近年来，无界分块算子矩阵引发了众多学者的浓厚兴趣与深入探讨^[1-3], 文献[4]研究了无界算子矩阵的可分解性问题. 文献[5]深入探讨了当次对角元素受限时，2×2无界算子矩阵的谱. 文献[6]聚焦力学领域, 具体讨论了其中一类特殊的4×4无界Hamilton算子矩阵, 揭示了其在力学应用中的独特作用.

拟谱凭借其稳定性特质, 已然成为学者们讨论的热点课题, 尤其在分块算子矩阵拟谱的研究方面, 取得了较好的结果^{[8, 9]}. 文献[10]深入剖析了2×2分块算子矩阵的本质拟谱.文献[13]得到了分块算子矩阵的各类本质拟谱与其内部元的关系. 文献[11-12]从扰动理论的独特视角出发, 提出了拟点谱、拟剩余谱和拟连续谱的概念, 并对拟点谱和拟连续谱进行了细致分类. 文献[14]深入探讨了在上三角扰动情形下, 对角分块算子矩阵的精细拟谱和固有拟谱.

近年来，学者们在谱理论的研究上已取得了显著进展, 然而, 对于四类拟点谱与两类拟剩余谱的研究甚少. 众所周知，拟点谱与拟剩余谱的细分研究在拟谱理论中占据重要的地位, 而拟谱是谱的拓展, 那么2×2分块对角无界算子矩阵点谱与剩余谱的精细刻画, 是否可以平行地推广到拟谱情形? 这是值得思考的问题. 文中, 通过具体的实例验证, 答案是否定的. 进一步, 在点谱与剩余谱描绘方法的启迪之下, 本文巧妙利用内部元的深刻信息, 对2×2无界算子矩阵在主对角扰动情形下的四类拟点谱与两类拟剩余谱的分布范围进行了详尽剖析, 并得到了精细的拟谱等式.

在本文的论述中，用$ \ X $来表示Hilbert空间. $ \ T^{\ast } $、$ \ \mathcal{D} \left ( T \right ) $、$ \ \mathcal{R} \left ( T \right ) $和$ \ \mathcal{N} \left ( T \right ) $分别表示$ \ X $中稠定线性算子$ \ T $的共轭算子、定义域、值域和零空间. $ \ \mathcal{B} \left ( X \right ) $表示$ \ X $上所有有界线性算子全体构成的集合, $ \ \mathcal{C} \left ( X \right ) $表示$ \ X $中所有稠定闭线性算子的集合. 尤为特别的是, 当$ \ T $为有界线性算子时, 总假设$ \ \mathcal{D} \left ( T\right )=X $.

定义2.1^[11]  设$ \ T\in \mathcal{C} \left ( X\right ) $.对于任意的$ \ \varepsilon > 0 $, 算子$ \ T $的拟谱$ \ \mathcal{\sigma _{\varepsilon } } \left ( T\right ) $和拟预解集$ \ \mathcal{\rho _{\varepsilon } } \left ( T\right ) $定义如下：

$ \mathcal{\sigma _{\varepsilon } } \left ( T\right )=\bigcup_{\underset{\parallel E\parallel< \varepsilon }{}}^{}\sigma \left (T+E \right )=\left \{ \lambda \in C: \text{存在} E\in \mathcal{B} \left ( X \right ), \parallel E\parallel <\varepsilon, \text{使得}\lambda \in \sigma \left (T+E \right ) \right \}, $

$ \mathcal{\rho _{\varepsilon } } \left ( T\right )=\bigcap _{\underset{\parallel E\parallel< \varepsilon }{}}^{}\rho\left (T+E \right )=\left \{ \lambda \in C: \text{对于任意的} E\in \mathcal{B} \left ( X \right ), \parallel E\parallel < \varepsilon, \text{都有}\lambda \in \rho \left (T+E \right ) \right \}. $

不难发现，有如下等式成立

$ \begin{equation} \notag \mathcal{\sigma _{\varepsilon } } \left ( T\right ) \cup \mathcal{\rho _{\varepsilon } } \left ( T\right )=C. \end{equation} $

定义2.2^[12]   记闭线性算子$ \ T $的拟点谱、拟剩余谱和拟连续谱为$ \mathcal{\sigma} _{\mathcal{\varepsilon}, p} \left ( T\right ) , \ \mathcal{\sigma} _{\mathcal{\varepsilon}, r} \left ( T\right ) $和$ \mathcal{\sigma} _{\mathcal{\varepsilon}, c} \left ( T\right ) $.

令$ \ T\in \mathcal{C} \left ( X\right ), E\in \mathcal{B}\left ( X\right ) $.对于任意的$ \ \varepsilon > 0 $, 定义

$ \mathcal{\sigma} _{\mathcal{\varepsilon}, p} \left ( T\right )=\bigcup_{\underset{\parallel E\parallel< \varepsilon }{}}^{}\sigma_{p } \left (T+E \right )=\left \{ \lambda \in C: \text{存在} E\in \mathcal{B} \left ( X \right ), \parallel E\parallel < \varepsilon, \text{使得}\lambda \in \sigma_{p} \left (T+E \right ) \right \}; $

$ \mathcal{\sigma} _{\mathcal{\varepsilon}, r} \left ( T\right )=\bigcup_{\underset{\parallel E\parallel<\varepsilon }{}}^{}\sigma_{r } \left (T+E \right )=\left \{ \lambda \in C: \text{存在} E\in \mathcal{B} \left ( X \right ), \parallel E\parallel <\varepsilon, \text{使得}\lambda \in \sigma_{r} \left (T+E \right ) \right \}; $

