Cowen和Douglas^[1]在Hilbert空间上引入并研究一类重要算子—Cowen-Douglas算子. Cowen-Douglas算子是根据全纯向量丛定义的, 这是首次应用复几何研究算子理论. 此后, Cowen-Douglas算子得到了广泛且深入的研究, 特别是这类算子的相似性与酉相似性取得了丰硕的成果^[2-10]. Ji等^[3]在Hilbert空间上引入了一类用上三角算子矩阵表示的Cowen-Douglas算子, 并研究其强不可约性. 本文作者^[9]在Banach空间上研究这类Cowen-Douglas算子的强不可约性. Hou和Ji^[2]在Hilbert空间上研究形如$ \left( \begin{array}{cc} T_1\; & A_{12}T_2-T_1A_{12} \\ 0\; & T_2 \end{array}\right ) $的$ 2\times 2 $上三角算子矩阵的酉相似性. 本文引入类似的$ n\times n $上三角算子矩阵, 并在Banach空间上研究其相似性, 在Hilbert空间上研究其酉相似性.

首先回顾一些符号和定义. 设$ X, Y $是Banach空间, 以$ B(X, Y) $表示从$ X $到$ Y $的所有有界线性算子全体, $ B(X, X) $简记为$ B(X) $. $ X $上的恒等算子记为$ I_X $, 简记为$ I $. 设$ T\in B(X, Y) $, 以$ \ker T $与$ \rm{ran}\mathit{T} $分别表示$ T $的零空间$ \ker T:=\{x\in X: Tx=0\} $与值域$ \rm{ran}\mathit{T}:=\{\mathit{T}\mathit{x}: \mathit{x}\in \mathit{X}\} $. $ X $的子集$ A $的线性张闭包记为$ \overline{\mbox{span}}A $. 设$ T\in B(X, Y) $, 若存在$ S\in B(Y, X) $, 使得$ TS=I_Y $, $ ST=I_X $, 则称$ T $可逆, 若只有$ TS=I_Y $成立, 则称$ T $右可逆. 分别以$ {\cal G}(X) $与$ {\cal G}_r(X) $表示$ B(X) $中的可逆算子全体与右可逆算子全体.

定义1.1   设$ X $是Banach空间, $ T, S\in B(X) $, 若存在可逆算子$ U\in B(X) $, 使得$ UT=SU $, 则称$ T $与$ S $相似.

定义1.2   设$ H $是Hilbert空间, $ T, S\in B(H) $, 若存在酉算子$ U\in B(H) $, 即$ UU^*=U^*U=I $, 使得$ UT=SU $, 则称$ T $与$ S $酉相似.

定义1.3   设$ X, Y $是Banach空间, $ T\in B(X) $, $ S\in B(Y) $, Rosenblum算子$ \tau_{T, S} $定义为

$\tau_{T, S}: B(Y, X)\rightarrow B(Y, X): \tau_{T, S}(A)=TA-AS, \ A\in B(Y, X)$.

定义1.4   设$ X $是Banach空间, $ \Omega $是复平面上的一个非空连通开集, $ n $是正整数, $ T\in B(X) $. 若对任意$ \omega\in \Omega $, 有$ \dim\ker (T-\omega)=n $, $ \mbox{ran} (T-\omega)=X $, 且$ \overline{\mbox{span}}\{\ker (T-\omega): \omega\in\Omega\}=X $, 则称$ T $是$ \Omega $上指标为$ n $的Cowen-Douglas算子, 记为$ T\in{\cal B}_n(\Omega)(X) $.

本文研究如下一类上三角算子矩阵.

定义1.5   设$ X $是Banach空间且有直和分解$ X=X_1\oplus X_2\oplus\cdots\oplus X_n $, 考虑如下类型的算子$ T\in B(X) $

$ T=\left( \begin{array}{cccc} T_1 &\; T_{12} \; & \cdots & \; T_{1n} \\ & T_2 & \ddots & \; \vdots \\ & & \ddots & \; T_{n-1, n}\\ 0 & & & \; T_n \\ \end{array}\right ), $

其中$ T_i\in B(X_i) $, $ 1\leq i\leq n $; $ T_{ij}=A_{ij}T_j-T_iA_{ij} $, $ A_{ij}\in B(X_j, X_i) $, $ 1\leq i<j\leq n $.

当$ T_{ij}=0 $ ($ 1\leq i<j\leq n $)时, 记上述算子$ T $为$ T=T_1\oplus T_2\oplus\cdots\oplus T_n $.

为了方便, 有时也记$ T_{i} $为$ T_{ii} $.

首先在Banach空间上研究定义1.5中算子的相似性.

定理2.1   设Banach空间$ X $与算子$ T\in B(X) $如定义1.5, 若$ A_{ij}A_{jk}=0 $, $ A_{ij}T_jA_{jk}=0 $ ($ 1\leq i<j<k\leq n $), 则$ T $与$ T_1\oplus T_2\oplus\cdots\oplus T_n $相似.

证   对$ 1\leq i<j<k\leq n $, 由于$ A_{ij}A_{jk}=0 $, $ A_{ij}T_jA_{jk}=0 $, 则

$ A_{ij}T_{jk}=A_{ij}A_{jk}T_k-A_{ij}T_jA_{jk}=0, $

$ T_{ij}-A_{ij}T_{j}=A_{ij}T_{j}-T_iA_{ij}-A_{ij}T_{j}=-T_iA_{ij}. $

记$ U={\left( \begin{array}{cccc} I & \; -A_{12} \; & \cdots & -A_{1n} \\ & I & \ddots & \vdots \\ & & \ddots & -A_{n-1, n} \\ 0 & & & I \\ \end{array}\right )}, $$ V={\left( \begin{array}{cccc} I &\; A_{12}\; & \cdots & A_{1n} \\ & I & \ddots & \vdots \\ & & \ddots & A_{n-1, n} \\ 0 & & & I \\ \end{array}\right )}, $则$ U, V\in B(X) $且

$UV={\left( \begin{array}{ccccc} I0 &\; -A_{12}A_{23} \; & \cdots & -{\sum\limits_{i=2}^{n-1}}A_{1i}A_{in} \\ I & 0 & \ddots & \vdots \\ & \ddots & \ddots & -A_{n-2, n-1}A_{n-1, n} \\ & & \ddots & 0 \\ 0 & & & I \\ \end{array}\right )}=I.$

同理可证$ VU=I $, 即$ U $可逆. 又由于

$ \begin{eqnarray*} UT&=&{\left( \begin{array}{ccccc} T_1 & \; T_{12}-A_{12}T_{2}\; & T_{13}-A_{12}T_{23}-A_{13}T_{3} & \cdots & T_{1n}-{\sum\limits_{i=2}^{n}}A_{1i}T_{in} \\ & T_2 & T_{23}-A_{23} & \ddots & \vdots \\ & & \ddots T_{3} & \ddots & T_{n-2, n}-A_{n-2, n-1}T_{n-1, n}-A_{n-2, n}T_{n} \\ & & & \ddots & T_{n-1, n}-A_{n-1, n}T_{n} \\ 0 & & & & T_n \\ \end{array}\right )}\\ &=&{\left( \begin{array}{ccccc} T_1 & \; -T_1A_{12} \; & -T_1A_{13} & \cdots & -T_1A_{1n} \\ & T_2 & -T_2A_{23} & \ddots & \vdots \\ & & \ddots & \ddots & -T_{n-2}A_{n-2, n} \\ & & & \ddots & -T_{n-1}A_{n-1, n} \\ 0 & & & & T_n \\ \end{array}\right )}=(T_1\oplus T_2\oplus\cdots\oplus T_n)U, \end{eqnarray*} $

则$ T $与$ T_1\oplus T_2\oplus\cdots\oplus T_n $相似. 证毕.

推论2.1   设Banach空间$ X $与算子$ T\in B(X) $如定义1.5, 且定义1.5中的$ n=2 $, 则$ T $与$ T_1\oplus T_2 $相似.

