Solving bi-objective bi-item solid transportation problem with fuzzy stochastic constraints involving normal distribution

1.   Introduction

Let $\mathcal{H}$ and $\mathcal{K}$ be complex infinite-dimensional separable Hilbert spaces, and let $\mathcal{B}(\mathcal{H}, \mathcal{K})$ (resp., $\mathcal{C}(\mathcal{H}, \mathcal{K})$, $\mathcal{C^{+}}(\mathcal{H}, \mathcal{K})$) be the set of all bounded (resp., closed, closable) operators from $\mathcal{H}$ to $\mathcal{K}$. If $\mathcal{K} = \mathcal{H}$, we use $\mathcal{B}(\mathcal{H})$, $\mathcal{C}(\mathcal{H})$, and $\mathcal{C^{+}}(\mathcal{H})$ as usual. The domain of $T\in\mathcal{C}(\mathcal{H})$ is denoted by $\mathcal D(T)$. A closed operator $T$ is said to be left (resp., right) Fredholm if its range $\mathcal R(T)$ is closed and $\alpha(T) < \infty$ (resp., $\beta(T) < \infty$), where $\alpha(T)$ and $\beta(T)$ denote the dimension of the null space $\mathcal N(T)$ and the quotient space $\mathcal H/\mathcal R(T)$, respectively. $T$ is said to be Fredholm if it is both left and right Fredholm. If $T$ is a left or right Fredholm operator, then we define the index of $T$ by $ind(T) = \alpha(T)-\beta(T)$. $T$ is called Weyl if it is a Fredholm operator of index zero. The left essential spectrum, right essential spectrum, essential spectrum, and Weyl spectrum of $T$ are, respectively, defined by

Iterates $T^{2}, T^{3}, \cdots$ of $T\in\mathcal{C}(\mathcal{H})$ are defined by $T^{n}x = T(T^{n-1}x)$ for $x\in\mathcal D(T^{n})$ ($n = 2, 3, \cdots$), where

It is easy to see that

If $n\geq0$, we follow the convention that $T^{0} = I$ (the identity operator on $\mathcal H$, with $\mathcal D(I) = \mathcal H$). Then $\mathcal N(T^{0}) = \{0\}$ and $\mathcal R(T^{0}) = \mathcal H$. It is also well known that if $\mathcal N(T^{k}) = \mathcal N(T^{k+1})$, then $\mathcal N(T^{n}) = \mathcal N(T^{k})$ for $n\geq k$. In this case, the smallest nonnegative integer $k$ such that $\mathcal N(T^{k}) = \mathcal N(T^{k+1})$ is called the ascent of $T$, and is denoted by $asc(T)$. If no such $k$ exists, we define $asc(T) = \infty$. Similarly, if $\mathcal R(T^{k}) = \mathcal R(T^{k+1})$, then $\mathcal R(T^{n}) = \mathcal R(T^{k})$ for $n\geq k$. Thus, we can analogously define the descent of $T$ denoted by $des(T)$, and define $des(T) = \infty$ if $R(T^{n+1})$ is always a proper subset of $R(T^{n})$. We call $T$ left (resp., right) Browder if it is left (resp., right) Fredholm and $asc(T) < \infty$ (resp., $des(T) < \infty$). The left Browder spectrum, right Browder spectrum, and Browder spectrum of $T$ are, respectively, defined by

We write $\sigma(T), \sigma_{ap}(T)$, and $\sigma_{\delta}(T)$ for the spectrum, approximate point spectrum, and defect spectrum of $T$, respectively.

