近年来,众多学者对算子代数上的保持问题进行了广泛深入的研究,并且取得了一系列的成果. 所谓保持问题,即研究保持算子某种特征、关系,函数以及集合等不变的映射问题,核心问题即是寻找发现算子代数同构(或反同构)的本质不变量(参见文献[1]). 其中,一类有趣的保持问题是以算子对的某种关系为不变量,刻画其映射特征,如文献[2, 3]中研究了保持两个算子乘积投影的线性或非线性映射,文献[4, 5]考虑了保持两个算子乘积幂等性的线性或非线性映射.算子束广泛应用于数学与物理研究领域,其中国内外学者对算子束进行了大量的研究,如文献[6, 7, 8, 9]. 同时,部分学者研究了算子束的保持问题,如文献[10, 11, 12].部分等距作为一类重要的算子类,文献[13, 14]研究了${\cal PI(H)}$上双边保持偏序和正交性的双射. 本文将结合部分等距和算子束,考虑算子代数上保持算子束部分等距的映射特征.

下面引入本文的一些概念和记号.设${\cal H}$是复Hilbert空间,$\dim {\cal H}$表示${\cal H}$的维数,${\cal B(H)}$表示${\cal H}$上全体有界线性算子构成的代数. ${\cal P(H)}$表示全体投影构成的集合. 算子$ A\in {\cal B(H)}$,${\cal N}(A)$表示算子$A$的零空间. 若$\| Ax\|=\| x\| $对任意的$x\in (\ker A)^{\perp}$成立,则称A为部分等距算子,即$A^{*}A$是投影,其中$(\ker A)^{\perp}$记作${\cal I}_{A}$称作A的始空间,ranA记作${\cal F}_{A}$称作A的终空间. ${\cal PI(H)}$表示全体部分等距构成的集合. 对任意的$x,y\in {\cal H}$,$\langle x,y \rangle $表示$x$和$y$的内积. $x\otimes y$表示 ${\cal H}$上的一秩算子,其中$(x\otimes y)z=\langle z,y\rangle x,\forall z\in {\cal H}$. 一秩算子$x\otimes y$是部分等距当且仅当$\| x\|\| y\|=1$. 对任意的$P,\ Q\in {\cal PI(H)}$,若$P| _{{\cal I}_{p}}=Q| _{{\cal I}_{p}}$,则称$P\leq Q$. 由$P| _{{\cal I}_{p}}=Q| _{{\cal I}_{p}}$,可得${\cal I}_{p}\subseteq {\cal I}_{Q},$ ${\cal F}_{P}\subseteq {\cal F}_{Q}$. 若$P^{*}Q=PQ^{*}=0$,则称部分等距算子$P$和$Q$正交,记作$P\perp Q$. 设$U$是${\cal H}$上的共轭线性双射,且对任意的$x,y\in{\cal H}$都有 $\langle Ux,Uy\rangle =\langle y,x\rangle $,则称$U$是共轭酉算子或反酉算子.在不引起混淆时,$I$表示任意Hilbert空间上的恒等算子. ${\Bbb T}=\{z\in {\Bbb C}:| z| =1\}$表示复平面上的单位圆周. 任意的$A,\ B\in {\cal B(H)},\ \lambda\in {\Bbb C}$,$A-\lambda B$称为$A,\ B$的算子束(pencil).

下面考虑${\cal B(H)}$上保持算子束部分等距的满射的特征.

设${\cal H}$是复Hilbert空间,$\Phi$是${\cal B(H)}$上的满射且保持算子束部分等距,即对任意$A,\ B\in {\cal B(H)}$,$\lambda\in\Bbb C$,$ A-\lambda B \in {\cal PI(H)}\Leftrightarrow\Phi(A)-\lambda \Phi(B)\in {\cal PI(H)}$. 下面给出本文的主要结论.