引理2.1   设Banach空间$ X $与算子$ T\in B(X) $如定义1.5, 满足$ \ker\tau_{T_i, T_j}=\{0\} $ ($ 1\leq j<i\leq n $). 若存在$ U=(U_{ij})_{n\times n}\in B(X) $, 使得$ UT=(T_1\oplus T_2\oplus\cdots\oplus T_n)U $, 则$ U_{ij} $满足

(1) $ U_{ij}=0 $ ($ 1\leq j<i\leq n $),

(2) $ U_{ii}\in \ker\tau_{T_i, T_i} $ ($ 1\leq i\leq n $),

(3) $ U_{i, i+1}+U_{ii}A_{i, i+1}\in \ker\tau_{T_i, T_{i+1}} $ ($ 1\leq i<n-1 $),

(4) $ {\sum\limits_{m=i+1}^{j-1}}U_{im}T_{mj}=\tau_{T_i, T_{j}}(U_{ij}+U_{ii}A_{ij}) $ ($ 1\leq i<j-1\leq n-1 $).

证   记$ UT=(T_1\oplus T_2\oplus\cdots\oplus T_n)U=(V_{ij})_{n\times n} $.

对$ 1<i\leq n $, 由于$ V_{i1}=U_{i1}T_1=T_{i}U_{i1} $, 则$ U_{i1}\in \ker\tau_{T_i, T_1}=\{0\} $, 即$ U_{i1}=0 $. 此时对$ 2<i\leq n $, 有$ V_{i2}=U_{i2}T_2=T_{i}U_{i2} $, 则$ U_{i2}\in \ker\tau_{T_i, T_2}=\{0\} $, 即$ U_{i2}=0 $. 递推可得: 对$ 1\leq j<i\leq n $, 有$ U_{ij}=0 $.

对$ 1\leq i\leq n $, 由于$ V_{ii}=U_{ii}T_i=T_{i}U_{ii} $, 则$ U_{ii}\in \ker\tau_{T_i, T_i} $.

对$ 1\leq i<j\leq n $, 有

$ V_{ij}={\sum\limits_{m=i}^{j}}U_{im}T_{mj}=T_{i}U_{ij}. $

特别地, 对$ 1\leq i<n-1 $, 有

$ V_{i, i+1}=U_{ii}T_{i, i+1}+U_{i, i+1}T_{i+1}=T_{i}U_{i, i+1}. $

由于

$ \begin{eqnarray*} U_{ii}T_{i, i+1}&=&U_{ii}(A_{i, i+1}T_{i+1}-T_{i}A_{i, i+1}) =U_{ii}A_{i, i+1}T_{i+1}-U_{ii}T_{i}A_{i, i+1} \\ &=&U_{ii}A_{i, i+1}T_{i+1}-T_{i}U_{ii}A_{i, i+1}, \end{eqnarray*} $

则

$ \begin{eqnarray*} (U_{ii}A_{i, i+1}+U_{i, i+1})T_{i+1}&=&U_{ii}A_{i, i+1}T_{i+1}+U_{i, i+1}T_{i+1} \\ &=&T_{i}U_{i, i+1}+T_{i}U_{ii}A_{i, i+1}=T_{i}(U_{i, i+1}+U_{ii}A_{i, i+1}). \end{eqnarray*} $

即$ U_{i, i+1}+U_{ii}A_{i, i+1}\in \ker\tau_{T_i, T_{i+1}} $.

对$ 1\leq i<j-1\leq n-1 $, 有

$ V_{ij}=U_{ii}T_{ij}+{\sum\limits_{m=i+1}^{j-1}}U_{im}T_{mj}+U_{ij}T_{j}=T_{i}U_{ij}. $

由于$ U_{ii}T_{ij}=U_{ii}(A_{ij}T_{j}-T_{i}A_{ij}) $$ =U_{ii}A_{ij}T_{j}-U_{ii}T_{i}A_{ij} $$ =U_{ii}A_{ij}T_{j}-T_{i}U_{ii}A_{ij} $, 则

$ \begin{eqnarray*} \sum\limits_{m=i+1}^{j-1}U_{im}T_{mj}&=&T_{i}U_{ij}+T_{i}U_{ii}A_{ij}-U_{ij}T_{j}-U_{ii}A_{ij}T_{j} \\ &=&T_{i}(U_{ij}+U_{ii}A_{ij})-(U_{ij}+U_{ii}A_{ij})T_{j}=\tau_{T_i, T_{j}}(U_{ij}+U_{ii}A_{ij}). \end{eqnarray*} $

证毕.

定理2.2   设Banach空间$ X $与算子$ T\in B(X) $如定义1.5, 满足$ \ker\tau_{T_i, T_j}=\{0\} $ ($ 1\leq j<i\leq n $), $ \ker\tau_{T_i, T_i}\bigcap {\cal G}_r(X_i)\subseteq {\cal G}(X_i) $ ($ 1\leq i\leq n $), 则$ T $与$ T_1\oplus T_2\oplus\cdots\oplus T_n $相似当且仅当存在$ U_{ij}\in B(X_j, X_i) $ ($ 1\leq i\leq j\leq n $), 满足

(1) $ U_{ii}\in \ker\tau_{T_i, T_i}\bigcap {\cal G}(X_i) $ ($ 1\leq i\leq n $),

(2) $ U_{i, i+1}+U_{ii}A_{i, i+1}\in \ker\tau_{T_i, T_{i+1}} $ ($ 1\leq i<n-1 $),

(3) $ {\sum\limits_{m=i+1}^{j-1}}U_{im}T_{mj}=\tau_{T_i, T_{j}}(U_{ij}+U_{ii}A_{ij}) $ ($ 1\leq i<j-1\leq n-1 $).

证   “$ \Rightarrow $”: 由$ T $与$ T_1\oplus T_2\oplus\cdots\oplus T_n $相似可知: 存在可逆算子$ U=(U_{ij})_{n\times n}\in B(X) $, 使得$ UT=(T_1\oplus T_2\oplus\cdots\oplus T_n)U $. 首先由引理2.1可得$ (2)\mbox{、}\ (3) $成立, 且

$ U_{ij}=0\ (1\leq j<i\leq n), \ \ U_{ii}\in \ker\tau_{T_i, T_i}\ (1\leq i\leq n). $

记$ U^{-1}=B=(B_{ij})_{n\times n}\in B(X) $, 则$ UB=BU=I $. 记$ UB=(W_{ij})_{n\times n}=I $, 则

$ I=W_{nn}=U_{nn}B_{nn}, $

即$ U_{nn}\in {\cal G}_r(X_n) $, 故$ U_{nn}\in \ker\tau_{T_n, T_n}\bigcap{\cal G}_r(X_n) $, 所以$ U_{nn}\in {\cal G}(X_n) $. 此时对$ 1\leq i\leq n-1 $, 有$ 0=W_{ni}=U_{nn}B_{ni}, $则$ B_{ni}=0 $. 故

$ I=W_{n-1, n-1}=U_{n-1, n-1}B_{n-1, n-1}+U_{n-1, n}B_{n, n-1}=U_{n-1, n-1}B_{n-1, n-1}, $

即$ U_{n-1, n-1}\in {\cal G}_r(X_{n-1}) $, 故$ U_{n-1, n-1}\in \ker\tau_{T_{n-1}, T_{n-1}}\bigcap{\cal G}_r(X_{n-1}) $, 所以$ U_{n-1, n-1}\in {\cal G}(X_{n-1}) $. 递推可得$ U_{ii}\in {\cal G}(X_i) $ ($ 1\leq i\leq n $). 即(1)成立.

"$ \Leftarrow $": 令$ U=(U_{ij})_{n\times n}\in B(X) $, 其中$ U_{ij}=0 $ ($ 1\leq j<i\leq n $), $ U_{ij} $满足条件(1)–(3) ($ 1\leq i\leq j\leq n $). 由于$ U_{ii}\in {\cal G}(X_i) $ ($ 1\leq i\leq n $), 则$ U $可逆.