For the spectrum and Browder spectrum, the perturbations of $2\times2$ bounded upper triangular operator matrix $M_{C} = \begin{bmatrix} A & C\\0 & B \\ \end{bmatrix}$, acting on a Hilbert or Banach space, have been studied by numerous authors (see, e.g., ^{[5,8,9,12,13,18]}). However, the characterizations of the unbounded case are still unknown. In this paper, the unbounded operator matrix

is considered, where $A\in \mathcal{C}(H)$, $D\in \mathcal{C}(K)$, and $B\in \mathcal{C}^{+}_{D}(\mathcal{K}, \mathcal{H}) = \{B\in\mathcal{C^{+}}(\mathcal{K}, \mathcal{H}): \mathcal{D}(B)\supset\mathcal{D}(D)\}$. It is obvious that $T_{B}$ is a closed operator matrix. Our main goal is to present some sufficient and necessary conditions for $T_{B}$ to be Browder (resp., invertible) for some closable operator $B\in \mathcal{C}^{+}_{D}(\mathcal{K}, \mathcal{H})$ applying a space decomposition technique. Further, the sufficient and necessary condition, which is completely described by the diagonal operators $A$ and $D$, for

is characterized, where $\sigma_{*}$ is the Browder spectrum (resp., spectrum). This is an extension of the results from ^[3,7,10,12]. In addition, this result is applied to a Hamiltonian operator matrix from elasticity theory.

Definition 1.1. ^[4] Let $T:\mathcal{D}(T)\subset \mathcal{H}\longrightarrow \mathcal{K}$ be a densely defined closed operator. If there is an operator $T^{\dagger}:\mathcal D(T^{\dagger})\subset \mathcal{K}\longrightarrow \mathcal{H}$ such that $\mathcal{D}(T^{\dagger}) = \mathcal{R}(T)\oplus\mathcal{R}(T)^{\perp}, \; \mathcal{N}(T^{\dagger}) = \mathcal{R}(T)^{\perp}$ and

then $T^{\dagger}$ is called the maximal Tseng inverse of $T$.

For the proof of the main results in the next sections, we need the following lemmas:

Lemma 1.1. ^[2] Let $T_{B} = \begin{bmatrix} A & B \\0 & D \\ \end{bmatrix}:\mathcal{D}(A)\oplus \mathcal{D}(D)\subset\mathcal{H}\oplus \mathcal{H}\longrightarrow \mathcal{H} \oplus \mathcal{H}$ be a closed operator matrix such that $A, D$ are closed operators with dense domains and $B$ is a closable operator. Then,

$(1)$ $\sigma_{le}(A)\cup\sigma_{re}(D) \subset \sigma_{e}(T_{B}) \subset\sigma_{e}(A)\cup \sigma_{e}(D).$

$(2)$ $\sigma_{w}(T_{B}) \subset\sigma_{w}(A)\cup \sigma_{w}(D).$

$(3)$ $\sigma_{b}(T_{B}) \subset\sigma_{b}(A)\cup \sigma_{b}(D).$

Lemma 1.2. ^[17] Let $T_{B} = \begin{bmatrix} A & B \\0 & D \\ \end{bmatrix}: \mathcal{D}(A)\oplus \mathcal{D}(D)\subset \mathcal{H}\oplus\mathcal{H}\longrightarrow \mathcal{H} \oplus \mathcal{H}$ be a closed operator matrix such that $A, D$ are closed operators with dense domains and $B$ is a closable operator. Then,

$(1)$ $asc(A)\leq asc(T_{B})\leq asc(A)+asc(D),$

$(2)$ $des(D)\leq des(T_{B})\leq des(A)+des(D).$

Lemma 1.3. ^[16] Let $T\in\mathcal{C}(H)$. Suppose that either $\alpha(T)$ or $\beta(T)$ is finite, and that $asc(T)$ is finite. Then, $\alpha(T)\leq\beta(T)$.

Lemma 1.4. ^[6] Let $T\in\mathcal{C}(H)$. Then, the following inequalities hold for any non negative integer $k$:

$(1)$ $\alpha(T^{k})\leq asc(T_{B})\cdot\alpha(T);$

$(2)$ $\beta(T^{k})\leq des(T_{B})\cdot\beta(T)$.

Lemma 1.5. ^[11] Suppose that the closed operator $T$ is a Fredholm operator and $B$ is $T$-compact. Then,

$(1)$ $T+B$ is a Fredholm operator,

$(2)$ $ind(T+B) = ind(T)$.

