算子矩阵是以线性算子为元素的矩阵,其研究在算子矩阵理论中非常广泛.无论从理论角度还是从实际应用角度来讲,算子矩阵理论的研究具有深远的意义.理论上,如果Hilbert空间$H$可以分为两个Hilbert空间$X$和$Y$的直和,即$H=X\oplus Y$,那么$H$上的任意有界线性算子${\cal A}$可以表示为$2\times2$算子矩阵

(1.1) $\begin{equation}\label{eq:0.1}{\cal A}=\left( \begin{array}{cc} A ~~& B \\ C ~~& D \\ \end{array} \right), \end{equation} $

其中$A\in B(X), B\in B(Y, X), C\in B(X, Y), D\in B(Y).$由此,通过研究子块算子的性质可以刻画算子矩阵${\cal A}$的性质.实际问题上,算子矩阵出现在很多数学物理问题中,如流体力学、弹性力学、电磁学以及量子力学等数学物理问题.我们知道这些问题一部分可以导入无穷维Hamilton系统,与此对应的算子矩阵就是Hamilton算子矩阵.它是一类广泛应用的$2\times2$算子矩阵(见文献[1-5]).因此,算子矩阵理论的研究非常重要.

通常算子矩阵的谱与其内部算子的谱有着紧密联系,算子矩阵的谱是否可以由其内部算子的谱完全刻画呢?很多学者对此做了大量工作,例如1994年杜鸿科教授(见文献[6])对上三角算子矩阵$M_{C}=\left(\begin{array}{cc} A ~~& C \\ 0 ~~& B \\ \end{array} \right)$进行研究,得到$\sigma(M_{C})\subset\sigma(A)\cup\sigma(B)$; 2000年Han等人在文献[7]中给出,当$\sigma_{ap}(A)\cap\sigma_{\delta}(B)=\emptyset$时, $\sigma(M_{C})=\sigma(A)\cup\sigma(B).$此外, Zguitti, Zerouali等学者给出近似点谱、本质谱、Weyl谱、Browder谱、本质近似点谱和Browder本质近似点谱的类似结果(见文献[8-13]).鉴于算子矩阵谱性质的研究现状,本文主要考虑一般$2\times2$有界算子矩阵${\cal A}=\left(\begin{array}{cc} A ~~& B \\ C ~~& D \\ \end{array} \right)$的本质谱、Weyl谱与其内部算子谱的关系,得到

$ \begin{eqnarray*}\sigma_{e}({\cal A})=\{\lambda\in{\Bbb C} : C-(D-\lambda)B^{-1}(A-\lambda) \hbox{不是Fredholm算子}\} \end{eqnarray*} $

的充分条件,进而得到一些相关结论.

本文中,以$X, Y$和$Z$表示复无穷维Hilbert空间, $X^{*}$表示$X$的共轭空间, $X\times Y$表示$X$与$Y$的乘积空间.对于$X$的子空间$U, V$,以$U\oplus V$表示$U, V$的正交直和. $B(X, Y)$表示$X$到$Y$的有界线性算子全体; $B(X)$简记$X$到$X$的有界线性算子全体.对于$T\in B(X)$, $T^{*}$表示$T$的共轭算子, $\rho(T)$表示$T$的预解集, $N(T)$与$R(T)$分别表示$T$的零空间与值域, $T$的零空间的维数表示为$\alpha(T)$, $T$的商空间$X/R(T)$的余维数表示为$\beta(T)$.如果$\alpha(T)$或$\beta(T)$有限,则称$ind(T)=\alpha(T)-\beta(T)$为$T$的指标. $T$的升指数表示为$asc(T)$,是指满足$N(T^{k})=N(T^{k+1})$的最小非负整数$k$,如果这样的$k$不存在,则记$asc(T)=\infty$; $T$的降指数记为$des(T)$,是指满足$R(T^{k})=R(T^{k+1})$的最小非负整数$k$,如果这样的$k$不存在,则记$des(T)=\infty$.如无特殊说明,本文的有界线性算子矩阵${\cal A}$都是指(1.1)式给出的算子矩阵${\cal A}=\left(\begin{array}{cc} A ~~& B \\ C ~~& D \\ \end{array} \right)$.

