算子矩阵的谱问题是近年来线性算子理论中较为活跃的研究课题之一, 许多学者对其进行过研究, 参见文献[1-4]及其所引文献.我们知道, 如果$M$是Hilbert空间$X$上线性算子$T$的不变子空间, 在空间分解$X=M\oplus M^{\bot}$下, 线性算子$T$有如下上三角矩阵表示:

$ T=\left( \begin{array}{cc} \ast & \ast \\ 0 & \ast \\ \end{array} \right): M\oplus M^{\bot}\rightarrow M\oplus M^{\bot}. $

国内外许多学者对$2\times 2$上三角算子矩阵的补问题和Weyl定理进行了研究, 取得了丰硕的研究成果^[5-8].近来, $3\times 3$算子矩阵的扰动问题引起了一些学者的兴趣, 文献[9]通过对角元的信息给出了所有以这些算子为对角元的算子矩阵的Weyl型定理, 文献[10]则给出了对角元给定的所有$3\times 3$算子矩阵的可能点谱、可能剩余谱、可能连续谱和可能谱的刻画.文献[11]研究了一类$3\times 3$算子矩阵的本质谱.我们考虑$3\times3$上三角算子矩阵的点谱和剩余谱, 推广了文献[12]的结论, 给出了一类无界$3\times3$上三角算子矩阵点谱和剩余谱的完全描述, 并且在$l^{2}\times l^{2}\times l^{2}$中构造了一些剩余谱非空的上三角算子矩阵实例, 为进一步研究$3\times3$上三角算子矩阵的谱理论提供了重要事实.

除非特别说明, 本文始终用$X$表示Hilbert空间.为了简洁, $X$和乘积空间$X\times X\times X$中的单位算子均用$I$表示.若$A$是Hilbert空间中的线性算子, 我们用$D(A)$、$R(A)$和$A^{\ast}$分别表示$A$的定义域、值域和共轭算子; $N(A)$表示$A$的零空间, 即$N(A)=\{x\in D(A)|Ax=0\}$. $W^{c}$表示集合$W$相对于复数域$\mathbb{C}$的补集, $\emptyset$表示空集.

定义2.1 若$W_{1}$和$W_{2}$为$X$的子集, 则$W_{1}+W_{2}$定义为

$W_{1}+W_{2}=\{x+y\in X|x\in W_{1}, y\in W_{2}\}, $

若$W_{1}, W_{2}$中有一个为$\emptyset$, 约定$W_{1}+W_{2}=\emptyset$.

定义2.2 ^[13] $X$中的线性算子$T$的谱$\sigma(T)$可分为三个互不相交的集合的并集, 即

$\sigma(T)=\sigma_{p}(T)\cup\sigma_{r}(T)\cup\sigma_{c}(T),$

其中

1) 称$\lambda$为$T$的点谱, 如果$\lambda I-T$不是单射.点谱的全体记为$\sigma_{p}(T)$, 即

$\sigma_{p}(T)=\{\lambda\in \mathbb{C}: \lambda I-T {\hbox{不是单射}}\};$

2) 称$\lambda$为$T$的剩余谱, 如果$\lambda I-T$是单射并且$\overline{R(\lambda I-T)}\neq X$.剩余谱的全体记为$\sigma_{r}(T)$, 即

$\sigma_{r}(T)=\{\lambda\in \mathbb{C}: \lambda I-T {\hbox{是单射}}, \overline{R(\lambda I-T)}\neq X\};$

3) 称$\lambda$为$T$的连续谱, 如果$\lambda I-T$是单射, $\overline{R(\lambda I-T)}= X$, 并且它的逆算子不连续.连续谱的全体记为$\sigma_{c}(T)$, 即

$\sigma_{c}(T)=\{\lambda\in \mathbb{C}: \lambda I-T{\hbox{是单射}}, \overline{R(\lambda I-T)}= X, {\hbox{但}}(\lambda I-T)^{-1}{\hbox{不连续}}\}.$

定理3.1 设$H= \left( \begin{array}{ccc} A & E & F \\ 0 & B & 0\\ 0 & 0 & C \\ \end{array} \right)$是Hilbert空间$X\times X\times X$中的稠定闭线性算子, $D(B)\subset D(E),D(C)\subset D(F)$, 则