$ \mathcal{\sigma} _{\mathcal{\varepsilon}, c} \left ( T\right )=\bigcup_{\underset{\parallel E\parallel< \varepsilon }{}}^{}\sigma_{c } \left (T+E \right )=\left \{ \lambda \in C: \text{存在} E\in \mathcal{B} \left ( X \right ), \parallel E\parallel <\varepsilon, \text{使得}\lambda \in \sigma_{c} \left (T+E \right ) \right \}. $

显而易见，以下等式是成立的

$ \begin{equation} \notag \mathcal{\sigma _{\varepsilon } } \left ( T\right ) = \mathcal{\sigma} _{\mathcal{\varepsilon}, p} \left ( T\right ) \cup \mathcal{\sigma} _{\mathcal{\varepsilon}, r} \left ( T\right )\cup \mathcal{\sigma} _{\mathcal{\varepsilon}, c} \left ( T\right ). \end{equation} $

在深入探讨值域的稠定性与闭性之后, 可以对$ \ X $中稠定闭线性算子$ \ T $的拟点谱与拟剩余谱进行更为精细的划分.

定义2.3^[13]   闭线性算子$ \ T $的拟点谱细分为四类, 分别用$ \ \mathcal{\sigma} _{\varepsilon, p, {1}} \left ( T\right ), \mathcal{\sigma} _{\varepsilon, p, {2}} \left ( T\right ), \mathcal{\sigma} _{\varepsilon, p, {3}} \left ( T\right ) $和$ \ \mathcal{\sigma} _{\varepsilon, p, {4}} \left ( T\right ) $来表示1-拟点谱、2-拟点谱、3-拟点谱和4-拟点谱. 拟剩余谱细分为两类, 分别用$ \ \mathcal{\sigma} _{\varepsilon, r, {1}} \left ( T\right ) $和$ \ \mathcal{\sigma} _{\varepsilon, r, {2}} \left ( T\right ) $来表示1-拟剩余谱和2-拟剩余谱. 假设$ \ T\in \mathcal{C} \left ( X\right ) $, $ \ \varepsilon > 0 $, 它们分别定义为

$ \begin{equation} \notag \mathcal{\sigma} _{\varepsilon, p, {1}} \left ( T\right )=\{\lambda\in\mathcal{C}:\text{存在}E\in\mathcal{B}\left( X\right ), \parallel E\parallel<\varepsilon, \lambda\in\sigma_{p} \left ( T+E \right ), \mathcal{R}\left ( T + E-\lambda I \right )=X\}; \end{equation} $

$ \begin{equation} \notag \begin{aligned} \mathcal{\sigma} _{\varepsilon, p, {2}} \left ( T\right ) = & \{\lambda\in\mathcal{C}:\text{存在}E\in\mathcal{B}\left( X\right ), \parallel E\parallel<\varepsilon, \lambda\in\sigma_{p} \left ( T+E \right ), \overline{\mathcal{R}\left ( T + E-\lambda I \right )}=X , \\ & \mathcal{R}\left ( T + E-\lambda I \right )\text{不闭} \}; \end{aligned} \end{equation} $

$ \begin{equation} \notag \begin{aligned} \mathcal{\sigma} _{\varepsilon, p, {3}} \left ( T\right )=& \{\lambda\in\mathcal{C}:\text{存在}E\in\mathcal{B}\left( X\right ), \parallel E\parallel<\varepsilon, \lambda\in\sigma_{p} \left ( T+E \right ), \overline{\mathcal{R}\left ( T + E-\lambda I \right )}\ne X , \\ & \mathcal{R}\left ( T + E-\lambda I \right )\text{闭} \}; \end{aligned} \end{equation} $

$ \begin{equation} \notag \begin{aligned} \mathcal{\sigma} _{\varepsilon, p, {4}} \left ( T\right )=&\{\lambda\in\mathcal{C}:\text{存在}E\in\mathcal{B}\left( X\right ), \parallel E\parallel<\varepsilon, \lambda\in\sigma_{p} \left ( T+E \right ), \overline{\mathcal{R}\left ( T + E-\lambda I \right )}\ne X , \\& \mathcal{R}\left ( T + E-\lambda I \right )\text{不闭} \}; \end{aligned} \end{equation} $

$ \begin{equation} \notag \mathcal{\sigma} _{\varepsilon, r, {1}} \left ( T\right )=\{\lambda\in\mathcal{C}:\text{存在}E\in\mathcal{B}\left( X\right ), \parallel E\parallel<\varepsilon, \lambda\in\sigma_{r} \left ( T+E \right ), \mathcal{R}\left ( T + E-\lambda I \right )\text{闭} \}; \end{equation} $

$ \begin{equation} \notag \mathcal{\sigma} _{\varepsilon, r, {2}} \left ( T\right )=\{\lambda\in\mathcal{C}:\text{存在}E\in\mathcal{B}\left( X\right ), \parallel E\parallel<\varepsilon, \lambda\in\sigma_{r} \left ( T+E \right ), \mathcal{R}\left ( T + E-\lambda I \right )\text{不闭} \}. \end{equation} $

定义2.4^[13]  记闭线性算子$ \ T $的拟近似点谱和拟压缩谱为$ \mathcal{\sigma} _{\varepsilon, ap} \left ( T\right ) $和$ \ \mathcal{\sigma} _{\varepsilon, com} \left ( T\right ) $, 具体为$ \sigma_{\varepsilon, ap}(T)=\{\lambda\in\mathbb{C}:\text {存在}E\in\mathcal{B}(X), \|E\|<\varepsilon, T + E-\lambda I\ \text{不是单射或} \ R(T + E-\lambda I) \text{不闭} \}; $ $ \sigma_{\varepsilon, com}(T)=\{\lambda\in\mathbb{C}:\text {存在}E\in\mathcal{B}(X), \|E\|<\varepsilon, \overline{R(T + E-\lambda I) }\ne X\}. $

