缺少了某些元素的算子矩阵称为缺项算子矩阵.算子矩阵的谱补问题旨在讨论缺项算子矩阵中所缺的元素对整个算子矩阵谱的影响, 在换位提升理论, 插值理论, 以及系统控制理论中具有重要应用.算子矩阵的谱扰动属于谱补问题的研究范畴, 它是当所缺的元素跑遍特定的集合时整个算子矩阵谱的稳定的组成部分.

设 $\mathcal H_{1}$, $\mathcal H_{2}$, $\mathcal H_{3}$为无穷维复可分Hilbert空间.以 $\mathcal{B}(\mathcal H_{i}, \mathcal H_{j})$表示从 $\mathcal H_{i}$到 $\mathcal H_{j}$的所有有界(线性)算子构成的集合, $\mathcal{B}(\mathcal H_{i}, \mathcal H_{i})$简记为 $\mathcal{B}(\mathcal H_{i})$, 其中 $i, j=1, 2, 3$.给定 $A\in\mathcal{B}(\mathcal H_{1}), B\in \mathcal{B}(\mathcal H_{2}), C\in \mathcal{B}(\mathcal H_{3})$, 记

$ M_{D, E, F}= \left(\begin{matrix} A&D&E\\ 0&B&F\\ 0&0&C \end{matrix}\right), $

其中

$ D\in\mathcal{B}(\mathcal H_{2}, \mathcal H_{1}), E\in\mathcal{B}(\mathcal H_{3}, \mathcal H_{1}), F\in\mathcal{B}(\mathcal H_{3}, \mathcal H_{2}) $

为待定的未知算子.显然, $M_{D, E, F}$是 $\mathcal H_{1}\oplus \mathcal H_{2}\oplus \mathcal H_{3}$上的缺项上三角算子矩阵.为叙述方便, 对于给定算子 $A\in \mathcal{B}(\mathcal H_{1}), B\in \mathcal{B}(\mathcal H_{2}), $以 $M_{D}$表示缺项 $2\times 2$阶上三角算子矩阵

$ M_{D}= \left(\begin{smallmatrix} A & D \\ 0 & B\end{smallmatrix}\right)\in \mathcal{B}( \mathcal H_{1}\oplus \mathcal H_{2}), $

其中 $ D\in\mathcal{B}(\mathcal H_{2}, \mathcal H_{1})$待定.

经过近二十年的积累, $2\times 2$阶上三角算子矩阵的谱扰动研究已日臻完善, 涌现出诸如谱, 点谱, 剩余谱, 连续谱, 近似点谱, 亏谱, 本质谱, Weyl谱, 以及Browder谱等的扰动结果^{[2, 3, 5-8, 10-18]}.例如, 文[15]研究了上三角算子矩阵 $M_{D}$的点谱和剩余谱扰动, 得到如下结果

$ \begin{eqnarray*}\label{eq0001} \bigcap\limits_{D\in{\cal B}({\cal H}_2, {\cal H}_1)}\sigma_{p, 1}(M_{D}) &=&\{\lambda\in\sigma_{p}(A): \overline{\mathcal{R}(A-\lambda)}=\mathcal{H}_1, \overline{\mathcal{R}(B-\lambda)}=\mathcal{H}_2\}\\ &&\cup\{\lambda\in\sigma_{p}(B): \mathcal{R}(A-\lambda)=\mathcal{H}_1, \overline{\mathcal{R}(B-\lambda)}=\mathcal{H}_2\}, \\ \bigcap\limits_{D\in{\cal B}({\cal H}_2, {\cal H}_1)} \sigma_{p, 2}(M_{D})&=&\{\lambda\in\sigma_{p}(A): \overline{\mathcal{R}(B-\lambda)}\neq\mathcal{H}_2\}\\ &&\cup\{\lambda\in\sigma_{p}(A): \mathcal{R}(B-\lambda)=\mathcal{H}_2, d(A-\lambda)>n(B-\lambda)\}\\ &&\cup\{\lambda\in\sigma_{p}(B): \mathcal{R}(A-\lambda)\mbox{闭}, d(A-\lambda) < n(B-\lambda),\\ \overline{\mathcal{R}(B-\lambda)}\neq\mathcal{H}_2\}, \\ \bigcap\limits_{D\in{\cal B}({\cal H}_2, {\cal H}_1)}\sigma_{r, 1}(M_{D})&=& (\sigma_{r, 1}(A)\cap\rho(B))\cup(\sigma_{r, 1}(A)\cap\sigma_{r, 1}(B))\\ &&\cup(\rho(A)\cap\sigma_{r, 1}(B)), \\ \bigcap\limits_{D\in{\cal B}({\cal H}_2, {\cal H}_1)}\sigma_{r, 2}(M_{D})&=& (\sigma_{r, 2}(A)\cap\sigma_r(B))\cup(\sigma_{c}(A)\cap\sigma_r(B))\cup(\sigma_{r, 2}(A)\cap\rho(B))\\ &&\cup(\rho(A)\cap\sigma_{r, 2}(B)) \cup\{\lambda\in\sigma_{r, 1}(A)\cap\sigma_{r, 2}(B): d(A-\lambda) < \infty\}. \end{eqnarray*} $

最近, 文[4, 18]分别研究了上三角算子矩阵 $M_{D, E, F}$的左(右)本质谱, 以及点谱, 剩余谱和连续谱的扰动; 文[9]给出了 $M_{D, E, F}$的点谱, 连续谱和剩余谱之并集的描述.本文在文[15]的基础上探讨了 $3\times 3$阶情形, 得到

$ \begin{array}{l} \bigcap\limits_{D, E, F}\sigma_{p, 1}(M_{D, E, F}), \ \bigcap\limits_{D, E, F}\sigma_{r, i}(M_{D, E, F})\quad(i=1, 2) \end{array} $

的完全描述.

对于有界算子 $T$, 分别以 $T^{*}$, $\mathcal{N}(T)$和 $\mathcal{R}(T)$表示 $ T$的共轭算子, 零空间和值域; 以 $n(T)$, $d(T)$分别表示 $ \mathcal{N}(T)$, $ \mathcal{N}(T^{*})$的维数, 即 $ n(T)=\dim \mathcal{N}(T)$, $d(T)=\dim \mathcal{N}(T^{*})$.下面给出文中涉及的一些基本概念和辅助引理.

定义1.1^[1]  设 $T$为Banach空间 $X$中的有界线性算子, 则 $T$的预解集 $\rho(T)$定义为

$ \rho(T)=\{\lambda \in \mathbb{C}: T-\lambda\, \textrm{具有定义于}\, X\, \textrm{上的有界逆}\}, $

并称集合 $\sigma(T)=\mathbb{C}\setminus \rho(T)$为 $T$的谱.