Lemma 1.6. ^[16] Let $T\in\mathcal{C}(H)$. Suppose that $asc(T)$ is finite, and that $\alpha(T) = \beta(T) < \infty$. Then, $des(T) = asc(T)$.

2.   Browder spectra of operator matrix

In this section, some sufficient and necessary conditions for $T_{B}$ to be Browder for some closable operator $B$ with $\mathcal D(B)\supset \mathcal D(D)$ are given and the set $\bigcap_{B\in \mathcal{C}^{+}_{D}(\mathcal{K, H})}\sigma_{b}(T_{B})$ is estimated.

Theorem 2.1. Let $A\in\mathcal{C}(\mathcal{H})$ and $D\in\mathcal{C}(\mathcal{K})$ be given operators with dense domains. If $\beta(A) = \infty$, then $T_{B}$ is left Browder for some $B\in\mathcal{C}_{D}^{+}(\mathcal{K}, \mathcal{H})$ if and only if $A$ is left Browder.

Proof. The necessity is obvious by Lemmas 1.1–1.3.

Now we verify the sufficiency. Let $p = asc(A) < \infty$, then $\alpha(A^{p})\leq p\cdot\alpha(A) < \infty$ from Lemma 1.4. Thus, we get $\dim(\mathcal{R}(A)+\mathcal{N}(A^{p}))^{\perp} = \infty$, since $\beta(A) = \infty$. It follows that there are two infinite dimensional subspaces $\Delta_{1}$ and $\Delta_{2}$ of $(\mathcal{R}(A)+\mathcal{N}(A^{p}))^{\perp}$ such that $(\mathcal{R}(A)+\mathcal{N}(A^{p}))^{\perp} = \Delta_{1}\oplus\Delta_{2}$. Define an operator $B:\mathcal{K}\rightarrow \mathcal{H}$ by

where $U:\mathcal{K}\rightarrow \Delta_{1}$ is a unitary operator. Then, we can claim that $T_{B}$ is left Browder.

First we prove that $\mathcal{N}(T_{B}) = \mathcal{N}(A)\oplus \{0\}$, which implies $\alpha(T_{B}) < \infty$. It is enough to verify that $\mathcal{N}(T_{B})\subset\mathcal{N}(A)\oplus \{0\}$. Let $\left(\begin{matrix}x\\y\\\end{matrix}\right)\in \mathcal{N}(T_{B})$, then $Ax+By = 0$ and $Dy = 0$. Thus, $Ax = -By\in \mathcal{R}(A)\cap\Delta_{1} = \{0\}$, and hence $x\in\mathcal{N}(A)$ and $By = 0$. From the definition of $B$, we get $y = 0$, which means $\mathcal{N}(T_{B})\subset\mathcal{N}(A)\oplus \{0\}$.

Next, we check that $\mathcal{R}(T_{B})$ is closed. We only need to check that the range of $\begin{bmatrix} U\\0\\D\\ \end{bmatrix}$ is closed since $\mathcal{R}(A)$ is closed. Suppose that

where $y_{n}\in \mathcal{D}(D)$. Then, $Uy_{n}\rightarrow y _{1}(n \rightarrow \infty)$ and $Dy_{n}\rightarrow y_{3}(n \rightarrow \infty)$. Thus, we have that $y_{n}\rightarrow U^{-1}y_{1}(n\rightarrow \infty)$ from $U$ is a unitary operator. We also obtain $U^{-1}y_{1}\in\mathcal{D}(D)$ and $DU^{-1}y_{1} = y _{3}$ since $D$ is a closed operator. Hence,

which implies $\mathcal{R}(T_{B})$ is closed.