定理3.1  令$ \ M=\begin{bmatrix} A &0 \\ 0&B\end{bmatrix} $: $ \mathcal{D} (A)\times \mathcal{D} (B)\to X\times X $是$ 2\times2 $分块对角算子矩阵, 则

$ \left (i\right )\; \; \sigma_{\varepsilon, p, 1}(M)\cap \left ( \rho _{\varepsilon } \left ( A \right )\cup \rho _{\varepsilon } \left (B \right ) \right )=\left ( \sigma _{\varepsilon, p, 1 } \left ( A \right )\cap \rho _{\varepsilon } \left (B \right ) \right ) \cup\left ( \rho _{\varepsilon } \left ( A \right )\cap \sigma _{\varepsilon , p, 1 } \left (B \right ) \right ); $

$ \left (ii\right )\; \; \sigma_{\varepsilon, p, 2}(M)\cap \left ( \rho _{\varepsilon } \left ( A \right )\cup \rho _{\varepsilon } \left (B \right ) \right )=\left ( \sigma _{\varepsilon, p, 2 } \left ( A \right )\cap \rho _{\varepsilon } \left (B \right ) \right ) \cup\left ( \rho _{\varepsilon } \left ( A \right )\cap \sigma _{\varepsilon , p, 2 } \left (B \right ) \right ); $

$ \left (iii\right )\; \; \sigma_{\varepsilon, p, 3}(M)\cap \left ( \rho _{\varepsilon } \left ( A \right )\cup \rho _{\varepsilon } \left (B \right ) \right )=\left (\sigma _{\varepsilon, p, 3 } \left ( A \right )\cap \rho _{\varepsilon } \left (B \right ) \right ) \cup\left ( \rho _{\varepsilon } \left ( A \right )\cap \sigma _{\varepsilon , p, 3 } \left (B \right ) \right ). $

证 (ⅰ) 首先证明必要性，假设$ \lambda \in \sigma_{\varepsilon, p, 1}(M)\cap \left ( \rho _{\varepsilon } \left ( A \right )\cup \rho _{\varepsilon } \left (B \right ) \right ) $, 分两种情形讨论.

情形1：当$ \lambda \in \sigma_{\varepsilon, p, 1}(T)\cap \rho _{\varepsilon } \left ( A \right ) $时，由$ \lambda \in \sigma_{\varepsilon, p, 1}(T) $可知，存在扰动算子

$ \begin{equation} \notag E=\begin{bmatrix}E_{1} & 0\\ 0 &E_{2} \end{bmatrix}\in \mathcal{B} \left ( X\oplus X \right ) \text{且} \parallel E\parallel < \varepsilon, \text{有} \end{equation} $

$ \begin{equation} \notag \mathcal{N} \left ( M+E-\lambda I \right )\ne \left \{ 0 \right \} \text{且}\; \mathcal{R} \left ( M+E-\lambda I \right )=X\oplus X , \end{equation} $

根据$ \mathcal{N} \left ( M+E-\lambda I \right )\ne \left \{ 0 \right \} $, 可以得到

$ \begin{equation} \notag \mathcal{N} \left ( A+E_{1} -\lambda I \right )\ne \left \{ 0 \right \} \text{或} \; \mathcal{N} \left ( B+E_{2} -\lambda I \right )\ne \left \{ 0 \right \}. \end{equation} $

又由$ \mathcal{R}(M+E-\lambda I)=\left\{\left(A+E_{1}-\lambda I\right) x \mid x \in \mathcal{D}(A)\right\} \times\left\{\left(B+E_{2}-\lambda I\right) y \mid y \in \mathcal{D}(B)\right\} $知

$ \begin{equation} \notag \mathcal{R} \left ( A+E_{1} -\lambda I \right )=X \text{且}\; \mathcal{R} \left ( B+E_{2} -\lambda I \right )=X. \end{equation} $

根据$ \lambda \in \rho_{\varepsilon }(A) $可知, 对任意的$ \ E_{1} \in \mathcal{B}\left ( X \right ) $, $ \parallel E_{1} \parallel < \varepsilon $都有

$ \begin{equation} \notag \mathcal{N} \left ( A+E_{1} -\lambda I \right )= \left \{ 0 \right \} \text{且}\; \mathcal{R} \left ( A+E_{1} -\lambda I \right )= X, \end{equation} $

那么

$ \begin{equation} \notag \mathcal{N} \left ( B+E_{2} -\lambda I \right )\ne \left \{ 0 \right \}. \end{equation} $

于是存在算子$ \ E_{2} \in \mathcal{B} \left ( X \right ) $, $ \parallel E_{2}\parallel < \varepsilon $, 使得$ \lambda \in \sigma _{p, 1} \left ( B+E_{2} \right ) $，即$ \lambda \in \sigma _{\varepsilon , p, 1} \left ( B \right ) $. 由$ \lambda \in \rho_{\varepsilon }(A) $，得到$ \lambda \in \rho _{\varepsilon } \left ( A \right )\cap\sigma_{\varepsilon, p, 1}(B) $.