由闭图象定理, $\lambda\in\rho(T)$当且仅当 $T-\lambda$是双射.显然, $\sigma(T)$可分成下述互不相交的组成部分 $\sigma(T)=\sigma_{p}(T)\cup\sigma_{r}(T)\cup\sigma_{c}(T), $其中

$ \begin{eqnarray*}&&\sigma_{p}(T)=\{\lambda\in\mathbb{C}:\ T-\lambda\, \mbox{不是单射}\}, \\ &&\sigma_{r}(T)=\{\lambda\in\mathbb{C}:\ T-\lambda\, \mbox{是单射且}\, \overline{\mathcal{R}(T-\lambda)}\neq X\}, \\ &&\sigma_{c}(T)=\{\lambda\in\mathbb{C}:\ T-\lambda\, \mbox{是单射}, \ \overline{\mathcal{R}(T-\lambda)}=X\, \mbox{且}\, \mathcal{R}(T-\lambda)\neq X\} \end{eqnarray*} $

分别为 $T$的点谱, 剩余谱和连续谱.类似于文[1], 可对点谱和剩余谱进一步细分为

$ \begin{eqnarray*} &&\sigma_{p, 1}(T)=\{\lambda\in\sigma_{p}(T):\ \overline{\mathcal{R}(T-\lambda)}= X\}, \\ &&\sigma_{p, 2}(T)=\{\lambda\in\sigma_{p}(T):\ \overline{\mathcal{R}(T-\lambda)}\neq X\}, \\ &&\sigma_{r, 1}(T)=\{\lambda\in\sigma_{r}(T):\ \mathcal{R}(T-\lambda)\, \mbox{闭}\}, \\ &&\sigma_{r, 2}(T)=\{\lambda\in\sigma_{r}(T):\ \mathcal{R}(T-\lambda)\, \mbox{不闭}\}. \end{eqnarray*} $

此外, 称集合

$ \sigma_{\delta}(T)=\{\lambda\in\mathbb{C}:\ \mathcal{R}(T-\lambda)\neq X\} $

为 $T$的亏谱.为叙述方便, 记

$ \begin{eqnarray*} \rho_l(T)&=&\{\lambda\in\mathbb{C}:\, T-\lambda\, \mbox{是左可逆的}\}, \\ \sigma_l(T)&=&\{\lambda\in\mathbb{C}:\, T-\lambda\, \mbox{不是左可逆的}\}, \\ \rho_m(T)&=&\{\lambda\in\mathbb{C}:\, T-\lambda\, \mbox{是Moore-Penrose 可逆的}\}, \\ \sigma_m(T)&=&\{\lambda\in\mathbb{C}:\, T-\lambda\, \mbox{不是Moore-Penrose 可逆的}\}, \\ \rho_{co}(T)&=&\{\lambda\in\mathbb{C}:\, \overline{\mathcal{R}(T-\lambda) }=X\}, \\ \sigma_{co}(T)&=&\{\lambda\in\mathbb{C}:\, \overline{\mathcal{R}(T-\lambda)} \neq X \}, \\ \rho_{in}(T)&=&\{\lambda\in\mathbb{C}:\, T-\lambda\, \mbox{是单射 }\}, \\ \rho_{\delta}(T)&=&\{\lambda\in\mathbb{C}:\, T-\lambda\, \mbox{是满射}\}. \end{eqnarray*} $

如所熟知, $\lambda\in\rho_{m}(T)$当且仅当 $\mathcal{R}(T- \lambda)$闭; $ \lambda\in\rho_{l}(T)$当且仅当 $T-\lambda$为具有闭值域的单射.显然, $\sigma_{p, 1}(T)=\sigma_{p}(T)\cap\rho_{\rm co}(T)$, $\sigma_{p, 2}(T)=\sigma_{p}(T)\cap\sigma_{\rm co}(T)$, $\sigma_{r, 1}(T)=\sigma_{r}(T)\cap\rho_{m}(T)$, $\sigma_{r, 2}(T)=\sigma_{r}(T)\cap\sigma_{m}(T)$.

引理1.1^[9]  设 $X$, $Y$为Banach空间, $T\in \mathcal{B}(X, Y)$.若 $\mathcal{R}(T)$不闭, 则存在无穷维子空间 $M\subset \overline{\mathcal{R}(T)}$使得

$ \overline{\mathcal{R}(T)}= \mathcal{R}(T)+M, \quad M\cap \mathcal{R}(T)=\{0\}. $

引理1.2  设 $X$, $Y$为Banach空间, $T\in \mathcal{B}(X, Y)$.下述结论显然成立.

(ⅰ)若 $S\in \mathcal{B}(X, Y)$为有限秩算子, 则 $\mathcal{R}(T)$闭当且仅当 $\mathcal{R}(T+S)$闭;

(ⅱ)若 $U\in \mathcal{B}(Y), V\in \mathcal{B}(X)$为可逆算子, 则 $\mathcal{R}(T)$闭当且仅当 $\mathcal{R}(UTV)$闭.

引理1.3  设 $A\in \mathcal{B}(\mathcal{H}_{1}), B\in \mathcal{B}(\mathcal{H}_{2}), C\in \mathcal{B}(\mathcal{H}_{3})$为给定算子, 则任给 $D, E, F$均有 $\lambda\in\rho_{\rm in}( M_{D, E, F})$的充要条件为 $\lambda\in\rho_{\rm in}(A)\cap\rho_{\rm in}(B)\cap\rho_{\rm in}(C).$

引理1.4  设 $A\in \mathcal{B}(\mathcal{H}_{1}), B\in \mathcal{B}(\mathcal{H}_{2}), C\in \mathcal{B}(\mathcal{H}_{3})$为给定算子, 则任给 $ D, E, F $均有 $\lambda\in\rho_{\rm co}( M_{D, E, F})$的充要条件为 $\lambda\in\rho_{\rm co}(A)\cap\rho_{\rm co}(B)\cap\rho_{\rm co}(C).$

引理1.1的证明见[9, 推论1];引理1.2是熟知的, 引理1.3, 1.4是显然的, 证明从略.

本节给出本文的主要结果和证明.为便于叙述, 以下记

$ \{e_{i}^{(1)}\}_{i=1}^{n(B-\lambda)}, \{e_{i}^{(2)}\}_{i=1}^{n(C-\lambda)}, \{f_{i}^{(1)}\}_{i=1}^{d(A-\lambda)}, \{f_{i}^{(2)}\}_{i=1}^{d(B-\lambda)}, \{h_{i}^{(1)}\}_{i=1}^{\infty}, \{h_{i}^{(2)}\}_{i=1}^{\infty} $

分别为 $\mathcal{N}(B-\lambda), \mathcal{N}(C-\lambda), \mathcal{R}(A-\lambda)^{\perp}, \mathcal{R}(B-\lambda)^{\perp}, \mathcal{H}_1, \mathcal{H}_2$的规范正交基, 其中 $A\in \mathcal{B}(\mathcal{H}_{1}), B\in \mathcal{B}(\mathcal{H}_{2}), C\in \mathcal{B}(\mathcal{H}_{3})$.