定义2.1   $T\in B(X)$,

(ⅰ)若$T$的值域是闭的且$\alpha(T) < \infty$,则称$T$为左Fredholm算子;

(ⅱ)若$T$的$\beta(T) < \infty$,则称$T$为右Fredholm算子;

(ⅲ)若$T$既是左Fredholm算子又是右Fredholm算子,则称$T$是Fredholm算子.

定义2.2   $T\in B(X)$,

(ⅰ)如果$T$是左Fredholm算子且$ind(T)\leq 0$,则称$T$为左Weyl算子;

(ⅱ)如果$T$是右Fredholm算子且$ind(T)\geq 0$,则称$T$为右Weyl算子;

(ⅲ)如果$T$既是左Weyl算子又是右Weyl算子,则称$T$为Weyl算子.

定义2.3   $T\in B(X)$,则$T$的左本质谱、右本质谱、本质谱和Weyl谱定义为

$\begin{eqnarray*}&&\sigma_{le}(T)=\{\lambda\in {\Bbb C}: T-\lambda I \hbox{不是左Fredholm算子}\};\\&&\sigma_{re}(T)=\{\lambda\in {\Bbb C}: T-\lambda I \hbox{不是右Fredholm算子}\};\\&&\sigma_{e}(T)=\{\lambda\in {\Bbb C}: T-\lambda I \hbox{不是Fredholm算子}\};\\&&\sigma_{w}(T)=\{\lambda\in {\Bbb C}: T-\lambda I \hbox{不是Weyl算子}\}.\end{eqnarray*}$

定义2.4  设$A\in B(X, Y)$, $X=N(A)\oplus X_{0}$, $Y=R(A)\oplus Y_{0}$,其中$X_{0}$和$Y_{0}$分别是$X$和$Y$的闭子空间.定义映射$\widetilde{A}:X_{0}\times Y_{0}\rightarrow Y$为

(2.1) $\begin{equation}\widetilde{A}(x_{0}, y_{0})=Ax_{0}+y_{0}, \end{equation}$

称$\widetilde{A}$为$A$生成的双射.

引理2.1^[14]  如果$A:X\rightarrow Y$和$B:Y\rightarrow Z$是Fredholm算子,那么$BA$也是Fredholm算子,且$ ind(BA)=ind(A)+ind(B).$

引理2.2^[14]  令$A:X\rightarrow Y$是Fredholm算子, $\widetilde{A}$是由$A$生成的双射.如果$B:X\rightarrow Y$是有界线性算子且$\|B\| < \|\tilde{A}^{-1}\|^{-1}$,那么$A+B$是Fredholm算子,并且

(ⅰ) $\alpha(A+B)\leq \alpha(A); $

(ⅱ) $\beta(A+B)\leq \beta(A); $

(ⅲ) $ind(A+B)=ind(A).$

引理2.3^[14]  如果$A:X\rightarrow Y$是Fredholm算子, $K:Y\rightarrow Z$是紧算子,那么$A+K$也是Fredholm算子,并且$ ind(A+K)=ind(A).$

首先,利用算子矩阵的二次补方法研究算子矩阵${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$的次对角元素,得到${\cal A}$是Fredholm算子的充分必要条件.

定理3.1^[14]  设${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$是${\cal H}_{1}\times {\cal H}_{2}$上的有界线性算子.若$B$有有界逆算子,则对$\lambda\in{\Bbb C}$,有

(ⅰ) ${\cal A}-\lambda$是左Fredholm算子当且仅当$C-(D-\lambda)B^{-1}(A-\lambda)$是左Fredholm算子;

(ⅱ) ${\cal A}-\lambda$是右Fredholm算子当且仅当$C-(D-\lambda)B^{-1}(A-\lambda)$是右Fredholm算子.

证  因为$B$有有界逆算子,所以我们可以得到

(3.1) $\begin{equation}\label{eq:0} {\cal A}-\lambda=\left( \begin{array}{cc} I~~& 0 \\ (D-\lambda)B^{-1} ~~& I \\ \end{array} \right) \left( \begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right) \left( \begin{array}{cc} I~~& 0 \\ B^{-1}(A-\lambda) ~~& I \\ \end{array} \right), \end{equation}$

其中$\lambda\in {\Bbb C}$, $T_{1}(\lambda)=C-(D-\lambda)B^{-1}(A-\lambda).$

注意到(3.1)式中第一和最后的算子矩阵是${\cal H}_{1}\times {\cal H}_{2}$上的双射,并且$B$有有界逆算子,因此${\cal A}-\lambda$是左Fredholm算子当且仅当$\left(\begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right)$是左Fredholm算子.

