is closed, the resolvent set of T is the set

$$\rho \left(T\right)=\{\lambda \in \mathbb{C}:{(T-\lambda I)}^{-1}\mathrm{is}\mathrm{everywhere}\mathrm{defined}\mathrm{and}\mathrm{single}\mathrm{valued}\}.$$

The spectrum of T is the set $\sigma \left(T\right):=\mathbb{C}\backslash \rho \left(T\right).$ For $T\in \mathcal{CR}(H,K),$$\sigma \left(T^{*}\right)=\{\overline{\lambda }\in \mathbb{C}:\lambda \in \sigma \left(T\right)\}$ (see ([1], Lemma 2.14)).

The point spectrum ${\sigma }_p\left(T\right)$, defect spectrum ${\sigma }_{\delta }\left(T\right)$, approximate point spectrum ${\sigma }_{ap}\left(T\right)$, essential spectrum ${\sigma }_e\left(T\right)$, left essential spectrum ${\sigma }_{le}\left(T\right)$, right essential spectrum ${\sigma }_{re}\left(T\right)$, Weyl spectrum ${\sigma }_w\left(T\right)$, essential approximate point spectrum ${\sigma }_{ea}\left(T\right)$, Browder spectrum ${\sigma }_b\left(T\right)$, Browder essential approximate point spectrum ${\sigma }_{ab}\left(T\right)$, and closed-range spectrum ${\sigma }_{cr}\left(T\right)$ of T are, respectively, defined by

$$\begin{array}{c}{\sigma }_p\left(T\right)=\{\lambda \in \mathbb{C}:T-\lambda I\mathrm{is}\mathrm{not}\mathrm{injective}\},\hfill \\ {\sigma }_{\delta }\left(T\right)=\{\lambda \in \mathbb{C}:R(T-\lambda I)\ne K\},\hfill \\ {\sigma }_{ap}\left(T\right)=\{\lambda \in \mathbb{C}:\mathrm{there}\mathrm{is}\mathrm{a}\mathrm{sequence}{\left\{x_n\right\}}_{n=1}^{\infty }\subseteq H\mathrm{such}\mathrm{that}\Vert x_n\Vert =1\hfill \\ \mathrm{for}\mathrm{all}n\in \mathbb{N},\mathrm{and}\Vert (T-\lambda I)x_n\Vert \to 0\mathrm{as}n\to \infty \},\hfill \\ {\sigma }_e\left(T\right)=\{\lambda \in \mathbb{C}:T-\lambda I\notin \mathsf{\Phi }(H\oplus K)\},\hfill \\ {\sigma }_{le}\left(T\right)=\{\lambda \in \mathbb{C}:T-\lambda I\notin {\mathsf{\Phi }}_{+}(H\oplus K)\},\hfill \\ {\sigma }_{re}\left(T\right)=\{\lambda \in \mathbb{C}:T-\lambda I\notin {\mathsf{\Phi }}_{-}(H\oplus K)\},\hfill \\ {\sigma }_w\left(T\right)=\{\lambda \in \mathbb{C}:T-\lambda I\notin {\mathsf{\Phi }}_0(H\oplus K)\},\hfill \\ {\sigma }_{ea}\left(T\right)=\{\lambda \in \mathbb{C}:T-\lambda I\notin {\mathsf{\Phi }}_{+}^{-}(H\oplus K)\},\hfill \\ {\sigma }_b\left(T\right)=\{\lambda \in \mathbb{C}:T-\lambda I\notin {\mathsf{\Phi }}_b(H\oplus K)\},\hfill \\ {\sigma }_{ab}\left(T\right)=\{\lambda \in \mathbb{C}:T-\lambda I\notin {\mathsf{\Phi }}_0^{-}(H\oplus K)\},\hfill \\ {\sigma }_{cr}\left(T\right)=\{\lambda \in \mathbb{C}:R(T-\lambda I)\ne \overline{R(T-\lambda I)}\}.\hfill \end{array}$$

Let $T\in \mathcal{BCR}\left(H\right)$ and $\lambda \in \mathbb{C}$. Then $\lambda \notin {\sigma }_{ap}\left(T\right)$ if and only if $N(T-\lambda I)=\left\{0\right\}$ and $R(T-\lambda I)$ is closed (see ([1], Theorem 2.10)).

Definition 1 

(see ([3], Definition 3.1)). Let $T\in \mathcal{L}\mathcal{R}\left(H\right)$. The local resolvent set ${\rho }_T\left(x\right)$ of T at $x\in H$ is defined as the set of all $\lambda \in \mathbb{C}$ such that there are an open neighborhood $U\left(\lambda \right)$ of λ and an analytic function $f:U\left(\lambda \right)\to D\left(T\right)$, which satisfies

$$(T-\lambda I)f\left(\lambda \right)=x+T\left(0\right),\lambda \in U\left(\lambda \right)$$

The local spectrum set ${\sigma }_T\left(x\right)$ of T at $x\in H$ is defined as ${\sigma }_T\left(x\right)=\mathbb{C}\backslash {\rho }_T\left(x\right)$.

Definition 2 

(see ([3], Definition 3.2)). Let $T\in \mathcal{L}\mathcal{R}\left(H\right)$, ${\lambda }_0\in \mathbb{C}$. If for every neighborhood $U\left({\lambda }_0\right)$ of ${\lambda }_0$ the only analytic function $f:U\left({\lambda }_0\right)\to H$ which satisfies Equation $(T-\lambda I)f\left(\lambda \right)=T\left(0\right)$ is the constant function $f\equiv 0$, then T is said to have the single-valued extension property at ${\lambda }_0\in \mathbb{C}$, abbreviated T has the SVEP at ${\lambda }_0$.

T is said to have the SVEP if T has the SVEP at any $\lambda \in \mathbb{C}$. Denote $S\left(T\right)$ the set of $\lambda \in \mathbb{C}$ for which T has no SVEP. Clearly, T has the SVEP if and only if $S\left(T\right)=\varnothing .$ Let $T\in \mathcal{L}\mathcal{R}\left(H\right)$ and $M\subseteq H$ be a subspace, then the relation $T_M$ is given by

$$\begin{array}{c}G\left(T_M\right):=\{(x,y)\in H\oplus H:y\in Tx+M\}.\hfill \end{array}$$

Write $T_{[\perp ]}=P_{T{\left(0\right)}^{\perp }}T:H\to T{\left(0\right)}^{\perp }$. It is clear that $T_{[\perp ]}$ is single-valued. For $A\in \mathcal{L}\mathcal{R}\left(H\right)$ and $C\in \mathcal{L}\mathcal{R}(K,H)$, the notation $\left(AC\right)$ is a row relation from $H\oplus K$ into H, i.e., $\left(AC\right)\left(\begin{array}{c}x\\ y\end{array}\right)=Ax+Cy$ for all $x\in D\left(A\right),y\in D\left(C\right)$.

The linear relation was introduced to consider adjoints of nondensely defined linear differential operators by J. von Neumann [4]. At present, the theoretical research has applications in many problems, for example, the study of some Cauchy problems is related to the theory of linear relations [5]; the fixed point theory of linear relations has applications in mathematical economics, optimal control, digital imaging, discontinuous differential equations, and game theory [6]. Recently, several researchers have been interested in the study of the spectral theory for relation matrices [1,7,8,9,10,11,12], of which article [12] shows that the spectral properties of the upper triangular relation matrix

$$M_C=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right):H\oplus K\to H\oplus K$$

are related to the multi-valued part $C\left(0\right)$ of relation $C\in \mathcal{BR}(K,H)$, and it obtains the inclusion ${\sigma }_{\star }\left(M_C\right)\subseteq {\sigma }_{\star }\left(A_{C\left(0\right)}\right)\cup {\sigma }_{\star }\left(B\right).$ In Section 2, we investigate the basic properties of relation T and upper triangular relation matrix $M_C$. Based on these properties, we obtain the spectral properties for $M_C$. In Section 3, based on the results in [12], we give the set

$$({\sigma }_{\star }\left(A_{C\left(0\right)}\right)\cup {\sigma }_{\star }\left(B\right))\backslash {\sigma }_{\star }\left(M_C\right)$$

under the local spectral theory, where ${\sigma }_{\star }\in \{\sigma ,{\sigma }_e,{\sigma }_w,{\sigma }_b\}$. In Section 4, we extend the results in [13] and explore how Browder’s theorem, a-Browder’s theorem, and Weyl’s theorem survive for upper triangular relation matrix $M_C$. Meanwhile, we give a new method for studying the Weylness of relation matrices where all internal elements are multi-valued linear operators.

2. Auxiliary Results

In this section, we collect some fundamental properties on linear operators, relations, and relation matrices, which will be used in later proofs. Let T be a linear relation in a linear space X; the singular chain manifold $R_c\left(T\right)$ of T is defined by $R_c\left(T\right)=R_0\left(T\right)\cap R_{\infty }\left(T\right)$, where $R_0\left(T\right)={\displaystyle \bigcup _{i=1}^{\infty }}N{\left(T\right)}^i$ and $R_{\infty }\left(T\right)={\displaystyle \bigcup _{i=1}^{\infty }}{T}^i\left(0\right)$. The lemma below is from Theorems 6.5, 6.13, and 6.11 and Corollary 6.7 of [14].

Lemma 1.

Assume that T is a linear relation in a linear space X:

(i) 

If $\alpha \left(T\right)<\infty$ and $R_c\left(T\right)=\left\{0\right\}$, then $n\left(T\right)\le d\left(T\right);$

(ii) 

If $\delta \left(T\right)<\infty$ and $D\left(T\right)=X$, then $d\left(T\right)\le n\left(T\right);$

(iii) 

If $\alpha \left(T\right)<\infty$, $n\left(T\right)=d\left(T\right)<\infty$ and $R_c\left(T\right)=\left\{0\right\}$, then $\alpha \left(T\right)=\delta \left(T\right)$;

(iv) 

If $X=R\left(T\right)+D\left(T^q\right)$ with $n\left(T\right)=d\left(T\right)<\infty$ and $q=\delta \left(T\right)<\infty$, then $R_c\left(T\right)=\left\{0\right\}$ and $\alpha \left(T\right)=\delta \left(T\right)$.

Lemma 2 

(see ([7], Lemma 4.2 and [9], Proposition 10)). Let $T\in \mathcal{B}\mathcal{C}\mathcal{R}\left(H\right)$:

(i) 

$T\in {\mathsf{\Phi }}_{+}\left(H\right)$ if and only if $Q_{T}T\in {\mathsf{\Phi }}_{+}(H,H/T\left(0\right))$, and $i\left(T\right)=i\left(Q_{T}T\right);$

(ii) 

$T\in {\mathsf{\Phi }}_{-}\left(H\right)$ if and only if $Q_{T}T\in {\mathsf{\Phi }}_{-}(H,H/T\left(0\right))$, and $i\left(T\right)=i\left(Q_{T}T\right);$

(iii) 

If $T\in {\mathsf{\Phi }}_{+}\left(H\right)$, then $\alpha \left(T\right)=\delta \left(T^{*}\right)$ and $\delta \left(T\right)=\alpha \left(T^{*}\right)$;

(iv) 

If $T\in {\mathsf{\Phi }}_{-}\left(H\right)$ and $\rho \left(T\right)\ne \varnothing$, then $\alpha \left(T\right)=\delta \left(T^{*}\right)$ and $\delta \left(T\right)=\alpha \left(T^{*}\right)$.

Lemma 3 

(see ([15], Proposition 2.1)). Let $T\in \mathcal{BCR}\left(H\right)$ with $\rho \left(T\right)\ne \varnothing$. If $\lambda \in \mathbb{C}$ such that $T-\lambda I\in {\mathsf{\Phi }}_{+}\left(H\right)\cup {\mathsf{\Phi }}_{-}\left(H\right)$, then

(i) 

T has the SVEP at λ if and only if $\alpha (T-\lambda I)<\infty$;

(ii) 

$T^{*}$ has the SVEP at λ if and only if $\delta (T-\lambda I)<\infty$.

Lemma 4 

(see ([16], Lemma 2.6)). Let T be a linear relation in a Banach space X. Then

(i) 

$R_c\left(T\right)=\left\{0\right\}$ if and only if $R_c(T-\lambda I)=\left\{0\right\}$ for any $\lambda \in \mathbb{C}$;

(ii) 

If $\rho \left(T\right)\ne \varnothing$, then $R_c\left(T\right)=\left\{0\right\}$.