定理2.1  设 $A\in \mathcal{B}(\mathcal{H}_{1}), B\in \mathcal{B}(\mathcal{H}_{2}), C\in \mathcal{B}(\mathcal{H}_{3})$为给定算子, 则

$ \begin{eqnarray}\label{eq01} \bigcap\limits_{D, E, F}\sigma_{p, 1}(M_{D, E, F})&=&(\sigma_{p, 1}(A)\cap\rho_{\rm co}(B)\cap\rho_{\rm co}(C))\cup (\rho_{\delta}(A)\cap\sigma_{p, 1}(B)\cap\rho_{\rm co}(C))\nonumber\\ &&\cup(\rho_{\delta}(A)\cap\rho_{\delta}(B)\cap\sigma_{p, 1}(C)). \end{eqnarray} $

(2.1)

证  记 $\Sigma=\bigcap\limits_{D, E, F}\sigma_{p, 1}(M_{D, E, F})$, $\Sigma_{1}, \Sigma_{2}, \Sigma_{3} $依次为式(2.1) 等号右端的集合.先证 $\Sigma\supset\Sigma_1\cup\Sigma_2\cup\Sigma_3$.设 $\lambda\in\Sigma_1\cup\Sigma_2\cup\Sigma_3$, 由引理1.4可知任给 $D, E, F$均有 $ \lambda\in\rho_{\rm co}(M_{D, E, F}) $, 下面只须证还有 $\lambda\in\sigma_{p}(M_{D, E, F})$.

若 $\lambda\in\Sigma_{1}$, 则 $\lambda\in\sigma_p(A)$, 显然任给 $D, E, F$均有 $\lambda\in\sigma_{p}(M_{D, E, F})$.若 $\lambda\in\Sigma_{2}\setminus\Sigma_{1}$, 则 $\lambda\in\rho(A)\cap\sigma_{p, 1} (B)\cap\rho_{\rm co}(C)$, 所以任给 $D, E, F$均存在可逆算子 $V$使得

$ \begin{pmatrix} A-\lambda & D & E\\ 0 & B-\lambda & F\\ 0 & 0 & C-\lambda \end{pmatrix}V=\begin{pmatrix} A-\lambda & 0 & 0\\ 0 & B-\lambda & F\\ 0 & 0 & C-\lambda \end{pmatrix}. $

注意到 $\lambda\in\sigma_{p}(B)$和 $V$的可逆性, 显然任给 $D, E, F$均有 $\lambda\in\sigma_{p}(M_{D, E, F})$.若 $\lambda\in\Sigma_{3}\setminus (\Sigma_{1}\cup\Sigma_{2})$, 则 $\lambda\in\rho(A)\cap\rho(B)\cap\sigma_{p, 1}(C)$, 同上讨论即可, 不再赘述.

下面证明 $\Sigma\subset\Sigma_1\cup\Sigma_2\cup\Sigma_3$.为此, 只须证若 $\lambda\notin\Sigma_1\cup\Sigma_2\cup\Sigma_3 $必有 $\lambda\notin\Sigma$, 可分以下4种情形讨论.

情形1   $\lambda\in\sigma_{co}(A) \cup\sigma_{co}(B)\cup\sigma_{co}(C) $.此时, 取 $D, E, F$均为零算子, 显然 $\overline{\mathcal{R}(M_{D, E, F}-\lambda)}\neq\mathcal{H}_{1}\oplus\mathcal{H}_{2}\oplus \mathcal{H}_{3}$, 因此 $\lambda\notin\Sigma.$

情形2   $\lambda\in\rho_{\rm co}(A) \cap\rho_{\rm co}(B)\cap\rho_{\rm co}(C)\cap\rho_{\rm in}(A)\cap\sigma_{m}(A) $.此时, 由引理1.1, 存在无穷维子空间 $ M\subset \overline{\mathcal{R}(A-\lambda)}$使得 $ M \cap \mathcal{R}(A-\lambda)= \{0\}$.记 $M$的某个含有无穷多元素的规范正交集为 $ \{g_{i}^{(1)}\}_{i=1}^{\infty}$.定义 $ F=0$,

$ \begin{eqnarray*} &&\begin{cases} D(e_{i}^{(1)})=g_{2i-1}^{(1)}, &i=1, 2, \cdots, n(B-\lambda), \\ D(y)=0, &y\in \mathcal{N}(B-\lambda)^{\perp}, \end{cases}\\ &&\begin{cases} E(e_{i}^{(2)})=g_{2i}^{(1)}, &i=1, 2, \cdots, n(C-\lambda), \\ E(y)=0, &y\in \mathcal{N}(C-\lambda)^{\perp}. \end{cases} \end{eqnarray*} $

则 $M_{D, E, F}-\lambda $为单射, 显然 $\lambda\notin\Sigma.$

情形3   $\lambda\in\rho_{\rm co}(A) \cap\rho_{\rm co}(B)\cap\rho_{\rm co}(C)\cap\rho_{\rm in}(A) \cap\rho_{\rm in}(B)\cap \sigma_{m}(B) $.此时, 类似于情形2, 存在无穷维子空间 $ N\subset \overline{\mathcal{R}(B-\lambda)}$使得 $ N \cap \mathcal{R}(B-\lambda)=\{0\}$.记 $N$的某个含有无穷多元素的规范正交集为 $ \{g_{i}^{(2)}\}_{i=1}^{\infty}$.取 $D, E$均为零算子, 并定义

$ \begin{cases} F(e_{i}^{(2)})=g_{i}^{(2)}, &i=1, 2, \cdots, n(C-\lambda), \\ F(y)=0, &y\in \mathcal{N}(C-\lambda)^{\perp}. \end{cases} $

显然 $M_{D, E, F}-\lambda $为单射, 因此 $\lambda\notin\Sigma .$

情形4   $\lambda\in\rho_{\rm co}(A) \cap\rho_{\rm co}(B)\cap\rho_{\rm co}(C)\cap\rho_{\rm in}(A) \cap\rho_{\rm in}(B)\cap\rho_{\rm in}(C) $.此时, 由引理1.3, 取 $D, E, F$均为零算子便有 $M_{D, E, F}-\lambda $为单射, 因此 $\lambda\notin\Sigma.$

定理2.2  设 $A\in \mathcal{B}(\mathcal{H}_{1}), B\in \mathcal{B}(\mathcal{H}_{2}), C\in \mathcal{B}(\mathcal{H}_{3})$为给定算子, 则

$ \begin{array}{l}\bigcap\limits_{D, E, F}\sigma_{p, 2}(M_{D, E, F})=\Sigma_{1}\cup\Sigma_{2}\cup\Sigma_{3}, \end{array} $