记$ UT=(V_{ij})_{n\times n} $, $ (T_1\oplus T_2\oplus\cdots\oplus T_n)U=(W_{ij})_{n\times n} $, 则

$ V_{ij}={\sum\limits_{m=1}^{j-1}}U_{im}T_{mj}+U_{ij}T_{j}, \ \ W_{ij}=T_iU_{ij}. $

由于$ U_{ij}=0 $ ($ 1\leq j<i\leq n $), 则当$ 1\leq j<i\leq n $时, 则$ V_{ij}=0=W_{ij} $. 当$ 1\leq i\leq j\leq n $时,

$ V_{ij}={\sum\limits_{m=i}^{j-1}}U_{im}T_{mj}+U_{ij}T_{j}, \ \ W_{ij}=T_iU_{ij}. $

当$ 1\leq i=j\leq n $时, 由于$ U_{ii}\in \ker\tau_{T_i, T_i} $, 即$ U_{ii}T_{i}=T_iU_{ii} $, 则$ V_{ij}=W_{ij} $. 当$ 1\leq i\leq n-1 $, $ j=i+1 $时, 由于$ U_{i, i+1}+U_{ii}A_{i, i+1}\in \ker\tau_{T_i, T_{i+1}} $, 则$ (U_{i, i+1}+U_{ii}A_{i, i+1})T_{i+1}=T_i(U_{i, i+1}+U_{ii}A_{i, i+1}) $, 结合$ U_{ii}T_{i}=T_iU_{ii} $可得

$ \begin{eqnarray*} V_{i, i+1}&=&U_{ii}T_{i, i+1}+U_{i, i+1}T_{i+1}=U_{ii}(A_{i, i+1}T_{i+1}-T_{i}A_{i, i+1})+U_{i, i+1}T_{i+1} \\ &=&U_{ii}A_{i, i+1}T_{i+1}-T_{i}U_{ii}A_{i, i+1}+U_{i, i+1}T_{i+1}=T_iU_{i, i+1}=W_{i, i+1}. \end{eqnarray*} $

当$ 1\leq i<j-1\leq n-1 $时, 由于$ {\sum\limits_{m=i+1}^{j-1}}U_{im}T_{mj}=\tau_{T_i, T_{j}}(U_{ij}+U_{ii}A_{ij}) $, 则

$ {\sum\limits_{m=i+1}^{j-1}}U_{im}T_{mj}=T_i(U_{ij}+U_{ii}A_{ij})-(U_{ij}+U_{ii}A_{ij})T_j. $

结合$ U_{ii}T_{i}=T_iU_{ii} $可得

$ \begin{eqnarray*} V_{ij}&=&U_{ii}T_{ij}+{\sum\limits_{m=i+1}^{j-1}}U_{im}T_{mj}+U_{ij}T_{j} \\ & =&U_{ii}(A_{ij}T_{j}-T_{i}A_{ij})+T_i(U_{ij}+U_{ii}A_{ij})-(U_{ij}+U_{ii}A_{ij})T_j+U_{ij}T_{j} \\ &=&-T_{i}U_{ii}A_{ij}+T_iU_{ij}+T_iU_{ii}A_{ij}=T_iU_{ij}=W_{ij}. \end{eqnarray*} $

综上, $ UT=(V_{ij})_{n\times n}=(W_{ij})_{n\times n}=(T_1\oplus T_2\oplus\cdots\oplus T_n)U $. 故$ T $与$ T_1\oplus T_2\oplus\cdots\oplus T_n $相似.

下面在Hilbert空间上研究一类更特殊的上三角算子矩阵(除了主对角线和副对角线上的算子可以不为零外, 其他位置上的算子均为零)的酉相似性. 只给出3阶算子矩阵和4阶算子矩阵的结果, $ n $阶算子矩阵也有类似的结果, 这里从略.

定理2.3   设$ H $是Hilbert空间且有直和分解$ H=H_1\oplus H_2\oplus H_3 $, $ T, S\in B(H) $有如下表示$ T={\left( \begin{array}{ccc} T_1 & 0 & A_{13}T_3-T_1A_{13} \\ 0 & T_2 & 0 \\ 0 & 0 & T_3 \end{array}\right )}, $$ S={\left( \begin{array}{ccc} S_1 & 0 & B_{13}S_3-S_1B_{13} \\ 0 & S_2 & 0 \\ 0 & 0 & S_3 \end{array}\right )}, $满足$ T_3\in {\cal B}_1(\Omega)(H_3) $, $ S_1\in {\cal B}_1(\Omega)(H_1) $, 且$ \ker \tau_{T_1, S_1}=\{0\} $, $ \ker \tau_{T_2, S_i}=\ker \tau_{S_2, T_i}=\{0\} $($ i=1, 3 $), $ \ker \tau_{S_3, T_3}=\{0\} $. 则$ T $与$ S $酉相似当且仅当存在酉算子$ U_{22}\in \ker \tau_{S_2, T_2} $以及可逆算子$ U_{13}\in B(H_3, H_1) $, $ U_{31}\in B(H_1, H_3) $满足

(1) $ I+A_{13}A_{13}^*=(U_{31}^*U_{31})^{-1} $, $ I+A_{13}^*A_{13}=(U_{13}^*U_{13})^{-1} $,

(2) $ U_{31}T_1 U_{31}^{-1}=S_3 $, $ (U_{13}^*)^{-1}T_3U_{13}^*=S_1 $,

(3) $ U_{13}A_{13}^*U_{31}^{-1}-B_{13}\in\ker \tau_{S_1, S_3} $.

证   "$ \Rightarrow $": 由于$ T $与$ S $酉相似, 则存在酉算子$ U=(U_{ij})_{3\times 3}\in B(H) $使得$ UT=SU $, 则$ TU^*=U^*UTU^*=U^*SUU^*=U^*S $. 即

$ {\left( \begin{array}{ccc} U_{11} & \; U_{12}\; & U_{13} \\ U_{21} & U_{22} & U_{23} \\ U_{31} & U_{32} & U_{33} \end{array}\right )} {\left( \begin{array}{ccc} T_1 & 0 & A_{13}T_3-T_1A_{13} \\ 0 & T_2 & 0 \\ 0 & 0 & T_3 \end{array}\right )} ={\left( \begin{array}{ccc} S_1 & 0 & B_{13}S_3-S_1B_{13} \\ 0 & S_2 & 0 \\ 0 & 0 & S_3 \end{array}\right )} {\left( \begin{array}{ccc} U_{11} &\; U_{12}\; & U_{13} \\ U_{21} & U_{22} & U_{23} \\ U_{31} & U_{32} & U_{33} \end{array}\right )} $

且

$ {\left( \begin{array}{ccc} T_1 & 0 & A_{13}T_3-T_1A_{13} \\ 0 & T_2 & 0 \\ 0 & 0 & T_3 \end{array}\right )} {\left( \begin{array}{ccc} U_{11}^* &\; U_{21}^* \; & U_{31}^* \\ U_{12}^* & U_{22}^* & U_{32}^* \\ U_{13}^* & U_{23}^* & U_{33}^* \end{array}\right )} ={\left( \begin{array}{ccc} U_{11}^* & \; U_{21}^* \; & U_{31}^* \\ U_{12}^* & U_{22}^* & U_{32}^* \\ U_{13}^* & U_{23}^* & U_{33}^* \end{array}\right )} {\left( \begin{array}{ccc} S_1 & 0 & B_{13}S_3-S_1B_{13} \\ 0 & S_2 & 0 \\ 0 & 0 & S_3 \end{array}\right )}. $

故

$ \begin{eqnarray*} &&{\left( \begin{array}{ccc} U_{11}T_1 &\; U_{12}T_2 \; & U_{11}(A_{13}T_3-T_1A_{13})+U_{13}T_3 \\ U_{21}T_1 & U_{22}T_2 & U_{21}(A_{13}T_3-T_1A_{13})+U_{23}T_3 \\ U_{31}T_1 & U_{32}T_2 & U_{31}(A_{13}T_3-T_1A_{13})+U_{33}T_3 \end{array}\right )}\\ &=&{\left( \begin{array}{ccc} S_1U_{11}\!+\!(B_{13}S_3\!-\!S_1B_{13})U_{31} & \; S_1U_{12}\!+\!(B_{13}S_3\!-\!S_1B_{13})U_{32} \; & S_1U_{13}\!+\!(B_{13}S_3\!-\!S_1B_{13})U_{33} \\ S_2U_{21} & S_2U_{22} & S_2U_{23} \\ S_3U_{31} & S_3U_{32} & S_3U_{33} \end{array}\right )} \end{eqnarray*} $