For upper triangular relation matrix $M_C$, we have the following Lemmas 5–10.

Lemma 5 

(see ([17], Proposition 2.7)). For $A\in \mathcal{LR}\left(H\right),B\in \mathcal{LR}\left(K\right)$, and $C\in \mathcal{LR}(K,H)$,

$$Q_{M_C}{M}_C=\left(\begin{array}{cc}{Q}_{\left(\begin{array}{cc}A& C\end{array}\right)}A& Q_{\left(\begin{array}{cc}A& C\end{array}\right)}C\\ 0& Q_{B}B\end{array}\right).$$

Remark 1.

For $A\in \mathcal{BR}\left(H\right),B\in \mathcal{BR}\left(K\right)$, and $C\in \mathcal{BR}(K,H)$ from Lemma 5, we can obtain that $M_C$ is closed if and only if $A\left(0\right)+C\left(0\right)$ and $B\left(0\right)$ are closed.

Lemma 6 

(see ([1], Theorem 2.12)). Let $A\in \mathcal{BCR}\left(H\right),B\in \mathcal{BCR}\left(K\right)$, and $C\in \mathcal{BCR}(K,H)$ with $A\left(0\right)+C\left(0\right)$ closed. Then the adjoint $M_C^{*}$ of $M_C$ is single-valued, and

$$M_C^{*}=\left(\begin{array}{cc}{A}^{*}& 0\\ C^{*}& B^{*}\end{array}\right):{(A\left(0\right)+C\left(0\right))}^{\perp }\oplus B{\left(0\right)}^{\perp }\to H\oplus K.$$

Lemma 7 

(see ([1], Theorem 3.2)). Let $A\in \mathcal{BCR}\left(H\right)$, $B\in \mathcal{BCR}\left(K\right)$. Then

$$\bigcap _{C\in \mathcal{B}(K,H)}\sigma \left(M_C\right)={\sigma }_{ap}\left(A\right)\cup {\sigma }_{\delta }\left(B\right)\cup \{\lambda \in \mathbb{C}:n(B-\lambda I)\ne d(A-\lambda I)\}.$$

Lemma 8 

(see ([17], Theorem 3.22)). For $A\in \mathcal{BR}\left(H\right)$ and $B\in \mathcal{BR}\left(K\right)$, there exists $C\in \mathcal{BR}(K,H)$ with $C\left(0\right)\subseteq A\left(0\right)$$(orC\in \mathcal{B}(K,H\left)\right)$ such that $M_C\in {\mathsf{\Phi }}_{+}^{-}(H\oplus K)$ if and only if $A\in {\mathsf{\Phi }}_{+}\left(H\right)$, $B\left(0\right)$ is closed, and

$$\left\{\begin{array}{c}n\left(B\right)<\infty andn\left(A\right)+n\left(B\right)\le d\left(A\right)+d\left(B\right),\hfill \\ orn\left(B\right)=d\left(A\right)=\infty ,\hfill & R\left(B\right)isclosed;\hfill \\ d\left(A\right)=\infty ,\hfill & R\left(B\right)isnotclosed.\hfill \end{array}\right.$$

Lemma 9 

(see ([17], Theorem 3.17)). Let $A\in \mathcal{BR}\left(H\right),B\in \mathcal{BR}\left(K\right)$. Then there is $C\in \mathcal{BR}(K,H)$ with $C\left(0\right)\subseteq A\left(0\right)$ such that $M_C\in {\mathsf{\Phi }}_0(H\oplus K)$ if and only if $A\in {\mathsf{\Phi }}_{+}\left(H\right),B\in {\mathsf{\Phi }}_{-}\left(K\right)$, and $n\left(A\right)+n\left(B\right)=d\left(A\right)+d\left(B\right).$

Lemma 10 

(see ([12], Theorem 3.1)). Let $A\in \mathcal{BR}\left(H\right),B\in \mathcal{BR}\left(K\right)$, and $C\in \mathcal{BR}(K,H)$. Then

(i) 

${\sigma }_e\left(M_C\right)\subseteq {\sigma }_e\left(A_{C\left(0\right)}\right)\cup {\sigma }_e\left(B\right);$

(ii) 

If $A\left(0\right)$ and $C\left(0\right)$ are closed, then ${\sigma }_{ab}\left(M_C\right)\subseteq {\sigma }_{ab}\left(A_{C\left(0\right)}\right)\cup {\sigma }_{ab}\left(B\right);$

(iii) 

If $A\left(0\right)$ and $C\left(0\right)$ are closed, then ${\sigma }_b\left(M_C\right)\subseteq {\sigma }_b\left(A_{C\left(0\right)}\right)\cup {\sigma }_b\left(B\right).$

Proposition 1.

Let $T\in \mathcal{BCR}\left(H\right)$ with $0\in \rho \left(T\right)$. Then $\sigma \left(T^{-1}\right)=\{{\displaystyle \frac{1}{\lambda }}:\lambda \in \sigma \left(T\right)\}$.

Proof. 

Note that $T-\lambda I=-\lambda T(T^{-1}-{\displaystyle \frac{1}{\lambda }}I)$ for any $\lambda \ne 0.$ Since $T\in \mathcal{BCR}\left(H\right)$ and $0\in \rho \left(T\right)$, we then know that $\lambda \in \sigma \left(T\right)$ if and only if ${\displaystyle \frac{1}{\lambda }}\in \sigma \left(T^{-1}\right).$ Hence, $\sigma \left(T^{-1}\right)=\{{\displaystyle \frac{1}{\lambda }}:\lambda \in \sigma \left(T\right)\}$. □

Proposition 2.

Let $T\in \mathcal{BCR}\left(H\right)$ with $R\left(T\right)$ closed. If $\rho \left(T\right)\ne \varnothing$ and $n\left(T\right)>d\left(T\right)$, then $0\in S\left(T\right)$.

Proof. 

Assume, to the contrary, that $0\notin S\left(T\right)$. Then T has the SVEP at 0. It is clear that $T\in {\mathsf{\Phi }}_{-}\left(H\right)$. So, from Lemma 3(i), we have $\alpha \left(T\right)<\infty .$ By Lemmas 1(i) and 4, $n\left(T\right)\le d\left(T\right)$, contradicting the assumption $n\left(T\right)>d\left(T\right)$. □

Proposition 3.

Let $A\in \mathcal{BR}\left(H\right)$, $B\in \mathcal{BR}\left(K\right)$, and $C\in \mathcal{BR}(K,H)$. If $M_C\in \mathsf{\Phi }(H\oplus K)$, then $A_{C\left(0\right)}\in {\mathsf{\Phi }}_{+}\left(H\right)$ and $B\in {\mathsf{\Phi }}_{-}\left(K\right)$.

Proof. 

Assume that $M_C\in \mathsf{\Phi }(H\oplus K)$. By Lemma 2(i),(ii), we have $Q_{M_X}{M}_X\in \mathsf{\Phi }(H\oplus K,(H\oplus K)/M_C\left(0\right))$, so then $Q_{\left(\begin{array}{cc}A& C\end{array}\right)}A\in {\mathsf{\Phi }}_{+}(H,H/(A\left(0\right)+C\left(0\right))),Q_{B}B\in {\mathsf{\Phi }}_{-}(K,K/B\left(0\right))$ according to Lemma 5. Again, from Lemma 2(ii), $B\in {\mathsf{\Phi }}_{-}\left(K\right)$. To complete the proof, it remains to be shown $A_{C\left(0\right)}\in {\mathsf{\Phi }}_{+}\left(H\right)$. The closedness of $M_C$ means that $A\left(0\right)+C\left(0\right)$ is closed and so $A_{C\left(0\right)}\left(0\right)$ is closed. It is clear that $\Vert A_{C\left(0\right)}\Vert \le \Vert A\Vert <\infty$, therefore $A_{C\left(0\right)}$ is closed. Note that

$$n\left(A_{C\left(0\right)}\right)=dim\{x\in H:Ax\subseteq A\left(0\right)+C\left(0\right)\}=n\left(Q_{\left(\begin{array}{cc}A& C\end{array}\right)}A\right)\le n\left(M_C\right)<\infty .$$

Moreover, it follows from $Q_{\left(\begin{array}{cc}A& C\end{array}\right)}A\in {\mathsf{\Phi }}_{+}(H,H/(A\left(0\right)+C\left(0\right)))$ that $R\left(Q_{\left(\begin{array}{cc}A& C\end{array}\right)}A\right)$ is closed, so

$$\begin{array}{c}R\left(Q_{\left(\begin{array}{cc}A& C\end{array}\right)}\left(AC-C\right)\right)=R\left(Q_{\left(\begin{array}{cc}A& C\end{array}\right)}A\right)+R\left(Q_{\left(\begin{array}{cc}A& C\end{array}\right)}(C-C)\right)\hfill \\ \hspace{0.17em}=R\left(Q_{\left(\begin{array}{cc}A& C\end{array}\right)}A\right)\hfill \end{array}$$

(1)

is closed. This together with the closedness of $A\left(0\right)+C\left(0\right)$ means that $R\left(A\right)+C\left(0\right)$ is closed, and so $R\left(A_{C\left(0\right)}\right)$ is closed. Hence, $A_{C\left(0\right)}\in {\mathsf{\Phi }}_{+}\left(H\right)$. □

Proposition 4.

Let $T\in \mathcal{BR}\left(H\right)$ and $S\in \mathcal{BR}(K,H)$. Then $Q_{\left(\begin{array}{cc}T& S\end{array}\right)}T$ and $T_{S\left(0\right)}$ are bounded. Moreover, if $T\left(0\right)+S\left(0\right)$ is closed, then $Q_{\left(\begin{array}{cc}T& S\end{array}\right)}T$ is invertible if and only if so is $T_{S\left(0\right)}$.

Proof. 

Note that

$$\Vert Q_{\left(\begin{array}{cc}T& S\end{array}\right)}Tx\Vert \le \Vert Q_{T}Tx\Vert =\Vert Tx\Vert ,x\in H,$$

and

$$\Vert T_{S\left(0\right)}x\Vert \le \Vert Tx\Vert ,x\in H.$$

It follows from the boundedness of T that both $Q_{\left(\begin{array}{cc}T& S\end{array}\right)}T$ and $T_{S\left(0\right)}$ are bounded linear relations. Assume that $T\left(0\right)+S\left(0\right)$ is closed. Then $n\left(Q_{\left(\begin{array}{cc}T& S\end{array}\right)}T\right)=\dim \{x\in H:Tx\subseteq T\left(0\right)+S\left(0\right)\}=n\left(T_{S\left(0\right)}\right).$ Clearly, $Q_{\left(\begin{array}{cc}T& S\end{array}\right)}T$ is surjective if and only if $T_{S\left(0\right)}$ is too. Hence, the conclusion is valid. □

Proposition 5.

Let $T\in \mathcal{BCR}\left(H\right)$. Then $S\left(T\right)\subseteq {\sigma }_p\left(T\right)$.

Proof. 

Assume that ${\lambda }_0\notin {\sigma }_p\left(T\right)$, so it suffices to prove that ${\lambda }_0\notin S\left(T\right)$, i.e., T has the SVEP at ${\lambda }_0.$ Let $f:U\left({\lambda }_0\right)\to H$ be an analytic function and $(T-\lambda I)f\left(\lambda \right)=T\left(0\right)$ for all $\lambda \in U\left({\lambda }_0\right).$ Note that f is an analytic function on $U\left({\lambda }_0\right)$. Then there is a sequence ${\left\{x_n\right\}}_{n=0}^{\infty }\subseteq H$ so that $f\left(\lambda \right)={\displaystyle \sum _{n=0}^{\infty }}{(\lambda -{\lambda }_0)}^n{x}_n$ for all $\lambda \in U\left({\lambda }_0\right)\backslash \left\{{\lambda }_0\right\}.$ The analyticity of $f\left(\lambda \right)$ shows that $f\left({\lambda }_0\right)=x_0,$ and then

$$(T-{\lambda }_{0}I)x_0=(T-{\lambda }_{0}I)f\left({\lambda }_0\right)=T\left(0\right),$$

which ensures $x_0\in N(T-{\lambda }_{0}I)=\left\{0\right\}$, and hence $x_0=0.$ Furthermore, this means that $(T-\lambda I)f\left(\lambda \right)=(T-\lambda I){\displaystyle \sum _{n=0}^{\infty }}{(\lambda -{\lambda }_0)}^n{x}_n=(\lambda -{\lambda }_0)(T-\lambda I){\displaystyle \sum _{n=0}^{\infty }}{(\lambda -{\lambda }_0)}^n{x}_{n+1}$ for $\lambda \in U\left({\lambda }_0\right)\backslash \left\{{\lambda }_0\right\}$, so

$$(T-\lambda I)\sum _{n=0}^{\infty }{(\lambda -{\lambda }_0)}^n{x}_{n+1}=T\left(0\right)$$

(2)

for all $\lambda \in U\left({\lambda }_0\right)\backslash \left\{{\lambda }_0\right\}$. Note that T is closed. Then it is not hard to see that equality (2) is valid for every $\lambda \in U\left({\lambda }_0\right)$ by continuity. Similar to the proof above, we can obtain that $x_1=0$. Then, for all integers $i\ge 2$, $x_i=0$ by iterating this procedure. Hence, $f\equiv 0$ on $U\left({\lambda }_0\right),$ which indicates that T has the SVEP at ${\lambda }_0.$ □

Proposition 6.