情形2：当$ \lambda \in \sigma_{\varepsilon, p, 1}(T)\cap \rho _{\varepsilon } \left ( B\right ) $时，同理可证$ \lambda \in \rho _{\varepsilon } \left ( B \right )\cap\sigma_{\varepsilon, p, 1}(A) $.

因此

$ \begin{equation} \notag \lambda \in \left ( \rho _{\varepsilon } \left (A \right )\cap\sigma _{\varepsilon, p, 1 } \left ( B \right ) \right ) \cup\left ( \rho _{\varepsilon } \left ( B \right )\cap \sigma_{\varepsilon , p, 1 } \left (A \right ) \right ). \end{equation} $

下面证明充分性: 假设$ \lambda \in \left ( \sigma _{\varepsilon, p, 1 } \left ( A \right )\cap \rho _{\varepsilon } \left (B \right ) \right ) \cup\left ( \rho _{\varepsilon } \left ( A \right )\cap \sigma _{\varepsilon , p, 1 } \left (B \right ) \right ) $,

当$ \lambda \in \left ( \sigma _{\varepsilon, p, 1 } \left ( A \right )\cap \rho _{\varepsilon } \left (B \right ) \right ) $时, 由$ \lambda \in \sigma _{\varepsilon, p, 1 } \left ( A \right ) $知, 存在$ \ E_{3} \in\mathcal{B}(X), \|E_{3} \|<\varepsilon $, 使得$ \lambda \in \sigma _{P, 1} \left ( A+E_{3} \right ) $即

$ \begin{equation} \notag \mathcal{N} \left ( A+E_{3} -\lambda I \right )\ne \left \{ 0 \right \} \text{且}\; \mathcal{R} \left ( A+E_{3} -\lambda I \right )= X. \end{equation} $

由$ \lambda \in \rho _{\varepsilon } \left ( B\right ) $可知，对任意的算子$ \ E_{4} \in\mathcal{B}(X), \|E_{4} \|<\varepsilon $, 有

$ \begin{equation} \notag \mathcal{N} \left ( B+E_{4} -\lambda I \right )= \left \{ 0 \right \} \text{且}\; \mathcal{R} \left ( B+E_{4} -\lambda I \right )= X. \end{equation} $

令$ \ E=\begin{bmatrix}E_{3} & 0\\ 0 &E_{4} \end{bmatrix}\in \mathcal{B} \left ( X \times X \right ), \parallel E\parallel=\max\{{\parallel E_{3}\parallel, \parallel E_{4}\parallel}\}<\varepsilon, $有

$ \begin{equation} \notag \mathcal{N} \left ( M+E-\lambda I \right )\ne \left \{ 0 \right \} \text{且}\; \mathcal{R} \left ( M+E-\lambda I \right )=X \times X , \end{equation} $

可得$ \ \lambda\in\sigma_{p, 1} \left ( M+E \right ), $根据定义4可得$ \ \lambda\in\mathcal{\sigma} _{\varepsilon, p, {1}} \left ( M \right ). $又由$ \ \lambda\in\mathcal{\rho _{\varepsilon } } \left ( B\right ), $故

$ \begin{equation} \notag \lambda\in\sigma_{\varepsilon, p, 1}(T)\cap \left ( \rho _{\varepsilon } \left ( A \right )\cup \rho _{\varepsilon } \left (B \right ) \right ). \end{equation} $

当$ \ \lambda\in\mathcal{\rho _{\varepsilon } } \left ( A\right )\cap \sigma_{\varepsilon, p, 1}(B) $时, 同理可得$ \ \lambda\in\sigma_{\varepsilon, p, 1}(T)\cap \left ( \rho _{\varepsilon } \left ( A \right )\cup \rho _{\varepsilon } \left (B \right ) \right ). $

综上所述,

$ \begin{equation} \notag \sigma_{\varepsilon, p, 1}(M)\cap \left ( \rho _{\varepsilon } \left ( A \right )\cup \rho _{\varepsilon } \left (B \right ) \right )=\left (\sigma _{\varepsilon, p, 1 } \left ( A \right )\cap \rho _{\varepsilon } \left (B \right ) \right ) \cup\left ( \rho _{\varepsilon } \left ( A \right )\cap \sigma _{\varepsilon , p, 1 } \left (B \right ) \right ). \end{equation} $

(ⅱ)和(ⅲ)的证明与(ⅰ)完全类似, 此处从略.

注3.1  在本文定理证明过程当中, 为了书写方便, 定理之间扰动算子$ \ E_{1}, E_{2}, E_{3} $和$ \ E_{4} $的取法不一致.

定理3.2  令$ \ M=\begin{bmatrix} A &0 \\ 0&B\end{bmatrix} $: $ \mathcal{D} (A)\times \mathcal{D} (B)\to X\times X $是$ 2\times2 $分块对角算子矩阵, 则

$ \begin{equation} \notag \sigma_{\varepsilon, p, 4}(T)= {{\Delta }_{41}}\cup {{\Delta }_{42}}\cup {{\Delta }_{43}}\cup {{\Delta }_{44}}, \end{equation} $

其中

$ \begin{equation} \notag {{\Delta }_{41}} = \sigma_{\varepsilon, p, 4}(A)\cup \sigma_{\varepsilon, p, 4}(B); \end{equation} $

$ \begin{equation} \notag {{\Delta }_{42}}={(\sigma_{\varepsilon, p, 1}(A)\cap \sigma_{\varepsilon, r, 2}(B))}\cup{(\sigma_{\varepsilon, p, 1}(B)\cap \sigma_{\varepsilon, r, 2}(A))}; \end{equation} $