$\begin{eqnarray*}\sigma_{p}(H)&=&\{\lambda\in C:\lambda\in\sigma_{p}(A)\}\\ &&\cup\{\lambda\in C:\lambda\in\sigma_{p}(B),R(E_{1})\cap R(\lambda I-A)\neq\emptyset\}\\ &&\cup\{\lambda\in C:\lambda\in\sigma_{p}(C),R(F_{1})\cap R(\lambda I-A)\neq\emptyset\}\\ &&\cup\{\lambda\in C:\lambda\in\sigma_{p}(B),\lambda\in\sigma_{p}(C),R(E_{1})\cap R(\lambda I-A) =\emptyset, \\ && R(F_{1})\cap R(\lambda I-A)=\emptyset, R(\lambda I-A)\cap(R(E_{1})+R(F_{1}))\neq\emptyset\},\end{eqnarray*}$

其中$E_{1}=E|_{N(\lambda I-B)\setminus\{0\}}$是$E$在$N(\lambda I-B)\setminus\{0\}$的限制, $F_{1}=F|_{N(\lambda I-C)\setminus\{0\}}$是$F$在$N(\lambda I-C)\setminus\{0\}$的限制.

证 由$D(B)\subset D(E),D(C)\subset D(F)$, 可知$D(H)=D(A)\times D(B)\times D(C).$为叙述简洁, 令

$\begin{eqnarray*} M&=&\{\lambda \in C :\lambda\in\sigma_{p}(B),R(E_{1})\cap R(\lambda I-A)\neq\emptyset\},\\ N&=&\{\lambda\in C:\lambda\in\sigma_{p}(C),R(F_{1})\cap R(\lambda I-A)\neq\emptyset\}, \\ Q&=&\{\lambda\in C:\lambda\in\sigma_{p}(B),\lambda\in\sigma_{p}(C),R(E_{1})\cap R(\lambda I-A)=\emptyset,R(F_{1})\cap R(\lambda I-A)=\emptyset,\\ && R(\lambda I-A)\cap(R(E_{1})+R(F_{1}))\neq\emptyset\}.\end{eqnarray*}$

先证$\sigma_{p}(A)\cup M\cup N\cup Q\subset\sigma_{p}(H)$.若$\lambda\in\sigma_{p}(A)$, 则存在$x_{0}\in D(A)$且$x_{0}\neq0$, 有$(\lambda I-A)x_{0}=0$.从而

$\left( \begin{array}{c} x_{0} \\ 0\\ 0 \\ \end{array} \right) \in D(H){\hbox{且}} \left( \begin{array}{c} x_{0} \\ 0\\ 0 \\ \end{array} \right)\neq0,$

满足如下的算子方程组

$\begin{align} \left\{ \begin{array}{c r} (\lambda I-A)x- Ey- Fz=0 ,\\ (\lambda I-B)y=0,\\ (\lambda I-C)z=0,\\ \end{array} \right. \end{align}$

(3.1)

故$\lambda\in\sigma_{p}(H)$.

若$\lambda\in M$, 则有$\lambda\in\sigma_{p}(B)$, 从而$N(\lambda I-B)\setminus\{0\}\neq\emptyset$, 又由$R(E_{1})\cap R(\lambda I-A)\neq\emptyset$, 则存在$x_{0}\in D(A)$和$y_{0}\in N(\lambda I-B)\setminus\{0\}$使得

$\begin{equation} E_{1}y_{0}=(\lambda I-A)x_{0}, \end{equation}$

(3.2)

从而$Ey_{0}=(\lambda I-A)x_{0}$, 又由$y_{0}\in N(\lambda I-B)\setminus\{0\}$, 自然有$y_{0}\neq0$且$(\lambda I-B)y_{0}=0$, 从而对上面存在的$x_{0},y_{0}$, 有$\left( \begin{array}{c} x_{0} \\ y_{0} \\ 0 \\ \end{array} \right) \in D(H)$且满足方程组$(3.1)$, 因此$\lambda\in\sigma_{p}(H)$.同理可证$\lambda\in N$时, $\lambda\in\sigma_{p}(H)$.

若$\lambda\in Q$, 则$\lambda\in\sigma_{p}(B), \lambda\in\sigma_{p}(C)$, 从而$N(\lambda I-B)\setminus\{0\}\neq\emptyset$, $N(\lambda I-C)\setminus\{0\}\neq\emptyset$, 又由$R(\lambda I-A)\cap(R(E_{1})+R(F_{1}))\neq\emptyset$, 故存在$x_{0}\in D(A)$, $y_{0}\in N(\lambda I-B)\setminus\{0\}$, $z_{0}\in N(\lambda I-C)\setminus\{0\}$, 使得

$\begin{equation} (\lambda I-A)x_{0}=E_{1}y_{0}+F_{1}z_{0}, \end{equation}$

(3.3)