Last, we verify that $\mathcal{N}(T_{B}^{p}) = \mathcal{N}(T_{B_{0}}^{p+1})$, which induces $asc(T_{B}) < \infty$. Let $\left(\begin{matrix}x\\y\\ \end{matrix}\right)\in \mathcal{N}(T_{B}^{p+1})$, then

Thus, $D^{p}y\in \mathcal{N}(D)\cap\mathcal{R}(D^{p})$ and $BD^{p}y\in(\mathcal{R}(A)+\mathcal{N}(A^{p}))^{\perp}\subset\mathcal{R}(A)^{\perp}$. This implies that $A^{p+1}x+A^{p}By+A^{p-1}BDy+\cdot\cdot\cdot+ABD^{p-1}y = -BD^{p}y\in\mathcal{R}(A) \cap(\mathcal{R}(A)+\mathcal{R}(A^{p}))^{\perp}\subset\mathcal{R}(A) \cap\mathcal{R}(A)^{\perp} = \{0\}$, and then

We can obtain $D^{p}y = 0$ by the definition of $B$. According to $A^{p+1}x+A^{p}By+\cdot\cdot\cdot+ABD^{p-1}y = A(A^{p}x+A^{p-1}By+ \cdot\cdot\cdot+BD^{p-1}y) = 0$, we have

Let $x_{1}: = A^{p}x+A^{p-1}By+\cdot\cdot\cdot+BD^{p-1}y$, then

It follows that $A^{p}x+A^{p-1}By+\cdots+ABD^{p-2}y-x_{1}$= $-BD^{p-1}y\in (\mathcal{R}(A)+\mathcal{N}(A))\cap(\mathcal{R}(A)+$$\mathcal{N}(A^{p}))^{\perp}\subset (\mathcal{R}(A)+\mathcal{N}(A^{p}))\cap(\mathcal{R}(A)+\mathcal{N}(A^{p}))^{\perp} = \{0\}$. Then, $A^{p}x+A^{p-1}By+\cdots+ABD^{p-2}y-x_{1} = -BD^{p-1}y = 0,$ i.e.,

Let $x_{2}: = A^{p-1}x+A^{p-2}By+A^{p-1}BDy+\cdot\cdot\cdot+BD^{p-2}y$, then $x_{2}\in\mathcal{N}(A^{2})$, as $Ax_{2} = x_{1}$ and $x_{1}\in\mathcal{N}(A)$. This implies $A^{p-1}x+A^{p-2}By+A^{p-3}BDy+\cdot\cdot\cdot+BD^{p-3}y-x_{2}$$= -BD^{p-2}y\in (\mathcal{R}(A)+\mathcal{N}(A^{2}))$$\cap(\mathcal{R}(A)+\mathcal{N}(A^{p}))^{\perp}\subset (\mathcal{R}(A)+\mathcal{N}(A^{p}))\cap(\mathcal{R}(A)+\mathcal{N}(A^{p}))^{\perp} = \{0\}$, and then

Continuing this process, there is $x_{p}\in\mathcal{N}(A^{p})$ such that

Thus, $x\in\mathcal{N}(A^{p+1}) = \mathcal{N}(A^{p})$. This induces that $\left(\begin{matrix}x\\y\\ \end{matrix}\right)\in \mathcal{N}(T^{p}_{B})$, which means that $\mathcal{N}(T_{B}^{p}) = \mathcal{N}(T_{B}^{p+1})$. □

Theorem 2.2. Let $A\in\mathcal{C}(\mathcal{H})$ and $D\in\mathcal{C}(\mathcal{K})$ be given operators with dense domains. If $\alpha(D) = \infty$, then $T_{B}$ is right Browder for some $B\in\mathcal{C}_{D}^{+}(\mathcal{K}, \mathcal{H})$ if and only if $D$ is right Browder.

Proof. If $T_{B}$ is right Browder for some $B\in\mathcal{C}_{D}^{+}(\mathcal{K}, \mathcal{H})$, then $D$ is right Fredholm and $des(D) < \infty$ by Lemmas 1.1 and 1.2. According to the assumption $\alpha(D) = \infty$, we can obtain $ind(D)\geq0$, which means $D$ is right Browder.