Let $T\in \mathcal{BCR}\left(H\right)$. Then $S\left(T\right)\cup {\sigma }_{su}\left(T\right)=\sigma \left(T\right).$

Proof. 

The inclusion $S\left(T\right)\cup {\sigma }_{su}\left(T\right)\subseteq \sigma \left(T\right)$ is clear by Proposition 5. Next we will show that the converse inclusion is also valid. Now suppose that ${\lambda }_0\notin S\left(T\right)\cup {\sigma }_{su}\left(T\right)$, so we need only to prove ${\lambda }_0\notin \sigma \left(T\right)$. Let $x_1\in N(T-{\lambda }_{0}I).$ Note that $T-{\lambda }_{0}I$ is surjective, which together with the closedness of T implies that there is ${\left\{x_n\right\}}_{n=0}^{\infty }$ and $\delta >0$ such that

$$(T-{\lambda }_{0}I)x_{n+1}=x_n+T\left(0\right),\Vert x_n\Vert \le {\delta }^n\Vert x_1\Vert ,$$

and it is clear that $x_0\in T\left(0\right).$ Take analytic function $f\left(\lambda \right)={\displaystyle \sum _{n=1}^{\infty }}{(\lambda -{\lambda }_0)}^{n-1}{x}_n$ for all $\lambda \in U({\lambda }_0,{\displaystyle \frac{1}{\delta }})$,

$$\begin{array}{c}(T-\lambda I)f\left(\lambda \right)=(T-{\lambda }_{0}I)f\left(\lambda \right)+({\lambda }_0-\lambda )f\left(\lambda \right)\hfill \\ ={\displaystyle \sum _{n=0}^{\infty }}{(\lambda -{\lambda }_0)}^n{x}_n-{\displaystyle \sum _{n=1}^{\infty }}{(\lambda -{\lambda }_0)}^n{x}_n+T\left(0\right)\hfill \\ =T\left(0\right).\hfill \end{array}$$

Since T has the SVEP at ${\lambda }_0$, we then have that $f\left(\lambda \right)\equiv 0$ for all $\lambda \in U({\lambda }_0,{\displaystyle \frac{1}{\delta }})$, and hence $x_1=0$, which means that $N(T-{\lambda }_{0}I)=\left\{0\right\}.$ Observe that $T-{\lambda }_{0}I$ is surjective. Thus, ${\lambda }_0\notin \sigma \left(T\right).$ □

Proposition 7.

Let $T\in \mathcal{BCR}\left(H\right)$. Then ${\sigma }_{su}\left(T\right)={\displaystyle \bigcup _{x\in H}}{\sigma }_T\left(x\right).$

Proof. 

Evidently, ${\sigma }_{su}\left(T\right)\subseteq {\displaystyle \bigcup _{x\in H}}{\sigma }_T\left(x\right).$ To complete the proof, it suffices to prove that the opposite inclusion is valid. We now suppose that ${\lambda }_0\notin {\sigma }_{su}\left(T\right)$, $x_0\in H.$ Note that $T-{\lambda }_{0}I$ is surjective. Then the closedness of T implies that there is ${\left\{x_n\right\}}_{n=0}^{\infty }$ and $\delta >0$ such that

$$\Vert x_n\Vert \le {\delta }^n\Vert x_1\Vert ,(T-{\lambda }_{0}I)x_{n+1}=x_n+T\left(0\right).$$

Take analytic function $f\left(\lambda \right)={\displaystyle \sum _{n=1}^{\infty }}{(\lambda -{\lambda }_0)}^{n-1}{x}_n$ for any $\lambda \in U({\lambda }_0,{\displaystyle \frac{1}{\delta }})$, so we have

$$\begin{array}{c}(T-\lambda I)f\left(\lambda \right)=(T-{\lambda }_{0}I)f\left(\lambda \right)+({\lambda }_0-\lambda )f\left(\lambda \right)\hfill \\ ={\displaystyle \sum _{n=0}^{\infty }}{(\lambda -{\lambda }_0)}^n{x}_n-{\displaystyle \sum _{n=1}^{\infty }}{(\lambda -{\lambda }_0)}^n{x}_n+T\left(0\right)\hfill \\ =x_0+T\left(0\right),\hfill \end{array}$$

which ensures that $\lambda \in {\rho }_T\left(x_0\right)$. Thus, ${\sigma }_{su}\left(T\right)\supseteq {\displaystyle \bigcup _{x\in H}}{\sigma }_T\left(x\right).$ □

Proposition 8.

Let $T\in \mathcal{BCR}(H,K)$, and $F\in \mathcal{B}(H,K)$ is finite rank. Then $R(T+F)$ is closed if and only if $R\left(T\right)$ is closed.

Proof. 

Sufficiency: The closedness of T implies that $T\left(0\right)$ is closed. Since $R\left(T\right)$ is closed and $R\left(T\right)=R\left(P_{T{\left(0\right)}^{\perp }}T\right)\oplus T\left(0\right)$, we know that $R\left(P_{T{\left(0\right)}^{\perp }}T\right)$ is closed. Hence $R(T+F)=R(P_{T{\left(0\right)}^{\perp }}T+P_{T{\left(0\right)}^{\perp }}F)\oplus T\left(0\right)$ is closed.

Necessity: Note that $R\left(T\right)=R(T+F-F)$. Then, from the sufficiency, the necessity follows immediately. □

Proposition 9.

Let $A\in \mathcal{BR}\left(H\right)$, $B\in \mathcal{BR}\left(K\right)$, and $C\in \mathcal{BR}(K,H)$. Then $S\left(A_{C\left(0\right)}\right)\subseteq S\left(M_C\right)$.

Proof. 

Assume that ${\lambda }_0\in S\left(A_{C\left(0\right)}\right)$. Then there are a neighborhood $U\left({\lambda }_0\right)$ of ${\lambda }_0$ and a nonzero analytic function $f:U\left({\lambda }_0\right)\to H$ such that $(A_{C\left(0\right)}-\lambda I)f\left(\lambda \right)=A\left(0\right)+C\left(0\right)$ for all $\lambda \in U\left({\lambda }_0\right).$ It is not hard to see that

$$(M_C-\lambda I)\left(\begin{array}{c}f\left(\lambda \right)\\ 0\end{array}\right)=\left(\begin{array}{c}(A-\lambda I)f\left(\lambda \right)+C\left(0\right)\\ (B-\lambda I)\left(0\right)\end{array}\right)=\left(\begin{array}{c}(A_{C\left(0\right)}-\lambda I)f\left(\lambda \right)\\ B\left(0\right)\end{array}\right)=\left(\begin{array}{c}A\left(0\right)+C\left(0\right)\\ B\left(0\right)\end{array}\right),$$

which means that ${\lambda }_0\in S\left(M_C\right)$ and hence $S\left(A_{C\left(0\right)}\right)\subseteq S\left(M_C\right)$. □

Proposition 10.

Let $A\in \mathcal{BR}\left(H\right),B\in \mathcal{BR}\left(K\right)$, and $C\in \mathcal{BR}(K,H)$. Then ${\sigma }_{A_{C\left(0\right)}}\left(x\right)\cup \sigma \left(B\right)={\sigma }_{M_C}(x\oplus 0)\cup S\left(B\right)$ for any $x\in H$.

Proof. 

It is clear that ${\sigma }_{A_{C\left(0\right)}}\left(x\right)\cup \sigma \left(B\right)\supseteq {\sigma }_{M_C}(x\oplus 0)\cup S\left(B\right).$ For the opposite inclusion, we assume that ${\lambda }_0\notin {\sigma }_{M_C}(x\oplus 0)\cup S\left(B\right).$ Then there are a neighborhood $U\left({\lambda }_0\right)$ and analytic functions $f_1:U\left({\lambda }_0\right)\to H$, $f_2:U\left({\lambda }_0\right)\to K$ such that

$$(A-\lambda I)f_1\left(\lambda \right)+C{f}_2\left(\lambda \right)=x+A\left(0\right)+C\left(0\right),(B-\lambda I)f_2\left(\lambda \right)=B\left(0\right)$$

for all $\lambda \in U\left({\lambda }_0\right).$ The fact that ${\lambda }_0\notin S\left(B\right)$ implies that $f_2\equiv 0$ on $U\left({\lambda }_0\right)$, and so $(A-\lambda I)f_1\left(\lambda \right)+C\left(0\right)=x+A\left(0\right)+C\left(0\right)$. This means that

$$(A_{C\left(0\right)}-\lambda I)f_1\left(\lambda \right)=x+A\left(0\right)+C\left(0\right)$$

for all $\lambda \in U\left({\lambda }_0\right).$ Therefore, ${\lambda }_0\notin {\sigma }_{A_{C\left(0\right)}}\left(x\right)$. □

Proposition 11.

Let $A\in \mathcal{BR}\left(H\right),B\in \mathcal{BCR}\left(K\right)$ and $C\in \mathcal{BR}(K,H)$ with $A\left(0\right)+C\left(0\right)$ closed. Then ${\sigma }_{su}\left(A_{C\left(0\right)}\right)\cup \sigma \left(B\right)={\sigma }_{su}\left(M_C\right)\cup S\left(B\right).$

Proof. 

Note that $A\in \mathcal{BR}\left(H\right)$. Then $\parallel A_{C\left(0\right)}\parallel \le \parallel A\parallel <\infty$ and so $A_{C\left(0\right)}$ is bounded, which together with the closedness of $A\left(0\right)+C\left(0\right)$ ensures that $A_{C\left(0\right)}$ is closed. It follows from Remark 1 that $M_C$ is closed. By virtue of Propositions 7 and 10, we know

$$\begin{array}{c}{\sigma }_{su}\left(A_{C\left(0\right)}\right)\cup \sigma \left(B\right)={\displaystyle \bigcup _{x\in H}}{\sigma }_{A_{C\left(0\right)}}\left(x\right)\cup \sigma \left(B\right)\hfill \\ \hspace{0.17em}={\displaystyle \bigcup _{x\in H}}{\sigma }_{M_C}(x\oplus 0)\cup S\left(B\right)\hfill \\ \hspace{0.17em}\subseteq {\displaystyle \bigcup _{x,y\in H}}{\sigma }_{M_C}(x\oplus y)\cup S\left(B\right)\hfill \\ \hspace{0.17em}={\sigma }_{su}\left(M_C\right)\cup S\left(B\right).\hfill \end{array}$$

Moreover, the converse inclusion ${\sigma }_{su}\left(A_{C\left(0\right)}\right)\cup \sigma \left(B\right)\supseteq {\sigma }_{su}\left(M_C\right)\cup S\left(B\right)$ is clear. Hence, the conclusion is valid. □

3. Spectral Properties of Relation Matrices

In this section, we mainly obtain the properties of spectra of upper triangular relation matrices, i.e., Theorems 1–5.

Theorem 1.

Let $A\in \mathcal{BR}\left(H\right),B\in \mathcal{BCR}\left(K\right)$, and $C\in \mathcal{BR}(K,H)$ with $A\left(0\right)+C\left(0\right)$ closed. Then $\sigma \left(M_C\right)\cup S\left(B\right)=\sigma \left(A_{C\left(0\right)}\right)\cup \sigma \left(B\right).$

Proof. 