其中

$ \begin{eqnarray*} \Sigma_{1}&=& (\sigma_{p}(A)\!\cap\! \sigma_{co}(C))\cup\{\lambda\!\in\!\sigma_{p}(A)\!\cap\! \rho_{\delta}(C)\!: d(B-\lambda)\!>\! n(C-\lambda) \}\\ &&\cup\{\lambda\!\in\!\sigma_{p, 2}(A)\!\cap\!\rho_{m}(B)\!\cap\!\rho_{\delta}(C)\!: d(A-\lambda)\!+\!d(B-\lambda)\!>\!n(B-\lambda)\!+\!n(C-\lambda)\}, \\ \Sigma_{2}&=& \{ \lambda\!\in\!\rho_{l}(A)\!\cap\! \sigma_{co}(C)\!: d(A-\lambda)\! < \!n(B-\lambda) \}\\ &&\cup\{\lambda\!\in\!\rho_{l}(A)\!\cap\!\sigma_{p, 2}(B)\!\cap\! \rho_{\delta}(C)\!: d(A-\lambda)\! < \!n(B-\lambda), d(B-\lambda)\!>\! n(C-\lambda)\}, \\ \Sigma_{3}&=& \{ \lambda\!\in\!\rho_{l}(A)\!\cap\!\rho_{m}(B)\!\cap\! \sigma_{p, 2}(C)\!: d(A-\lambda)\!+\!d(B-\lambda)\! < \!n(B-\lambda)\!+\!n(C-\lambda) \}. \end{eqnarray*} $

证  记 $\Sigma=\bigcap\limits_{D, E, F}\sigma_{p, 2}(M_{D, E, F})$, $\Sigma_{11}, \Sigma_{12}, \Sigma_{13} $依次为 $\Sigma_{1}$等号右端的集合.先证 $\Sigma\supset\Sigma_1\cup\Sigma_2\cup\Sigma_3$.

若 $\lambda\in\Sigma_{11}$, 则 $\lambda\in\sigma_p(A)$且 $\overline{\mathcal{R}(C-\lambda)}\neq\mathcal{H}_3$.显然 $\lambda\in\Sigma$.现设 $\lambda\in\Sigma_{12}$.由 $\lambda\in\rho_\delta(C)$可知任给 $D, E, F$算子矩阵 $M _{D, E, F}-\lambda$均有如下分块表示

$ \begin{array}{l} M_{D, E, F}-\lambda =\left(\begin{smallmatrix} A-\lambda & D & E_{1} & E_{2}\\ 0 & (B-\lambda)_1 & F_{11} & F_{12}\\ 0 & 0 & F_{21} & F_{22}\\ 0 & 0 & (C-\lambda)_{1} & 0 \end{smallmatrix}\right):\, \left(\begin{smallmatrix} {\cal H}_{1}\\ {\cal H}_{2}\\ \mathcal{N}(C-\lambda)^{\perp}\\ \mathcal{N}(C-\lambda)\end{smallmatrix}\right) \to\left(\begin{smallmatrix} {\cal H}_{1}\\ \overline{\mathcal{R}(B-\lambda)}\\ \mathcal{R}(B-\lambda)^{\perp}\\ {\cal H}_{3} \end{smallmatrix}\right), \end{array} $

其中 $(C-\lambda)_1$为可逆算子.因此存在可逆算子 $U$使得

$ U\left(\begin{smallmatrix} A-\lambda & D & E_{1} & E_{2}\\ 0 & (B-\lambda)_1 & F_{11} & F_{12}\\ 0 & 0 & F_{21} & F_{22}\\ 0 & 0 & (C-\lambda)_{1} & 0 \end{smallmatrix}\right)=\left(\begin{smallmatrix} A-\lambda & D & 0 & E_{2}\\ 0 & (B-\lambda)_1 & 0 & F_{12}\\ 0 & 0 & 0 & F_{22}\\ 0 & 0 & (C-\lambda)_{1} & 0 \end{smallmatrix}\right). $

注意到 $d(B-\lambda)> n(C-\lambda)$可知无论如何选取 $D, E, F$均有 $F_{22} $具有闭值域且不是满射, 进而由 $U$的可逆性得到 $\lambda\in\sigma_{co} (M_{D, E, F})$.再由 $\lambda\in\sigma_{p}(A)$, 显然 $\lambda\in\Sigma$.以下假设 $\lambda\in\Sigma_{13}$.由于 $\lambda\in\rho_\delta(C)\cap \rho_m(B)$, 所以任给 $D, E, F$均有

$ \begin{equation*}\label{equc} M_{D, E, F}-\lambda =\left(\begin{smallmatrix} (A-\lambda)_1 & D_{11} & D_{12} & E_{11} & E_{12}\\ 0 & D_{21} & D_{22} & E_{21} & E_{22}\\ 0 & (B-\lambda)_1 & 0 & F_{11}& F_{12}\\ 0 & 0 & 0 & F_{21}& F_{22}\\ 0 & 0 & 0 & (C-\lambda)_1 &0 \end{smallmatrix}\right): \left(\begin{smallmatrix} {\cal H}_{1}\\ \mathcal{N}(B-\lambda)^\perp\\ \mathcal{N}(B-\lambda)\\ \mathcal{N}(C-\lambda)^\perp\\ \mathcal{N}(C-\lambda) \end{smallmatrix}\right) \to\left(\begin{smallmatrix} \overline{\mathcal{R}(A-\lambda)}\\ \mathcal{R}(A-\lambda)^\perp\\ \mathcal{R}(B-\lambda)\\ \mathcal{R}(B-\lambda)^\perp\\ {\cal H}_3 \end{smallmatrix}\right), \end{equation*} $

其中 $(B-\lambda)_1$, $(C-\lambda)_1$为可逆算子.因此存在可逆算子 $U$使得

$ U (M_{D, E, F}-\lambda)=\left(\begin{smallmatrix} (A-\lambda)_1 & 0 & D_{12} & 0 & E_{12}^{\prime}\\ 0 & 0 & D_{22} & 0 & E_{22}^{\prime}\\ 0 & (B-\lambda)_1 & 0 & 0& F_{12}\\ 0 & 0 & 0 & 0 & F_{22}\\ 0 & 0 & 0 & (C-\lambda)_1 &0 \end{smallmatrix}\right), $

其中 $ E_{12}^{\prime}=E_{12}-D_{11}(B-\lambda)_{1}^{-1}F_{12}, E_{22}^{\prime}=E_{22}-D_{21}(B-\lambda)_{1}^{-1}F_{12}.$由

$ d(A-\lambda)+d(B-\lambda)> n(B-\lambda)+ n(C-\lambda) $

易知无论如何选取 $D, E, F$均有 $ \left(\begin{smallmatrix} D_{22}&E_{22}^{\prime}\\ 0&F_{22} \end{smallmatrix}\right) $具有闭值域且不是满射, 结合 $\lambda\in\sigma_{p}(A)$便有 $\lambda\in\Sigma$.于是 $\Sigma\supset\Sigma_1$.