且

$ \begin{eqnarray*} &&{\left( \begin{array}{ccc} T_1U_{11}^*\!+\!(A_{13}T_3\!-\!T_1A_{13})U_{13}^* &\; T_1U_{21}^*\!+\!(A_{13}T_3\!-\!T_1A_{13})U_{23}^* \; & T_1U_{31}^*\!+\!(A_{13}T_3\!-\!T_1A_{13})U_{33}^* \\ T_2U_{12}^* & T_2U_{22}^* & T_2U_{32}^* \\ T_3U_{13}^* & T_3U_{23}^* & T_3U_{33}^* \end{array}\right )}\\ &=&{\left( \begin{array}{ccc} U_{11}^*S_1 & \; U_{21}^*S_2\; & U_{11}^*(B_{13}S_3-S_1B_{13})+U_{31}^*S_3 \\ U_{12}^*S_1 & U_{22}^*S_2 & U_{12}^*(B_{13}S_3-S_1B_{13})+U_{32}^*S_3 \\ U_{13}^*S_1 & U_{23}^*S_2 & U_{13}^*(B_{13}S_3-S_1B_{13})+U_{33}^*S_3 \end{array}\right )}. \end{eqnarray*} $

由上两式可得

(2.1) $ \begin{equation} U_{11}T_1=S_1U_{11}+(B_{13}S_3-S_1B_{13})U_{31}, \end{equation} $

(2.2) $ \begin{equation} U_{21}T_1=S_2U_{21}, \ \ T_2U_{12}^*=U_{12}^*S_1, \ \ U_{22}T_2=S_2U_{22}, \end{equation} $

(2.3) $ \begin{equation} U_{21}(A_{13}T_3-T_1A_{13})+U_{23}T_3=S_2U_{23}, \ \ T_2U_{32}^*=U_{12}^*(B_{13}S_3-S_1B_{13})+U_{32}^*S_3, \end{equation} $

(2.4) $ \begin{equation} U_{31}T_1=S_3U_{31}, \ \ U_{31}(A_{13}T_3-T_1A_{13})+U_{33}T_3=S_3U_{33}, \end{equation} $

(2.5) $ \begin{equation} T_1U_{11}^*+(A_{13}T_3-T_1A_{13})U_{13}^*=U_{11}^*S_1, \ \ T_3U_{13}^*=U_{13}^*S_1. \end{equation} $

由(2.2)式可得$ U_{22}\in \ker \tau_{S_2, T_2} $. 由于$ \ker \tau_{S_2, T_1}=\ker \tau_{T_2, S_1}=\{0\} $, 由(2.2)式可得$ U_{21}=U_{12}^*=0 $. 此时(2.3)式化为$ U_{23}T_3=S_2U_{23} $, $ T_2U_{32}^*=U_{32}^*S_3 $. 由于$ \ker \tau_{S_2, T_3}=\ker \tau_{T_2, S_3}=\{0\} $, 则$ U_{23}=U_{32}^*=0 $, 故$ U_{12}=0 $, $ U_{32}=0 $. 由(2.4)式和(2.5)式可得

$ U_{31}A_{13}T_3-S_3U_{31}A_{13}+U_{33}T_3=U_{31}A_{13}T_3-U_{31}T_1A_{13}+U_{33}T_3=S_3U_{33}, $

$ T_1U_{11}^*+A_{13}U_{13}^*S_1-T_1A_{13}U_{13}^*=T_1U_{11}^*+A_{13}T_3U_{13}^*-T_1A_{13}U_{13}^*=U_{11}^*S_1, $

则

$ (U_{31}A_{13}+U_{33})T_3=U_{31}A_{13}T_3+U_{33}T_3=S_3U_{33}+S_3U_{31}A_{13}=S_3(U_{33}+U_{31}A_{13}), $

$ T_1(U_{11}^*-A_{13}U_{13}^*)=T_1U_{11}^*-T_1A_{13}U_{13}^*=U_{11}^*S_1-A_{13}U_{13}^*S_1=(U_{11}^*-A_{13}U_{13}^*)S_1. $

由于$ \ker \tau_{S_3, T_3}=\{0\} $, $ \ker \tau_{T_1, S_1}=\{0\} $, 则$ U_{33}+U_{31}A_{13}=0 $, $ U_{11}^*-A_{13}U_{13}^*=0 $, 即$ U_{33}=-U_{31}A_{13} $, $ U_{11}^*=A_{13}U_{13}^* $, 则$ U_{11}=U_{13}A_{13}^* $. 此时

$ U={\left( \begin{array}{ccc} U_{13}A_{13}^* & 0 & U_{13} \\ 0 &\; U_{22} \; & 0 \\ U_{31} & 0 & -U_{31}A_{13} \end{array}\right )}, {\quad} U^*={\left( \begin{array}{ccc} A_{13}U_{13}^* & 0 & U_{31}^* \\ 0 &\; U_{22}^*\; & 0 \\ U_{13}^* & 0 & -A_{13}^*U_{31}^* \end{array}\right )}, $

则

$ UU^*=(U_{13}A_{13}^*A_{13}U_{13}^*+U_{13}U_{13}^*)\oplus(U_{22}U_{22}^*)\oplus(U_{31}U_{31}^*+U_{31}A_{13}A_{13}^*U_{31}^*), $

$ U^*U={\left( \begin{array}{ccc} A_{13}U_{13}^*U_{13}A_{13}^*+U_{31}^*U_{31} & 0 & A_{13}U_{13}^*U_{13}-U_{31}^*U_{31}A_{13} \\ 0 &\; U_{22}^*U_{22} \; & 0 \\ U_{13}^*U_{13}A_{13}^*-A_{13}^*U_{31}^*U_{31} & 0 & U_{13}^*U_{13}+A_{13}^*U_{31}^*U_{31}A_{13} \end{array}\right )}. $

由于$ UU^*=U^*U=I $, 则$ U_{22}U_{22}^*=U_{22}^*U_{22}=I $, 即$ U_{22} $是酉算子, 且有如下等式

(2.6) $ \begin{equation} U_{13}(I+A_{13}^*A_{13})U_{13}^*=U_{13}A_{13}^*A_{13}U_{13}^*+U_{13}U_{13}^*=I, \end{equation} $

(2.7) $ \begin{equation} A_{13}U_{13}^*U_{13}A_{13}^*+U_{31}^*U_{31}=I, \ U_{13}^*U_{13}+A_{13}^*U_{31}^*U_{31}A_{13}=I, \ A_{13}U_{13}^*U_{13}=U_{31}^*U_{31}A_{13}. \end{equation} $

由(2.6)式可得$ U_{13}(I+A_{13}^*A_{13}) $是满射. 由于$ T_3\in {\cal B}_1(\Omega)(H_3) $, $ S_1\in {\cal B}_1(\Omega)(H_1) $, 由(2.5)式与文献[9, 命题2.3]可得$ {\overline{\rm ran}}U_{13}^*=H_3 $, 进而由(2.6)式可得$ U_{13}(I+A_{13}^*A_{13}) $是单射(事实上, 若有$ y\in H_3 $, 使得$ U_{13}(I+A_{13}^*A_{13})y=0 $, 则存在$ x_n\in H_1 $, 使得$ U_{13}^*x_n\rightarrow y $, 故$ x_n=U_{13}(I+A_{13}^*A_{13})U_{13}^*x_n\rightarrow U_{13}(I+A_{13}^*A_{13})y=0 $, 所以$ U_{13}^*x_n\rightarrow 0 $, 因此$ y=0 $). 故$ U_{13}(I+A_{13}^*A_{13}) $可逆. 因此$ U_{13}^*=(U_{13}(I+A_{13}^*A_{13}))^{-1} $可逆, 故$ U_{13} $可逆. 所以$ I+A_{13}^*A_{13} $可逆. 由于$ \sigma(A_{13}^*A_{13})\backslash\{0\}=\sigma(A_{13}A_{13}^*)\backslash\{0\} $, 则$ I+A_{13}A_{13}^* $可逆. 由(2.7)式可得