We first prove $\sigma \left(M_C\right)\subseteq \sigma \left(A_{C\left(0\right)}\right)\cup \sigma \left(B\right).$ Let $\lambda \notin \sigma \left(A_{C\left(0\right)}\right)\cup \sigma \left(B\right).$ It suffices to show that $\lambda \notin \sigma \left(M_C\right),$ i.e., $M_C-\lambda I$ is invertible. The invertibility of $B-\lambda I$ means that $Q_B(B-\lambda I)$ is invertible. It follows from the invertibility of $A_{C\left(0\right)}-\lambda I$ that $Q_{\left(\begin{array}{cc}A& C\end{array}\right)}(A-\lambda I)$ is invertible by Proposition 4. Note that $Q_{\left(\begin{array}{cc}A& C\end{array}\right)}C$ is bounded. According to Lemma 5,

$$Q_{M_C}(M_C-\lambda I)=\left(\begin{array}{cc}{Q}_{\left(\begin{array}{cc}A& C\end{array}\right)}(A-\lambda I)& Q_{\left(\begin{array}{cc}A& C\end{array}\right)}C\\ 0& Q_B(B-\lambda I)\end{array}\right),$$

and $Q_{M_C}(M_C-\lambda I)$ is a single-valued relation. From this, we then have that $Q_{M_C}(M_C-\lambda I)$ is invertible, and hence $M_C-\lambda I$ is invertible, which shows that $\sigma \left(M_C\right)\subseteq \sigma \left(A_{C\left(0\right)}\right)\cup \sigma \left(B\right).$ By Proposition 5, $\sigma \left(M_C\right)\cup S\left(B\right)\subseteq \sigma \left(A_{C\left(0\right)}\right)\cup \sigma \left(B\right)$. In addition, the opposite inclusion can be obtained according to Propositions 9 and 11. Hence, $\sigma \left(M_C\right)\cup S\left(B\right)=\sigma \left(A_{C\left(0\right)}\right)\cup \sigma \left(B\right).$ □

The following theorem is a generalization of ([18], Theorem 2.5).

Theorem 2.

Let $A\in \mathcal{BCR}\left(H\right),B\in \mathcal{BCR}\left(K\right)$, and $C\in \mathcal{BCR}(K,H)$ with $A\left(0\right)+C\left(0\right)$ closed. Then $\sigma \left(M_C\right)\cup (S\left(B\right)\cap \{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\})=\sigma \left(A_{C\left(0\right)}\right)\cup \sigma \left(B\right).$

Proof. 

We first show that $\sigma \left(M_C\right)\cup \{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\}=\sigma \left(A_{C\left(0\right)}\right)\cup \sigma \left(B\right)$. By virtue of Lemma 6 and Theorem 1, we know

$$\sigma \left(M_C^{*}\right)\cup S\left(A^{*}{\mid }_{{(A\left(0\right)+C\left(0\right))}^{\perp }}\right)=\sigma \left(B^{*}\right)\cup \sigma \left(A^{*}{\mid }_{{(A\left(0\right)+C\left(0\right))}^{\perp }}\right).$$

We claim that $A^{*}{\mid }_{{(A\left(0\right)+C\left(0\right))}^{\perp }}={\left(A_{C\left(0\right)}\right)}^{*}.$ Indeed, it is not hard to see that $D{\left(A_{C\left(0\right)}\right)}^{*}={(A\left(0\right)+C\left(0\right))}^{\perp }=D\left(A^{*}{\mid }_{{(A\left(0\right)+C\left(0\right))}^{\perp }}\right)$ and

$$\langle {\left(A_{C\left(0\right)}\right)}^{*}x,y\rangle =\langle x,\left(A_{C\left(0\right)}\right)y\rangle =\langle x,Ay\rangle =\langle A^{*}x,y\rangle$$

for $x\in {(A\left(0\right)+C\left(0\right))}^{\perp }$ and $y\in H$. Since $x\in {(A\left(0\right)+C\left(0\right))}^{\perp }$, $A^{*}{\mid }_{{(A\left(0\right)+C\left(0\right))}^{\perp }}={\left(A_{C\left(0\right)}\right)}^{*}$. Therefore, $\sigma \left(M_C\right)\cup \{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\}=\sigma \left(A_{C\left(0\right)}\right)\cup \sigma \left(B\right)$. Note that $\sigma \left(M_C\right)\cup S\left(B\right)=\sigma \left(A_{C\left(0\right)}\right)\cup \sigma \left(B\right)$ by Theorem 1. Hence, $\sigma \left(M_C\right)\cup (S\left(B\right)\cap \{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\})=\sigma \left(A_{C\left(0\right)}\right)\cup \sigma \left(B\right).$ □

Next we give an example to illustrate the result above.

Example 1.

Let $A,B,C\in \mathcal{BCR}\left({\ell }^2\left(\mathbb{N}\right)\right)$ be given by

$$\begin{array}{c}Ax=(0,x_1,x_2,x_3,x_4\cdots ),Bx=(0,x_2,0,x_4,0,x_6\cdots ),\hfill \\ Cx=\left\{\right(a,0,0,0,0\cdots ):a\in \mathbb{N}\}\hfill \end{array}$$

for all $x=(x_1,x_2,x_3,\cdots )\in {\ell }^2\left(\mathbb{N}\right)$, respectively. Then $\sigma \left(M_C\right)=\left\{0\right\}\cup \{\lambda \in \mathbb{C}:|\lambda |\ge 1\}.$

Note that $\sigma \left(A\right)=\{\lambda \in \mathbb{C}:|\lambda |\le 1\},\sigma \left(B\right)=\{0,1\}.$ We claim $\sigma \left(A_{C\left(0\right)}\right)=\{\lambda \in \mathbb{C}:|\lambda |\ge 1\}.$ Indeed, let T be the left shift operator on ${\ell }^2\left(\mathbb{N}\right)$ defined by $Tx=(x_2,x_3,x_4,x_5,\cdots )$ for all $x=(x_1,x_2,x_3,\cdots )\in {\ell }^2\left(\mathbb{N}\right)$, so it is easy to see that $A_{C\left(0\right)}=T^{-1}$ and $0\in \rho \left(T^{-1}\right)$, which together with Proposition 1 and $\sigma \left(T\right)=\{\lambda \in \mathbb{C}:|\lambda |\le 1\}$ imply that $\sigma \left(A_{C\left(0\right)}\right)=\{\lambda \in \mathbb{C}:|\lambda |\ge 1\}.$ Since $S\left(B\right)=\varnothing$ and $A\left(0\right)+C\left(0\right)$ is closed, then $\sigma \left(M_C\right)=\sigma \left(A_{C\left(0\right)}\right)\cup \sigma \left(B\right)=\left\{0\right\}\cup \{\lambda \in \mathbb{C}:|\lambda |\ge 1\}$ by Theorem 2,

Moreover, observing that $n\left(M_C\right)=\{x\in H:Ax\subseteq A\left(0\right)+C\left(0\right)\}\oplus N\left(B\right)=N\left(A_{C\left(0\right)}\right)\oplus N\left(B\right)$ and $R\left(M_C\right)=R\left(A_{C\left(0\right)}\right)\oplus R\left(B\right)$. Then we can know that $\sigma \left(M_C\right)=\sigma \left(A_{C\left(0\right)}\right)\cup \sigma \left(B\right)=\left\{0\right\}\cup \{\lambda \in \mathbb{C}:|\lambda |\ge 1\}.$

Note that $\sigma \left(M_C\right)\ne \sigma \left(A\right)\cup \sigma \left(B\right)$. This shows that the spectral properties of upper triangular relation matrices are related to the multi-valued part $C\left(0\right)$ of relation $C\in \mathcal{BR}(K,H)$.

Theorem 3.

Let $A\in \mathcal{BR}\left(H\right),B\in \mathcal{BR}\left(K\right)$, and $C\in \mathcal{BR}(K,H)$ with $\rho \left(A_{C\left(0\right)}\right)\ne \varnothing$ and $\rho \left(B\right)\ne \varnothing$. Then ${\sigma }_e\left(M_C\right)\cup (S\left(B\right)\cap \{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\})={\sigma }_e\left(A_{C\left(0\right)}\right)\cup {\sigma }_e\left(B\right)\cup (S\left(B\right)\cap \{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\}).$

Proof. 

Assume that $\lambda \in ({\sigma }_e\left(A_{C\left(0\right)}\right)\cup {\sigma }_e\left(B\right))\backslash {\sigma }_e\left(M_C\right).$ Then $M_C-\lambda I\in \mathsf{\Phi }(H\oplus K)$. By Proposition 3, we have $A_{C\left(0\right)}-\lambda I\in {\mathsf{\Phi }}_{+}\left(H\right)$ and $B-\lambda I\in {\mathsf{\Phi }}_{-}\left(K\right).$ Note that ${\sigma }_e\left(M_C\right)\subseteq {\sigma }_e\left(A_{C\left(0\right)}\right)\cup {\sigma }_e\left(B\right)$ by Lemma 10(i). Then, to finish the proof, it is sufficient to show $\lambda \in S\left(B\right)\cap \{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\}).$ On the contrary, suppose $\lambda \notin S\left(B\right)\cap \{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\}),$ then there are two cases to consider.

Case 1: If $\lambda \notin \{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\}),$ i.e., $\overline{\lambda }\notin S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)$. It is clear that ${\left(A_{C\left(0\right)}\right)}^{*}-\overline{\lambda }I={(A_{C\left(0\right)}-\lambda I)}^{*}\in {\mathsf{\Phi }}_{-}\left(H\right).$ In view of $\rho \left(A_{C\left(0\right)}\right)\ne \varnothing$, we can see that $\rho \left({(A_{C\left(0\right)}-\lambda I)}^{*}\right)\ne \varnothing$. Then, from Proposition 2, $n\left({(A_{C\left(0\right)}-\lambda I)}^{*}\right)\le d\left({(A_{C\left(0\right)}-\lambda I)}^{*}\right)$ and so $d(A_{C\left(0\right)}-\lambda I)\le n(A_{C\left(0\right)}-\lambda I)<\infty$, which means that $A_{C\left(0\right)}-\lambda I\in \mathsf{\Phi }\left(H\right)$. Moreover, from equality (1), we have

$$d\left(Q_{\left(\begin{array}{cc}A& C\end{array}\right)}(A-\lambda I)\right)=d\left(Q_{\left(\begin{array}{cc}A& C\end{array}\right)}\left(A-\lambda IC-C\right)\right)=d(A_{C\left(0\right)}-\lambda I)<\infty .$$

The fact that $M_C-\lambda I\in \mathsf{\Phi }(H\oplus K)$ implies $Q_{M_C}(M_C-\lambda I)\in \mathsf{\Phi }(H\oplus K,H\oplus K/M_C\left(0\right))$, and hence, $Q_{\left(\begin{array}{cc}A& C\end{array}\right)}(A-\lambda I)\in {\mathsf{\Phi }}_{+}(H,H/(A\left(0\right)+C\left(0\right)))$ according to Lemma 5, so $Q_{\left(\begin{array}{cc}A& C\end{array}\right)}(A-\lambda I)\in \mathsf{\Phi }(H,H/(A\left(0\right)+C\left(0\right)))$. Again, by Lemma 5, $Q_B(B-\lambda I)\in \mathsf{\Phi }(K,K/B\left(0\right)),$ so $B-\lambda I\in \mathsf{\Phi }\left(K\right),$ which contradicts the hypothesis $\lambda \in {\sigma }_e\left(A_{C\left(0\right)}\right)\cup {\sigma }_e\left(B\right).$

Case 2: If $\lambda \notin S\left(B\right),$ then $n(B-\lambda I)\le d(B-\lambda I)<\infty$ according to Proposition 2, so $B-\lambda I\in \mathsf{\Phi }\left(K\right).$ Similar to the proof above, we can obtain that $A_{C\left(0\right)}-\lambda I\in \mathsf{\Phi }\left(H\right),$ which contradicts the hypothesis $\lambda \in {\sigma }_e\left(A_{C\left(0\right)}\right)\cup {\sigma }_e\left(B\right).$ □

Theorem 4.

Let $A\in \mathcal{BR}\left(H\right),B\in \mathcal{BR}\left(K\right)$, and $C\in \mathcal{BR}(K,H)$ with $\rho \left(A_{C\left(0\right)}\right)\ne \varnothing$ and $\rho \left(B\right)\ne \varnothing$. Then

$$\begin{array}{c}({\sigma }_w\left(A_{C\left(0\right)}\right)\cup {\sigma }_w\left(B\right))\backslash {\sigma }_w\left(M_C\right)\subseteq (S\left(B\right)\cap \{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\})\hfill \\ \cup (S\left(A_{C\left(0\right)}\right)\cap \{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left(B^{*}\right)\}).\hfill \end{array}$$

Proof. 