定理2.1 设${\cal H}$是复Hilbert空间,$\Phi$是${\cal B(H)}$上的满射,则对任意的$A,\ B\in{\cal B(H)}$,$\lambda\in{\Bbb C}$,$\Phi$满足

$\label{eq:a1} A-\lambda B\in {\cal PI(H)}\Leftrightarrow\Phi(A) -\lambda\Phi(B)\in{\cal PI(H)}$

(2.1)

的充要条件是存在${\cal H}$上的两个酉算子$U,\ V$使得$\Phi(X)=UXV,\ \forall X\in {\cal B(H)}$或存在${\cal H}$上的两个共轭酉算子$U,\ V$ 使得$\Phi(X)=UX^{*}V,\ \forall X\in {\cal B(H)}$.

为了证明该定理,我们先给出部分等距$A,\ B$满足$ A\leq B $的一个充要条件.

命题2.2 设$A,\ B\in{\cal PI(H)}$,则 $ A\leq B $当且仅当$ B-2A\in {\cal PI(H)}$.

证 设 $ A\leq B$,则$ B-A\in {\cal PI(H)}.$ 又因为$A\perp B-A,$ 所以$(B-A)-A=B-2A\in {\cal PI(H)}.$ 反之,用反证法,存在$x\in {\cal I}_A$,使得 $Ax\not=Bx$. 注意到 $B-2A$,$B,\ A$都是部分等距算子,

$$\|x\|\geq \|Bx-2Ax\|\geq 2\|Ax\|-\|Bx\|\geq \|x\|, $$ 因此,$\|Bx-2Ax\|=\|x\|$. 由于$\|Ax\|=\|x\|$,$\|Bx\|\leq \|x\|$,故 $\|Bx\|=\|x\|$. $$4\|Ax\|^2+\|Bx\|^2-2\| Ax,Bx\|-2\| Bx,Ax\|=\|x\|^2. $$ 我们有$4\|Ax\|^2=2\| Ax,Bx\|+2\| Bx,Ax\| $,利用Schwatz不等式,得到 $Ax=Bx$,矛盾.

引理2.3 设$\Phi$是${\cal B(H)}$上的满射,且保持算子束部分等距,则有

(1) $\Phi(0)=0$且 $\Phi$是单射;

(2) $\Phi$是齐次的;

(3) $\Phi$ 双边保持部分等距的正交性和偏序;

(4) $\Phi$双边保持一秩算子.

证 首先证明一个充要条件,即$I-B,\ I+B \in {\cal PI(H)}$当且仅当$B=0$. 事实上,若$B=0$,则显然有$I\pm B\in {\cal PI(H)}$; 反之,设$I\pm B\in {\cal PI(H)}$,则在空间分解${\cal H}={\cal I}_{I-B}\oplus {\cal N}(I-B)$下,若${\cal N}(I-B)\neq \{0\}$,则存在单位向量$x\in{\cal N}(I-B)$,使得 $(I-B)x=0$,于是,$\|(I+B)x\|=\| 2x\|=2,$ 矛盾. 因此,$I-B$为等距算子. 同理,$I+B$也是等距算子. 于是有

\begin{eqnarray*} \left\{ \begin{array}{lll} (I-B)^{*}(I-B)&=&I-B-B^{*}+B^{*}B=I. \\ (I+B)^{*}(I+B)&=&I+B+B^{*}+B^{*}B=I. \end{array} \right. \end{eqnarray*} 因此,$ B^{*}B=-(B+B^{*})=B+B^{*}$, 因此,$B=0$.

(1) 对于任意$ \lambda \in {\Bbb C}$,由于$0-\lambda 0=0 \in{\cal PI(H)},$ 因此$\Phi(0)- \lambda \Phi(0)=(1-\lambda)\Phi(0) \in{\cal PI(H)}$,事实上,$\lambda$是任意的,因此,$\Phi(0)=0.$ 进而可知,$\Phi$双边保持部分等距.