又因为$B$是有界算子且有有界逆算子,所以$B$和$B^{^{-1}}$都是闭算子,从而$R(B)$是闭的.因此$R\left(\begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array}\right)$是闭的当且仅当$R(T_{1}(\lambda))$是闭的, $\alpha\left(\begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array}\right) < \infty$当且仅当$\alpha(T_{1}(\lambda)) < \infty.$所以(ⅰ)成立.

同理, ${\cal A}-\lambda$是右Fredholm算子当且仅当$\left(\begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right)$是右Fredholm算子.又因为

$ \begin{eqnarray*}\beta\left( \begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right) <\infty\Leftrightarrow \beta(T_{1}(\lambda) <\infty, \end{eqnarray*} $

所以(ⅱ)成立.结论证毕.

注3.1  设${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$是${\cal H}_{1}\times {\cal H}_{2}$上的有界线性算子, $C$有有界逆算子.则对$\lambda\in{\Bbb C}$,

(ⅰ) ${\cal A}-\lambda$是左Fredholm算子当且仅当$B-(A-\lambda)C^{-1}(D-\lambda)$是左Fredholm算子;

(ⅱ) ${\cal A}-\lambda$是右Fredholm算子当且仅当$B-(A-\lambda)C^{-1}(D-\lambda)$是右Fredholm算子.

推论3.1  设${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$是${\cal H}_{1}\times {\cal H}_{2}$上的有界线性算子, $B$有有界逆算子, $C$是Fredholm算子, $\widetilde{C}$是$C$生成的双射.若$\|(D-\lambda)B^{-1}(A-\lambda)\|\leq\|\widetilde{C}^{-1}\|^{-1}$,则

(ⅰ) ${\cal A}-\lambda$是Fredholm算子;

(ⅱ) $ind({\cal A}-\lambda)=ind C$;

(ⅲ)若$C$是Weyl算子,则${\cal A}-\lambda$是Weyl算子.

证  对$\lambda\in {\Bbb C}$,根据定理$3.1$知${\cal A}-\lambda$是Fredholm算子当且仅当$C-(D-\lambda)B^{-1}(A-\lambda)$是Fredholm算子.因为

(3.2) $\begin{equation}\|(D-\lambda)B^{-1}(A-\lambda)\|\leq\|\widetilde{C}^{-1}\|^{-1}, \end{equation} $

所以根据引理$2.2$知$C-(D-\lambda)B^{-1}(A-\lambda)$是Fredholm算子,并且

(3.3) $\begin{equation} \label{eq:t2}ind\left(C-(D-\lambda)B^{-1}(A-\lambda)\right)=ind(C).\end{equation}$

因此(ⅰ)成立.

因为$B$有有界逆算子,所以

(3.4) $\begin{equation}\label{eq:t1} {\cal A}-\lambda=\left( \begin{array}{cc} I~~& 0 \\ (D-\lambda)B^{-1} ~~& I \\ \end{array} \right) \left( \begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right) \left( \begin{array}{cc} I~~& 0 \\ B^{-1}(A-\lambda) ~~& I \\ \end{array} \right), \end{equation}$

其中$\lambda\in {\Bbb C}$, $T_{1}(\lambda)=C-(D-\lambda)B^{-1}(A-\lambda).$

注意到(3.4)式中第一和最后的算子矩阵是${\cal H}_{1}\times {\cal H}_{2}$上的双射,因此

(3.5) $ \begin{equation}ind\left({\cal A}-\lambda\right)= ind\left(C-(D-\lambda)B^{-1}(A-\lambda)\right). \end{equation} $

从而由(3.3式知(ⅱ)成立.

(ⅲ)因为$C$是Weyl算子,所以$B$和$C$都是Fredholm算子,并且

(3.6) $ \begin{equation}ind(B)=ind(C)=0. \end{equation}$

由(ⅰ)和(ⅱ)知${\cal A}-\lambda$是Fredholm算子,并且

(3.7) $ \begin{equation}ind({\cal A}-\lambda)=0. \end{equation} $

因此(ⅲ)成立.结论证毕.