Suppose that $\lambda \in {\sigma }_w\left(A_{C\left(0\right)}\right)\cup {\sigma }_w\left(B\right)\backslash {\sigma }_w\left(M_C\right)$, so we have $M_C-\lambda I\in \mathsf{\Phi }(H\oplus K)$. Let $\lambda \notin S\left(B\right)\cap \{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\}.$ Then to complete the proof, it suffices to show that $\lambda \in S\left(A_{C\left(0\right)}\right)\cap \{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left(B^{*}\right)\}$. Note that $Q_{\left(\begin{array}{cc}A& C\end{array}\right)}(A-\lambda I)\in \mathsf{\Phi }(H,H/(A\left(0\right)+C\left(0\right)))$ and $Q_B(B-\lambda I)\in \mathsf{\Phi }(K,K/B\left(0\right))$ by the proof of Theorem 3. By virtue of Lemma 5, $i\left(Q_{M_C}(M_C-\lambda I)\right)=i\left(Q_{\left(\begin{array}{cc}A& C\end{array}\right)}(A-\lambda I)\right)+i\left(Q_B(B-\lambda I)\right)=0.$ Note that

$$\begin{array}{c}d(A_{C\left(0\right)}-\lambda I)=d\left(Q_{\left(\begin{array}{cc}A& C\end{array}\right)}\left(A-\lambda IC-C\right)\right)=d\left(Q_{\left(\begin{array}{cc}A& C\end{array}\right)}(A-\lambda I)\right)<\infty \hfill \end{array}$$

and

$$\begin{array}{c}n(A_{C\left(0\right)}-\lambda I)=dim\{x\in H:(A-\lambda I)x\subseteq A\left(0\right)+C\left(0\right)\}\hfill \\ =n\left(Q_{\left(\begin{array}{cc}A& C\end{array}\right)}(A-\lambda I)\right)<\infty .\hfill \end{array}$$

Then $i(A_{C\left(0\right)}-\lambda I)=i\left(Q_{\left(\begin{array}{cc}A& C\end{array}\right)}(A-\lambda I)\right)$, which together with Lemma 2(i),(ii) means that $i(M_C-\lambda I)=i(A_{C\left(0\right)}-\lambda I)+i(B-\lambda I)=0.$ Observe that $\lambda \in {\sigma }_w\left(A_{C\left(0\right)}\right)\cup {\sigma }_w\left(B\right)$. If $i(A_{C\left(0\right)}-\lambda I)>0$, then $i(B-\lambda I)<0$. By Proposition 2, we have

$$\lambda \in S\left(A_{C\left(0\right)}\right)\cap \{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left(B^{*}\right)\}).$$

However, if $i(A_{C\left(0\right)}-\lambda I)<0$, then similarly, we have $\lambda \in S\left(B\right)\cap \{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\}$, contradicting the assumption $\lambda \notin S\left(B\right)\cap \{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\}$. Hence, $\lambda \in S\left(A_{C\left(0\right)}\right)\cap \{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left(B^{*}\right)\}).$ □

Theorems 3 and 4 extend the result of ([19], Theorems 3.2 and 3.3) to the case of a linear relation. The next theorem is a generalization of ([20], Theorem 2.3).

Theorem 5.

Let $A\in \mathcal{BCR}\left(H\right),B\in \mathcal{BCR}\left(K\right)$, and $C\in \mathcal{BCR}(K,H)$ with $\rho \left(A_{C\left(0\right)}\right)\ne \varphi$ and $\rho \left(B\right)\ne \varphi$. Then

$$\begin{array}{c}{\sigma }_b\left(M_C\right)\cup (\{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\}\cap S\left(B\right))={\sigma }_b\left(A_{C\left(0\right)}\right)\cup {\sigma }_b\left(B\right).\hfill \end{array}$$

Proof. 

We first show that

$$\begin{array}{c}{\sigma }_b\left(M_C\right)\cup (\{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\}\cap S\left(B\right))\subseteq {\sigma }_b\left(A_{C\left(0\right)}\right)\cup {\sigma }_b\left(B\right).\hfill \end{array}$$

It suffices to prove that $\{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\}\cap S\left(B\right))\subseteq {\sigma }_b(A_{C\left(0\right)}\cup {\sigma }_b\left(B\right)$ by Lemma 10(iii). We now suppose $\lambda \notin {\sigma }_b\left(A_{C\left(0\right)}\right)\cup {\sigma }_b\left(B\right),$ so then $A_{C\left(0\right)}-\lambda I\in \mathsf{\Phi }\left(H\right)$, $B-\lambda I\in \mathsf{\Phi }\left(K\right),$$\alpha \left({(A_{C\left(0\right)}-\lambda I)}^{*}\right)=\delta (A_{C\left(0\right)}-\lambda I)<\infty$, and $\alpha (B-\lambda I)<\infty$. By virtue of Lemma 3, $\overline{\lambda }\notin S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)$ and $\lambda \notin S\left(B\right)$, which means that $\{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\}\cap S\left(B\right)\subseteq {\sigma }_b\left(A_{C\left(0\right)}\right)\cup {\sigma }_b\left(B\right)$.

For the reverse inclusion, let $\lambda \notin {\sigma }_b\left(M_C\right)\cup (\{\overline{\lambda }\in \mathbb{C}:\lambda \in S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)\}\cap S\left(B\right))$, so it suffices to prove $\lambda \notin {\sigma }_b\left(A_{C\left(0\right)}\right)\cup {\sigma }_b\left(B\right).$ Evidently, $M_C-\lambda I\in {\mathsf{\Phi }}_b(H\oplus K)$, which together with Proposition 3 means that $A_{C\left(0\right)}-\lambda I\in {\mathsf{\Phi }}_{+}\left(H\right)$ and $B-\lambda I\in {\mathsf{\Phi }}_{-}\left(K\right)$, and according to Lemmas 2(iii), (iv) and 6,

$$\begin{array}{c}\delta (B-\lambda I)=\alpha \left({(B-\lambda I)}^{*}\right)<\alpha \left({(M_C-\lambda I)}^{*}\right)=\delta (M_C-\lambda I)<\infty ,\hfill \\ \delta \left({(A-\lambda I)}^{*}{|}_{{(A\left(0\right)+C\left(0\right))}^{\perp }}\right)<\delta \left({(M_C-\lambda I)}^{*}\right)=\alpha (M_C-\lambda I)<\infty .\hfill \end{array}$$

It follows from the proof of Theorem 2 that ${(A-\lambda I)}^{*}{\mid }_{{(A\left(0\right)+C\left(0\right))}^{\perp }}={(A_{C\left(0\right)}-\lambda I)}^{*}$, then $\alpha (A_{C\left(0\right)}-\lambda I)<\infty$ by Lemma 2(iii). Note that $M_C-\lambda I\in {\mathsf{\Phi }}_b(H\oplus K)$. Then $A\left(0\right)+C\left(0\right)$ is closed. It is clear that $R\left(Q_{\left(AC\right)}(A-\lambda I)\right)=R\left(Q_{\left(AC\right)}(A-\lambda IC-C)\right)$, so

$$\begin{array}{c}\beta \left(Q_{\left(AC\right)}(A-\lambda I)\right)=\beta \left(Q_{\left(AC\right)}(A-\lambda IC-C)\right)\hfill \\ =\beta (A-\lambda IC-C)\hfill \\ =\beta (A_{C\left(0\right)}-\lambda I).\hfill \end{array}$$

It is not hard to see that $n\left(Q_{\left(AC\right)}(A-\lambda I)\right)=n(A_{C\left(0\right)}-\lambda I),$ so $i\left(Q_{\left(AC\right)}(A-\lambda I)\right)=i(A_{C\left(0\right)}-\lambda I).$ There are two case as follows.

Case 1: If $\overline{\lambda }\notin S\left({\left(A_{C\left(0\right)}\right)}^{*}\right)$, by Proposition 2, $n\left({(A_{C\left(0\right)}-\lambda I)}^{*}\right)\le d\left({(A_{C\left(0\right)}-\lambda I)}^{*}\right)$, and so $d(A_{C\left(0\right)}-\lambda I))\le n(A_{C\left(0\right)}-\lambda I)$, which together with $\alpha (A_{C\left(0\right)}-\lambda I)<\infty$ implies that $i(A_{C\left(0\right)}-\lambda I)=0$ according to Lemmas 1(i) and 4; thus, $i\left(Q_{\left(AC\right)}(A-\lambda I)\right)=0$. Then, from $M_C-\lambda I\in {\mathsf{\Phi }}_b(H\oplus K)$, we have $Q_{\left(AC\right)}(A-\lambda I)\in \mathsf{\Phi }(H,H/(A\left(0\right)+C\left(0\right))).$ Since $i(Q_{M_X}{M}_X-\lambda I)=i(M_C-\lambda I)=0,$ then Lemma 5 show that $i\left(Q_{\left(AC\right)}(A-\lambda I)\right)+i\left(Q_B(B-\lambda I)\right)=0$, which together with $i\left(Q_{\left(AC\right)}(A-\lambda I)\right)=i(A_{C\left(0\right)}-\lambda I)$ and $i\left(Q_B(B-\lambda I)\right)=i(B-\lambda I)$ means that

$$i(A_{C\left(0\right)}-\lambda I)+i(B-\lambda I)=0.$$

It follows from $i(A_{C\left(0\right)}-\lambda I)+i(B-\lambda I)=0$ that $i(B-\lambda I)=0$. By Lemma 1(iii), (iv), $\alpha (A_{C\left(0\right)}-\lambda I)=\delta (A_{C\left(0\right)}-\lambda I)<\infty$ and $\alpha (B-\lambda I)=\delta (B-\lambda I)<\infty$. Hence, $\lambda \notin {\sigma }_b\left(A_{C\left(0\right)}\right)\cup {\sigma }_b\left(B\right).$

Case 2: If $\lambda \notin S\left(B\right)$, similar to the proof of case 1, we can obtain $\lambda \notin {\sigma }_b\left(A_{C\left(0\right)}\right)\cup {\sigma }_b\left(B\right).$

Therefore, the reverse inclusion is valid. □

4. Weyl-Type Theorems of Relation Matrices

In this section, we mainly study Browder’s theorem, a-Browder’s theorem, and Weyl’s theorem for upper triangular relation matrices, i.e., Theorems 7–10. As their corollaries, some related properties are also characterized. In the sequel, we write ${\overline{\sigma }}_{\star }\left(T\right)=\{\overline{\lambda }\in \mathbb{C}:\lambda \in {\sigma }_{\star }\left(T\right)\}$, ${\pi }_0\left(T\right)=\{\lambda \in iso\sigma \left(T\right):0<n(T-\lambda I)<\infty \}$, and $p\left(T\right)=\{\lambda \in \mathbb{C}:\alpha (T-\lambda I)=\delta (T-\lambda I)<\infty \},$ where $iso\sigma \left(T\right)$ is the isolated point of $\sigma \left(T\right).$

For $T\in \mathcal{BCR}\left(H\right)$, if ${\sigma }_w\left(T\right)={\sigma }_b\left(T\right)$, we say Browder’s theorem holds for T, while if ${\sigma }_{ea}\left(T\right)={\sigma }_{ab}\left(T\right)$, we say a-Browder’s theorem holds for T; T satisfies Weyl’s theorem if $\sigma \left(T\right)\backslash {\sigma }_w\left(T\right)={\pi }_0\left(T\right)$.

Lemma 11 

(see ([21], Theorem 4.1)). Let $T\in \mathcal{BCR}\left(H\right)$ with $\rho \left(T\right)\ne \varnothing$. If Weyl’s theorem holds for T, then ${\sigma }_w\left(T\right)={\sigma }_b\left(T\right)$.

Lemma 12 

(see ([22], Theorem 4.2)). Let $T\in \mathcal{BCR}\left(H\right)$ with $\rho \left(T\right)\ne \varnothing$. If $\lambda \in \sigma \left(T\right)$ and $\alpha (T-\lambda I)=\delta (T-\lambda I)<\infty$, then λ is an isolated point of $\sigma \left(T\right)$.

Proposition 12.