$ U_{31}^*U_{31}(I+A_{13}A_{13}^*)=U_{31}^*U_{31}A_{13}A_{13}^*+U_{31}^*U_{31}=A_{13}U_{13}^*U_{13}A_{13}^*+U_{31}^*U_{31}=I, $

$ (I+A_{13}^*A_{13})U_{13}^*U_{13}=U_{13}^*U_{13}+A_{13}^*A_{13}U_{13}^*U_{13}=U_{13}^*U_{13}+A_{13}^*U_{31}^*U_{31}A_{13}=I, $

则$ U_{31}^*U_{31} $可逆, 由于$ U_{31}(I+A_{13}A_{13}^*)U_{31}^*=U_{31}U_{31}^*+U_{31}A_{13}A^*_{13}U_{31}^*=I, $故$ U_{31}^* $与$ U_{31} $均可逆, 且有

$ I+A_{13}A_{13}^*=(U_{31}^*U_{31})^{-1}, \ \ I+A_{13}^*A_{13}=(U_{13}^*U_{13})^{-1}. $

再由(2.4)式和(2.5)式可得

$ U_{31}T_1 U_{31}^{-1}=S_3, \ \ (U_{13}^*)^{-1}T_3U_{13}^*=S_1. $

由(2.1)式和(2.4)式可得

$ \begin{eqnarray*} U_{11}U_{31}^{-1}S_3U_{31}-B_{13}S_3U_{31}&=&U_{11}U_{31}^{-1}U_{31}T_1-B_{13}S_3U_{31}=U_{11}T_1-B_{13}S_3U_{31} \\ &=&S_1U_{11}-S_1B_{13}U_{31}. \end{eqnarray*} $

上式两边同时右乘$ U_{31}^{-1} $可得

$ (U_{11}U_{31}^{-1}-B_{13})S_3=U_{11}U_{31}^{-1}S_3-B_{13}S_3=S_1U_{11}U_{31}^{-1}-S_1B_{13} =S_1(U_{11}U_{31}^{-1}-B_{13}). $

即

$ U_{13}A_{13}^*U_{31}^{-1}-B_{13}=U_{11}U_{31}^{-1}-B_{13}\in\ker \tau_{S_1, S_3}. $

"$ \Leftarrow $": 令$ U={\left( \begin{array}{ccc} U_{13}A_{13}^* & 0 & U_{13} \\ 0 & U_{22} & 0 \\ U_{31} & 0 & -U_{31}A_{13} \end{array}\right )}, $则$ U^*={\left( \begin{array}{ccc} A_{13}U_{13}^* & 0 & U_{31}^* \\ 0 & U_{22}^* & 0 \\ U_{13}^* & 0 & -A_{13}^*U_{31}^* \end{array}\right )}. $

由(1)可得$ I+A_{13}A_{13}^*=U_{31}^{-1}(U_{31}^*)^{-1} $, $ I+A_{13}^*A_{13}=U_{13}^{-1}(U_{13}^*)^{-1} $, 则

(2.8) $ \begin{equation} U_{13}A_{13}^*A_{13}U_{13}^*+U_{13}U_{13}^*=U_{13}(I+A_{13}^*A_{13})U_{13}^*=I, \end{equation} $

$ U_{31}U_{31}^*+U_{31}A_{13}A_{13}^*U_{31}^*=U_{31}(I+A_{13}A_{13}^*)U_{31}^*=I. $

由于$ (I+A_{13}A_{13}^*)A_{13}=A_{13}(I+A_{13}^*A_{13}) $, 则$ A_{13}(I+A_{13}^*A_{13})^{-1}=(I+A_{13}A_{13}^*)^{-1}A_{13} $. 由(1)可得

$ A_{13}U_{13}^*U_{13}=U_{31}^*U_{31}A_{13}, $

则

$ U_{13}^*U_{13}A_{13}^*=A_{13}^*U_{31}^*U_{31}. $

故

$ A_{13}U_{13}^*U_{13}A_{13}^*+U_{31}^*U_{31}=U_{31}^*U_{31}A_{13}A_{13}^*+U_{31}^*U_{31}=U_{31}^*U_{31}(I+A_{13}A_{13}^*)=I, $

$ U_{13}^*U_{13}+A_{13}^*U_{31}^*U_{31}A_{13}=U_{13}^*U_{13}+U_{13}^*U_{13}A_{13}^*A_{13}=U_{13}^*U_{13}(I+A_{13}^*A_{13})=I. $

又由于$ U_{22} $是酉算子, 即$ U_{22}U_{22}^*=U_{22}^*U_{22}=I $, 则

$ UU^*=(U_{13}A_{13}^*A_{13}U_{13}^*+U_{13}U_{13}^*)\oplus(U_{22}U_{22}^*)\oplus(U_{31}U_{31}^*+U_{31}A_{13}A_{13}^*U_{31}^*)=I, $

$ U^*U={\left( \begin{array}{ccc} A_{13}U_{13}^*U_{13}A_{13}^*+U_{31}^*U_{31} & 0 & A_{13}U_{13}^*U_{13}-U_{31}^*U_{31}A_{13} \\ 0 &\; U_{22}^*U_{22}\; & 0 \\ U_{13}^*U_{13}A_{13}^*-A_{13}^*U_{31}^*U_{31} & 0 & U_{13}^*U_{13}+A_{13}^*U_{31}^*U_{31}A_{13} \end{array}\right )}=I. $

所以$ U $是酉算子. 此时

$ UT={\left( \begin{array}{ccc} U_{13}A_{13}^*T_1 & 0 & U_{13}A_{13}^*(A_{13}T_3-T_1A_{13})+U_{13}T_3 \\ 0 & \; U_{22}T_2\; & 0 \\ U_{31}T_1 & 0 & U_{31}(A_{13}T_3-T_1A_{13})-U_{31}A_{13}T_3 \end{array}\right )}, $

$ SU={\left( \begin{array}{ccc} S_1U_{13}A_{13}^*+(B_{13}S_3-S_1B_{13})U_{31} & 0 & S_1U_{13}-(B_{13}S_3-S_1B_{13})U_{31}A_{13} \\ 0 & \; S_2U_{22} \; & 0 \\ S_3U_{31} & 0 & -S_3U_{31}A_{13} \end{array}\right )}. $

由于$ U_{22}\in \ker \tau_{S_2, T_2} $, 则

$ U_{22}T_2=S_2U_{22}. $

由(2)可得

$ U_{31}T_1=S_3U_{31}, $

$ U_{31}(A_{13}T_3-T_1A_{13})-U_{31}A_{13}T_3=U_{31}A_{13}T_3-U_{31}T_1A_{13}-U_{31}A_{13}T_3=-S_3U_{31}A_{13}. $

由(3)可得$ (U_{13}A_{13}^*U_{31}^{-1}-B_{13})S_3=S_1(U_{13}A_{13}^*U_{31}^{-1}-B_{13}) $, 又$ T_1U_{31}^{-1}=U_{31}^{-1}S_3 $, 则

$ U_{13}A_{13}^*T_1U_{31}^{-1}-B_{13}S_3=U_{13}A_{13}^*U_{31}^{-1}S_3-B_{13}S_3=S_1U_{13}A_{13}^*U_{31}^{-1}-S_1B_{13}. $

上式两边同时右乘$ U_{31} $可得

$ U_{13}A_{13}^*T_1-B_{13}S_3U_{31}=S_1U_{13}A_{13}^*-S_1B_{13}U_{31}, $

即

$ U_{13}A_{13}^*T_1=S_1U_{13}A_{13}^*+B_{13}S_3U_{31}-S_1B_{13}U_{31}=S_1U_{13}A_{13}^*+(B_{13}S_3-S_1B_{13})U_{31}. $

由上式可得

(2.9) $ \begin{equation} (B_{13}S_3-S_1B_{13})U_{31}=U_{13}A_{13}^*T_1-S_1U_{13}A_{13}^*. \end{equation} $