$ \begin{equation} \notag {{\Delta }_{43}}={(\sigma_{\varepsilon, p, 2}(A)\cap \sigma_{\varepsilon, com}(B))}\cup{(\sigma_{\varepsilon, p, 2}(B)\cap \sigma_{\varepsilon, com}(A))}; \end{equation} $

$ \begin{equation} \notag {{\Delta }_{44}}={(\sigma_{\varepsilon, p, 3}(A)\cap \sigma_{\varepsilon, cr}(B))}\cup{(\sigma_{\varepsilon, p, 3}(B)\cap \sigma_{\varepsilon, cr}(A))}. \end{equation} $

其中

$ \begin{equation} \notag \sigma_{\varepsilon, cr}(T)=\{\lambda\in\mathbb{C}:\text {存在}E\in\mathcal{B}(X), \|E\|<\varepsilon, \text {使得}{R(T + E-\lambda I) }\text {不闭}\}. \end{equation} $

证  首先证明充分性, 即$ \ {{\Delta }_{41}}\cup {{\Delta }_{42}}\cup {{\Delta }_{43}}\cup {{\Delta }_{44}}\subseteq {{\sigma }_{\varepsilon, P, 4}}(M). $

(ⅰ) 假设$ \ \lambda \in{{\Delta }_{41}}, $即$ \ \lambda \in\sigma_{\varepsilon, p, 4}(A)\cup \sigma_{\varepsilon, p, 4}(B) $, 根据定义3可知, 存在算子$ {{E}_{1}}\in \mathcal{B}(X) $, $ {{E}_{\text{2}}}\in \mathcal{B}(X) $. 且$ \| {{E}_{1}}\| <\varepsilon, \| {{E}_{2}}\| <\varepsilon $, 有

$ \begin{equation*} \lambda \in {{\sigma }_{p, 4}}(A+{{E}_{1}})\cup {{\sigma }_{p, 4}}(B+{{E}_{2}}), \end{equation*} $

即$ A+{{E}_{1}}-\lambda I $不是单射, $ \overline{\mathsf{\mathcal{R}}(A\text{+}{{E}_{1}}-\lambda I)}\ne X $, 且$ \mathsf{\mathcal{R}}(A+{{E}_{1}}-\lambda I) $不闭, $ B+{{E}_{2}}-\lambda I $不是单射, $ \overline{\mathsf{\mathcal{R}}(B+{{E}_{2}}-\lambda I)}\ne X, $且$ \mathsf{\mathcal{R}}(B+{{E}_{2}}-\lambda I) $不是闭集.

令$ E\text{=}\left[ \begin{matrix} {{E}_{1}} & 0 \\ 0 & {{E}_{2}} \\ \end{matrix} \right]\in \mathcal{B}(X\times X) $, $ \| E\| =\max \{\| {{E}_{1}}\|, \| {{E}_{2}}\| \}<\varepsilon $, 得到$ M+E-\lambda I $不是单射, $ \overline{\mathsf{\mathcal{R}}(M+E-\lambda I)}\ne X, $且$ \mathsf{\mathcal{R}}(M+E-\lambda I) $不是闭集, 故$ \lambda \in {{\sigma }_{p, 4}}(M+E), $则$ \lambda \in {{\sigma }_{\varepsilon, p, 4}}(M) $.

(ⅱ) 下面证明$ \ {{\Delta }_{42}}\subseteq {{\sigma }_{\varepsilon, P, 4}}(M). $假设$ \lambda \in ({{\sigma }_{\varepsilon, p, 1}}(A)\cap {{\sigma }_{\varepsilon, r, 2}}(B)) $, 根据定义3可知, 存在算子$ {{E}_{3}}\in \mathcal{B}(X) $, $ {{E}_{\text{4}}}\in \mathcal{B}(X) $. 且$ \| {{E}_{3}}\| <\varepsilon, \| {{E}_{4}}\| <\varepsilon $, 令$ \widehat{E}\text{=}\left[ \begin{matrix} {{E}_{3}} & 0 \\ 0 & {{E}_{4}} \\ \end{matrix} \right]\in \mathcal{B}(X\times X) $, $ \| \widehat{E}\| =\max \{\| {{E}_{3}}\|, \| {{E}_{4}}\| \}<\varepsilon $, 得到$ M+\widehat{E}-\lambda I $不是单射, $ \overline{\mathsf{\mathcal{R}}(M+\widehat{E}-\lambda I)}\ne X $, 且$ \mathsf{\mathcal{R}}(M+\widehat{E}-\lambda I) $不是闭集, 故$ \lambda \in {{\sigma }_{p, 4}}(M+\widehat{E}) $, 即$ \lambda \in {{\sigma }_{\varepsilon, p, 4}}(M) $.

当$ \lambda \in ({{\sigma }_{\varepsilon, p, 1}}(B)\cap {{\sigma }_{\varepsilon, r, 2}}(A)) $时, 同理可证$ \lambda \in {{\sigma }_{\varepsilon, p, 4}}(M) $.