Let $T\in \mathcal{BCR}\left(H\right)$ with $\rho \left(T\right)\ne \varnothing$. Then $\sigma \left(T\right)\backslash {\sigma }_b\left(T\right)\subseteq {\pi }_0\left(T\right).$

Proof. 

Let $\lambda \in \sigma \left(T\right)\backslash {\sigma }_b\left(T\right)$. Then $\lambda \in \sigma \left(T\right)$, and $T-\lambda I$ is a Browder relation. It is clear that $0<n(T-\lambda I)<\infty$. By Lemma 12, we can obtain that $\lambda \in iso\sigma \left(T\right).$ Hence, $\sigma \left(T\right)\backslash {\sigma }_b\left(T\right)\subseteq {\pi }_0\left(T\right).$ □

Theorem 6.

Let $A\in \mathcal{BCR}\left(H\right),B\in \mathcal{BCR}\left(K\right)$ with $\rho \left(A\right)\ne \varnothing$ and $\rho \left(B\right)\ne \varnothing$. Then

(i) 

${\sigma }_w\left(M_0\right)={\sigma }_b\left(M_0\right)$ if and only if A and B have the SVEP on $\{\lambda \in \mathbb{C}:A-\lambda I\in \mathsf{\Phi }(H),B-\lambda I\in \mathsf{\Phi }(K),i(A-\lambda I)+i(B-\lambda I)=0\};$

(ii) 

${\sigma }_{ea}\left(M_0\right)={\sigma }_{ab}\left(M_0\right)$ if and only if A and B have the SVEP on $\{\lambda \in \mathbb{C}:A-\lambda I\in {\mathsf{\Phi }}_{+}\left(H\right),B-\lambda I\in {\mathsf{\Phi }}_{+}\left(K\right),i(A-\lambda I)+i(B-\lambda I)\le 0\}.$

Proof. 

(i) Let ${\sigma }_w\left(M_0\right)={\sigma }_b\left(M_0\right)$. Then it is clear that

$${\sigma }_w\left(M_0\right)={\sigma }_b\left(A\right)\cup {\sigma }_b\left(B\right).$$

Let $\lambda \in \{\lambda \in \mathbb{C}:A-\lambda I\in \mathsf{\Phi }(H),B-\lambda I\in \mathsf{\Phi }(K),i(A-\lambda I)+i(B-\lambda I)=0\},$ that is, $\lambda \notin {\sigma }_w\left(M_0\right)$. It follows from ${\sigma }_w\left(M_0\right)={\sigma }_b\left(A\right)\cup {\sigma }_b\left(B\right)$ that $A-\lambda I\in \mathsf{\Phi }\left(H\right),B-\lambda I\in \mathsf{\Phi }\left(K\right),$ and $\alpha (A-\lambda I)<\infty ,\alpha (B-\lambda I)<\infty$. By Lemma 3, A and B have the SVEP on $\{\lambda \in \mathbb{C}:A-\lambda I\in \mathsf{\Phi }(H),B-\lambda I\in \mathsf{\Phi }(K),i(A-\lambda I)+i(B-\lambda I)=0\}.$

Conversely, since ${\sigma }_w\left(M_0\right)\subseteq {\sigma }_b\left(M_0\right)$, we only need to show that ${\sigma }_b\left(M_0\right)\subseteq {\sigma }_w\left(M_0\right)$. Assume that $\lambda \notin {\sigma }_w\left(M_0\right).$ Then $A-\lambda I\in \mathsf{\Phi }\left(H\right),B-\lambda I\in \mathsf{\Phi }\left(K\right),i(A-\lambda I)+i(B-\lambda I)=0$. Note that A and B have the SVEP at $\lambda$. Then $i(A-\lambda I)\le 0$ and $i(B-\lambda I)\le 0$ according to Proposition 2, which together with $i(A-\lambda I)+i(B-\lambda I)=0$ implies that

$$i(A-\lambda I)=i(B-\lambda I)=0.$$

Since A and B have the SVEP at $\lambda$, $\alpha (A-\lambda I)<\infty$ and $\alpha (B-\lambda I)<\infty$. By virtue of Lemmas 1(iii) and 4, $\alpha (A-\lambda I)=\delta (A-\lambda I)<\infty$ and $\alpha (B-\lambda I)=\delta (B-\lambda I)<\infty$, which means that $\lambda \notin {\sigma }_b\left(M_0\right).$ Hence, ${\sigma }_w\left(M_0\right)={\sigma }_b\left(M_0\right).$

(ii) Let ${\sigma }_{ea}\left(M_0\right)={\sigma }_{ab}\left(M_0\right)$. Then we can know ${\sigma }_{ea}\left(M_0\right)={\sigma }_{ab}\left(A\right)\cup {\sigma }_{ab}\left(B\right).$ Assume that $\lambda \in \{\lambda \in \mathbb{C}:A-\lambda I\in {\mathsf{\Phi }}_{+}\left(H\right),B-\lambda I\in {\mathsf{\Phi }}_{+}\left(K\right),i(A-\lambda I)+i(A-\lambda I)\le 0\},$ that is, $\lambda \notin {\sigma }_{ea}\left(M_0\right)$. Then ${\sigma }_{ea}\left(M_0\right)={\sigma }_{ab}\left(A\right)\cup {\sigma }_{ab}\left(B\right)$ implies that $\lambda \notin {\sigma }_{ab}\left(A\right)\cup {\sigma }_{ab}\left(B\right)$ and so $A-\lambda I\in {\mathsf{\Phi }}_{+}\left(H\right),B-\lambda I\in {\mathsf{\Phi }}_{+}\left(K\right),$$\alpha (A-\lambda I)<\infty$, and $\alpha (B-\lambda I)<\infty$. By Lemma 3, A and B have the SVEP on $\{\lambda \in \mathbb{C}:A-\lambda I\in {\mathsf{\Phi }}_{+}\left(H\right),B-\lambda I\in {\mathsf{\Phi }}_{+}\left(K\right),i(A-\lambda I)+i(A-\lambda I)$$\le 0\}.$

Conversely, since ${\sigma }_{ea}\left(M_0\right)\subseteq {\sigma }_{ab}\left(M_0\right)$, we only need to show ${\sigma }_{ab}\left(M_0\right)\subseteq {\sigma }_{ea}\left(M_0\right)$. Suppose $\lambda \notin {\sigma }_{ea}\left(M_0\right).$ Then $A-\lambda I\in {\mathsf{\Phi }}_{+}\left(H\right),B-\lambda I\in {\mathsf{\Phi }}_{+}\left(K\right),i(A-\lambda I)+i(B-\lambda I)\le 0$. Since A and B have the SVEP at $\lambda$, $i(A-\lambda I)\le 0$ and $i(B-\lambda I)\le 0$ according to Proposition 2, then $i(A-\lambda I)=i(B-\lambda I)=0$. By Lemma 3, we can learn that $\alpha (A-\lambda I)<\infty$ and $\alpha (B-\lambda I)<\infty$, which means that ${\sigma }_{ab}\left(M_0\right)\subseteq {\sigma }_{ea}\left(M_0\right)$. Hence ${\sigma }_{ab}\left(M_0\right)={\sigma }_{ea}\left(M_0\right)$. □

Remark 2.

(i) Clearly, Theorem 6 implies that Browder’s theorem holds for $M_0$ if and only if A and B have the SVEP at $\lambda \notin {\sigma }_w\left(M_0\right)$, and a-Browder’s theorem holds for $M_0$ if and only if A and B have the SVEP at $\lambda \notin {\sigma }_{ea}\left(M_0\right)$.

(ii) For $M_C=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right)$ and $\left(\begin{array}{cc}{A}_{C\left(0\right)}& C_{[\perp ]}\\ 0& B\end{array}\right)$, where $A\in \mathcal{BR}\left(H\right),B\in \mathcal{BR}\left(K\right),C\in \mathcal{BCR}(K,H)$, then clearly, $C_{[\perp ]}=P_{C{\left(0\right)}^{\perp }}C$ is single-valued and bounded. Note that

$$G\left(M_C\right)=G\left(\left(\begin{array}{cc}{A}_{C\left(0\right)}& C_{[\perp ]}\\ 0& B\end{array}\right)\right),$$

so $M_C$ satisfies Weyl’s theorem if and only if $\left(\begin{array}{cc}{A}_{C\left(0\right)}& C_{[\perp ]}\\ 0& B\end{array}\right)$ satisfies Weyl’s theorem. This will simplify the study of the Weylness of relation matrices with all internal elements being multi-valued operators.

Theorem 7.

Let $A\in \mathcal{BCR}\left(H\right),B\in \mathcal{BCR}\left(K\right),C\in \mathcal{BCR}(K,H)$ with $\rho \left(A_{C\left(0\right)}\right)\ne \varnothing$, $\rho \left(B\right)\ne \varnothing$, and $A\left(0\right)+C\left(0\right)$ closed. If $\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ satisfies Browder’s theorem and at least one of the following conditions holds:

(i) 

$A_{C\left(0\right)}$ has the SVEP at $\lambda \in {\sigma }_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\backslash {\sigma }_{le}\left(A_{C\left(0\right)}\right)$ and B has the SVEP at $\mu \in {\sigma }_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\backslash {\sigma }_{re}\left(B\right);$

(ii) 

$A_{C\left(0\right)}$ has the SVEP at $\lambda \in {\sigma }_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\backslash {\sigma }_{le}\left(A_{C\left(0\right)}\right)$ and ${\left(A_{C\left(0\right)}\right)}^{*}$ has the SVEP at $\mu \in {\overline{\sigma }}_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\backslash {\overline{\sigma }}_{le}\left(A_{C\left(0\right)}\right);$

(iii) 

${\left(A_{C\left(0\right)}\right)}^{*}$ has the SVEP at $\lambda \in {\overline{\sigma }}_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\backslash {\overline{\sigma }}_{le}\left(A_{C\left(0\right)}\right)$ and $B^{*}$ has the SVEP at $\mu \in {\overline{\sigma }}_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\backslash {\overline{\sigma }}_{re}\left(B\right),$

then $M_C$ satisfies Browder’s theorem.

Proof. 

Let Browder’s theorem holds for $\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$. Note that

$$\parallel A_{C\left(0\right)}x\parallel = \parallel Q_{\left(AC\right)}{A}_{C\left(0\right)}x\parallel = \parallel Q_{\left(AC\right)}Ax\parallel \le \parallel Q_{A}Ax\parallel = \parallel Ax\parallel ,x\in H.$$

Then the boundedness of A means that $A_{C\left(0\right)}$ is bounded, which together with the closedness of $A\left(0\right)+C\left(0\right)$ shows that $A_{C\left(0\right)}$ is closed, and so $\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ is closed since B is closed. From this, we have that $R\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ is closed if and only if $R{\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)}^{*}$ is closed. It follows from $N{\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)}^{*}={\left(R\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\right)}^{\perp }$ and $N\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)={\left(R{\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)}^{*}\right)}^{\perp }$ that

$${\sigma }_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)={\overline{\sigma }}_w{\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)}^{*}.$$

According to Lemma 2(iii), we can obtain that ${\sigma }_b\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)={\overline{\sigma }}_b{\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)}^{*}.$ Then Browder’s theorem holds for $\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ if and only if Browder’s theorem holds for ${\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)}^{*}$. By virtue of Theorem 6, $A_{C\left(0\right)}$ and B have the SVEP at $\lambda \notin {\sigma }_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right),$ and ${\left(A_{C\left(0\right)}\right)}^{*}$ and $B^{*}$ have the SVEP at $\lambda \notin {\overline{\sigma }}_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ according to Lemma 6. Note that ${\sigma }_{le}\left(A_{C\left(0\right)}\right)\subseteq {\sigma }_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ and ${\sigma }_{re}\left(B\right)\subseteq {\sigma }_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right),$ then hypotheses (i), (ii), and (iii) mean, respectively, that

(i)’

$A_{C\left(0\right)}$ has the SVEP at $\lambda \notin {\sigma }_{le}\left(A_{C\left(0\right)}\right)$ and B has the SVEP at $\mu \notin {\sigma }_{re}\left(B\right);$

(ii)’

$A_{C\left(0\right)}$ has the SVEP at $\lambda \notin {\sigma }_{le}\left(A_{C\left(0\right)}\right)$ and ${\left(A_{C\left(0\right)}\right)}^{*}$ has the SVEP at $\mu \notin {\overline{\sigma }}_{le}\left(A_{C\left(0\right)}\right);$