设 $\lambda\in\Sigma_{2} $.由于 $\lambda\in\rho_{l}(A) $, 所以任给 $D, E, F$均有

$ M_{D, E, F}-\lambda=\left(\begin{smallmatrix} (A-\lambda)_1 & D_{11} & D_{12} & E_{1}\\ 0 & D_{21} & D_{22} & E_{2}\\ 0 & (B-\lambda)_1 & 0 & F\\ 0 & 0 & 0 & C-\lambda \end{smallmatrix}\right):\, \left(\begin{smallmatrix} \mathcal{H}_{1}\\ \mathcal{N}(B-\lambda)^\perp\\ \mathcal{N}(B-\lambda)\\ \mathcal{H}_{3}\end{smallmatrix}\right) \to\left(\begin{smallmatrix} \mathcal{R}(A-\lambda)\\ \mathcal{R}(A-\lambda)^\perp\\ \mathcal{H}_{2}\\ \mathcal{H}_{3} \end{smallmatrix}\right), $

其中 $(A-\lambda)_1$为可逆算子.因此存在可逆算子 $V$使得

$ \begin{equation}\label{equ2} (M_{D, E, F}-\lambda)V= \left(\begin{smallmatrix} (A-\lambda)_1 & 0 & 0 & 0\\ 0& D_{21} & D_{22} & E_{2}\\ 0& (B-\lambda)_1 &0 & F\\ 0 & 0 &0& C-\lambda \end{smallmatrix}\right). \end{equation} $

(2.2)

注意到 $n(B-\lambda)> d(A-\lambda)$, 显然无论如何选取 $D, E, F$均有 $D_{22}$不是单射, 进而由式(2.2) 便知 $\lambda\in\sigma_{p}(M_{D, E, F})$.这样, 若 $\lambda\in\sigma_{co}(C), $则任给 $D, E, F$均有 $\lambda\in\sigma_{co}(M_{D, E, F})$, 显然 $\lambda\in\Sigma$; 若 $\lambda\in\rho_{\delta}(C)$且 $d(B-\lambda)>n(C-\lambda)$, 则由 $\Sigma_{12}$的证明可知任给 $D, E, F$均有 $\lambda\in\sigma_{co}(M_{D, E, F})$, 从而 $\lambda\in\Sigma$.于是 $\Sigma\supset\Sigma_2$.

现设 $\lambda\in\Sigma_{3}$.由 $\lambda\in\rho_l(A)\cap \rho_m(B)$可知任给 $D, E, F$均有

$ \begin{equation*} M_{D, E, F}-\lambda =\left(\begin{smallmatrix} (A-\lambda)_1 & D_{11} & D_{12} & E_{11} & E_{12}\\ 0 & D_{21} & D_{22} & E_{21} & E_{22}\\ 0 & (B-\lambda)_1 & 0 & F_{11}& F_{12}\\ 0 & 0 & 0 & F_{21}& F_{22}\\ 0 & 0 & 0 & (C-\lambda)_1 &0 \end{smallmatrix}\right): \left(\begin{smallmatrix} {\cal H}_{1}\\ \mathcal{N}(B-\lambda)^\perp\\ \mathcal{N}(B-\lambda)\\ \mathcal{N}(C-\lambda)^\perp\\ \mathcal{N}(C-\lambda) \end{smallmatrix}\right) \to\left(\begin{smallmatrix} \mathcal{R}(A-\lambda)\\ \mathcal{R}(A-\lambda)^\perp\\ \mathcal{R}(B-\lambda)\\ \mathcal{R}(B-\lambda)^\perp\\ {\cal H}_{3} \end{smallmatrix}\right), \end{equation*} $

其中 $(A-\lambda)_1$, $(B-\lambda)_1$均为可逆算子, $(C-\lambda)_1$为单射.因此存在可逆算子 $U, V$使得

$ \begin{equation}\label{equ3} U(M_{D, E, F}-\lambda)V= \left(\begin{smallmatrix} (A-\lambda)_1 & 0 & 0 & 0&0\\ 0& 0 & D_{22}&E_{21}^{\prime} & E_{22}^{\prime}\\ 0& (B-\lambda)_1 &0 &0&0\\ 0 & 0 &0& F_{21} & F_{22}\\ 0 & 0 &0& C-\lambda & 0 \end{smallmatrix}\right), \end{equation} $

(2.3)

其中 $ E_{21}^{\prime}= E_{21}-D_{21}(B-\lambda)_{1}^{-1}F_{11}, E_{22}^{\prime}= E_{22}-D_{21}(B-\lambda)_{1}^{-1}F_{12}$.由

$ d(A-\lambda)+ d(B-\lambda) < n(B-\lambda)+n(C-\lambda) $

易知无论如何选取 $D, E, F$均有 $ \left(\begin{smallmatrix} D_{22} & E_{22}^{\prime}\\ 0 & F_{22} \end{smallmatrix}\right) $不是单射, 进而由式(2.3) 可有 $\lambda\in\sigma_{p}(M_{D, E, F})$.注意到 $\lambda\in \sigma_{p, 2}(C)$, 则任给 $D, E, F$均有 $\lambda\in\sigma_{co}(M_{D, E, F})$, 显然 $\lambda\in\Sigma$.于是 $\Sigma\supset\Sigma_3$.综上可知 $\Sigma\supset\Sigma_1\cup\Sigma_2\cup\Sigma_3$.

下面证明相反的包含关系.为此只须证若 $\lambda\notin\Sigma_1\cup\Sigma_2\cup\Sigma_3 $必有 $\lambda\notin\Sigma$, 可分以下6种情形讨论.

情形1   $\lambda\in\rho_{\rm in}(A)\cap\sigma_{m}(A)$.此时, 取定理2.1中情形2的 $D, E, F$便有 $M_{D, E, F}-\lambda $为单射, 因此 $\lambda\notin\Sigma$.

情形2   $\lambda\in\rho_{l}(A)\cap \sigma_{m}(B)$且 $d(A-\lambda)\geq n(B-\lambda)$.此时, 取定理2.1中情形3的集合 $N$, $\{g_{i}^{(2)}\}_{i=1}^{\infty}$和算子 $E, F$, 并定义

$ \begin{cases} D(e_{i}^{(1)})=f_{i}^{(1)}, &i=1, 2, \cdots, n(B-\lambda), \\ D(y)=0, &y\in \mathcal{N}(B-\lambda)^{\perp}. \end{cases} $

则 $M_{D, E, F}-\lambda $为单射, 显然 $\lambda\notin\Sigma$.