推论3.2  设${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$是${\cal H}_{1}\times {\cal H}_{2}$上的有界线性算子, $B$有有界逆算子,则

(ⅰ) ${\cal A}-\lambda$是左Weyl算子当且仅当$C-(D-\lambda)B^{-1}(A-\lambda)$是左Weyl算子;

(ⅱ) ${\cal A}-\lambda$是右Weyl算子当且仅当$C-(D-\lambda)B^{-1}(A-\lambda)$是右Weyl算子.

证  令$T_{1}(\lambda)=C-(D-\lambda)B^{-1}(A-\lambda)$.

因为$B$有有界逆算子,所以$\alpha(B)=\beta(B)=0$,因此

(3.8) $\begin{equation}ind({\cal A}-\lambda)=ind\left( \begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right)=ind(T_{1}(\lambda)).\end{equation}$

根据定理$3.1$知(ⅰ)成立. (ⅱ)的证明与(ⅰ)相似.

注3.2  设${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$是${\cal H}_{1}\times {\cal H}_{2}$上的有界线性算子, $C$有有界逆算子,则

(ⅰ) ${\cal A}-\lambda$是左Weyl算子当且仅当$B-(A-\lambda)C^{-1}(D-\lambda)$是左Weyl算子;

(ⅱ) ${\cal A}-\lambda$是右Weyl算子当且仅当$B-(A-\lambda)C^{-1}(D-\lambda)$是右Weyl算子.

定理3.2  设${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$是${\cal H}_{1}\times {\cal H}_{2}$上的有界线性算子.若$B$有有界逆算子,则

(3.9) $\begin{equation}\label{eq:01}\sigma_{e}({\cal A})=\{\lambda\in{\Bbb C} : C-(D-\lambda)B^{-1}(A-\lambda) \hbox{不是Fredholm算子}\}; \end{equation} $

(3.10) $\begin{equation}\label{eq:02}\sigma_{w}({\cal A})=\{\lambda\in{\Bbb C} : C-(D-\lambda)B^{-1}(A-\lambda) \hbox{不是Weyl算子}\}. \end{equation} $

证  因为$B$有有界逆算子,所以我们可以得到

(3.11) $\begin{equation}\label{eq:3} {\cal A}-\lambda=\left( \begin{array}{cc} I~~& 0 \\ (D-\lambda)B^{-1} ~~& I \\ \end{array} \right) \left( \begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right) \left( \begin{array}{cc} I~~& 0 \\ B^{-1}(A-\lambda) ~~& I \\ \end{array} \right), \end{equation}$

其中$\lambda\in {\Bbb C}$, $T_{1}(\lambda)=C-(D-\lambda)B^{-1}(A-\lambda).$

注意到(3.11)式中第一和最后的算子矩阵是${\cal H}_{1}\times {\cal H}_{2}$上的双射,并且$B$有有界逆算子,因此

(3.12) $\begin{equation}\alpha({\cal A}-\lambda)=\alpha(\left( \begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right))=\alpha\left(T_{1}(\lambda)\right), \end{equation}$

(3.13) $ \begin{equation}\beta({\cal A}-\lambda)=\beta(\left( \begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right))=\beta(T_{1}(\lambda)).\end{equation}$

根据定理3.1知${\cal A}-\lambda$是Fredholm算子当且仅当$T_{1}(\lambda)$是Fredholm算子,从而(3.9)式成立.

由推论3.2知${\cal A}-\lambda$是Weyl算子当且仅当$T_{1}(\lambda)$是Weyl算子,从而(3.10)式成立.结论证毕.

注3.3  设${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$是${\cal H}_{1}\times {\cal H}_{2}$上的有界线性算子, $C$有有界逆算子,则

(3.14)$ \begin{equation}\label{eq:03}\sigma_{e}({\cal A})=\{\lambda\in{\Bbb C} : B-(A-\lambda)C^{-1}(D-\lambda) \hbox{不是Fredholm算子}\}; \end{equation} $

(3.15)$ \begin{equation}\label{eq:04}\sigma_{w}({\cal A})=\{\lambda\in{\Bbb C} : B-(A-\lambda)C^{-1}(D-\lambda) \hbox{不是Weyl算子}\}. \end{equation} $

定理3.3  设${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$是${\cal H}_{1}\times{\cal H}_{2}$上的有界线性算子.若$B$有有界逆算子, $C$是紧算子,则