设任意的$A,B\in {\cal B(H)}$且$\Phi(A)=\Phi(B)$. 由$(I-A)+A\in {\cal PI(H)}$知 $\Phi(I-A)+\Phi(A)\in {\cal PI(H)}$. 从而$\Phi(I-A)+\Phi(B)\in {\cal PI(H)}$. 于是,$I-A+B\in {\cal PI(H)}$. 由$(I-B)+B\in {\cal PI(H)}$知 $\Phi(I-B)+\Phi(B)\in {\cal PI(H)}$. 从而$\Phi(I-B)+\Phi(A)\in {\cal PI(H)}$. 于是,$I-B+A\in {\cal PI(H)}$. 由上述充要条件可知,$A=B$,因此,$\Phi$是单射.

(2) 由于$\Phi$是${\cal B(H)}$上的满射,因而存在$X\in{\cal B(H)}$,使得$\Phi(X)=\lambda \Phi(A)$,于是有 $\Phi(I-X)+\lambda \Phi(A)\in {\cal PI(H)}$以及$ \Phi(I+X)-\lambda \Phi(A)\in PI(H)$,因此,$I-X+\lambda A\in {\cal PI(H)}$ 以及$I+X-\lambda A\in {\cal PI(H)}$. 由上述充要条件可得 $X-\lambda A=0$,即 X=$\lambda A$. 因此,$\Phi(\lambda A)=\lambda \Phi (A)$.

(3) 由文献[14,定理2]的证明过程可知,对于任意的$A,\ B\in {\cal PI(H)},\ A+B,\ A-B\in {\cal PI(H)}$当且仅当$A\perp B.$ 又因为$\Phi$双边保持部分等距,且有

$$A \bot B \Leftrightarrow A\pm B\in{\cal PI(H)}\Leftrightarrow\Phi(A)\pm \Phi(B)\in {\cal PI(H)}\Leftrightarrow\Phi(A)\bot \Phi (B). $$ 因此,$\Phi:{\cal PI(H)}\rightarrow{\cal PI(H)}$ 双边保部分等距的正交性.

另外,对于任意的$A,\ B\in {\cal PI(H)},$ 若$A\leq B$,则$ B-2A\in {\cal PI(H)}$. 于是,$\Phi(B)-2 \Phi (A)\in {\cal PI(H)}$. 因此,由命题2.2可知 $\Phi (A)\leq \Phi(B)$,由于$\Phi^{-1}$与$\Phi$性质相同,因此 $\Phi$双边保部分等距的偏序.

(4) 对于任意的一秩算子$x\otimes y,$ $\frac{x\otimes y}{\| x\| \| y\|}$为一秩部分等距,由(1),(2)可知

$$\Phi (x\otimes y)=\Phi \bigg(\| x\| \| y\|\frac{x\otimes y}{\| x\| \| y\|}\bigg)=\| x\| \| y\| \Phi\bigg(\frac{x\otimes y}{\| x\| \| y\|}\bigg). $$ 由文献[13,定理1]的证明过程可知 $\Phi$双边保持部分等距的秩,所以$\Phi (x\otimes y)$是一秩的. 由于$\Phi^{-1}$与$\Phi$性质相同,因此,$\Phi$双边保持一秩算子.

定理2.1的证明 充分性显然,故只需证明必要性. 由引理2.3可知,$\Phi$在${\cal PI(H)}$上的限制 $\Phi|_{{\cal PI(H)}}:{\cal PI(H)}\rightarrow{\cal PI(H)}$是一个双边保正交性和偏序的双射,因此,由文献[14,定理1]可知,当dim${\cal H}\geq3$时,对任意的$R\in{\cal PI(H)}$,有下述四种形式之一成立:

(1) 存在${\cal H}$上的两个酉算子$U,\ V$使得$\Phi (R)=URV,\ \ \ \forall R\in {\cal PI(H)}$;

(2) 存在${\cal H}$上的两个酉算子$U,\ V$使得$\Phi (R)=UR^{*}V,\ \ \ \forall R\in {\cal PI(H)}$;

(3) 存在${\cal H}$上的两个共轭酉算子$U,\ V$使得$\Phi (R)=URV,\ \ \ \forall R\in {\cal PI(H)}$;