情形3   $\lambda\in\rho_{l}(A)\cap \rho_{m}(B)$, $d(A-\lambda)\geq n(B-\lambda) $且 $d(A-\lambda)+ d(B-\lambda)\geq n(B-\lambda)+n(C-\lambda)$.此时, 定义

$ \begin{cases} D(e_{i}^{(1)})=f_{i}^{(1)}, &i=1, 2, \cdots, n(B-\lambda), \\ D(y)=0, &y\in \mathcal{N}(B-\lambda)^{\perp}. \end{cases} $

若 $d(B-\lambda)\geq n(C-\lambda)$, 则定义 $E=0$,

$ \begin{cases} F(e_{i}^{(2)})=f_{i}^{(2)}, &i=1, 2, \cdots, n(B-\lambda), \\ F(y)=0, &y\in \mathcal{N}(B-\lambda)^{\perp}; \end{cases} $

若 $d(B-\lambda) < n(C-\lambda)$, 由 $d(A-\lambda)+ d(B-\lambda)\geq n(B-\lambda)+n(C-\lambda)$可知

$ d(A-\lambda)-n(B-\lambda)\geq n(C-\lambda)-d(B-\lambda). $

则定义

$ \begin{eqnarray*} &&\begin{cases} E(e_{d(B-\lambda)+i}^{(2)})=f_{n(B-\lambda)+i}^{(1)}, &i=1, 2, \cdots, n(C-\lambda)-d(B-\lambda), \\ E(y)=0, &y\perp\{e_{i}^{(2)}\}_{i=d(B-\lambda)+1}^{n(C-\lambda)-d(B-\lambda)}, \end{cases}\\ &&\begin{cases} F(e_{i}^{(2)})=f_{i}^{(2)}, &i=1, 2, \cdots, d(B-\lambda), \\ F(y)=0, &y\perp\{e_{i}^{(2)}\}_{i=1}^{d(B-\lambda)}. \end{cases} \end{eqnarray*} $

容易验证 $M_{D, E, F}-\lambda $为单射, 显然 $\lambda\notin\Sigma$.

情形4   $\lambda\in\rho_{\rm co}(C)\cap\sigma_{m}(C)$.此时, 注意到 $\lambda\in\sigma_{m}(C)$当且仅当 $\overline{ \lambda}\in\sigma_{m}(C^{*})$.由引理1.1, 存在无穷维子空间 $Z\subset \overline{\mathcal{R}(C^{*}-\overline{\lambda})}=\mathcal{N}(C-\lambda)^{\perp} $使得 $ Z\cap \mathcal{R}(C^{*}-\overline{\lambda})= \{0\}$.记 $Z$的某个含有无穷多元素的规范正交集为 $ \{g_{i}^{(3)}\}_{i=1}^{\infty}$.定义 $ D=0$,

$ \begin{eqnarray*} &&\begin{cases} E(g_{i}^{(3)})=h_{i}^{(1)}, &i=1, 2, \cdots, \\ E(y)=0, &y\perp\{g_{i}^{(3)}\}_{i=1}^{\infty}, \end{cases}\\ &&\begin{cases} F(g_{i}^{(3)})=h_{i}^{(2)}, &i=1, 2, \cdots, \\ F(y)=0, &y\perp\{g_{i}^{(3)}\}_{i=1}^{\infty}, \end{cases} \end{eqnarray*} $

则 $E^*\in\mathcal{B}(\mathcal{H}_1, \mathcal{H}_3), F^*\in\mathcal{B}(\mathcal{H}_2, \mathcal{H}_3)$均为单射, 而且 $\mathcal{R}(E^*)=\mathcal{R}(F^*)=\mbox{span}\{g_{i}^{(3)}:i=1, 2, \cdots\}$.注意到 $ Z\cap \mathcal{R}(C^{*}-\overline{\lambda})= \{0\}$和 $\lambda\in\rho_{\rm co}(C)$, 所以立即得到 $M_{D, E, F}^*-\overline{\lambda}$是单射.即 $M_{D, E, F}-\lambda $具有稠值域, 从而 $ \lambda\notin\Sigma$.

情形5   $\lambda\in\sigma_{m}(B)\cap \rho_{\delta}(C)$且 $d(B-\lambda)\leq n(C-\lambda)$.此时, 类似于情形4, 存在无穷维子空间 $W\subset \overline{\mathcal{R}(B^{*}-\overline{\lambda})}=\mathcal{N}(B-\lambda)^{\perp} $使得 $W \cap \mathcal{R}(B^{*}-\overline{\lambda}) =\{0\}$.记 $W$的某个含有无穷多元素的规范正交集为 $ \{g_{i}^{(4)}\}_{i=1}^{\infty}$.定义 $ E=0$,

$ \begin{eqnarray*} &&\begin{cases} D(g_{i}^{(4)})=h_{i}^{(1)}, &i=1, 2, \cdots, \\ D(y)=0, &y\perp\{g_{i}^{(4)}\}_{i=1}^{\infty}, \end{cases}\\ &&\begin{cases} F(e_{i}^{(2)})=f_{i}^{(2)}, &i=1, 2, \cdots, d(B-\lambda), \\ F(y)=0, &y\perp\{e_{i}^{(2)}\}_{i=1}^{d(B-\lambda)}. \end{cases} \end{eqnarray*} $

显然, $M_{D, E, F}-\lambda $具有稠值域, 因此 $\lambda\notin\Sigma$.

情形6   $\lambda\in\rho_{m}(B)\cap\rho_{\delta}(C)$, $d(B-\lambda)\leq n(C-\lambda) $且 $d(A-\lambda)+d(B-\lambda)\leq n(B-\lambda)+ n(C-\lambda)$.此时, 定义

$ \begin{cases} F(e_{i}^{(2)})=f_{i}^{(2)}, &i=1, 2, \cdots, d(B-\lambda), \\ F(y)=0, &y\perp\{e_{i}^{(2)}\}_{i=1}^{d(B-\lambda)}. \end{cases} $

若 $d(A-\lambda)\leq n(B-\lambda)$, 则定义 $E=0$,

$ \begin{cases} D(e_{i}^{(1)})=f_{i}^{(1)}, &i=1, 2, \cdots, d(A-\lambda), \\ D(y)=0, &y\perp\{e_{i}^{(1)}\}_{i=1}^{d(A-\lambda)}; \end{cases} $

否则注意到 $ d(A-\lambda)-n(B-\lambda)\leq n(C-\lambda)-d(B-\lambda), $定义

$ \begin{eqnarray*} &&\begin{cases} D(e_{i}^{(1)})=f_{i}^{(1)}, &i=1, 2, \cdots, n(B-\lambda), \\ D(y)=0, &y\in \mathcal{N}(B-\lambda)^{\perp}, \end{cases}\\ &&\begin{cases} E(e_{d(B-\lambda)+i}^{(2)})=f_{n(B-\lambda)+i}^{(1)}, &i=1, 2, \cdots, d(A-\lambda)-n(B-\lambda), \\ E(y)=0, &y\perp\{e_{i}^{(2)}\}_{i=1}^{d(A-\lambda)-n(B-\lambda)}. \end{cases} \end{eqnarray*} $

不难看出 $M_{D, E, F}-\lambda$是满射, 从而 $\lambda\notin\Sigma$.