从而$(\lambda I-A)x_{0}-Ey_{0}-Fz_{0}=0$, 又由$y_{0}\in N(\lambda I-B)\setminus\{0\}$, $z_{0}\in N(\lambda I-C)\setminus\{0\}$, 自然有$(\lambda I-B)y_{0}=0,(\lambda I-C)z_{0}=0$, 即$\left( \begin{array}{c} x_{0} \\ y_{0} \\ z_{0}\\ \end{array} \right) \in D(H)$且$\left( \begin{array}{c} x_{0}\\ y_{0} \\ z_{0}\\ \end{array} \right)\neq0$满足方程组$(3.1)$, 即$\lambda\in \sigma_{p}(H)$.进而证明了$\sigma_{p}(A)\cup M\cup N\cup Q\subset \sigma_{p}(H)$.

下面证明$\sigma_{p}(H)\subset \sigma_{p}(A)\cup M\cup N\cup Q $.对于任意$\lambda\in \sigma_{p}(H)$, 存在

$\left( \begin{array}{c} x_{0} \\ y_{0} \\ z_{0} \\ \end{array} \right)\in D(H){\hbox{ 且}}\left( \begin{array}{c} x_{0} \\ y_{0} \\ z_{0} \\ \end{array} \right)\neq0, $

满足方程组$(3.1)$.以下分四种情况讨论:

1) 若$y_{0}=0, z_{0}=0$, 则必有$x_{0}\neq0$, 此时方程$(\lambda I-A)x_{0}-Ey_{0}-Fz_{0}=0$变为$(\lambda I-A)x_{0}=0$, 这意味着$\lambda\in\sigma_{p}(A)$.

2) 若$y_{0}=0, z_{0}\neq0$, 则由$(\lambda I-C)z_{0}=0$可得$\lambda\in\sigma_{p}(C)$且$z_{0}\in N(\lambda I-C)\setminus\{0\}$, 此时方程组的第一个方程变为$(\lambda I-A)x_{0}-Fz_{0}=0$, 从而$(\lambda I-A)x_{0}=F_{1}z_{0}$, 即

$R(\lambda I-A)\cap R(F_{1})\neq\emptyset,$

故$\lambda\in N$.

3) 若$y_{0}\neq0, z_{0}=0$, 类似前面的证明可知$\lambda\in\sigma_{p}(B)$且$R(E_{1})\cap R(\lambda I-A)\neq\emptyset$, 因此$\lambda\in M$.

4) 若$y_{0}\neq0, z_{0}\neq0$, 则$\lambda\in\sigma_{p}(B)$, $\lambda\in\sigma_{p}(C)$, 并且有$y_{0}\in N(\lambda I-B)\setminus\{0\}$, $z_{0}\in N(\lambda I-C)\setminus\{0\}$, 且$(\lambda I-A)x_{0}=E_{1}y_{0}+F_{1}z_{0}$, 从而

$R(\lambda I-A)\cap(R(E_{1})+R(F_{1}))\neq\emptyset.$

若$E_{1}y_{0}\in R(\lambda I-A)$或$F_{1}z_{0}\in R(\lambda I-A)$, 则$\lambda\in M\cap N$, 从而$\lambda\in M\cup N$.若$E_{1}y_{0}\overline{\in}R(\lambda I-A)$且$F_{1}z_{0}\overline{\in}R(\lambda I-A)$, 即$R(\lambda I-A)\cap R(E_{1})=\emptyset$, $R(\lambda I-A)\cap R(F_{1})=\emptyset$, 此时$\lambda\in Q$.

综上所述$\lambda\in\sigma_{p}(H)$, 则$\lambda\in\sigma_{p}(A)\cup M\cup N\cup Q$.证毕.

注3.1  定理3.1中的结论$\sigma_{p}(H)$可描述为七个互不相交的集合之并:

$\begin{eqnarray*}\sigma_{p}(H)&=&\{\lambda\in C:\lambda\in\sigma_{p}(A), \lambda\overline{\in}\sigma_{p}(B), \lambda\overline{\in}\sigma_{p}(C)\}\\ &&\cup\{\lambda\in C:\lambda\in\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\overline{\in}\sigma_{p}(C), R(\lambda I-A)\cap R(E_{1})\neq\emptyset\}\\ &&\cup\{\lambda\in C:\lambda\in\sigma_{p}(A), \lambda\overline{\in}\sigma_{p}(B), \lambda\in\sigma_{p}(C), R(\lambda I-A)\cap R(F_{1})\neq\emptyset\}\\ &&\cup\{\lambda\in C:\lambda\in\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\in\sigma_{p}(C), R(\lambda I-A)\cap(R(E_{1})+R(F_{1}))\neq\emptyset\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\overline{\in}\sigma_{p}(C), R(\lambda I-A)\cap R(E_{1})\neq\emptyset\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\overline{\in}\sigma_{p}(B), \lambda\in\sigma_{p}(C), R(\lambda I-A)\cap R(F_{1})\neq\emptyset\}\\ && \cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\in\sigma_{p}(C), R(\lambda I-A)\cap(R(E_{1})+R(F_{1}))\neq\emptyset\}.\end{eqnarray*}$