(4) 存在${\cal H}$上的两个共轭酉算子$U,\ V$使得$\Phi (R)=UR^{*}V,\ \ \ \forall R\in {\cal PI(H)}.$

由引理2.3可知,$\Phi$是齐次的,因而上述(2),(3)不成立. 由文献[15,推论 2.3]可知 ${\cal B(H)}$中的任一有界线性算子都可以表示成有限个投影的线性组合. 投影是特殊的部分等距. 若记$\Omega_{m}=\{\Sigma_{k=1}^{m}\lambda_{k}R_{k}:\ \lambda_{k}\in {\Bbb C},\ R_{k}\in {\cal PI(H)}\}.$ 则${\cal B(H)}$中任一算子都是$\Omega_{m}$中之一. 下面对其它两种情形分别进行讨论.

情形1 存在酉算子$U,\ V:H\rightarrow H$,使得对于任意$R\in {\cal PI(H)}$,有$\Phi(R)=URV,$ 定义映射$\Psi:{\cal B(H)}\rightarrow{\cal B(H)}$,$\Psi(A)=U^{*}\Phi(A)V^{*}$,则

$$A-\lambda B \in{\cal PI(H)}\Leftrightarrow\Psi(A)-\lambda \Psi(B) \in {\cal PI(H)}. $$ 且对于任意$R\in{\cal PI(H)}$,有$\Psi(R)=R$. 下面用数学归纳法证明对任意A$\in \Omega_{m}$,有$\Psi(A)=A$.

当$m=1$时,由引理2.3可知,对任意$\lambda_{1} \in {\Bbb C}$及$R_{1}\in {\cal PI(H)}$,有

$$\Psi(\lambda_{1} R_{1})=\lambda_{1}\Psi(R_{1})=\lambda_{1} R_{1}. $$ 假设对任意A$\in \Omega_{k-1}$,有$\Psi(A)=A$,即$\Psi(\Sigma_{i=1}^{k-1}\lambda_{i}R_{i})=\Sigma_{i=1}^{k-1}\lambda_{i}R_{i},\forall \lambda_{i}\in {\Bbb C},R_{i}\in{\cal PI(H)}.$

当$m=k$时,记A($\lambda$)=$R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k},\ \ (\forall \lambda \in {\Bbb C})$. 若$\lambda=0$,则A($\lambda$)$\in \Omega_{k-1},$ 从而由归纳假设可得 $\Psi(A(\lambda ))=A(\lambda).$ 若$\lambda \neq 0$,则由 $R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k}-(\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k})\in {\cal PI(H)}$ 可得

\[Q(\lambda ) = \Psi ({R_1} + \Sigma _{i = 2}^{k - 1}{\lambda _i}{R_i} + \lambda {R_k}) - \Psi (\Sigma _{i = 2}^{k - 1}{\lambda _i}{R_i} + \lambda {R_k}) \in PI(H).\]

(2.2)

由于$\Sigma_{i=2}^{k-1}\lambda_{i}R_{i} +\lambda R_{k} \in \Omega_{k-1},$ 从而由归纳假设和(2.2)式可得

\[\Psi ({R_1} + \Sigma _{i = 2}^{k - 1}{\lambda _i}{R_i} + \lambda {R_k}) = Q(\lambda ) + \Sigma _{i = 2}^{k - 1}{\lambda _i}{R_i} + \lambda {R_k}.\]

(2.3)

事实上, $$\lambda^{-1}(R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k})-\lambda^{-1}(R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i})\in {\cal PI(H)}, $$ 由$\Psi$的齐次性可得

\[D(\lambda ) = {\lambda ^{ - 1}}\Psi ({R_1} + \Sigma _{i = 2}^{k - 1}{\lambda _i}{R_i} + \lambda {R_k}) - {\lambda ^{ - 1}}\Psi ({R_1} + \Sigma _{i = 2}^{k - 1}{\lambda _i}{R_i}) \in PI(H).\]

(2.4)