定理2.3  设 $A\in \mathcal{B}({\cal H}_{1}), B\in \mathcal{B}({\cal H}_{2}), C\in \mathcal{B}({\cal H}_{3})$为给定算子, 则

$ \begin{eqnarray}\label{eq003} \bigcap\limits_{D, E, F}\sigma_{r, 1}(M_{D, E, F}) &=&(\sigma_{r, 1}(A)\cap\rho(B)\cap\rho(C)) \cup (\rho_{l}(A)\cap\sigma_{r, 1}(B)\cap\rho(C))\nonumber\\ &&\cup (\rho_{l}(A)\cap\rho_{l}(B)\cap\sigma_{r, 1}(C)). \end{eqnarray} $

(2.4)

证  记 $\Sigma=\bigcap\limits_{D, E, F}\sigma_{r, 1}(M_{D, E, F})$, $\Sigma_{1}, \Sigma_{2}, \Sigma_{3} $依次为式(2.4) 右端的集合.先证 $\Sigma\supset\Sigma_1\cup\Sigma_2\cup\Sigma_3$.设 $\lambda\in\Sigma_1\cup\Sigma_2\cup\Sigma_3$, 显然 $\lambda\in\rho_l(A)\cap\rho_l(B)\cap\rho_l(C)$, 所以任给 $D, E, F$均有

$ M_{D, E, F}-\lambda=\left(\begin{smallmatrix} (A-\lambda)_1 & D_{1} & E_{1}\\ 0 & D_{2} & E_{2}\\ 0 & (B-\lambda)_1 & F_{1}\\ 0 & 0 & F_{2}\\0 & 0 & (C-\lambda)_1 \end{smallmatrix}\right):\, \left(\begin{smallmatrix} \mathcal{H}_{1}\\ \mathcal{H}_{2}\\ \mathcal{H}_{3} \end{smallmatrix}\right) \to\left(\begin{smallmatrix} \mathcal{R}(A-\lambda)\\ \mathcal{R}(A-\lambda)^\perp\\ \mathcal{R}(B-\lambda)\\ \mathcal{R}(B-\lambda)^\perp\\ \mathcal{R}(C-\lambda)\\ \mathcal{R}(C-\lambda)^\perp\\ \end{smallmatrix}\right), $

其中 $(A-\lambda)_1$, $(B-\lambda)_1$, $(C-\lambda)_1$均为可逆算子.因此存在可逆算子 $U$使得

$ \begin{equation}\label{eq004} U(M_{D, E, F}-\lambda)=\left(\begin{smallmatrix} (A-\lambda)_1 & 0 & 0\\ 0 & 0 &0\\ 0 & (B-\lambda)_1 & 0\\ 0 & 0 & 0\\0&0& (C-\lambda)_1 \end{smallmatrix}\right). \end{equation} $

(2.5)

注意到 $d(A-\lambda), d(B-\lambda), d(C-\lambda)$至少有一个大于零, 由式(2.5) 可知无论如何选取 $D, E, F $均有 $M_{D, E, F}- \lambda$左可逆且 $ \overline{\mathcal{R}(M_{D, E, F}- \lambda)}\neq \mathcal{H}_{1}\oplus \mathcal{H}_{2}\oplus \mathcal{H}_{3}$.因此 $\lambda\in\Sigma. $于是 $\Sigma\supset\Sigma_1\cup\Sigma_2\cup\Sigma_3$.

再证相反的包含关系.本文断言: $\lambda\in\Sigma $蕴含 $\lambda\in\rho_l(A)\cap\rho_l(B)\cap\rho_l(C)$.若不然, 取 $D, E, F$皆为零算子便有 $\lambda\in \sigma _{l}(M_{D, E, F})$, 这与 $\lambda\in\Sigma$矛盾.因此只须证若 $\lambda\in(\rho_l(A)\cap\rho_l(B)\cap\rho_l(C))\setminus(\Sigma_1\cup\Sigma_2\cup\Sigma_3) $便有 $\lambda\notin\Sigma$.这是显然的, 因为 $\lambda\in\Sigma_1\cup\Sigma_2\cup\Sigma_3$蕴含 $\lambda\in\rho_l(A)\cap\rho_l(B)\cap\rho_l(C)$且 $ A-\lambda, B-\lambda, C-\lambda$至少有一个值域不稠, 从而若 $\lambda\in(\rho_l(A)\cap\rho_l(B)\cap\rho_l(C)) \setminus(\Sigma_1\cup\Sigma_2\cup\Sigma_3)$便有 $\lambda\in\rho(A)\cap\rho(B)\cap\rho(C)$.

定理2.4  设 $A\in \mathcal{B}(H_{1}), B\in \mathcal{B}(H_{2}), C\in \mathcal{B}(H_{3})$为给定算子, 则

$ \begin{array}{c} \bigcap\limits_{D, E, F} \sigma_{r, 2}(M_{D, E, F})=\Sigma_{1}\cup\Sigma_{2}, \end{array} $

其中

$ \begin{eqnarray*} \Sigma_{1}&=&(( \sigma_{c}(A)\cup\sigma_{r, 2}(A))\cap\sigma_{r}(B)\cap\rho(C)) \cup(\sigma_{r, 2}(A)\cap\rho(B)\cap\rho(C))\\ &&\cup\{\lambda\in \rho_{\rm in}(A)\cap\sigma_{r, 2}(B)\cap\rho(C):d(A-\lambda) < \infty\}, \\ \Sigma_{2}&=&(( \sigma_{c}(A)\cup\sigma_{r, 2}(A))\cap \rho_{\rm in}(B)\cap\sigma_{r}(C))\\ &&\cup\{\lambda\in \rho_{\rm in}(A)\cap(\sigma_{c}(B)\cup\sigma_{r, 2}(B)) \cap\sigma_{r}(C): d(A-\lambda) < \infty\}\\ &&\cup\{\lambda\in \rho_{\rm in}(A)\cap\rho_{\rm in}(B)\cap\sigma_{r, 2}(C): d(A-\lambda) < \infty, d(B-\lambda) < \infty\}. \end{eqnarray*} $

证  记 $\Sigma=\bigcap\limits_{D, E, F}\sigma_{r, 2}(M_{D, E, F}) .$先证 $\Sigma\supset\Sigma_{1}\cup\Sigma_{2}$.设 $\lambda\in\Sigma_{1}\cup\Sigma_{2}$.因为 $\lambda\in\Sigma_{2}$蕴含 $\lambda\in\sigma_{r}(C), $显然任给 $D, E, F$均有 $\Sigma_{2}\subset \sigma_{co}(M_{D, E, F})$; 注意到 $\lambda\in\Sigma_{1}$蕴含 $\lambda\in(\sigma_r(B)\cap\rho(C))\cup(\sigma_r(A)\cap\rho(B)\cap\rho(C))$, 容易证明任给 $D, E, F$均有 $\Sigma_{1}\subset\sigma_{co}(M_{D, E, F})$.此外, 由引理1.3易知任给 $D, E, F$均有 $M_{D, E, F}-\lambda$为单射.以下只须证还有 $\lambda\in\sigma_{m}(M_{D, E, F})$.