Now, we verify the reverse implication. Denote $q = des(D)$. By Lemma 1.4, we have $\beta(D^{q})\leq q\cdot\beta(D) < \infty$, and then $\mathcal{R}(D^{q})$ is closed. Thus, $\mathcal{K} = \mathcal{R}(D^{q}) \oplus \mathcal{R}(D^{q})^{\perp}$. This includes $\mathcal{N}(D) = [\mathcal{N}(D)\cap\mathcal{R}(D^{q})]\oplus [\mathcal{N}(D)\cap\mathcal{R}(D^{q})^{\perp}]$. According to the assumption $\dim(\mathcal{N}(D)) = \infty$ and $\dim(\mathcal{N}(D)\cap\mathcal{R}(D^{q})^{\perp}) < \infty$, we know that $\dim(\mathcal{N}(D)\cap\mathcal{R}(D^{q})) = \infty$. Then, there exist two infinite dimensional subspaces $\Omega_{1}$ and $\Omega_{2}$ such that $\mathcal{N}(D)\cap\mathcal{R}(D^{q}) = \Omega_{1}\oplus \Omega_{2}$. Define the operator $B_{0}$ by

where $U:\Omega_{1}\rightarrow \mathcal{H}$ is a unitary operator. Then, we obtain that $\mathcal{R}(T_{B}) = \begin{bmatrix} \mathcal{H}\\ \mathcal{R}(D)\\ \end{bmatrix}$ is closed, $\beta(T_{B}) = \beta(D) < \infty$, and $\alpha(T_{B}) = \alpha(A)+\dim\Omega_{2}+\dim(\mathcal{N}(D)\cap\mathcal{R}(D^{q})^{\perp}) = \infty$. Thus, we also get $des(T_{B}) < \infty$ by $des(D) = q < \infty$. This means $T_{B}$ is right Browder. □

Remark 2.1. Let $A\in\mathcal{C}(\mathcal{H})$ and $D\in\mathcal{C}(\mathcal{K})$ be given operators with dense domains. If $T_{B}$ is Browder for some $B\in\mathcal{C}_{D}^{+}(\mathcal{K}, \mathcal{H})$, then $\alpha(D) < \infty$ if and only if $\beta(A) < \infty$. In fact, if $T_{B}$ is Browder for some $B\in\mathcal{C}_{D}^{+}(\mathcal{K}, \mathcal{H})$, then $A$ is left Fredholm with finite ascent and $D$ is right Fredholm with finite descent, by Lemmas 1.1 and 1.2. It follows that $T_{B}$ admits the following representation:

where $D_{1}:\mathcal{D}(D)\rightarrow \mathcal{R}(D)$ is invertible and $A_{1}:\mathcal{D}(A)\rightarrow \mathcal{R}(A_{1})( = \mathcal{R}(A))$ is an operator with closed range. Let $A^{\dagger}_{1}:\mathcal{R}(A)\rightarrow \mathcal{H}$ be the maximal Tseng inverse of $A_{1}$, then $A_{1}A^{\dagger}_{1} = I_{\mathcal{R}(A)}$. Set $P = \begin{bmatrix} I & 0 & -B_{12}D_{1}^{-1} & 0\\0&I & -B_{22}D_{1}^{-1} & 0\\0 & 0&I & 0\\0 & 0 & 0&I\\ \end{bmatrix}$, $Q = \begin{bmatrix} I & -A_{1}^{\dagger}B_{11} & 0\\0&I & 0\\0 & 0&I\\ \end{bmatrix}$, then

Thus, $ind(T_{B}) = ind(T_{B_{21}})$ since $P$ and $Q$ are both injective. Assume that $\alpha(D) < \infty$ (resp., $\beta(A) < \infty$), then $B_{21}$ is a compact operator. By Lemma 1.5, we also obtain $ind(T_{B}) = ind(A)+ind(D) = 0$. Therefore, $\alpha(D) < \infty$ if and only if $\beta(A) < \infty$.