$ \sigma_{e}({\cal A})\subseteq(\sigma_{e}(A)\cup\sigma_{e}(D)), \sigma_{w}({\cal A})\subseteq(\sigma_{w}(A)\cup\sigma_{w}(D)). $

证  若$\lambda\notin \sigma_{e}(A)\cup\sigma_{e}(D)$,则$(D-\lambda)$和$(A-\lambda)$都是Fredholm算子.因为$B$有有界逆算子,所以$B$是Fredholm算子.根据引理2.1知, $(D-\lambda)B^{-1}(A-\lambda)$是Fredholm算子.因为$C$是紧算子,所以根据引理2.2知$T_{1}(\lambda)$是Fredholm算子.由定理3.2知$\lambda\notin \sigma_{e}({\cal A}).$

根据引理2.3,我们得到

(3.16) $ \begin{equation}ind({\cal A}-\lambda)=ind(\left( \begin{array}{cc} 0~~&B \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right))=ind(T_{1}(\lambda))=ind(A-\lambda)+ind(D-\lambda).\end{equation}$

若$\lambda\notin \sigma_{w}(A)\cup\sigma_{w}(D)$,则$\lambda\notin \sigma_{e}({\cal A})$且$ind(A-\lambda)=ind(D-\lambda)=0.$因此$\lambda\notin \sigma_{w}({\cal A}).$结论证毕.

推论3.3  设${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$是${\cal H}_{1}\times{\cal H}_{2}$上的有界线性算子,若$C$有有界逆算子, $B$是紧算子,则

$\sigma_{e}({\cal A})\subseteq(\sigma_{e}(A)\cup\sigma_{e}(D)), \sigma_{w}({\cal A})\subseteq(\sigma_{w}(A)\cup\sigma_{w}(D)).$

例3.1  考虑有界线性算子矩阵

${\cal A}=\left( \begin{array}{cc} A ~~& I \\ C~~&D \\ \end{array} \right), $

其中$I$表示单位算子, $A, B, C\in L(l^{2})$分别定义为

$A(x_{1}, x_{2}, x_{3}, \cdots)=(0, x_{1}, 0, \frac{1}{2}x_{2}, 0, \frac{1}{3}x_{3}, \cdots), $

$C(x_{1}, x_{2}, x_{3}, \cdots)=(0, x_{2}, 0, x_{4}, 0, x_{6}, \cdots), $

$D(x_{1}, x_{2}, x_{3}, \cdots)=(0, 0, x_{2}, 0, x_{3}, 0, x_{4}, \cdots).$

显然,对$\lambda\in {\Bbb C}$,有

$ \begin{eqnarray*}{\cal A}-\lambda=\left( \begin{array}{cc} I~~& 0 \\ (D-\lambda) & I \\ \end{array} \right) \left( \begin{array}{cc} 0 & I \\ T_{1}(\lambda) ~ & 0 \\ \end{array} \right) \left( \begin{array}{cc} I~~& 0 \\ (A-\lambda) & I \\ \end{array} \right), \end{eqnarray*} $

其中$T_{1}(\lambda)=C-(D-\lambda)(A-\lambda)$.根据定理3.2知

(3.17) $ \begin{equation}\label{eq:3.1} \sigma_{e}({\cal A})=\{\lambda\in{\Bbb C} : C-(D-\lambda)(A-\lambda) \hbox{不是Fredholm算子}\}, \end{equation}$

(3.18) $ \begin{equation}\label{eq:3.2} \sigma_{w}({\cal A})=\{\lambda\in{\Bbb C} : C-(D-\lambda)(A-\lambda) \hbox{不是Weyl算子}\}. \end{equation}$

另一方面(参考文献[10]),通过直接计算得到

$\sigma_{p+}(A)\setminus\sigma_{le}(A)=\emptyset, \sigma_{p\infty}(A^{*})^{*}\setminus\sigma_{re}(A^{*})^{*}=\emptyset, $

并且

$\sigma_{p+}(D)\setminus\sigma_{re}(D)=\emptyset, \sigma_{p\infty}(D)\setminus\sigma_{re}(D)=\emptyset, \sigma_{p\infty}(D)=\{0, 1\}, $

其中

$\sigma_{p+}(A)=\{\lambda\in\sigma_{p}(A): \alpha(A-\lambda I)>\beta(A-\lambda I)\}, $

$ \sigma_{p\infty}(A)=\{\lambda\in\sigma_{p}(A): \alpha(A-\lambda I)=\infty\}. $

从而(3.17)式和(3.18)式成立.