由定理3.1证明中的(3.2) 和(3.3) 两式可看出, 若去掉定理3.1中条件$D(B)\subset D(E), D(C)\subset D(F)$, 有如下结论:

推论3.1 设$X$是Hilbert空间, $H= \left( \begin{array}{ccc} A & E & F \\ 0 & B & 0\\ 0 & 0 & C \\ \end{array} \right)$是$X\times X\times X$中的稠定闭线性算子, $D(H)=D(A)\times (D(E)\cap D(B))\times (D(F)\cap D(C))$, 则

$\begin{eqnarray*}\sigma_{p}(H)&=&\{\lambda\in C:\lambda\in\sigma_{p}(A)\}\\ && \cup\{\lambda\in C:\lambda\in\sigma_{p}(B), R(E_{2})\cap R(\lambda I-A)\neq\emptyset\}\\ && \cup\{\lambda\in C:\lambda\in\sigma_{p}(C), R(F_{2})\cap R(\lambda I-A)\neq\emptyset\}\\ && \cup\{\lambda\in C:\lambda\in\sigma_{p}(B), \lambda\in\sigma_{p}(C), R(E_{2})\cap R(\lambda I-A) =\emptyset, \\ && R(F_{2})\cap R(\lambda I-A)=\emptyset, R(\lambda I-A)\cap(R(E_{2})+R(F_{2}))\neq\emptyset\}, \end{eqnarray*}$

其中$E_{2}=E|_{(N(\lambda I-B)\cap D(E))\setminus\{0\}}$是$E$在$(N(\lambda I-B)\cap D(E))\setminus\{0\}$的限制,

$F_{2}=F|_{(N(\lambda I-C)\cap D(F))\setminus\{0\}}$

是$F$在$(N(\lambda I-C)\cap D(F))\setminus\{0\}$的限制.

定理3.2 设$X$是Hilbert空间, $H= \left( \begin{array}{ccc} A & E & F \\ 0 & B & 0\\ 0 & 0 & C \\ \end{array} \right)$是$X\times X\times X$中的稠定闭线性算子, $D(B)\subset D(E), D(C)\subset D(F)$, 则

$\begin{eqnarray*}&& \sigma_{r}(H)=\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\overline{\in}\sigma_{p}(B), \lambda\overline{\in}\sigma_{p}(C), \overline{R(\lambda I-A)+R(E_{3})+R(F_{3})}\neq X\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\overline{\in}\sigma_{p}(C), \lambda\in\sigma_{r}(B)\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\overline{\in}\sigma_{p}(B), \lambda\in\sigma_{r}(C)\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\overline{\in}\sigma_{p}(B), \lambda\in\sigma_{p}(C),\\ && R(\lambda I-A)\cap R(F_{1})=\emptyset, \overline{R(\lambda I-A)+R(E_{3})+R(F_{3})} \neq X\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\in\sigma_{r}(B), \lambda\in\sigma_{p}(C), R(\lambda I-A)\cap R(F_{1})=\emptyset\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\overline{\in}\sigma_{p}(B), \lambda\in\sigma_{p}(C), R(\lambda I-A)\cap R(F_{1})=\emptyset, \overline{R(\lambda I-C)}\neq X\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\overline{\in}\sigma_{p}(C), \\ && R(\lambda I-A)\cap R(E_{1})=\emptyset, \overline{R(\lambda I-A)+R(E_{3})+R(F_{3})}\neq X\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\overline{\in}\sigma_{p}(C),\\ && R(\lambda I-A)\cap R(E_{1})=\emptyset, \overline{R(\lambda I-B)}\neq X\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\in\sigma_{r}(C), R(\lambda I-A)\cap R(E_{1})=\emptyset\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\in\sigma_{p}(C), R(\lambda I-A)\cap R(E_{1})=\emptyset,\\ && R(\lambda I-A)\cap R(F_{1})=\emptyset, R(\lambda I-A)\cap(R(E_{1})+R(F_{1}))=\emptyset,\\ && \overline{R(\lambda I-A)+R(E_{3})+R(F_{3})}\neq X\}\\ && \cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\in\sigma_{p}(C), R(\lambda I-A)\cap R(E_{1})=\emptyset,\\ && R(\lambda I-A)\cap R(F_{1})=\emptyset, R(\lambda I-A)\cap(R(E_{1})+R(F_{1}))=\emptyset, \overline{R(\lambda I-B)}\neq X\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\in\sigma_{p}(C), R(\lambda I-A)\cap R(E_{1})=\emptyset, \\ && R(\lambda I-A)\cap R(F_{1})=\emptyset, R(\lambda I-A)\cap(R(E_{1})+R(F_{1}))=\emptyset, \overline{R(\lambda I-C)}\neq X\}, \end{eqnarray*}$