(ⅲ) 同理可证$ \ {{\Delta }_{43}}\subseteq {{\sigma }_{\varepsilon, p, 4}}(M), {{\Delta }_{44}}\subseteq {{\sigma }_{\varepsilon, p, 4}}(M). $

因此

$ \begin{equation*} {{\Delta }_{41}}\cup {{\Delta }_{42}}\cup {{\Delta }_{43}}\cup {{\Delta }_{44}}\subseteq {{\sigma }_{\varepsilon, p, 4}}(M). \end{equation*} $

下面证明必要性, 即$ \ {{\sigma }_{\varepsilon, p, 4}}(M)\subseteq{{\Delta }_{41}}\cup {{\Delta }_{42}}\cup {{\Delta }_{43}}\cup {{\Delta }_{44}} . $

假设$ \ \lambda \in{{\sigma }_{\varepsilon, p, 4}}(M) $, 根据$ \ {{\sigma }_{\varepsilon, p, 4}}(M)\subseteq{{\sigma }_{ p}}(M) $可知$ \ \lambda \in{{\sigma }_{p, 4}}(M) , $则存在$ \ \widetilde{E}\text{=}\left[ \begin{matrix} {{E}_{5}} & 0 \\ 0 & {{E}_{6}} \\ \end{matrix} \right]\in \mathcal{B}(X \times X) $, 且$ \| \widehat{E}\| <\varepsilon $, $ \ \lambda \in{{\sigma }_{p}}(M+\widetilde{E}) $, 即$ \ \lambda \in{{\sigma }_{p}}(A+{E}_{5})\cup{{\sigma }_{p}}(B+{E}_{6}) $, 进一步，得到以下四种情形:

情形1: 当$ \lambda \in {{\sigma }_{p, 4}}(A+{{E}_{5}}) $或$ \lambda \in {{\sigma }_{p, 4}}(B+{{E}_{6}}) $时, 即存在$ {{E}_{5}}\in \mathcal{B}(X) $, $ {{E}_{6}}\in \mathcal{B}(X) $, 且$ \| {{E}_{5}}\| <\varepsilon, \| {{E}_{6}}\| <\varepsilon $, 有

$ \begin{equation*} \lambda \in {{\sigma }_{\varepsilon, p, 4}}(A)\cup {{\sigma }_{\varepsilon, p, 4}}(B), \end{equation*} $

因此$ \lambda \in {{\Delta }_{41. }} $

情形2: 当$ \lambda \in {{\sigma }_{p, 1}}(A+{{E}_{5}}) $, 即$ \lambda \in {{\sigma }_{\varepsilon, p, 1}}(A) $时, 由于$ \overline{\mathsf{\mathcal{R}}(M+\widetilde{E}-\lambda I)}\ne X $且$ \mathsf{\mathcal{R}}(M+\widetilde{E}-\lambda I) $不是闭集, 故$ \overline{\mathsf{\mathcal{R}}(B+{{E}_{6}}-\lambda I)}\ne X $且$ \mathsf{\mathcal{R}}(B+{{E}_{6}}-\lambda I) $不是闭集. 当$ B+{{E}_{6}}-\lambda I $是单射时, 有$ \lambda \in {{\sigma }_{r, 2}}(B+{{E}_{6}}) $, 进而$ \lambda \in {{\sigma }_{\varepsilon, r, 2}}(B) $, 当$ B+{{E}_{6}}-\lambda I $不是单射时, $ \lambda \in {{\sigma }_{p, 4}}(B+{{E}_{6}}) $, 进而$ \lambda \in {{\sigma }_{\varepsilon, p, 4}}(B) $.

因此,

$ \begin{equation*} \lambda \in {{\sigma }_{\varepsilon, p, 1}}(A)\cap ({{\sigma }_{\varepsilon, r, 2}}(B)\cup {{\sigma }_{\varepsilon, p, 4}}(B)). \end{equation*} $

同理可证当$ \lambda \in {{\sigma }_{p, 1}}(B+{{E}_{6}}) $, 即$ \lambda \in {{\sigma }_{\varepsilon, p, 1}}(B) $时可以得到

$ \begin{equation*} \lambda \in {{\sigma }_{\varepsilon, p, 1}}(B)\cap ({{\sigma }_{\varepsilon, r, 2}}(A)\cup {{\sigma }_{\varepsilon, p, 4}}(A)). \end{equation*} $

因此当$ \lambda \in ({{\sigma }_{\varepsilon, p, 1}}(A)\cap {{\sigma }_{\varepsilon, r, 2}}(B))\cup ({{\sigma }_{\varepsilon, p, 1}}(B)\cap {{\sigma }_{\varepsilon, r, 2}}(A)) $时, 可以得到$ \lambda \in {{\Delta }_{42}} $, 当$ \lambda \in ({{\sigma }_{\varepsilon, p, 1}}(A)\cap {{\sigma }_{\varepsilon, p, 4}}(B))\cup ({{\sigma }_{\varepsilon, p, 1}}(B)\cap {{\sigma }_{\varepsilon, p, 4}}(A)), $此时$ \lambda \in {{\Delta }_{41. }} $

情形3:同理可证当$ \lambda \in {{\sigma }_{p, 2}}(A+{{E}_{5}})\cup {{\sigma }_{p, 2}}(B+{{E}_{6}}) $时$ \lambda \in {{\Delta }_{43}} $.

情形4:同理可证当$ \lambda \in {{\sigma }_{p, 3}}(A+{{E}_{5}})\cup {{\sigma }_{p, 3}}(B+{{E}_{6}}) $, 此时$ \lambda \in {{\Delta }_{44}} $.

因此

$ \begin{equation*} {{\sigma }_{\varepsilon, p, 4}}(M)\subseteq{{\Delta }_{41}}\cup {{\Delta }_{42}}\cup {{\Delta }_{43}}\cup {{\Delta }_{44}} . \end{equation*} $

综上所述

$ \begin{equation*} {{\sigma }_{\varepsilon, p, 4}}(M) = {{\Delta }_{41}}\cup {{\Delta }_{42}}\cup {{\Delta }_{43}}\cup {{\Delta }_{44}}. \end{equation*} $

对角算子矩阵剩余谱的细分，有如下结论^[7].