由(2.8)式可得

$ \begin{eqnarray*} (U_{13}A_{13}^*A_{13}+U_{13})U_{13}^*S_1 &=&(U_{13}A_{13}^*A_{13}U_{13}^*+U_{13}U_{13}^*)S_1=S_1 \\ &=&S_1(U_{13}A_{13}^*A_{13}U_{13}^*+U_{13}U_{13}^*)=S_1(U_{13}A_{13}^*A_{13}+U_{13})U_{13}^*. \end{eqnarray*} $

上式两边同时右乘$ (U_{13}^*)^{-1} $, 结合(1)可得

$ \begin{eqnarray*} U_{13}A_{13}^*A_{13}T_3+U_{13}T_3 &=&(U_{13}A_{13}^*A_{13}+U_{13})T_3=(U_{13}A_{13}^*A_{13}+U_{13})U_{13}^*S_1(U_{13}^*)^{-1} \\ &=&S_1(U_{13}A_{13}^*A_{13}+U_{13})=S_1U_{13}A_{13}^*A_{13}+S_1U_{13}. \end{eqnarray*} $

上式两边同时减去$ U_{13}A_{13}^*T_{1}A_{13} $, 结合(2.9)式可得

$ \begin{eqnarray*} U_{13}A_{13}^*(A_{13}T_3-T_1A_{13})+U_{13}T_3 &=&U_{13}A_{13}^*A_{13}T_3-U_{13}A_{13}^*T_{1}A_{13}+U_{13}T_3 \\ &=&S_1U_{13}A_{13}^*A_{13}+S_1U_{13}-U_{13}A_{13}^*T_{1}A_{13}\\ &=&S_1U_{13}-(U_{13}A_{13}^*T_{1}-S_1U_{13}A_{13}^*)A_{13} \\ &=&S_1U_{13}-(B_{13}S_3-S_1B_{13})U_{31}A_{13}. \end{eqnarray*} $

综上可得$ UT=SU $, 故$ T $与$ S $酉相似.证毕.

定理2.4   设$ H $是Hilbert空间且有直和分解$ H=H_1\oplus H_2\oplus H_3\oplus H_4 $, $ T, S\in B(H) $有如下表示

$ T={\left( \begin{array}{cccc} T_1 & 0 & 0 & A_{14}T_4-T_1A_{14} \\ 0 & T_2 & A_{23}T_3-T_3A_{23} & 0 \\ 0 & 0 & T_{3} & 0 \\ 0 & 0 & 0 & T_{4} \end{array}\right )}, \ S={\left( \begin{array}{cccc} S_1 & 0 & 0 & B_{14}S_4-S_1B_{14} \\ 0 & S_2 & B_{23}S_3-S_3B_{23} & 0 \\ 0 & 0 & S_{3} & 0 \\ 0 & 0 & 0 & S_{4} \end{array}\right )}, $

满足$ T_i\in {\cal B}_1(\Omega)(H_i) $ ($ i=3, 4 $), $ S_i\in {\cal B}_1(\Omega)(H_i) $ ($ i=1, 2 $), 且$ \ker \tau_{T_i, S_i}=\{0\} $ ($ i=1, 2 $), $ \ker \tau_{T_3, S_i}=\ker \tau_{S_3, T_i}=\{0\} $ ($ i=1, 4 $), $ \ker \tau_{T_i, S_2}=\ker \tau_{S_i, T_2}=\{0\} $ ($ i=1, 4 $), $ \ker \tau_{S_i, T_i}=\{0\} $ ($ i=3, 4 $). 则$ T $与$ S $酉相似当且仅当存在可逆算子$ U_{ij}\in B(H_j, H_i) $ ($ ij=14, 23, 41, 32 $), 满足

(1) $ I+A_{ij}A_{ij}^*=(U_{ji}^*U_{ji})^{-1} $, $ I+A_{ij}^*A_{ij}=(U_{ij}^*U_{ij})^{-1} $ ($ ij=14, 23 $),

(2) $ U_{ji}T_iU_{ji}^{-1}=S_j $ ($ ji=41, 32 $), $ (U_{ij}^*)^{-1}T_jU_{ij}^*=S_i $ ($ ij=14, 23 $),

(3) $ U_{ij}A_{ij}^*U_{ji}^{-1}-B_{ij}\in\ker \tau_{S_i, S_j} $ ($ ij=14, 23 $).

证   记$ T_{14}=A_{14}T_4-T_1A_{14} $, $ T_{23}=A_{23}T_3-T_3A_{23} $, $ S_{14}=B_{14}S_4-S_1B_{14} $, $ S_{23}=B_{23}S_3-S_3B_{23} $.

"$ \Rightarrow $": 由于$ T $与$ S $酉相似, 则存在酉算子$ U=(U_{ij})_{4\times 4}\in B(H) $使得$ UT=SU $, 则$ TU^*=U^*UTU^*=U^*SUU^*=U^*S $. 即

$ {\left( \begin{array}{cccc} U_{11} & U_{12} &\; U_{13} \; &\; U_{14} \\ U_{21} & U_{22} & U_{23} &\; U_{24} \\ U_{31} & U_{32} & U_{33} &\; U_{34} \\ U_{41} & U_{42} & U_{43} & \; U_{44} \end{array}\right )} {\left( \begin{array}{cccc} T_1 & 0 & 0 & T_{14} \\ 0 & T_2 & T_{23} & 0 \\ 0 & 0 & T_{3} & 0 \\ 0 & 0 & 0 & T_{4} \end{array}\right )} ={\left( \begin{array}{cccc} S_1 & 0 & 0 & S_{14} \\ 0 & S_2 & S_{23} & 0 \\ 0 & 0 & S_{3} & 0 \\ 0 & 0 & 0 & S_{4} \end{array}\right )} {\left( \begin{array}{cccc} U_{11} & \; U_{12}\; & U_{13} &\; U_{14} \\ U_{21} & U_{22} & U_{23} & \; U_{24} \\ U_{31} & U_{32} & U_{33} & \; U_{34} \\ U_{41} & U_{42} & U_{43} & \; U_{44} \end{array}\right )} $

且

$ {\left( \begin{array}{cccc} T_1 & 0 & 0 & T_{14} \\ 0 & T_2 & T_{23} & 0 \\ 0 & 0 & T_{3} & 0 \\ 0 & 0 & 0 & T_{4} \end{array}\right )} {\left( \begin{array}{cccc} U_{11}^* &\; U_{21}^* \; & U_{31}^* & \; U_{41}^* \\ U_{12}^* & U_{22}^* & U_{32}^* &\; U_{42}^* \\ U_{13}^* & U_{23}^* & U_{33}^* & \; U_{43}^* \\ U_{14}^* & U_{24}^* & U_{34}^* & \; U_{44}^* \end{array}\right )} ={\left( \begin{array}{cccc}U_{11}^* & U_{21}^* & U_{31}^* & U_{41}^* \\ U_{12}^* & \; U_{22}^* \; & U_{32}^* & \; U_{42}^* \\ U_{13}^* & U_{23}^* & U_{33}^* & \; U_{43}^* \\ U_{14}^* & U_{24}^* & U_{34}^* & \; U_{44}^* \end{array}\right )} {\left( \begin{array}{cccc} S_1 & 0 & 0 & \; S_{14} \\ 0 & S_2 & S_{23} & 0 \\ 0 & 0 & S_{3} & 0 \\ 0 & 0 & 0 & S_{4} \end{array}\right )}. $