其中$E_{3}=E|_{D(B)}$是$E$在$D(B)$的限制, $F_{3}=F|_{D(C)}$是$F$在$D(C)$的限制. $E_{1}$、$F_{1}$同定理3.1.

证 为简便计, 令

$\begin{eqnarray*}&& S_{1}=\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\overline{\in}\sigma_{p}(B), \lambda\overline{\in}\sigma_{p}(C)\},\\ && S_{2}=\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\overline{\in}\sigma_{p}(B), \lambda\in\sigma_{p}(C), R(\lambda I-A)\cap R(F_{1})=\emptyset\},\\ && S_{3}=\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\overline{\in}\sigma_{p}(C), R(\lambda I-A)\cap R(E_{1})=\emptyset\},\\ && S_{4}=\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\in\sigma_{p}(C), R(\lambda I-A)\cap R(E_{1})=\emptyset,\\ && R(\lambda I-A)\cap R(F_{1})=\emptyset, R(\lambda I-A)\cap(R(E_{1})+R(F_{1}))=\emptyset\}.\end{eqnarray*}$

根据定理3.1, 可知$(\sigma_{p}(H))^{c}=S_{1}\cup S_{2}\cup S_{3}\cup S_{4}$.注意到$\overline{R(\lambda I-H)}\neq X\times X\times X$当且仅当

$\overline{R(\lambda I-A)+R(E_{3})+R(F_{3})}\neq X,{\hbox{ 或}}\overline{R(\lambda I-B)}\neq X, {\hbox{或}}\overline{R(\lambda I-C)}\neq X.$

令

$ M=\{\lambda\in C:\overline{R(\lambda I-A)+R(E_{3})+R(F_{3})}\neq X, {\hbox{或}}\overline{R(\lambda I-B)}\neq X, {\hbox{或}} \overline{R(\lambda I-C)}\neq X\}.$

根据剩余谱的定义, 有

$\lambda\in\sigma_{r}(H)\Leftrightarrow\lambda\in(\sigma_{p}(H))^{c}\cap M\Leftrightarrow\lambda\in(S_{1}\cup S_{2}\cup S_{3}\cup S_{4})\cap M,$

利用结合的运算律, 便证明了定理3.2.

同推论3.1, 若去掉定理3.2中关于定义域的限制条件$D(B)\subset D(E), D(C)\subset D(F)$, 则有如下的结论:

推论3.2 设$X$是Hilbert空间, $H= \left( \begin{array}{ccc} A & E & F \\ 0 & B & 0\\ 0 & 0 & C \\ \end{array} \right)$是$X\times X\times X$中的稠定闭线性算子, $D(H)=D(A)\times (D(E)\cap D(B))\times (D(F)\cap D(C))$, 则