下面,我们研究算子矩阵${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$的主对角元素的性质.讨论主对角元素本质谱、Weyl谱与算子矩阵${\cal A}$的本质谱、Weyl谱的关系.

定理4.1  设${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$是${\cal H}_{1}\times{\cal H}_{2}$上的有界线性算子,则对任意$\lambda\in\rho(D)$,

(ⅰ) ${\cal A}-\lambda$是左Fredholm算子当且仅当$A-\lambda-B(D-\lambda)^{-1}C$是左Fredholm算子;

(ⅱ) ${\cal A}-\lambda$是右Fredholm算子当且仅当$A-\lambda-B(D-\lambda)^{-1}C$是右Fredholm算子.

证  对任意$\lambda\in\rho(D)$,我们有

(4.1) $\begin{equation}\label{eq:4.01} {\cal A}-\lambda =\left( \begin{array}{cc} I~~ & B(D-\lambda)^{-1} \\ 0 ~~& I \\ \end{array} \right) \left( \begin{array}{cc} S_{1}(\lambda) ~~& 0 \\ 0 ~~& D-\lambda \\ \end{array} \right) \left( \begin{array}{cc} I~~& 0 \\ (D-\lambda)^{-1}C~~& I \\ \end{array} \right), \end{equation}$

其中$S_{1}(\lambda)=A-\lambda-B(D-\lambda)^{-1}C.$

注意到在(4.1)式中,第一和最后一个块算子矩阵都是有界算子且有有界逆算子.因此${\cal A}-\lambda$是左Fredholm算子当且仅当$\left(\begin{array}{cc} S_{1}(\lambda) ~ & 0 \\ 0 ~~& D-\lambda \\ \end{array} \right)$是左Fredholm算子.

因为$D-\lambda$是双射,所以$R(D-\lambda)$是闭集.因此$R(\left(\begin{array}{cc} S_{1}(\lambda) ~~& 0 \\ 0 ~~& D-\lambda \\ \end{array} \right))$是闭集当且仅当$R(S_{1}(\lambda))$是闭集,并且$\alpha(\left(\begin{array}{cc} S_{1}(\lambda) ~~& 0 \\ 0 ~~& D-\lambda \\ \end{array} \right)) < \infty$当且仅当$\alpha(S_{1}(\lambda)) < \infty.$从而(ⅰ)成立.

同理, ${\cal A}-\lambda$是右Fredholm算子当且仅当$\left(\begin{array}{cc} S_{1}(\lambda) ~ & 0 \\ 0 ~~& D-\lambda \\ \end{array} \right)$是右Fredholm算子.因为$\beta(D-\lambda)=0$,所以$\beta(\left(\begin{array}{cc} S_{1}(\lambda) ~~& 0 \\ 0~~ & D-\lambda \\ \end{array} \right)) < \infty$当且仅当$\beta(S_{1}(\lambda)) < \infty$.从而(ⅱ)成立.结论证毕.

注4.1  设${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$是${\cal H}_{1}\times{\cal H}_{2}$上的有界线性算子,则对任意$\lambda\in\rho(A)$,

(ⅰ)${\cal A}-\lambda$是左Fredholm算子当且仅当$D-\lambda-C(A-\lambda)^{-1}B$是左Fredholm算子;

(ⅱ)${\cal A}-\lambda$是右Fredholm算子当且仅当$D-\lambda-C(A-\lambda)^{-1}B$是右Fredholm算子.

推论4.1  设${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$是${\cal H}_{1}\times{\cal H}_{2}$上的有界线性算子.

(ⅰ)若$A-\lambda$是Fredholm算子, $\widetilde{A-\lambda}$是$A-\lambda$生成的双射, $\lambda\in\rho(D)$,且$\|B(D-\lambda)^{-1}C\|\leq\|(\widetilde{A-\lambda})^{-1}\|^{-1}$,则${\cal A}-\lambda$是Fredholm算子;

(ⅱ)若$\lambda\in(\rho(A)\cap\rho(D))$,且$\|B(D-\lambda)^{-1}C\|\leq\|(\widetilde{A-\lambda})^{-1}\|^{-1}$,则${\cal A}-\lambda$是Fredholm算子.