由于$R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}\in \Omega_{k-1},$ 从而由归纳假设和(2.4)式可得

\[\Psi ({R_1} + \Sigma _{i = 2}^{k - 1}{\lambda _i}{R_i} + \lambda {R_k}) = {R_1} + \Sigma _{i = 2}^{k - 1}{\lambda _i}{R_i} + \lambda D(\lambda ).\]

(2.5)

由于$\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda D(\lambda )\in \Omega_{k-1},$ 则由归纳假设,$\Psi(\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda D(\lambda ))=\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda D(\lambda).$ 从而由(2.5)式可知 $$\Psi(R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k})-\Psi(\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda D(\lambda ))=R_{1}\in {\cal PI(H)}. $$ 这说明

\[{R_1} + \lambda ({R_k} - D(\lambda )) = ({R_1} + \Sigma _{i = 2}^{k - 1}{\lambda _i}{R_i} + \lambda {R_k}) - (\Sigma _{i = 2}^{k - 1}{\lambda _i}{R_i} + \lambda D(\lambda )) \in PI(H).\]

(2.6)

由(2.3),(2.5)式有

\[\lambda ({R_k} - D(\lambda )) = {R_1} - Q(\lambda ),\]

(2.7)

代入(2.6)式可得,$2R_{1}-Q(\lambda)\in {\cal PI(H)}.$ 由命题2.2可得,$R_{1}\leq Q(\lambda),$ 令$Q(\lambda)=R_{1}\oplus Q_{1}(\lambda),$ 其中$ Q_{1}(\lambda)\in {\cal PI(H)}.$ 由(2.3)式可得 $$\Psi(R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k})-(\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k})=Q(\lambda), $$ 因而 $$\lambda ^{-1}\Psi(R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k})-\lambda ^{-1}(\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+Q(\lambda ))=R_{k}. $$ 由$\Psi$的齐次性及(2.1)式可得 $$C(\lambda)=\lambda ^{-1}(R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k})-\lambda ^{-1}(\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+Q(\lambda))\in {\cal PI(H)}. $$ 因而

\[\lambda C(\lambda ) = {R_1} + \lambda {R_k} - Q(\lambda ).\]

(2.8)

由于$Q(\lambda)=R_{1}\oplus Q_{1}(\lambda)$,那么

\[{Q_1}(\lambda ) = \lambda ({R_k} - C(\lambda )).\]

(2.9)

将(2.3)式代入(2.8)式可得 $$\lambda C(\lambda )=R_{1}+\lambda R_{k}-(\Psi(R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k})-\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}-\lambda R_{k}). $$ 从而有 $$\lambda C(\lambda )=R_{1}+2\lambda R_{k}-(\Psi(R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k})-\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}). $$ 将(2.5)式代入上式可得$R_{1}+2\lambda R_{k}-(R_{1}+\lambda D(\lambda))=\lambda C(\lambda).$ 因此,$2 R_{k}-D(\lambda)=C(\lambda),$ 其中 $R_{k},\ D(\lambda),\ C(\lambda)\in {\cal PI(H)}.$ 由命题2.2可得 $R_{k} \leq D(\lambda).$ 令$D(\lambda)=R_{k} \oplus D_{1}(\lambda),$ 又因为$ Q(\lambda )=R_{1}\oplus Q_{1}(\lambda )$ 且 $D_{1}(\lambda),\ Q_{1}(\lambda )\in {\cal PI(H)},$ 所以将上式代入(2.7)式可得 $\lambda D_{1}(\lambda)=Q_{1}(\lambda)$.