$\begin{eqnarray*}&& \sigma_{r}(H)=\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\overline{\in}\sigma_{p}(B), \lambda\overline{\in}\sigma_{p}(C), \overline{R(\lambda I-A)+R(E_{4})+R(F_{4})}\neq X\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\overline{\in}\sigma_{p}(B), \lambda\overline{\in}\sigma_{p}(C), \overline{R(\lambda I-B_{1})}\neq X\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\overline{\in}\sigma_{p}(B), \lambda\overline{\in}\sigma_{p}(C)\overline{R(\lambda I-C_{1})}\neq X\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\overline{\in}\sigma_{p}(B), \lambda\in\sigma_{p}(C),\\ && R(\lambda I-A)\cap R(F_{2})=\emptyset, \overline{R(\lambda I-A)+R(E_{4})+R(F_{4})}\neq X\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\overline{\in}\sigma_{p}(B), \lambda\in\sigma_{p}(C), R(\lambda I-A)\cap R(F_{2})=\emptyset, \overline{R(\lambda I-B_{1})}\neq X\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\overline{\in}\sigma_{p}(B), \lambda\in\sigma_{p}(C), R(\lambda I-A)\cap R(F_{2})=\emptyset, \overline{R(\lambda I-C_{1})}\neq X\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\overline{\in}\sigma_{p}(C), R(\lambda I-A)\cap R(E_{2})=\emptyset,\\ && \overline{R(\lambda I-A)+R(E_{4})+R(F_{4})}\neq X\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\overline{\in}\sigma_{p}(C), R(\lambda I-A)\cap R(E_{2})=\emptyset, \overline{R(\lambda I-B_{1})}\neq X\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\overline{\in}\sigma_{p}(C), R(\lambda I-A)\cap R(E_{2})=\emptyset, \overline{R(\lambda I-C_{1})}\neq X\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\in\sigma_{p}(C), R(\lambda I-A)\cap R(E_{2})=\emptyset, R(\lambda I-A)\cap R(F_{2})=\emptyset, \\ &&R(\lambda I-A)\cap(R(E_{2})+R(F_{2}))=\emptyset, \overline{R(\lambda I-A)+R(E_{4})+R(F_{4})}\neq X\}\\ &&\cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\in\sigma_{p}(C), R(\lambda I-A)\cap R(E_{2})=\emptyset, R(\lambda I-A)\cap R(F_{2})=\emptyset, \\ && R(\lambda I-A)\cap(R(E_{2})+R(F_{2}))=\emptyset, \overline{R(\lambda I-B_{1})}\neq X\}\\ && \cup\{\lambda\in C:\lambda\overline{\in}\sigma_{p}(A), \lambda\in\sigma_{p}(B), \lambda\in\sigma_{p}(C), R(\lambda I-A)\cap R(E_{2})=\emptyset, R(\lambda I-A)\cap R(F_{2})=\emptyset, \\ && R(\lambda I-A)\cap(R(E_{2})+R(F_{2}))=\emptyset, \overline{R(\lambda I-C_{1})}\neq X\}, \end{eqnarray*}$

其中$E_{4}=E|_{D(B)\cap D(E)}$是$E$在$D(B)\cap D(E)$的限制, $F_{4}=F|_{D(C)\cap D(F)}$是$F$在$D(C)\cap D(F)$的限制, $ B_{1}=B|_{D(B)\cap D(E)}$是$B$在$D(B)\cap D(E)$的限制, $C_{1}=C|_{D(C)\cap D(F)}$是$C$在$D(C)\cap D(F)$的限制. $E_{2}$、$F_{2}$定义同推论3.1.

注3.2 若$A, B, C, E, F$为$X$中的有界线性算子, 则$D(B)\subset D(E), D(C)\subset D(F)$自然成立, 故前文所有结论对于$X\times X\times X$上的有界算子矩阵$H$均适用.

本节举例说明上三角算子矩阵$H$的剩余谱可以非空, 进而验证了定理3.2的有效性.

例1 设$X=l^{2}(0, \infty)$, 对于任意$x\in l^{2}$, $x=(x_{1}, x_{2}, x_{3}, x_{4},\cdots)$, 令

$\begin{align*} &Ax=(x_{1}+2x_{2}, x_{1}, x_{2}, x_{3}, \cdots),\\ &Bx=(2x_{2}, x_{1}, x_{2}, x_{3}, \cdots),\\ &Cx=(0, x_{1}, x_{2}, x_{3},\cdots),\\ &Ex=(4x_{1}, 2x_{1}, x_{2}, x_{3}, \cdots),\\ &Fx=(2x_{1}, x_{1}, x_{2}, x_{3}, \cdots). \end{align*}$

记

$H=\left( \begin{array}{lll} A & E & F \\ 0 &B & 0 \\ 0 &0& C\\ \end{array} \right),$

则$-1\in\sigma_{r}(H)$, 从而$\sigma_{r}(H)$非空.

证 经计算易知$-1\overline{\in}\sigma_{p}(A)$, $-1\overline{\in}\sigma_{p}(B)$, $-1\overline{\in}\sigma_{p}(C)$, $\forall x, y, z\in X$, 有

$\begin{eqnarray*}&&(-I-A)x-E_{3}y-F_{3}z \\ &=&(-2x_{1}-2x_{2}-4y_{1}-2z_{1}, -x_{1}-x_{2}-2y_{1}-z_{1},\\ &&-x_{3}-x_{2}-y_{2}-z_{2}, -x_{4}-x_{3}-y_{3}-z_{3}, \cdots).\end{eqnarray*}$

令

$M=\{x:x=(x_{1}, x_{2}, \cdots)\in l^{2}| x_{1}=2x_{2}\}.$

易证$M$是$X$的真闭子空间, 又$R(-I-A)+R(E_{3})+R(F_{3})$包含于$M$中, 从而

$\overline{R(-I-A)+R(E_{3})+R(F_{3})}\neq X.$

综上, $-1\overline{\in}\sigma_{p}(A)$, $-1\overline{\in}\sigma_{p}(B)$, $-1\overline{\in}\sigma_{p}(C)$, $\overline{R(-I-A)+R(E_{3})+R(F_{3})}\neq X$.由定理3.2可知, $-1\in\sigma_{r}(H)$, 即$\sigma_{r}(H)$非空.