(iii)’

${\left(A_{C\left(0\right)}\right)}^{*}$ has the SVEP at $\lambda \notin {\overline{\sigma }}_{le}\left(A_{C\left(0\right)}\right)$ and $B^{*}$ has the SVEP at $\mu \notin {\overline{\sigma }}_{re}\left(B\right).$

It is clear that ${\sigma }_w\left(M_C\right)\subseteq {\sigma }_b\left(M_C\right)$, so to finish the proof, we have to show ${\sigma }_b\left(M_C\right)\subseteq {\sigma }_w\left(M_C\right)$. Let $\lambda \notin {\sigma }_w\left(M_C\right),$ so then $M_C-\lambda I\in {\mathsf{\Phi }}_0(H\oplus K)$. It follows from $G\left(M_C\right)=G\left(\left(\begin{array}{cc}{A}_{C\left(0\right)}& C_{[\perp ]}\\ 0& B\end{array}\right)\right)$ and Lemma 9 that $\lambda \notin {\sigma }_{le}\left(A_{C\left(0\right)}\right)\cup {\sigma }_{re}\left(B\right)$ and

$$n(A_{C\left(0\right)}-\lambda I)+n(B-\lambda I)=d(A_{C\left(0\right)}-\lambda I)+d(B-\lambda I).$$

By Proposition 2, $i(A_{C\left(0\right)}-\lambda I)\le 0$ and $i(B-\lambda I)\le 0$ if (i)’ holds, $i(A_{C\left(0\right)}-\lambda I)=0$ if (ii)’ holds, and $i(A_{C\left(0\right)}-\lambda I)\ge 0$ and $i(B-\lambda I)\ge 0$ if (iii)’ holds. Then $n(A_{C\left(0\right)}-\lambda I)+n(B-\lambda I)=d(A_{C\left(0\right)}-\lambda I)+d(B-\lambda I)$ means that

$$i(A_{C\left(0\right)}-\lambda I)=i(B-\lambda I)=0,$$

and so $\lambda \notin {\sigma }_w\left(A_{C\left(0\right)}\right)\cup {\sigma }_w\left(B\right)$. Note that ${\sigma }_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\subseteq {\sigma }_w\left(A_{C\left(0\right)}\right)\cup {\sigma }_w\left(B\right).$ Then $\lambda \notin {\sigma }_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$, which means that $A_{C\left(0\right)}$ and B have the SVEP at $\lambda$ according to Remark 2(i). By virtue of Lemmas 3 and 1(iii), $\alpha \left(A_{C\left(0\right)}\right)=\delta \left(A_{C\left(0\right)}\right)<\infty$ and $\alpha \left(B\right)=\delta \left(B\right)<\infty .$ Then $\lambda \notin {\sigma }_w\left(A_{C\left(0\right)}\right)\cup {\sigma }_w\left(B\right)$ shows that $\lambda \notin {\sigma }_b\left(A_{C\left(0\right)}\right)\cup {\sigma }_b\left(B\right),$ hence $\lambda \notin {\sigma }_b\left(M_C\right)$ according to Lemma 10(iii). □

Corollary 1.

Let $A\in \mathcal{BCR}\left(H\right),B\in \mathcal{BCR}\left(K\right)$ with $\rho \left(A\right)\ne \varnothing$ and $\rho \left(B\right)\ne \varnothing$. If $\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)$ satisfies Browder’s theorem and at least one of the following conditions holds:

(i) 

A has the SVEP at $\lambda \in {\sigma }_w\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)\backslash {\sigma }_{le}\left(A\right)$ and B has the SVEP at $\mu \in {\sigma }_w\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)\backslash {\sigma }_{re}\left(B\right);$

(ii) 

A has the SVEP at $\lambda \in {\sigma }_w\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)\backslash {\sigma }_{le}\left(A\right)$ and $A^{*}$ has the SVEP at $\mu \in {\overline{\sigma }}_w\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)\backslash {\overline{\sigma }}_{le}\left(A\right);$

(iii) 

$A^{*}$ has the SVEP at $\lambda \in {\overline{\sigma }}_w\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)\backslash {\overline{\sigma }}_{le}\left(A\right)$ and $B^{*}$ has the SVEP at $\mu \in {\overline{\sigma }}_w\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)\backslash {\overline{\sigma }}_{re}\left(B\right),$

then $M_C$ satisfies Browder’s theorem for all $C\in \mathcal{B}(K,H)$.

Corollary 2.

Let $A\in \mathcal{BCR}\left(H\right),B\in \mathcal{BCR}\left(K\right),C\in \mathcal{BCR}(K,H)$ with $\rho \left(A_{C\left(0\right)}\right)\ne \varnothing$, $\rho \left(B\right)\ne \varnothing$, and $A\left(0\right)+C\left(0\right)$ closed. If $A_{C\left(0\right)}$ and B have the SVEP, and $\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ satisfies Browder’s theorem, then $M_C$ satisfies Browder’s theorem.

Corollary 3.

Let $A\in \mathcal{BCR}\left(H\right),B\in \mathcal{BCR}\left(K\right)$ with $\rho \left(A\right)\ne \varnothing$ and $\rho \left(B\right)\ne \varnothing$. If A and B have the SVEP, and $\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)$ satisfies Browder’s theorem, then $M_C$ satisfies Browder’s theorem for all $C\in \mathcal{B}(K,H)$.

Theorem 8.

Let $A\in \mathcal{BCR}\left(H\right),B\in \mathcal{BCR}\left(K\right),C\in \mathcal{BCR}(K,H)$ with $\rho \left(A_{C\left(0\right)}\right)\ne \varnothing$, $\rho \left(B\right)\ne \varnothing$, and $A\left(0\right)+C\left(0\right)$ closed. If $\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ satisfies a-Browder’s theorem and at least one of the following conditions holds:

(i) 

$A_{C\left(0\right)}$ has the SVEP at $\lambda \in {\sigma }_{ea}\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\backslash {\sigma }_{le}\left(A_{C\left(0\right)}\right)$ and ${\left(A_{C\left(0\right)}\right)}^{*}$ has the SVEP at $\mu \in {\overline{\sigma }}_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\backslash {\overline{\sigma }}_{le}\left(A_{C\left(0\right)}\right);$

(ii) 

${\left(A_{C\left(0\right)}\right)}^{*}$ has the SVEP at $\lambda \in {\overline{\sigma }}_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\backslash {\overline{\sigma }}_{le}\left(A_{C\left(0\right)}\right)$ and $B^{*}$ has the SVEP at $\mu \in {\overline{\sigma }}_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\backslash {\overline{\sigma }}_{le}\left(B\right),$

then $M_C$ satisfies a-Browder’s theorem.

Proof. 

Let a-Browder’s theorem hold for $\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$. Then it is easy to see that Browder’s theorem holds for $\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$. From the proof of Theorem 7, we can know that ${\left(A_{C\left(0\right)}\right)}^{*}$ has the SVEP at $\lambda \notin {\overline{\sigma }}_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$, which together with Remark 2(i) shows that hypotheses (i) and (ii) mean, respectively, that

(i)’

$A_{C\left(0\right)}$ has the SVEP at $\lambda \notin {\sigma }_{le}\left(A_{C\left(0\right)}\right)$ and ${\left(A_{C\left(0\right)}\right)}^{*}$ has the SVEP at $\mu \notin {\overline{\sigma }}_{le}\left(A_{C\left(0\right)}\right);$

(ii)’

${\left(A_{C\left(0\right)}\right)}^{*}$ has the SVEP at $\lambda \notin {\overline{\sigma }}_{le}\left(A_{C\left(0\right)}\right)$ and $B^{*}$ has the SVEP at $\mu \notin {\overline{\sigma }}_{le}\left(B\right).$

Clearly, ${\sigma }_{ea}\left(M_C\right)\subseteq {\sigma }_{ab}\left(M_C\right)$. To finish the proof, we need to verify that ${\sigma }_{ab}\left(M_C\right)\subseteq {\sigma }_{ea}\left(M_C\right)$. Let $\lambda \notin {\sigma }_{ea}\left(M_C\right),$ so we have $M_C-\lambda I\in {\mathsf{\Phi }}_{+}^{-}(H\oplus K)$, which implies that

$$\lambda \notin {\sigma }_{le}\left(A_{C\left(0\right)}\right)$$

by Remark 2(ii) and Lemma 8. Similar to the proof of Theorem 7, we have that $i(A_{C\left(0\right)}-\lambda I)=0$ if (i)’ holds and $i(A_{C\left(0\right)}-\lambda I)\ge 0$ and $i(B-\lambda I)\ge 0$ if (ii)’ holds. Let either (i)’or (ii)’ hold, so we know that

$$\lambda \notin {\sigma }_{le}\left(B\right)$$

and $i(A_{C\left(0\right)}-\lambda I)+i(B-\lambda I)\le 0$ by Lemma 8, which means $\lambda \notin {\sigma }_{ea}\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$. By Remark 2(i), $A_{C\left(0\right)}$ and B have the SVEP at $\lambda .$ It follows from Lemma 3 that

$$\alpha (A_{C\left(0\right)}-\lambda I)<\infty ,\alpha (B-\lambda I)<\infty .$$

So $\lambda \notin {\sigma }_{ab}\left(A_{C\left(0\right)}\right)\cup {\sigma }_{ab}\left(B\right)$, which means $\lambda \notin {\sigma }_{ab}\left(M_C\right)$ by Lemma 10(ii). Hence, ${\sigma }_{ab}\left(M_C\right)={\sigma }_{ea}\left(M_C\right),$ i.e., a-Browder’s theorem holds for $M_C$. □

Corollary 4.

Let $A\in \mathcal{BCR}\left(H\right),B\in \mathcal{BCR}\left(K\right)$ with $\rho \left(A\right)\ne \varnothing$ and $\rho \left(B\right)\ne \varnothing$. If $\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)$ satisfies a-Browder’s theorem and at least one of the following conditions holds:

(i) 

A has the SVEP at $\lambda \in {\sigma }_{ea}\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)\backslash {\sigma }_{le}\left(A\right)$ and $A^{*}$ has the SVEP at $\mu \in {\overline{\sigma }}_w\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)\backslash {\overline{\sigma }}_{le}\left(A\right);$

(ii) 

$A^{*}$ has the SVEP at $\lambda \in {\overline{\sigma }}_w\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)\backslash {\overline{\sigma }}_{le}\left(A\right)$ and $B^{*}$ has the SVEP at $\mu \in {\overline{\sigma }}_w\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)\backslash {\overline{\sigma }}_{le}\left(B\right)$,

then $M_C$ satisfies a-Browder’s theorem for all $C\in \mathcal{B}(K,H)$.

Applying Theorems 7 and 8 yields the following result for upper triangular relation matrices.

Theorem 9.

Let $A\in \mathcal{BCR}\left(H\right),B\in \mathcal{BCR}\left(K\right),C\in \mathcal{BCR}(K,H)$ with $\rho \left(A_{C\left(0\right)}\right)\ne \varnothing$, $\rho \left(B\right)\ne \varnothing$, and $A\left(0\right)+C\left(0\right)$ closed. If $\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ satisfies Browder’s (a-Browder’s) theorem, and either ${\left(A_{C\left(0\right)}\right)}^{*}$ and A have the SVEP or ${\left(A_{C\left(0\right)}\right)}^{*}$ and $B^{*}$ have the SVEP, then $M_C$ satisfies Browder’s (a-Browder’s) theorem.

Corollary 5.

Let $A\in \mathcal{BCR}\left(H\right),B\in \mathcal{BCR}\left(K\right)$ with $\rho \left(A\right)\ne \varnothing$ and $\rho \left(B\right)\ne \varnothing$. If $\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)$ satisfies Browder’s (a-Browder’s) theorem, and either $A^{*}$ and A have the SVEP or $A^{*}$ and $B^{*}$ have the SVEP, then $M_C$ satisfies Browder’s (a-Browder’s) theorem for all $C\in \mathcal{B}(K,H)$.

Theorem 10.

Let $A\in \mathcal{BCR}\left(H\right),B\in \mathcal{BCR}\left(K\right),C\in \mathcal{BCR}(K,H)$ with $\rho \left(A_{C\left(0\right)}\right)\ne \varnothing$, $\rho \left(B\right)\ne \varnothing$, and $A\left(0\right)+C\left(0\right)$ closed. If at least one of the hypotheses (i), (ii), and (iii) of Theorem 7 is satisfied, and $\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ satisfies Weyl’s theorem, then $M_C$ satisfies Weyl’s theorem if and only if ${\pi }_0\left(M_C\right)\subseteq p\left(A_{C\left(0\right)}\right)$.