命题4.1  令$ \ M=\begin{bmatrix} A &0 \\ 0&B\end{bmatrix} $: $ \mathcal{D} (A)\times \mathcal{D} (B)\to X\times X $是$ 2\times2 $分块对角算子矩阵,

$ \begin{equation} \notag (i)\; \; \sigma_{r, 1}(M)=(\sigma_{r, 1}(A)\setminus \sigma_{ap}(B))\cup(\sigma_{r, 1}(A)\setminus\sigma_{ap}(B)), \; \; (ii)\; \; \sigma_{r, 2}(M)={{\Delta }_{1}}\cup{{\Delta }_{2}}, \end{equation} $

其中

$ \begin{equation} \notag {{\Delta }_{1}}=(\sigma_{r, 1}(A)\cap \sigma_{c}(B))\cup(\sigma_{r, 2}(A)\setminus\sigma_{p}(B)), {{\Delta }_{2}}=(\sigma_{r, 1}(B)\cap \sigma_{c}(A))\cup(\sigma_{r, 2}(B)\setminus\sigma_{p}(A)). \end{equation} $

关于两类拟剩余谱的探讨, 若将命题4.1直接迁移到拟谱的情形, 其包含关系$ \ ({\sigma_{\varepsilon, r, 1}}(A)\setminus {\sigma_{\varepsilon, ap}}(B))\cup({\sigma_{\varepsilon, r, 1}}(A)\setminus {\sigma_{\varepsilon, ap}(B)})\subseteq{\sigma_{\varepsilon, r, 1}}(M) $成立, 然而，反包含关系在此并不成立, 我们可以通过以下实例来加以验证.

例4.1  设$ \ X=l^{2}(0, \infty), $对于任意的$ \ x\in l^{2}, x=(x_{1}, x_{2}, x_{3}\ldots) $, 令$ Ax=(0, x_{1}, x_{2}, x_{3}\ldots), $且

$ \begin{equation*} T\text{=}\left[ \begin{matrix} {A} & 0 \\ 0 & {A} \\ \end{matrix} \right]. \end{equation*} $

取有界扰动算子$ {{E}_{1}} $

$ \begin{equation*} {{E}_{1}}x=(\frac{\varepsilon }{2}{{x}_{1}}, 0, 0, 0\cdots ), \end{equation*} $

经过计算$ \| {{E}_{1}}\| <\varepsilon $, $ (A+{{E}_{1}}-\frac{1}{2}I)x=(\frac{\varepsilon }{2}{{x}_{1}}, {{x}_{1}}-\frac{1}{2}{{x}_{2}}, {{x}_{2}}-\frac{1}{2}{{x}_{3}}\cdots ), $令$ (A+{{E}_{1}}-\frac{1}{2}I)x=0 $, 显然$ x=0 $, 则$ A+{{E}_{1}}-\frac{1}{2}I $是单射.

$ \begin{equation*} ({{A}^{*}}+{{{E}_{1}}^*}-\frac{1}{2}I)x=({{x}_{2}}+\frac{\varepsilon -1}{2}{{x}_{1}}, {{x}_{3}}-\frac{1}{2}{{x}_{2}}, {{x}_{4}}-\frac{1}{2}{{x}_{3}}\cdots ), \end{equation*} $

令$ ({{A}^{*}}+{{{E}_{1}}^*}-\frac{1}{2}I)x=0 $, $ {{x}_{1}}=c(c\ne 0) $, 可得$ x=(c, -\frac{\varepsilon -1}{2}c, -\frac{\varepsilon -1}{4}c, -\frac{\varepsilon -1}{8}c\cdots )\in {{l}^{2}} $, 即$ \mathsf{\mathcal{N}}({{{E}_{1}}^*}-\frac{1}{2}I)\ne 0 $, 由$ \mathsf{\mathcal{N}}({{A}^{*}}+{{E}_{1}}*-\frac{1}{2}I)={{\overline{\mathsf{\mathcal{R}}(A+{{E}_{1}}-\frac{1}{2}I)}}^{\bot }} $可知

$ \begin{equation*} \overline{\mathsf{\mathcal{R}}(A+{{E}_{1}}-\frac{1}{2}I)}\ne X. \end{equation*} $

下面证明$ \mathsf{\mathcal{R}}(A+{{E}_{1}}-\frac{1}{2}I) $是闭集. 事实上, 对于任意序列$ \{{{y}_{n}}\}\subset R(A+{{E}_{1}}-\frac{1}{2}I) $, 令

$ \begin{equation*} {{y}_{n}}=\{{{y}_{n1, }}{{y}_{n2}}, {{y}_{n3}}\cdots \}, {{y}_{0}}=\{{{y}_{01, }}{{y}_{02}}, {{y}_{03}}\cdots \}, {{y}_{0}}\in {{l}^{2}}, \end{equation*} $

且$ {{y}_{n}}\to {{y}_{0}}(n\to \infty ) $, $ {{x}_{n}}=\{{{x}_{n1, }}{{x}_{n2}}, {{x}_{n3}}\cdots \}, {{x}_{0}}=\{{{x}_{01, }}{{x}_{02}}, {{x}_{03}}\cdots \}. $令$ (A+{{E}_{1}}-\frac{1}{2}I){{x}_{n}}={{y}_{n}} $, 取$ {{x}_{n}}=\{\frac{2}{\varepsilon -1}{{y}_{n1, }}2{{y}_{n2}}, 2{{y}_{n3}}\cdots \} $, 经验证$ {{x}_{n}}\in {{l}^{2}} $.