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

$\begin{align*} &Ax=(2x_{2}, x_{1}, x_{2}, x_{3}, \cdots),\\ &Bx=(x_{1}+2x_{2}, x_{1}, x_{2}, x_{3}, \cdots),\\ &Cx=(0, x_{1}, x_{2}, x_{3}, \cdots),\\ &Ex=(2x_{1}+x_{2}, x_{1}, x_{2}, x_{3}, \cdots),\\ &Fx=(x_{1}+x_{2}, x_{1}, x_{2}, x_{3}, \cdots) \end{align*}$

且

$H=\left( \begin{array}{lll} A & E & F \\ 0 &B & 0 \\ 0 &0& C \end{array} \right).$

经计算易知$-1\overline{\in}\sigma_{p}(A)$, $-1\in\sigma_{r}(B)$, $-1\overline{\in}\sigma_{p}(C)$, 由定理3.2可知, $-1\in\sigma_{r}(H)$, 于是$\sigma_{r}(H)$非空.

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

$\begin{align*} &Ax=( 2x_{2}, x_{1}, x_{2}, x_{3}, \cdots),\\ &Bx=(0, x_{1}, x_{2}, x_{3}, \cdots),\\ &Cx=(x_{2}, 2x_{1}+x_{3}, x_{4}, x_{5}, \cdots),\\ &Ex=(0, 0, 0, x_{1}, \cdots),\\ &Fx=(0, 0, x_{1}, x_{2}, \dots). \end{align*}$

通过计算, 有$ C=A^{\ast}$.令

$H=\left( \begin{array}{lll} A & E & F \\ 0 &B & 0 \\ 0 &0& C \end{array} \right),$

则$\sigma_{r}(H)$非空.

证 经计算可知, $0\overline{\in}\sigma_{p}(A)$, $0\overline{\in}\sigma_{p}(B)$, $0\in\sigma_{p}(C)$且

$N(\lambda I-C)\setminus\{0\}=\{(m_{1}, 0, -2 m_{1}, 0, 0, \cdots)^{T}\in X| m_{1}\neq0\}.$

于是$F_{1}m =(0, 0, m_{1}, 0, -2 m_{1}, \cdots)$, 由于$m_{1}\neq0$, 故$R(-A)\cap R(F_{1})=\emptyset$.对于任意$x, y, z\in X$, 有

$-Ax-E_{3}y-F_{3}z=(-2x_{2}, -x_{1}, -x_{2}-z_{1}, -x_{3}-y_{1}-z_{2}, \cdots),$

若$-x_{2}-z_{1}=0 $, 显然$\overline{R(-A)+R(E_{3})+R(F_{3})}\neq X$.若$-x_{2}-z_{1}\neq0$, 取

$q=(m_{1}, 0, \frac{2\overline{x_{2}}m_{1}}{\overline{-x_{2}-z_{1}} }, 0, 0, \cdots), q\neq0, $

进而$-Ax-E_{3}y-F_{3}z$与非零元$q$正交, 因此$q \in (R(-A)+R(E_{3})+R(F_{3}))^{\perp}$, 所以$\overline{R(-A)+R(E_{3})+R(F_{3})}\neq X $.

综上所述,

$0\overline{\in}\sigma_{p}(A), 0\overline{\in}\sigma_{p}(B), 0\in\sigma_{p}(C), R(-A)\cap R(F_{1})=\emptyset, \overline{R(-A)+R(E_{3})+R(F_{3})}\neq X .$

由定理3.2可知$0\in\sigma_{r}(H)$, 即$\sigma_{r}(H)$非空.

例4 设$X=l^{2}(0, \infty)$, 对于任意$x\in l^{2}$, $x=(x_{1}, x_{2}, x_{3}, \cdots)$, 令

$\begin{align*} &Ax=(2x_{2}, x_{1}, x_{2}, x_{3}, \cdots),\\ &Bx=(0, x_{1}, x_{2}, x_{3}, \cdots),\\ &Cx=(0, x_{2}, x_{3}, x_{4}, \cdots),\\ &Ex=(2x_{1}, x_{1}, x_{2}, x_{3}, \cdots),\\ &Fx=(0, 0, x_{1}, x_{2}, \cdots) \end{align*}$

且

$H=\left( \begin{array}{lll} A & E & F \\ 0 &B & 0 \\ 0 &0& C \end{array} \right),$

则$\sigma_{r}(H)$非空.