The next result is given under the hypothesis that $\beta(A)$ and $\alpha(D)$ are both finite.

Theorem 2.3. Let $A\in\mathcal{C}(\mathcal{H})$ and $D\in\mathcal{C}(\mathcal{K})$ be given operators with dense domains. Suppose that $\beta(A) < \infty$ and $\alpha(D) < \infty$. Then, $T_{B}$ is Browder for some $B\in\mathcal{C}^{+}_{D}(\mathcal{K}, \mathcal{H})$ if and only if $A$ is left Browder, $D$ is right Browder, and $\alpha(A)+\alpha(D) = \beta(A)+\beta(D)$.

Proof. Necessity. If $T_{B}$ is Browder for some $B\in\mathcal{C}^{+}_{D}(\mathcal{K}, \mathcal{H})$, then $A$ is left Fredholm, $D$ is right Fredholm, $asc(A) < \infty$, and $des(D) < \infty$ by Lemmas 1.1 and 1.2. This means $A$ is left Browder and $D$ is right Browder. We can also get $\alpha(A)+\alpha(D) = \beta(A)+\beta(D)$ by Remark 2.1.

Sufficiency. There are two cases to consider.

Case Ⅰ: Assume that $\alpha(D)\leq\beta(A)$, then we can define a unitary operator $J_{1}:\mathcal{N}(D)\rightarrow \mathcal{M}_{1}$, where $\mathcal{M}_{1}\subset\mathcal{R}(A)^{\perp}$ and $\dim(\mathcal{M}_{1}) = \alpha(D)$. Set

We have $\alpha(T_{B}) = \alpha(A)$ and $\beta(T_{B}) = \beta(D)+\beta(A)-\alpha(D)$. Then, $asc(T_{B}) = asc(A) < \infty$ and $\alpha(T_{B}) = \beta(T_{B}) < \infty$ by the assumption $\alpha(A)+\alpha(D) = \beta(A)+\beta(D)$. Thus, $des(T_{B}) < \infty$ from Lemma 1.6, which means $T_{B}$ is Browder.

Case Ⅱ: Suppose that $\alpha(D) > \beta(A)$. Define a unitary operator $J_{2}:\mathcal{M}_{2}\rightarrow \mathcal{R}(A)^{\perp}$, where $\mathcal{M}_{2}\subset\mathcal{N}(D)$ and $\dim(\mathcal{M}_{2}) = \beta(A)$. Let

We get $\alpha(T_{B}) = \alpha(A)+\alpha(D)-\beta(A)$ and $\beta(T_{B}) = \beta(D)$. Then, $des(T_{B}) = des(D) < \infty$ and $\alpha(T_{B}) = \beta(T_{B}) < \infty$ by the assumption $\alpha(A)+\alpha(D) = \beta(A)+\beta(D)$. Hence, $\alpha(T_{B}^{*}) = \beta(T_{B}^{*}) < \infty$ and $asc(T_{B}^{*}) = des(T_{B}) < \infty$ by [1, Proposition 3.1]. Therefore, we get $asc(T_{B}) = asc(T_{B}^{**}) = des(T_{B}^{*})<\infty$ by Lemma 1.6. This means $T_{B}$ is Browder. □

According to Theorems 2.1–2.3 and Remark 2.1, we obtain the theorem below, which is the main result of this section.

Theorem 2.4. Let $A\in\mathcal{C}(\mathcal{H})$ and $D\in\mathcal{C}(\mathcal{K})$ be given operators with dense domains. Then, $T_{B}$ is Browder for some $B\in\mathcal{C}^{+}_{D}(\mathcal{K}, \mathcal{H})$ if and only if $A$ is left Browder, $D$ is right Browder, and $\alpha(A)+\alpha(D) = \beta(A)+\beta(D)$.