证  (ⅰ)对$\lambda\in\rho(D)$,根据定理4.1知${\cal A}-\lambda$是Fredholm算子当且仅当$A-\lambda-B(D-\lambda)^{-1}C$是Fredholm算子.因为$A-\lambda$是Fredholm算子,并且$\|B(D-\lambda)^{-1}C\|\leq\|(\widetilde{A-\lambda})^{-1}\|^{-1}$,所以由引理2.2知$A-\lambda-B(D-\lambda)^{-1}C$是Fredholm算子.因此${\cal A}-\lambda$是Fredholm算子.

(ⅱ)对$\lambda\in(\rho(A)\cap\rho(D))$,则$A-\lambda$是双射.因此

(4.2) $\begin{equation}(\widetilde{A-\lambda})(x_{0}, y_{0})=Ax_{0}, \end{equation} $

(4.3) $\begin{equation}\|(\widetilde{A-\lambda})^{-1}\|=\|(A-\lambda)^{-1}\|. \end{equation} $

从而

(4.4) $\begin{equation}\|B(D-\lambda)^{-1}C\|\leq\|(A-\lambda)^{-1}\|^{-1} =\|(\widetilde{A-\lambda})^{-1}\|^{-1}. \end{equation}$

由(ⅰ)知${\cal A}-\lambda$是Fredholm算子.结论证毕.

注4.2  设${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$是${\cal H}_{1}\times{\cal H}_{2}$上的有界线性算子.

(ⅰ)若$\lambda\in\rho(A)$, $D-\lambda$是Fredholm算子, $\widetilde{D-\lambda}$是$D-\lambda$生成的双射,且$\|C(A-\lambda)^{-1}B\|\leq\|(\widetilde{D-\lambda})^{-1}\|^{-1}$,则${\cal A}-\lambda$是Fredholm算子;

(ⅱ)若$\lambda\in(\rho(A)\cap\rho(D))$,且$\|C(A-\lambda)^{-1}B\|\leq\|(\widetilde{D-\lambda})^{-1}\|^{-1}$,则${\cal A}-\lambda$是Fredholm算子.

推论4.2  设${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$是${\cal H}_{1}\times{\cal H}_{2}$上的有界线性算子,则对任意$\lambda\in \rho(D)$,

(ⅰ) ${\cal A}-\lambda$是左Weyl算子当且仅当$A-\lambda-B(D-\lambda)^{-1}C$是左Weyl算子;

(ⅱ) ${\cal A}-\lambda$是右Weyl算子当且仅当$A-\lambda-B(D-\lambda)^{-1}C$是右Weyl算子.

证  对任意$\lambda\in\rho(D)$,有

(4.5) $\begin{equation}\label{eq:4.02} {\cal A}-\lambda =\left( \begin{array}{cc} I~~ & B(D-\lambda)^{-1} \\ 0~~ & I \\ \end{array} \right) \left( \begin{array}{cc} S_{1}(\lambda) ~ & 0 \\ 0~~ & D-\lambda \\ \end{array} \right) \left( \begin{array}{cc} I~~& 0 \\ (D-\lambda)^{-1}C~~& I \\ \end{array} \right), \end{equation} $

其中$S_{1}(\lambda)=A-\lambda-B(D-\lambda)^{-1}C$.根据定理4.1知${\cal A}-\lambda$是左Fredholm算子当且仅当$A-\lambda-B(D-\lambda)^{-1}C$是左Fredholm算子.

注意到(4.5)式中第一和最后一个算子矩阵是双射.因此

(4.6) $\begin{equation}ind({\cal A}-\lambda)=ind \left( \begin{array}{cc} S_{1}(\lambda) ~~& 0 \\ 0 ~~& D-\lambda \\ \end{array} \right)=ind(S_{1}(\lambda))+ind(D-\lambda)).\end{equation} $

因为$\lambda\in\rho(D)$,并且$ind(D-\lambda)=0$,所以

(4.7) $\begin{equation}ind({\cal A}-\lambda)=ind(S_{1}(\lambda)). \end{equation}$

从而(ⅰ)成立. (ⅱ)的证明与(ⅰ)相似.结论证毕.