故

$ \begin{eqnarray*} &&{\left( \begin{array}{cccc} U_{11}T_1 & \; U_{12}T_2 \; & U_{12}T_{23}+U_{13}T_3 & \; U_{11}T_{14}+U_{14}T_4 \\ U_{21}T_1 & U_{22}T_2 & U_{22}T_{23}+U_{23}T_3 & \; U_{21}T_{14}+U_{24}T_4 \\ U_{31}T_1 & U_{32}T_2 & U_{32}T_{23}+U_{33}T_3 & \; U_{31}T_{14}+U_{34}T_4 \\ U_{41}T_1 & U_{42}T_2 & U_{42}T_{23}+U_{43}T_3 & \; U_{41}T_{14}+U_{44}T_4 \end{array}\right )}\\ &=&{\left( \begin{array}{cccc} S_1U_{11}+S_{14}U_{41} &\; S_1U_{12}+S_{14}U_{42} \; & S_1U_{13}+S_{14}U_{43} &\; S_1U_{14}+S_{14}U_{44} \\ S_2U_{21}+S_{23}U_{31} & S_2U_{22}+S_{23}U_{32} & S_2U_{23}+S_{23}U_{33} & \; S_2U_{24}+S_{23}U_{34} \\ S_3U_{31} & S_3U_{32} & S_3U_{33} & \; S_3U_{34} \\ S_4U_{41} & S_4U_{42} & S_4U_{43} & \; S_4U_{44} \end{array}\right )} \end{eqnarray*} $

且

$ \begin{eqnarray*} &&\left( \begin{array}{cccc} T_1U_{11}^*+T_{14}U_{14}^* &\; T_1U_{21}^*+T_{14}U_{24}^* \; & T_1U_{31}^*+T_{14}U_{34}^* & \; T_1U_{41}^*+T_{14}U_{44}^* \\ T_2U_{12}^*+T_{23}U_{13}^* & T_2U_{22}^*+T_{23}U_{23}^* & T_2U_{32}^*+T_{23}U_{33}^* & \; T_2U_{42}^*+T_{23}U_{43}^* \\ T_3U_{13}^* & T_3U_{23}^* & T_3U_{33}^* &\; T_3U_{43}^* \\ T_4U_{14}^* & T_4U_{24}^* & T_4U_{34}^* &\; T_4U_{44}^* \end{array}\right ) \\ &=&\left( \begin{array}{cccc} U_{11}^*S_1 &\; U_{21}^*S_2 \; & U_{21}^*S_{23}+U_{31}^*S_3 & \; U_{11}^*S_{14}+U_{41}^*S_4 \\ U_{12}^*S_1 & U_{22}^*S_2 & U_{22}^*S_{23}+U_{32}^*S_3 & \; U_{12}^*S_{14}+U_{42}^*S_4 \\ U_{13}^*S_1 & U_{23}^*S_2 & U_{23}^*S_{23}+U_{33}^*S_3 & \; U_{13}^*S_{14}+U_{43}^*S_4 \\ U_{14}^*S_1 & U_{24}^*S_2 & U_{24}^*S_{23}+U_{34}^*S_3 & \; U_{14}^*S_{14}+U_{44}^*S_4 \end{array}\right ). \end{eqnarray*} $

由上两式可得

(2.10) $ \begin{equation} U_{31}T_1=S_3U_{31}, \ \ T_3U_{13}^*=U_{13}^*S_1, \ \ U_{42}T_2=S_4U_{42}, \ \ T_4U_{24}^*=U_{24}^*S_2, \end{equation} $

(2.11) $ \begin{equation} U_{12}T_2=S_1U_{12}+S_{14}U_{42}, \ \ T_1U_{21}^*+T_{14}U_{24}^*=U_{21}^*S_2, \end{equation} $

(2.12) $ \begin{equation} U_{31}T_{14}+U_{34}T_4=S_3U_{34}, \ \ T_3U_{43}^*=U_{13}^*S_{14}+U_{43}^*S_4, \end{equation} $

(2.13) $ \begin{equation} U_{41}T_1=S_4U_{41}, \ \ U_{41}T_{14}+U_{44}T_4=S_4U_{44}, \end{equation} $

(2.14) $ \begin{equation} T_1U_{11}^*+T_{14}U_{14}^*=U_{11}^*S_1, \ \ T_4U_{14}^*=U_{14}^*S_1, \end{equation} $

(2.15) $ \begin{equation} U_{32}T_2=S_3U_{32}, \ \ U_{32}T_{23}+U_{33}T_3=S_3U_{33}, \end{equation} $

(2.16) $ \begin{equation} T_2U_{22}^*+T_{23}U_{23}^*=U_{22}^*S_2, \ \ T_3U_{23}^*=U_{23}^*S_2, \end{equation} $

(2.17) $ \begin{equation} U_{11}T_1=S_1U_{11}+S_{14}U_{41}\ \ U_{22}T_2=S_2U_{22}+S_{23}U_{32}. \end{equation} $

由于$ \ker \tau_{S_3, T_1}=\ker \tau_{T_3, S_1}=\{0\} $, $ \ker \tau_{S_4, T_2}=\ker \tau_{T_4, S_2}=\{0\} $, 由(2.10)式可得$ U_{31}=U_{13}^*=0 $, $ U_{42}=U_{24}^*=0 $. 此时(2.11)式和(2.12)式分别化为$ U_{12}T_2=S_1U_{12} $, $ T_1U_{21}^*=U_{21}^*S_2 $和$ U_{34}T_4=S_3U_{34} $, $ T_3U_{43}^*=U_{43}^*S_4 $. 由于$ \ker \tau_{S_1, T_2}=\ker \tau_{T_1, S_2}=\{0\} $, $ \ker \tau_{S_3, T_4}=\ker \tau_{T_3, S_4}=\{0\} $, 则$ U_{12}=U_{21}^*=0 $, $ U_{34}=U_{43}^*=0 $, 故$ U_{13}=0 $, $ U_{24}=0 $, $ U_{21}=0 $, $ U_{43}=0 $. 由(2.13)–(2.16)式, 及$ \ker \tau_{T_i, S_i}=\{0\} $ ($ i=1, 2 $), $ \ker \tau_{S_i, T_i}=\{0\} $ ($ i=3, 4 $), 类似定理2.3的证明可证得$ U_{11}=U_{14}A_{14}^* $, $ U_{22}=U_{23}A_{23}^* $, $ U_{33}=-U_{32}A_{23} $, $ U_{44}=-U_{41}A_{14} $. 此时

$ U={\left( \begin{array}{cccc} U_{14}A_{14}^* & 0 & 0 & U_{14} \\ 0 & U_{23}A_{23}^* & U_{23} & 0 \\ 0 & U_{32} & -U_{32}A_{23} & 0 \\ U_{41} & 0 & 0 & -U_{41}A_{14} \end{array}\right )}, \ U^*={\left( \begin{array}{cccc} A_{14}U_{14}^* & 0 & 0 & U_{41}^* \\ 0 & A_{23}U_{23}^* & U_{32}^* & 0 \\ 0 & U_{23}^* & -A_{23}^*U_{32}^* & 0 \\ U_{14}^* & 0 & 0 & -A_{14}^*U_{41}^* \end{array}\right )}, $

则

$ UU^*=P_{14}\oplus P_{23}\oplus Q_{32}\oplus Q_{41}, $

$ U^*U= \left( \begin{array}{cccc} \begin{array}{c} A_{14}U_{14}^*U_{14}A_{14}^*\\ +U_{41}^*U_{41} \end{array} & 0 \ & 0 & \begin{array}{c}A_{14}U_{14}^*U_{14}\\ -U_{41}^*U_{41}A_{14} \end{array} \\ 0 & \begin{array}{c} A_{23}U_{23}^*U_{23}A_{23}^*\\ +U_{32}^*U_{32} \end{array} \ & \begin{array}{c} A_{23}U_{23}^*U_{23}\\ -U_{32}^*U_{32}A_{23} \end{array} & 0 \\ 0 &\begin{array}{c} U_{23}^*U_{23}A_{23}^* \\ -A_{23}^*U_{32}^*U_{32}\end{array} \ & \begin{array}{c} U_{23}^*U_{23}\\ +A_{23}^*U_{32}^*U_{32}A_{23}\end{array} & 0 \\ \begin{array}{c} U_{14}^*U_{14}A_{14}^*\\ -A_{14}^*U_{41}^*U_{41} \end{array} & 0 \ & 0 & \begin{array}{c} U_{14}^*U_{14}\\ +A_{14}^*U_{41}^*U_{41}A_{14} \end{array} \end{array}\right ), $