Proof. 

Sufficiency: Assume that $\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ satisfies Weyl’s theorem, so we have that $\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ satisfies Browder’s theorem according to Lemma 11, which together with one of the hypotheses (i)–(iii) of Theorem 7 means that $M_C$ satisfies Browder’s theorem, and ${\sigma }_w\left(M_C\right)={\sigma }_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ by the proof of Theorem 7. We claim that $\sigma \left(M_C\right)=\sigma \left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right).$ Indeed, from the proof of Theorem 7, we have that ${\left(A_{C\left(0\right)}\right)}^{*}$ has the SVEP at $\mu \notin {\overline{\sigma }}_{le}\left(A_{C\left(0\right)}\right)$ or B has the SVEP at $\lambda \notin {\sigma }_{re}\left(B\right)$. Let $\lambda \notin \sigma \left(M_C\right)$. Since $G\left(M_C\right)=G\left(\left(\begin{array}{cc}{A}_{C\left(0\right)}& C_{[\perp ]}\\ 0& B\end{array}\right)\right),$$\lambda \notin {\sigma }_{ap}\left(A_{C\left(0\right)}\right)\cap {\sigma }_{\delta }\left(B\right)$ according to Lemma 7, and hence $A_{C\left(0\right)}-\lambda I$ is injective, $R(A_{C\left(0\right)}-\lambda I)$ is closed, and $B-\lambda I$ is surjective, then

$$\lambda \notin {\sigma }_{le}\left(A_{C\left(0\right)}\right)\cup {\sigma }_{re}\left(B\right).$$

By virtue of Proposition 2, we can obtain that $\lambda \notin \sigma \left(A_{\left(C\right(0\left)\right)}\right)\cup \sigma \left(B\right)$, and then $\lambda \notin \sigma \left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$, which implies that $\sigma \left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\subseteq \sigma \left(M_C\right).$ In addition, $\sigma \left(M_C\right)\subseteq \sigma \left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ is clear. Hence,

$$\sigma \left(M_C\right)=\sigma \left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right).$$

Since $\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ satisfies Weyl’s theorem and ${\sigma }_w\left(M_C\right)={\sigma }_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$, we have

$$\sigma \left(M_C\right)\backslash {\sigma }_w\left(M_C\right)=\sigma \left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\backslash {\sigma }_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)={\pi }_0\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right).$$

Note that $M_C$ satisfies Browder’s theorem. So, applying Proposition 12, we have that $\sigma \left(M_C\right)\backslash {\sigma }_w\left(M_C\right)=\sigma \left(M_C\right)\backslash {\sigma }_b\left(M_C\right)\subseteq {\pi }_0\left(M_C\right)$ and hence ${\pi }_0\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\subseteq {\pi }_0\left(M_C\right).$ To prove the sufficiency, it suffices to show ${\pi }_0\left(M_C\right)\subseteq {\pi }_0\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right).$ Assume that $\lambda \in {\pi }_0\left(M_C\right).$ Then the expression $\sigma \left(M_C\right)=\sigma \left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ indicates $\lambda \in iso\sigma \left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right).$ Evidently, $n(M_C-\lambda I)>0$ means that $n(\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)-\lambda I)>0$, and $n(M_C-\lambda I)<\infty$ means that $n(A_{C\left(0\right)}-\lambda I)<\infty$. Next we prove that $n(B-\lambda I)<\infty .$ On the contrary, let $n(B-\lambda I)=\infty .$ As a relation from $H\oplus N(B-\lambda I)\oplus N{(B-\lambda I)}^{\perp }$ to $H\oplus K$, the relation $\left(\begin{array}{cc}{A}_{C\left(0\right)}-\lambda I& C_{[\perp ]}\\ 0& B-\lambda I\end{array}\right)$ admits the block expression

$$\left(\begin{array}{ccc}{A}_{C\left(0\right)}-\lambda I& {\left(C_{[\perp ]}\right)}_1& {\left(C_{[\perp ]}\right)}_2\\ 0& B-B& {(B-\lambda I)}_1\end{array}\right).$$

There are two cases to consider here.

Case 1: dim$\left(R\left({\left(C_{[\perp ]}\right)}_1\right)\right)<\infty .$ It follows from $n(B-\lambda I)=\infty$ that $n\left({\left(C_{[\perp ]}\right)}_1\right)=\infty ,$ and $\left\{0\right\}\oplus N\left({\left(C_{[\perp ]}\right)}_1\right)\subseteq N\left(\begin{array}{cc}{A}_{C\left(0\right)}-\lambda I& C_{[\perp ]}\\ 0& B-\lambda I\end{array}\right)=N(M_C-\lambda I),$ so $n(M_C-\lambda I)=\infty$, which is a contradiction.

Case 2: dim$\left(R\left({\left(C_{[\perp ]}\right)}_1\right)\right)=\infty .$ Note that $\lambda \in iso\sigma \left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)=iso(\sigma \left(A_{C\left(0\right)}\right)\cup \sigma \left(B\right).$ Then $\lambda \in \rho \left(A_{C\left(0\right)}\right)\cup iso\sigma \left(A_{C\left(0\right)}\right),$ which together with ${\pi }_0\left(M_C\right)\subseteq p\left(A_{C\left(0\right)}\right)$ and $n(A_{C\left(0\right)}-\lambda I)<\infty$ indicates that $\beta (A_{C\left(0\right)}-\lambda I)<\infty .$ Then dim$(R\left({\left(C_{[\perp ]}\right)}_1\right)\cap R(A_{C\left(0\right)}-\lambda I))=\infty$ and so

$$n(M_C-\lambda I)\ge \dim \{(x,y)\in H\oplus K:0\in (A_{C\left(0\right)}-\lambda I)x+{\left(C_{[\perp ]}\right)}_{1}y\}=\infty ,$$

which is a contradiction. Hence, $n(B-\lambda I)<\infty .$ This, together with $n(\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)-\lambda I)>0$ and $n(A_{C\left(0\right)}-\lambda I)<\infty$, shows that $\lambda \in {\pi }_0\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right),$ which means that ${\pi }_0\left(M_C\right)\subseteq {\pi }_0\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right).$ Therefore, ${\pi }_0\left(M_C\right)={\pi }_0\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)=\sigma \left(M_C\right)\backslash {\sigma }_w\left(M_C\right),$ i.e., $M_C$ satisfies Weyl’s theorem.

Necessity: Assume that $M_C$ and $\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ satisfy Weyl’s theorem, and we have from the proof above that

$$\begin{array}{c}{\pi }_0\left(M_C\right)=\sigma \left(M_C\right)\backslash {\sigma }_w\left(M_C\right)=\sigma \left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\backslash {\sigma }_w\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\hfill \\ =\sigma \left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)\backslash {\sigma }_b\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right).\hfill \end{array}$$

Let $\lambda \in {\pi }_0\left(M_C\right),$ then $\alpha (\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)-\lambda I)=\delta (\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)-\lambda I)<\infty$ and $\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)-\lambda I\in \mathsf{\Phi }(H\oplus K)$, so we can easily obtain that $\alpha \left(A_{C\left(0\right)}\right)=\delta \left(A_{C\left(0\right)}\right)<\infty$, i.e., ${\pi }_0\left(M_C\right)\subseteq p\left(A_{C\left(0\right)}\right)$. □

From the proof of Theorem 10, we have the following theorem.

Theorem 11.

Let $A\in \mathcal{BCR}\left(H\right),B\in \mathcal{BCR}\left(K\right),C\in \mathcal{BCR}(K,H)$ with $\rho \left(A_{C\left(0\right)}\right)\ne \varnothing$, $\rho \left(B\right)\ne \varnothing$, and $A\left(0\right)+C\left(0\right)$ closed. If at least one of the hypotheses (i), (ii), and (iii) of Theorem 7 is satisfied, and $\left(\begin{array}{cc}{A}_{C\left(0\right)}& 0\\ 0& B\end{array}\right)$ satisfies Weyl’s theorem, then $M_C$ satisfies Weyl’s theorem if and only if $\delta (A_{C\left(0\right)}-\lambda I)<\infty$ for all $\lambda \in {\pi }_0\left(M_C\right)$.

Corollary 6.

Let $A\in \mathcal{BCR}\left(H\right),B\in \mathcal{BCR}\left(K\right),C\in \mathcal{B}(K,H)$ with $\rho \left(A\right)\ne \varnothing$ and $\rho \left(B\right)\ne \varnothing$. If at least one of the hypotheses (i), (ii), and (iii) of Corollary 1 is satisfied, and $\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)$ satisfies Weyl’s theorem, then $M_C$ satisfies Weyl’s theorem if and only if ${\pi }_0\left(M_C\right)\subseteq p\left(A\right)$.

Corollary 7.

Let $A\in \mathcal{BCR}\left(H\right),B\in \mathcal{BCR}\left(K\right),C\in \mathcal{B}(K,H)$ with $\rho \left(A\right)\ne \varnothing$ and $\rho \left(B\right)\ne \varnothing$. If at least one of the hypotheses (i), (ii), and (iii) of Corollary 1 is satisfied, and $\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)$ satisfies Weyl’s theorem, then $M_C$ satisfies Weyl’s theorem if and only if $\delta (A-\lambda I)<\infty$ for all $\lambda \in {\pi }_0\left(M_C\right)$.

We end this section with an example to illustrate the previous result.

Example 2.

Let $A\in \mathcal{BCR}\left({\ell }^2\right)$, $B\in \mathcal{BCR}\left({\ell }^2\right)$, and $C\in \mathcal{B}\left({\ell }^2\right)$ be given by

$$\begin{array}{c}Ax=(0,x_1,0,{\displaystyle \frac{1}{2}}{x}_2,0,{\displaystyle \frac{1}{3}}{x}_3,0,{\displaystyle \frac{1}{4}}{x}_4,\cdots ),Bx=(0,x_2,0,x_4,0,x_6\cdots ),\hfill \\ Cx=(0,0,x_2,0,x_3,0,x_4,\cdots )\hfill \end{array}$$

for any $x=(x_1,x_2,x_3,\cdots )\in {\ell }^2$, respectively.

Note that $\rho \left(A\right)=\mathbb{C}\backslash \left\{0\right\}\ne \varnothing$, $\rho \left(B\right)=\mathbb{C}\backslash \{0,1\}\ne \varnothing$, A and B have the SVEP. Since $\sigma \left(M_0\right)={\sigma }_w\left(M_0\right)=\{0,1\}$, ${\pi }_0\left(M_0\right)=\varnothing$, $M_0$ satisfies Weyl’s theorem. However, ${\pi }_0\left(M_C\right)=\left\{0\right\}$ and $\delta \left(A\right)=\infty$, so $M_C$ does not satisfy Weyl’s theorem according to Corollary 7. Indeed, $\sigma \left(M_0\right)={\sigma }_w\left(M_0\right)=\{0,1\}$, ${\pi }_0\left(M_0\right)=\left\{0\right\}$, which means that $M_C$ does not satisfy Weyl’s theorem.

5. Conclusions

Certain spectral properties and Weyl-type theorems of upper triangular relation matrix $M_C:=\left(\begin{array}{cc}A& C\\ 0& B\end{array}\right)$ in infinite dimensional complex separable Hilbert spaces are considered in this paper. Specifically, the defect set $({\sigma }_{\star }\left(A_{C\left(0\right)}\right)\cup {\sigma }_{\star }\left(B\right))\backslash {\sigma }_{\star }\left(M_C\right)$ is characterized for given relations $A\in \mathcal{BR}\left(H\right)$, $B\in \mathcal{BR}\left(K\right)$, and $C\in \mathcal{BR}(K,H)$, where ${\sigma }_{\star }$ is chosen from the spectrum, the essential spectrum, the Weyl spectrum, and the Browder spectrum, and we give an example to illustrate the result. Additionally, under certain conditions, we show that relation matrices satisfy Browder- and Weyl-type theorems. Our results extend those of the single-valued operator theory in a Hilbert space by using the properties of the multi-valued parts and the induced quotient map and space decomposition methods. As for future directions, there is a lack of specific examples for some of our main conclusions, and we are going to address this shortcoming in a forthcoming paper.