若$| \lambda | \neq 1$,则$Q_{1}(\lambda)=D_{1}(\lambda)=0.$ 此时,由(2.3)式可得

$$\Psi(R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k})=R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k}. $$

若$| \lambda | = 1,$ 由(2.3)式可得

\[\Psi ({R_1} + \Sigma _{i = 2}^{k - 1}{\lambda _i}{R_i} + \lambda {R_k}) = {R_1} + {Q_1}(\lambda ) + \Sigma _{i = 2}^{k - 1}{\lambda _i}{R_i} + \lambda {R_k}.\]

(2.10)

$\forall \mu \in {\Bbb C},$ 且$| \mu | \neq 1.$ 由于 $$(R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k})-(R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\mu R_{k})=(\lambda -\mu)R_{k}. $$ 因而$\frac{1}{\lambda -\mu}(R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k})-\frac{1}{\lambda -\mu}(R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\mu R_{k})=R_{k}\in {\cal PI(H)}.$ 那么,$\Psi(\frac{1}{\lambda -\mu}(R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k}))-\Psi(\frac{1}{\lambda -\mu}(R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\mu R_{k}))\in{\cal PI(H)}.$ 由$\Psi$的齐次性,则有 $$S(\lambda)=\frac{1}{\lambda -\mu}\Psi(R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k})-\frac{1}{\lambda -\mu}(R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\mu R_{k})\in{\cal PI(H)}. $$ 那么,

\[\Psi ({R_1} + \Sigma _{i = 2}^{k - 1}{\lambda _i}{R_i} + \lambda {R_k}) = {R_1} + \Sigma _{i = 2}^{k - 1}{\lambda _i}{R_i} + \mu {R_k} + (\lambda - \mu )S(\lambda ).\]

(2.11)

由(2.10),(2.11)式可得$R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\lambda R_{k}+Q_{1}(\lambda )=R_{1}+\Sigma_{i=2}^{k-1}\lambda_{i}R_{i}+\mu R_{k}+(\lambda-\mu)S(\lambda).$ 因此,$Q_{1}(\lambda )=(\mu-\lambda)(R_{k}-S(\lambda)).$ 事实上,由于$\| R_{k}-S(\lambda)\| \leq 2$且$\mu$是任意的,因此 $Q_{1}(\lambda )=0.$

综上可知,$Q(\lambda )=R_{1}$.

所以由(2.3)式可得

\[\Psi ({R_1} + \Sigma _{i = 2}^{k - 1}{\lambda _i}{R_i} + \lambda {R_k}) = {R_1} + \Sigma _{i = 2}^{k - 1}{\lambda _i}{R_i} + \lambda {R_k}.\]

(2.12)

任意$ A=\Sigma_{i=1}^{k}\lambda_{i}R_{i}\in \Omega _{k}$,若$\lambda_{1}=0,$ 则$A\in \Omega _{k-1}.$ 从而$\Psi(A)=A.$ 若$\lambda_{1}\neq0$,则$A=\lambda_{1}(R_{1}+\Sigma_{i=2}^{k}\lambda_{1}^{-1}\lambda_{i}R_{i}).$ 从而由$\Psi$的齐次性和(2.12)式有 $$\Psi (A)=\lambda_{1}\Psi(R_{1}+\Sigma_{i=2}^{k}\lambda_{1}^{-1}\lambda_{i}R_{i})=\lambda_{1}(R_{1}+\Sigma_{i=2}^{k}\lambda_{1}^{-1}\lambda_{i}R_{i})=A. $$ 因此,对于任意正整数$m$及$A\in \Omega_{m}$,有$\Psi(A)=A$,从而有$\Phi(A)=UAV$. 由于${\cal B(H)}$中任一算子都是$\Omega_{m}$中之一,于是对于任意$A\in{\cal B(H)}$,有$\Phi(A)=UAV$.

情形2 存在共轭酉算子$U,\ V:\ H\rightarrow H$,使得对于任意$R\in {\cal PI(H)}$,有$\Phi(R)=UR^{*}V,$ 类似地,定义映射$\Psi:{\cal B(H)}\rightarrow{\cal B(H)},\ \ \Psi(A)=U^{*}\Phi(A)V^{*}$,则