其中$ P_{ij}=U_{ij}A_{ij}^*A_{ij}U_{ij}^*+U_{ij}U_{ij}^* $ ($ ij=14, 23 $), $ Q_{ij}=U_{ij}U_{ij}^*+U_{ij}A_{ji}A_{ji}^*U_{ij}^* $ ($ ij=32, 41 $.) 由于$ UU^*=U^*U=I $, 有如下等式

(2.18) $ \begin{equation} U_{14}(I+A_{14}^*A_{14})U_{14}^*=P_{14}=I, \ \ U_{23}(I+A_{23}^*A_{23})U_{23}^*=P_{23}=I, \end{equation} $

(2.19) $ \begin{equation} A_{14}U_{14}^*U_{14}A_{14}^*+U_{41}^*U_{41}=I, \ U_{14}^*U_{14}+A_{14}^*U_{41}^*U_{41}A_{14}=I, \ A_{14}U_{14}^*U_{14}=U_{41}^*U_{41}A_{14}, \end{equation} $

(2.20) $ \begin{equation} A_{23}U_{23}^*U_{23}A_{23}^*+U_{32}^*U_{32}=I, \ U_{23}^*U_{23}+A_{23}^*U_{32}^*U_{32}A_{23}=I, \ A_{23}U_{23}^*U_{23}=U_{32}^*U_{32}A_{23}. \end{equation} $

由(2.13)–(2.20)式, 及$ T_i\in {\cal B}_1(\Omega)(H_i) $ ($ i=3, 4 $), $ S_i\in {\cal B}_1(\Omega)(H_i) $ ($ i=1, 2 $), 类似定理2.3的证明可证得$ U_{ij} $ ($ ij=14, 23, 41, 32 $)可逆, 且满足

(1) $ I+A_{ij}A_{ij}^*=(U_{ji}^*U_{ji})^{-1} $, $ I+A_{ij}^*A_{ij}=(U_{ij}^*U_{ij})^{-1} $ ($ ij=14, 23 $),

(2) $ U_{ji}T_iU_{ji}^{-1}=S_j $ ($ ji=41, 32 $), $ (U_{ij}^*)^{-1}T_jU_{ij}^*=S_i $ ($ ij=14, 23 $),

(3) $ U_{ij}A_{ij}^*U_{ji}^{-1}-B_{ij}\in\ker \tau_{S_i, S_j} $ ($ ij=14, 23 $).

"$ \Leftarrow $": 令

$ U={\left( \begin{array}{cccc} U_{14}A_{14}^* & 0 & 0 & U_{14} \\ 0 & U_{23}A_{23}^* & U_{23} & 0 \\ 0 & U_{32} & -U_{32}A_{23} & 0 \\ U_{41} & 0 & 0 & -U_{41}A_{14} \end{array}\right )}, $

则

$ U^*={\left( \begin{array}{cccc} A_{14}U_{14}^* & 0 & 0 & U_{41}^* \\ 0 & A_{23}U_{23}^* & U_{32}^* & 0 \\ 0 & U_{23}^* & -A_{23}^*U_{32}^* & 0 \\ U_{14}^* & 0 & 0 & -A_{14}^*U_{41}^* \end{array}\right )}. $

类似定理2.3的证明可证得

$ P_{14}=U_{14}A_{14}^*A_{14}U_{14}^*+U_{14}U_{14}^*=I, \ \ Q_{41}=U_{41}U_{41}^*+U_{41}A_{14}A_{14}^*U_{41}^*=I, $

$ P_{23}=U_{23}A_{23}^*A_{23}U_{23}^*+U_{23}U_{23}^*=I, \ \ Q_{32}=U_{32}U_{32}^*+U_{32}A_{23}A_{23}^*U_{32}^*=I, $

$ A_{14}U_{14}^*U_{14}A_{14}^*+U_{41}^*U_{41}=I, \ \ U_{14}^*U_{14}+A_{14}^*U_{41}^*U_{41}A_{14}=I, $

$ A_{14}U_{14}^*U_{14}=U_{41}^*U_{41}A_{14}, \ \ U_{14}^*U_{14}A_{14}^*=A_{14}^*U_{41}^*U_{41}, $

$ A_{23}U_{23}^*U_{23}A_{23}^*+U_{32}^*U_{23}=I, \ \ U_{23}^*U_{23}+A_{23}^*U_{32}^*U_{32}A_{23}=I, $

$ A_{23}U_{23}^*U_{23}=U_{32}^*U_{32}A_{23}, \ \ U_{23}^*U_{23}A_{23}^*=A_{23}^*U_{32}^*U_{32}. $

故

$ UU^*=P_{14}\oplus P_{23}\oplus Q_{32}\oplus Q_{41}=I, $

$ U^*U= \left( \begin{array}{cccc} \begin{array}{c} A_{14}U_{14}^*U_{14}A_{14}^*\\ +U_{41}^*U_{41} \end{array} & 0 \ & 0 & \begin{array}{c} A_{14}U_{14}^*U_{14}\\ -U_{41}^*U_{41}A_{14} \end{array} \\ 0 & \begin{array}{c} A_{23}U_{23}^*U_{23}A_{23}^*\\ +U_{32}^*U_{32}\end{array} \ & \begin{array}{c} A_{23}U_{23}^*U_{23}\\ -U_{32}^*U_{32}A_{23} \end{array} & 0 \\ 0 & \begin{array}{c}U_{23}^*U_{23}A_{23}^*\\ -A_{23}^*U_{32}^*U_{32}\end{array}\ & \begin{array}{c} U_{23}^*U_{23}\\ +A_{23}^*U_{32}^*U_{32}A_{23}\end{array} & 0 \\ \begin{array}{c} U_{14}^*U_{14}A_{14}^*\\ -A_{14}^*U_{41}^*U_{41} \end{array} & 0 \ & 0 &\begin{array}{c} U_{14}^*U_{14}\\ +A_{14}^*U_{41}^*U_{41}A_{14}\end{array} \end{array}\right ) =I. $

所以$ U $是酉算子. 此时

$ UT={\left( \begin{array}{cccc} U_{14}A_{14}^*T_1 & 0 & 0 & U_{14}A_{14}^*T_{14}+U_{14}T_4 \\ 0 &\; \; U_{23}A_{23}^*T_2\; \; & U_{23}A_{23}^*T_{23}+U_{23}T_3 & 0 \\ 0 & U_{32}T_2 & U_{32}T_{23}-U_{32}A_{23}T_3 & 0 \\ U_{41}T_1 & 0 & 0 & U_{41}T_{14}-U_{41}A_{14}T_4 \end{array}\right )}, $

$ SU={\left( \begin{array}{cccc} S_1U_{14}A_{14}^*+S_{14}U_{41} & 0 & 0 & S_1U_{14}-S_{14}U_{41}A_{14} \\ 0 & \; S_2U_{23}A_{23}^*+S_{23}U_{32} \; & S_2U_{23}-S_{23}U_{32}A_{23} & 0 \\ 0 & S_3U_{32} & -S_3U_{32}A_{23} & 0 \\ S_4U_{41} & 0 & 0 & -S_4U_{41}A_{14} \end{array}\right )}. $

类似定理2.3的证明可证得

$ U_{41}T_{14}-U_{41}A_{14}T_4=-S_4U_{41}A_{14}, \ \ U_{14}A_{14}^*T_1=S_1U_{14}A_{14}^*+S_{14}U_{41}, $

$ U_{41}T_1=S_4U_{41}, \ \ U_{14}A_{14}^*T_{14}+U_{14}T_4=S_1U_{14}-S_{14}U_{41}A_{14}, $

$ U_{32}T_{23}-U_{32}A_{23}T_3=-S_3U_{32}A_{23}, \ \ U_{23}A_{23}^*T_2=S_2U_{23}A_{23}^*+S_{23}U_{32}, $

$ U_{32}T_2=S_3U_{32}, \ \ U_{23}A_{23}^*T_{23}+U_{23}T_3=S_2U_{23}-S_{23}U_{32}A_{23}, $

则$ UT=SU $, 故$ T $与$ S $酉相似.证毕.