证 首先经计算, 我们有$0\overline{\in}\sigma_{p}(A)$, $0\overline{\in}\sigma_{p}(B)$, $0\in\sigma_{p}(C)$且

$N(-C)\setminus\{0\}=\{m=(m_{1}, 0, 0, \cdots)^{T}\in X| m_{1}\neq0\}.$

于是$F_{1}m =(0, 0, m_{1}, 0, 0, \cdots)$, 进而$R(-A)\cap R(F_{1})=\emptyset$, 又因为$\overline{R(-C)}\neq X$.

综上$0\overline{\in}\sigma_{p}(A)$, $0\overline{\in}\sigma_{p}(B)$, $0\in\sigma_{p}(C)$, $R(-A)\cap R(F_{1})=\emptyset$, $\overline{R(-C)}\neq X$, 由定理3.2可知, $0\in\sigma_{r}(H)$, 即$\sigma_{r}(H)$非空.

例5 设$X=l^{2}(0, \infty)$, 对于任意$x\in l^{2}$, $x=(x_{1}, x_{2}, x_{3}, \cdots)$, 令

$\begin{align*} &Ax=(3x_{1}, 2x_{1}, x_{1}, x_{2}, \cdots),\\ &Bx=(x_{1}+x_{2}, 2x_{1}+x_{3}, x_{4}, x_{5}, \cdots),\\ &Cx=(x_{2}, 2x_{1}+x_{3}, x_{4}, x_{5}, \cdots),\\ &Ex=(0, x_{1}, 0, x_{2}, x_{3}, \cdots),\\ &Fx=(0, 0, x_{1}, x_{2}, \cdots) \end{align*}$

且

$H=\left( \begin{array}{lll} A & E & F \\ 0 &B & 0 \\ 0 &0& C \end{array} \right),$

则$\sigma_{r}(H)$非空.

证 经计算易知$0\overline{\in}\sigma_{p}(A)$, $0\in\sigma_{p}(B)$, $0\in\sigma_{p}(C)$且

$\begin{eqnarray*}&& N(C)\setminus\{0\}=\{ m=(m_{1}, 0, -2 m_{1}, 0, 0, \cdots)^{T}\in X| m_{1}\neq0\},\\ && N(B)\setminus\{0\}=\{ n=(n_{1}, -n_{1}, -2 n_{1}, 0, 0, \cdots)\in X| n_{1}\neq0\}.\end{eqnarray*}$

于是$F_{1}m =(0, 0, m_{1}, 0, -2m_{1}, \cdots)$, 进而$R(-A)\cap R(F_{1})=\emptyset$.同理$R(-A)\cap R(E_{1})=\emptyset$.注意到$E_{1}n+F_{1}m=(0, n_{1}, m_{1}, -n_{1}, -2m_{1}-2 n_{1}, \cdots)$, 有$R(-A)\cap( R(E_{1})+R(F_{1}))=\emptyset$.

对于任意$x, y, z\in X$, 可得

$-Ax-E_{3}y-F_{3}z=(-3x_{1}, -2x_{1}-y_{1}, -x_{1}-z_{1}, -x_{2}-y_{2}-z_{2}, \cdots).$

若$ -x_{1}-z_{1}=0$, 则$\overline{R(-A)+R(E_{3})+R(F_{3})}\neq X $.若$ -x_{1}-z_{1}\neq0$, 则取

$q=(m_{1}, 0, \frac{3\overline{x_{1}}m_{1}}{\overline{-x_{1}-z_{1}} }, 0, 0, \cdots)\neq 0,$

由于$-Ax-E_{3}y-F_{3}z$与$q$正交, 所以, $\overline{R(-A)+R(E_{3})+R(F_{3})}\neq X $.综上$0\overline{\in}\sigma_{p}(A)$, $0\in\sigma_{p}(B)$, $0\in\sigma_{p}(C)$, $R(-A)\cap R(F_{1})=\emptyset$, $R(-A)\cap R(E_{1})=\emptyset$, $R(-A)\cap( R(E_{1})+R(F_{1})=\emptyset$, $\overline{R(-A)+R(E_{3})+R(F_{3})}\neq X $.由定理3.2可知, $0\in\sigma_{r}(H)$, 即$\sigma_{r}(H)$非空.