$$A-\lambda B \in{\cal PI(H)}\Leftrightarrow \Psi(A)-\overline{\lambda} \Psi(B) \in{\cal PI(H)}. $$ 且$\Psi$是共轭齐次的双射,即对于任意$A\in{\cal B(H)}$及$\lambda \in {\Bbb C},$ 有$\Psi(\lambda A)=\overline{\lambda}\Psi(A)$,由$\Psi$的定义,则对于任意$R\in{\cal PI(H)}$,有$\Psi(R)=R^{*}$. 因此与情形1类似,用数学归纳法可得,对任意$A=\Sigma_{i=1}^{m}\lambda_{i}R_{i}\in \Omega_{m},$ 有 $$\Psi(A)=\Psi (\Sigma_{i=1}^{m}\lambda_{i}R_{i})=\Sigma_{i=1}^{m}\overline{\lambda_{i}}R_{i}^{*}=A^{*}. $$ 因此,$\Phi (A)=UA^{*}V.$ 由于${\cal B(H)}$中任一算子都是$\Omega_{m}$中之一,于是对于任意$A\in {\cal B(H)}$,有$\Phi (A)=UA^{*}V$.

当dim${\cal H}=1$时,显然成立. 下面证明dim${\cal H}=2$的情形. 设$E_{ij}\in M_{2}({\Bbb C})$表示$(i,\ j)$位置为1,其余为0的2阶矩阵,对于任意的$A\in M_{2}({\Bbb C}),\ A^{t}$表示矩阵$A$的转置.

第一步: 存在酉算子$U_{1},\ V_{1}\in M_{2}({\Bbb C})$,使得$\Phi (E_{11})=U_{1}E_{11}V_{1}$.

由于$E_{11}$是部分等距,因而$\Phi(E_{11})$是部分等距. 由于$\Phi$是单射,因而$\Phi(E_{11})\neq0,$ 又由于$\Phi$双边保一秩,故$\Phi (E_{11})$是一秩部分等距,故存在两个酉算子$U_1,V_1\in M_{2}({\Bbb C})$,使得$\Phi (E_{11})=U_1E_{11}V_1$.

定义$\Phi_{1}(A)=U_{1}^*\Phi (A)V_{1}^*,$ $A\in M_2({\Bbb C})$,则$\Phi_1$保持算子束部分等距,此时,$\Phi_{1}(E_{11})=E_{11}$,且$\Phi_1$和$\Phi$有相同的性质.

第二步: 存在酉算子$U_2,V_2\in M_{2}({\Bbb C})$,使得$\Phi_1(E_{11})=U_2E_{11}V_2=E_{11}$,$\Phi_1(E_{22})=U_2E_{22}V_2$.

由于$E_{11} \perp E_{22}$,且$\Phi_1$双边保部分等距的正交性,因而$\Phi_1(E_{11}) \perp \Phi_1(E_{22})$. 设$\Phi_{1}(E_{22})=\left[{\begin{array}{*{20}{c}} a_{11}~&a_{12}\\ a_{21}~&a_{22}\\ \end{array}} \right]$,于是有 $\Phi_{1}(E_{22})E_{11}=0$,$E_{11}\Phi_{1}(E_{22})=0$. 计算可得$a_{11}=a_{12}=a_{21}=0.$ 又因为$\Phi_{1}$双边保部分等距,因而 $a_{22}\in {\Bbb T}.$ 令$U_{2}=E_{11}+a_{22}E_{22},$ $V_{2}=I$,则有 $\Phi_1(E_{11})=U_{2}E_{11}V_{2}=E_{11},$ $\Phi_1(E_{22})=U_{2}E_{22}V_{2}.$

设$\Phi_2(A)=U_{2}^*\Phi_1(A)V_2^*,$ $A\in M_2({\Bbb C})$,则有$\Phi_2(E_{ii})=E_{ii}(i=1,2)$. 由文献[13,定理1]的证明过程可知,$\Phi$在${\cal PI(H)}$上正交可和. 因此,由引理2.3知,$\forall \alpha,\ \beta \in {\Bbb T}$,$\Phi_2(\alpha E_{11}+\beta E_{22})= \alpha E_{11}+\beta E_{22}$.