Immediately, we get the following corollary, in which the set $\bigcap_{B\in \mathcal{C}^{+}_{D}(\mathcal{K, H})}\sigma_{b}(T_{B})$ is estimated:

Corollary 2.1. Let $A\in\mathcal{C}(\mathcal{H})$ and $D\in\mathcal{C}(\mathcal{K})$ be given operators with dense domains. Then,

Remark 2.2. Theorem 2.4 and Corollary 2.1 extend the results of bounded case in ^[5,18].

Theorem 2.5. Let $A\in\mathcal{C}(\mathcal{H})$ and $D\in\mathcal{C}(\mathcal{K})$ be given operators with dense domains. Then,

for every $B\in\mathcal{C}^{+}_{D}(\mathcal{K}, \mathcal{H})$, if and only if $\lambda\in\sigma_{asc}(D)\setminus(\sigma_{lb}(A)\cup\sigma_{rb}(D))$ implies $\alpha(A-\lambda I)+\alpha(D-\lambda I)\neq\beta(A-\lambda I)+\beta(D-\lambda I),$ where $\sigma_{asc}(D) = \{\lambda\in \mathbb{C}:asc(D-\lambda I) = \infty\}$.

Proof. Equation (2.1) holds for every $B\in \mathcal{C}^{+}_{D}(\mathcal{K, H})$ if and only if

by [2, Theorem 3.3]. That is,

This induces

□

3.   Spectra of operator matrix

Theorem 3.1. Let $A\in\mathcal{C}(\mathcal{H})$ and $D\in\mathcal{C}(\mathcal{K})$ with dense domains. Then, $T_{B}$ is right invertible for some $B\in\mathcal{C}^{+}_{D}(\mathcal{K}, \mathcal{H})$ if and only if $D$ is right invertible and $\alpha(D)\geq\beta(A)$.

Proof. Necessity. If $T_{B}$ is right invertible for some $B\in\mathcal{C}^{+}_{D}(\mathcal{K}, \mathcal{H})$, then $D$ is right invertible and $T_{B}$ has the following representation:

Obviously, $D_{1}:\mathcal{N}(D)^{\perp}\cap\mathcal{D}(A)\rightarrow \mathcal{K}$ is a bijection. Set $P = \begin{bmatrix} I & 0 & -B_{12}D_{1}^{-1}\\0&I & -B_{22}D_{1}^{-1}\\0 & 0&I\\ \end{bmatrix}$, then

Thus, $\mathcal{R}(\widehat{T}_{B}) = \mathcal{H} \oplus \mathcal{K}$ since $P$ is bijective, and hence $\alpha(D)\geq\beta(A)$.

Sufficiency. There are two cases to consider.

Case Ⅰ: Let $\alpha(D) = \infty$, then we can define a Unitary operator $U:\mathcal{N}(D)\rightarrow \mathcal{H}$. Set

Clearly, $T_{B}$ is right invertible.

Cases Ⅱ: Assume that $\alpha(D) < \infty$, then we can define the right invertible operator $R:\mathcal{N}(D)\rightarrow \mathcal{R}(A)^{\perp}$. Set

Clearly, $T_{B}$ is right invertible. □

From the above theorem and Theorems 5.2.1, 5.2.3 of ^[15], we can obtain the next results immediately.

Theorem 3.2. Let $A\in\mathcal{C}(\mathcal{H})$ and $D\in\mathcal{C}(\mathcal{K})$ with dense domains. Then, $T_{B}$ is invertible for some $B\in\mathcal{C}^{+}_{D}(\mathcal{K}, \mathcal{H})$ if and only if $A$ is left invertible, $D$ is right invertible, and $\alpha(D)\geq\beta(A)$.

Corollary 3.1. Let $A\in\mathcal{C}(\mathcal{H})$ and $D\in\mathcal{C}(\mathcal{K})$ with dense domains. Then,

Remark 3.1. Theorem 3.2 and Corollary 3.1 are also valid for bounded operator matrix $B\in\mathcal{B}(\mathcal{K}, \mathcal{H})$. These conclusions extend the results in ^[8,9,12,13].

From Corollary 3.1, we have the following theorem, which extends the results of ^[3,7,10,12].