注4.3  设${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$是${\cal H}_{1}\times{\cal H}_{2}$上的有界线性算子,则对任意$\lambda\in \rho(A)$,

(ⅰ) ${\cal A}-\lambda$是左Weyl算子当且仅当$D-\lambda-C(A-\lambda)^{-1}B$是左Weyl算子;

(ⅱ) ${\cal A}-\lambda$是右Weyl算子当且仅当$D-\lambda-C(A-\lambda)^{-1}B$是右Weyl算子.

定理4.2  设${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$是${\cal H}_{1}\times{\cal H}_{2}$上的有界线性算子,那么

(ⅰ) $\rho(D)\cap\sigma_{e}({\cal A})=\rho(D)\cap\{\lambda\in{\Bbb C} : A-\lambda-B(D-\lambda)^{-1}C$不是Fredholm算子};

(ⅱ) $\rho(D)\cap\sigma_{w}({\cal A})=\rho(D)\cap\{\lambda\in{\Bbb C} : A-\lambda-B(D-\lambda)^{-1}C$不是Weyl算子}.

证  对$\lambda\in(\rho(D)\cap\sigma_{e}({\cal A}))$,得到

(4.8) $\begin{equation}\label{eq:4.4} {\cal A}-\lambda =\left( \begin{array}{cc} I ~~& B(D-\lambda)^{-1} \\ 0 ~~& I \\ \end{array} \right) \left( \begin{array}{cc} S_{1}(\lambda) ~~& 0 \\ 0 ~~& D-\lambda \\ \end{array} \right) \left( \begin{array}{cc} I~~& 0 \\ (D-\lambda)^{-1}C~~& I \\ \end{array} \right), \end{equation}$

其中$S_{1}(\lambda)=A-\lambda-B(D-\lambda)^{-1}C$.

利用定理4.1知${\cal A}-\lambda$是Fredholm算子当且仅当$S_{1}(\lambda)$是Fredholm算子.因此(ⅰ)成立.

因为$\lambda\in\rho(D)$,所以

(4.9) $\begin{equation}ind({\cal A}-\lambda)=ind(S_{1}(\lambda)). \end{equation}$

因此(ⅱ)成立.结论证毕.

注4.4  设${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \\ \end{array} \right)$是${\cal H}_{1}\times{\cal H}_{2}$上的有界线性算子,那么

(ⅰ) $\rho(A)\cap\sigma_{e}({\cal A})=\rho(A)\cap\{\lambda\in{\Bbb C} : D-\lambda-C(A-\lambda)^{-1}B$不是Fredholm算子};

(ⅱ) $\rho(A)\cap\sigma_{w}({\cal A})=\rho(A)\cap\{\lambda\in{\Bbb C} : D-\lambda-C(A-\lambda)^{-1}B$不是Weyl算子}.

例4.1  设$A, B, C, D\in B(l^{2})$定义为

$A(x_{1}, x_{2}, x_{3}, \cdots)=(0, 0, 0, \frac{1}{2}x_{2}, 0, \frac{1}{3}x_{3}, \cdots), $

$B(x_{1}, x_{2}, x_{3}, \cdots)=(x_{1}, 0, x_{2}, 0, x_{3}\cdots), $

$C(x_{1}, x_{2}, x_{3}, \cdots)=(\alpha_{1}x_{1}, \alpha_{2}x_{2}, \alpha_{3}x_{3}, \cdots), $

$D(x_{1}, x_{2}, x_{3}, \cdots)=(0, x_{2}, 0, x_{4}, 0, x_{6}, \cdots), $

其中$\lim \alpha_{n}=0.$考虑有界线性算子矩阵${\cal A}=\left(\begin{array}{cc} A~~&B \\ C~~&D \end{array} \right).$

直接计算得到$\sigma(A)=\sigma_{e}(A)=\sigma_{w}(A)=\{0\}$, $\sigma(D)=\sigma_{e}(D)=\sigma_{w}(D)=\{0, 1\}$.因为

(4.10) $\begin{equation}\lim\alpha_{n}=0, \end{equation} $

所以$C(A-\lambda)^{-1}B$是紧算子,从而

(4.11) $\begin{equation}\{\lambda\in{\Bbb C} : D-\lambda-C(A-\lambda)^{-1}B不是{\rm{Fredholm}}算子\}=\{1\}. \end{equation}$

根据注4.4知$\sigma_{e}({\cal A})=\sigma_{w}({\cal A})=\{1\}$.