根据$ {{y}_{n}}\to {{y}_{0}}(n\to \infty ) $, 可以得到$ {{x}_{n}}\to {{x}_{0}}(n\to \infty ) $, 从而

$ \begin{equation*} {{x}_{0}}=\{\frac{2}{\varepsilon -1}{{y}_{01, }}2{{y}_{02}}, 2{{y}_{03}}\cdots \}\in {{l}^{2}}, \end{equation*} $

且$ (A+{{E}_{1}}-\frac{1}{2}I){{x}_{0}}={{y}_{0}} $, 即$ \mathsf{\mathcal{R}}(A+{{E}_{1}}-\frac{1}{2}I) $是闭集.

经过以上的分析可知, 存在$ {{E}_{1}}\in \mathcal{B}(X) $, $ \| {{E}_{1}}\| <\varepsilon, $ $ \frac{1}{2}\in {{\sigma }_{r, 1}}(A+{{E}_{1}}) $, 即$ \frac{1}{2}\in {{\sigma }_{\varepsilon, r, 1}}(A) $.

下面证明$ \lambda \in {{\sigma }_{\varepsilon, ap}}(A) $, 事实上, 取有界扰动算子$ {{E}_{2}} $

$ \begin{equation*} {{E}_{2}}x=(-\frac{1}{2}{{x}_{1}}, -{{x}_{1}}, 0, 0\cdots ), \end{equation*} $

经过计算$ \| {{E}_{2}}\| <\frac{\sqrt{5}}{2} $, $ (A+{{E}_{2}}-\frac{1}{2}I)x=(0, -\frac{1}{2}{{x}_{2}}, -{{x}_{2}}-\frac{1}{2}{{x}_{3}}, -{{x}_{3}}-\frac{1}{2}{{x}_{4}}\cdots ), $显然$ A+{{E}_{2}}-\frac{1}{2}I $不是单射, 故当$ \varepsilon >\frac{\sqrt{5}}{2} $时, 根据定义5可知, $ \frac{1}{2}\in {{\sigma }_{\varepsilon, ap}}(A) $.

因此, 当$ \varepsilon >\frac{\sqrt{5}}{2} $时, $ \frac{1}{2}\in ({{\sigma }_{\varepsilon, r, 1}}(A)\cap {{\sigma }_{\varepsilon, ap}}(A)). $

令

$ \begin{equation*} E=\left[ \begin{matrix} {{E}_{1}} & 0 \\ 0 & {{E}_{1}} \\ \end{matrix} \right]\in \mathcal{B}(X\oplus X), \end{equation*} $

易知$ \| E\| =\max \{\| {{E}_{1}}\|, \| {{E}_{3}}\| \}<\varepsilon, $得到$ T+E-\frac{1}{2}I $是单射, $ \overline{R(T+E-\frac{1}{2}I)}\ne X $, 且$ R(T+E-\frac{1}{2}I) $是闭集, 故$ \frac{1}{2}\in {{\sigma }_{\varepsilon, r, 1}}(T) $.

综上所述

$ \begin{equation*} {\sigma_{\varepsilon, r, 1}}(T)\nsubseteq({\sigma_{\varepsilon, r, 1}}(A)\setminus {\sigma_{\varepsilon, ap}}(B))\cup({\sigma_{\varepsilon, r, 1}}(B)\setminus {\sigma_{\varepsilon, ap}}(A)). \end{equation*} $

经过细致的剖析，我们得到如下结论.

定理4.1   令$ \ M=\begin{bmatrix} A &0 \\ 0&B\end{bmatrix} $: $ \mathcal{D} (A)\times \mathcal{D} (B)\to X\times X $是$ 2\times2 $分块对角算子矩阵, 则

$ \begin{equation*} \left (i\right )\sigma_{\varepsilon, r, 1}(M)\cap \left ( \rho _{\varepsilon } \left ( A \right )\cup \rho _{\varepsilon } \left (B \right ) \right )=\left ( \sigma _{\varepsilon, r, 1 } \left ( B \right )\cap \rho _{\varepsilon } \left (A \right ) \right ) \cup\left ( \rho _{\varepsilon } \left ( B \right )\cap \sigma _{\varepsilon , r, 1 } \left (A \right ) \right ); \end{equation*} $

$ \begin{equation*} \left (ii\right )\sigma_{\varepsilon, r, 2}(M)\cap \left ( \rho _{\varepsilon } \left ( A \right )\cup \rho _{\varepsilon } \left (B \right ) \right )=\left ( \sigma _{\varepsilon, r, 2 } \left ( B \right )\cap \rho _{\varepsilon } \left (A \right ) \right ) \cup\left ( \rho _{\varepsilon } \left ( B \right )\cap \sigma _{\varepsilon , r, 2 } \left (A \right ) \right ). \end{equation*} $

证  与定理3.1的证明类似, 此处从略.

本文聚焦2×2无界对角算子矩阵, 在对角的形式扰动下, 对其拟谱进行了精细的刻画. 通过运用对角元素所蕴含的信息，成功得到一系列拟谱等式, 我们将其与谱的相关结论进行了比较, 同时辅以具体的实例进行了详尽的阐释.

在深入探讨无界对角算子矩阵的拟谱特性时, 我们不难发现, 这一领域的精细刻画充满了复杂性, 尤其当扰动算子的形式复杂时, 更需要进一步研究. 充分挖掘算子的特性, 开展分类研究, 比如进一步探讨正规算子, 亚正规算子和无穷维Hamilton算子, 这些都可以进行深入研究.