Theorem 3.3. Let $A\in\mathcal{C}(\mathcal{H})$ and $D\in\mathcal{C}(\mathcal{K})$ be given operators with dense domains. Then,

for every $B\in \mathcal{C}^{+}_{D}(\mathcal{K, H})$ if and only if

where $\sigma_{p, 1}(D) = \{\lambda\in\mathbb{C}:\mathcal{N}(D-\lambda I)\neq\{0\}, \mathcal{N}(D-\lambda I) = \mathcal{K}\}$ and $\sigma_{r, 1}(A) = \{\lambda\in\mathbb{C}:\mathcal{N}(A-\lambda I) = \{0\}$, $\mathcal{R}(A-\lambda I) = \overline{\mathcal{R}(A-\lambda I)}\neq\mathcal{H}\}$.

Proof. Equation (3.1) holds for every $B\in \mathcal{C}^{+}_{D}(\mathcal{K, H})$ if and only if

by [7, Corollary 2]. That is,

This induces

by Corollary 3.1, $\sigma_{r, 1}(A)\cap\sigma_{ap}(A) = \emptyset$, and $\sigma_{p, 1}(D)\cap\sigma_{\delta}(D) = \emptyset$. □

Applying Theorems 2.5 and 3.3 for the Hamiltonian operator matrix, we obtain the next result.

Theorem 3.4. Let $H = \begin{bmatrix} A&B\\0 & -A^{*}\\\end{bmatrix}: \mathcal{D}(A)\oplus\mathcal{D}(A^{*})\subset\mathcal{H}\oplus\mathcal{K}\rightarrow \mathcal{H}\oplus\mathcal{K}$ be a Hamiltonian operator matrix, where $A$ is a densely defined closed operator and $B$ is a self adjoint operator. Then, the following assertions hold:

$(1)$ For every $C\in \mathcal{C}^{+}_{A^{*}}(\mathcal{K, H})$,

if and only if

$(2)$ For every $C\in \mathcal{C}^{+}_{A^{*}}(\mathcal{K, H})$,

if and only if

In particular, if $\sigma_{r, 1}(A)$ does not include symmetric points about the imaginary axis, then (3.3) holds.

Example 3.1. Consider the plate bending equation in the domain $\{(x, y): 0 < x < 1, 0 < y < 1\}$

with boundary conditions

Set

then the equation can be written as the Hamiltonian system (see ^[19])

and the corresponding Hamiltonian operator matrix is given by

with domain $\mathcal{D}(A)\oplus \mathcal{D}(A^{*})\subset \mathcal{H}\oplus \mathcal{H}$, where $\mathcal H = L_{2}(0, 1)\oplus L_{2}(0, 1)$, $\mathcal A = AC[0, 1]$, and

By a simple calculation, we get $\sigma_{r, 1}(A) = \emptyset$ and $\sigma(A) = \{k\pi:k = \pm1, \pm2, ...\}.$ By Theorem 3.4, we get

On the other hand, we can easily calculate that

4.   Conclusions

Certain spectral and Browder spectral properties of closed operator matrix $T_{B} = \begin{bmatrix}A & B\\ 0 & D \\\end{bmatrix}$ in a Hilbert space are considered in this paper. Specifically, some sufficient and necessary conditions are described under which $T_{B}$ is Browder (resp., invertible) for some closable operator $B$ with $\mathcal D(B)\supset \mathcal D(D)$. Additionally, a sufficient and necessary condition is obtained under which the Browder spectrum (resp., spectrum) of $T_{B}$ coincides with the union of the Browder spectrum (resp., spectrum) of its diagonal entries. These results also hold for bounded operator matrix. As an application, the corresponding properties of the Hamiltonian operator matrix from elasticity theory are given.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

We are grateful to the referees for their valuable comments on this paper.

This work is supported by the Basic Scientific Research Funds of Subordinate Universities of Inner Mongolia (Grant No. ZSLJ202213).

Conflict of interest

No potential conflict of interest was reported by the authors.