现记 $\Sigma_{i}$的表达式中等号右端的集合依次为 $\Sigma_{i1}, \Sigma_{i2}, \Sigma_{i3}\ (i=1, 2)$.若 $\lambda\in\Sigma_{11}\cup\Sigma_{12}\cup\Sigma_{21}, $则必有 $\lambda\in\sigma_{m}(A).$显然任给 $D, E, F$均有 $\lambda\in\sigma_{m}(M_{D, E, F}).$若 $\lambda\in\Sigma_{13}, $则任给 $D, E, F$均有

$ M_{D, E, F}-\lambda=\left(\begin{smallmatrix} (A-\lambda)_1 & D_{1} & E_{1}\\ 0 & D_{2} & E_{2}\\ 0 & (B-\lambda)_1 & F_{1}\\ 0 & 0 & F_{2}\\0 & 0 & C-\lambda \end{smallmatrix}\right):\, \left(\begin{smallmatrix} \mathcal{H}_{1}\\ \mathcal{H}_{2}\\ \mathcal{H}_{3} \end{smallmatrix}\right) \to\left(\begin{smallmatrix} \overline{\mathcal{R}(A-\lambda)}\\ \mathcal{R}(A-\lambda)^\perp\\ \overline{\mathcal{R}(B-\lambda)}\\ \mathcal{R}(B-\lambda)^\perp\\ \mathcal{H}_3 \end{smallmatrix}\right), $

其中 $C-\lambda$可逆.因此存在可逆算子 $U$使得

$ U(M_{D, E, F}-\lambda)=\left(\begin{smallmatrix} (A-\lambda)_1 & D_{1} & 0\\ 0 & D_{2} & 0\\ 0 & (B-\lambda)_1 & 0\\ 0 & 0 & 0\\0 & 0 & C-\lambda \end{smallmatrix}\right), $

结合 $d(A-\lambda) < \infty$, 应用引理1.2可有 $\lambda\in\sigma_{m}(M_{D, E, F})$当且仅当

$ \left(\begin{smallmatrix} (A-\lambda)_1 & D_{1} & 0\\ 0 & 0 & 0\\ 0 & (B-\lambda)_1 & 0\\ 0 & 0 & 0\\0 & 0 & C-\lambda \end{smallmatrix}\right) $

的值域不闭.这样若 $\lambda\in\sigma_{m}(A), $同前述讨论可知, 任给 $D, E, F$均有 $\lambda\in \sigma_{m}(M_{D, E, F});$若 $\lambda\notin\sigma_{m}(A)$, 则 $(A-\lambda)_1$为可逆算子, 注意到 $\lambda\in\sigma_{r, 2}(B)$容易得到任给 $D, E, F$均有 $\lambda\in\sigma_{m}(M_{D, E, F})$.类似地, 容易证明若 $\lambda\in\Sigma_{22}\cup\Sigma_{23}, $则任给 $D, E, F$仍有 $\lambda\in\sigma_{m}(M_{D, E, F}).$

再证 $\Sigma\subset\Sigma_{1}\cup\Sigma_{2}$.注意到任给 $D, E, F$均有 $ \sigma_{r, 2}(M_{D, E, F})\subset\sigma_{m}(M_{D, E, F})$.所以要证 $\lambda\notin\Sigma_{1} \cup\Sigma_{2}$蕴含 $\lambda\notin \Sigma$, 只须考虑如下3种情形即可.

情形1   $\lambda\in\sigma_{c}(C)$, $d(A-\lambda) < \infty$且 $d(B-\lambda) < \infty$.此时, 注意到 $\lambda\in\sigma_{m}(C)$, 取形如定理2.2中情形4的 $D, E, F$可知 $M_{D, E, F}-\lambda $具有稠值域, 显然 $\lambda\notin\Sigma$.

情形2 $\lambda\in\sigma_{r, 1}(A)\cap\sigma_{c}(B)\cap\rho(C)$且 $d(A- \lambda) < \infty$.此时, 取 $E, F$均为零算子并定义 $D(h_{i}^{(2)})=h_{i}^{(1)}$, 显然 $M_{D, E, F}-\lambda $具有稠值域, 从而 $ \lambda\notin\Sigma$.

情形3   $\lambda\in(\rho(A)\cap\sigma_{c}(B)\cap\rho(C)) \cup(\sigma_{c}(A)\cap(\rho(B)\cup\sigma_{c}(B))\cap\rho(C)) \cup(\sigma_{c}(A)\cap \sigma_{c}(B)\cap\rho(C))$.此时, 取 $D, E, F$均为零算子, 显然 $M_{D, E, F}-\lambda $具有稠值域, 因此 $ \lambda\notin\Sigma$.

注  由上述定理, 文[15]的结果可描述为

$ \begin{eqnarray*} \bigcap\limits_{D\in{\cal B}({\cal H}_2, {\cal H}_1)}\sigma_{p, 1}(M_{D}) &=&(\sigma_{p, 1}(A)\cap\rho_{\rm co}(B))\cup(\rho_{\delta}(A)\cap\sigma_{p, 1}(B)), \\ \bigcap\limits_{D\in{\cal B}({\cal H}_2, {\cal H}_1)} \sigma_{p, 2}(M_{D})&=&(\sigma_{p}(A)\cap\sigma_{co}(B))\\ &&\cup\{\lambda\in\sigma_{p, 2}(A)\cap\rho_{\delta}(B): d(A-\lambda)>n(B-\lambda)\}\\ &&\cup\{\lambda\in\rho_{l}(A)\cap\sigma_{p, 2}(B):d(A-\lambda) < n(B-\lambda))\}, \\ \bigcap\limits_{D\in{\cal B}({\cal H}_2, {\cal H}_1)}\sigma_{r, 1}(M_{D})&=& (\sigma_{r, 1}(A)\cap\rho(B))\cup(\rho_{l}(A)\cap\sigma_{r, 1}(B)), \\ \bigcap\limits_{D\in{\cal B}({\cal H}_2, {\cal H}_1)}\sigma_{r, 2}(M_{D})&=& ((\sigma_{r, 2}(A)\cup\sigma_{c}(A))\cap\sigma_r(B))\cup(\sigma_{r, 2}(A)\cap\rho(B))\\ &&\cup\{\lambda\in\rho_{\rm in}(A)\cap\sigma_{r, 2}(B): d(A-\lambda) < \infty\}. \end{eqnarray*} $