第三步: 存在酉算子$U_3,V_3\in M_{2}({\Bbb C})$,使得$\Phi_{2}(E_{ii})=U_{3}\Phi_{2}(E_{ii})V_{3}=E_{ii}(i=1,2)$,$\Phi_{2}(E_{12})=U_{3}E_{12}V_{3}$,或$\Phi_{2}(E_{12})=U_{3}E_{21}V_{3}$.

设$\Phi_{2}(E_{12})=\left[{\begin{array}{*{20}{c}} a_{11}~&a_{12}\\ a_{21}~&a_{22}\\ \end{array}} \right]$. 由于$\frac{1}{\sqrt{1+| \lambda| ^{2}}}(\lambda E_{11}+E_{12})$是部分等距,因而$\frac{1}{\sqrt{1+| \lambda| ^{2}}}(\lambda E_{11}+\Phi_{2} (E_{12}))$是部分等距. 因而

\begin{eqnarray*} &&\bigg[\frac{1}{\sqrt{1+| \lambda| ^{2}}}(\lambda E_{11}+\Phi_{2} (E_{12}))\bigg]\bigg[\frac{1}{\sqrt{1+| \lambda| ^{2}}}(\lambda E_{11}+\Phi_{2} (E_{12}))\bigg]^{*}\\&=&\frac{1}{1+| \lambda| ^{2}}\left[{\begin{array}{*{20}{c}} | \lambda+a_{11}| ^{2}+| a_{12}| ^{2}~&(\lambda+a_{11})\overline{a_{21}}+a_{12}\overline{a_{22}}\\ (\overline{\lambda+a_{11}})a_{21}+a_{22}\overline{a_{12}}~&| a_{21}| ^{2}+| a_{22}| ^{2} \end{array}} \right] \end{eqnarray*} 是投影. 因而 \begin{eqnarray*} \frac{1}{1+| \lambda| ^{2}}(| \lambda+a_{11}| ^{2}+| a_{12}| ^{2})=\frac{| \lambda| ^{2}+| a_{11}| ^{2}+| a_{12}| ^{2}+\lambda \overline{a_{11}}+\overline{\lambda}a_{11}}{1+| \lambda| ^{2}}\leq 1. \end{eqnarray*} 对上式取$\lambda =0,$ 可得$| a_{11}| ^{2}+| a_{12}| ^{2}\leq 1$,又由于$\lambda$是任意的,故矛盾. 因此,$a_{11}=0$. 同理,$\frac{1}{\sqrt{1+| \lambda| ^{2}}}(\lambda E_{22}+E_{12})$是部分等距,可得$a_{22}=0$. 又由于$\Phi_{2}$双边保持一秩,因而$a_{21}=0$,$a_{12} \in {\Bbb T}$或$a_{12}=0$,$a_{21} \in {\Bbb T}$. 假设第一种情形成立,令 $U_{3}=E_{11}+\overline{a_{12}}E_{22}$,$V_{3}=E_{11}+a_{12}E_{22}$,则 $\Phi_{2}(E_{ii})=U_{3}\Phi_{2}(E_{ii})V_{3}=E_{ii}(i=1,\ 2)$,$\Phi_{2}(E_{12})=U_{3}E_{12}V_{3}.$ 若第二种情形成立,则类似地可得 $\Phi_{2}(E_{12})=U_{3}E_{21}V_{3}.$

下面假设第一种情形成立,设$\Phi_3(A)=U_{3}^*\Phi_2(A)V_3^*,$ $A\in M_2({\Bbb C})$,则有$\Phi_3(E_{ii})=E_{ii}(i=1,2)$,$\Phi_{3}(E_{12})=E_{12}$.

由于$E_{21}\perp E_{12},$ 因而$\Phi_{3}(E_{21})\perp\Phi_{3}(E_{12})$. 因此,$\Phi_{3}(E_{21})=a_{21}E_{21}(a_{21}\in {\Bbb T})$. 由于$\Phi$在${\cal PI(H)}$上正交可和,因而$\Phi_{3}(E_{12}+E_{21})=E_{12}+a_{21}E_{21}$. 由于$\frac{1}{2}I+\frac{1}{2}\Phi_{3}(E_{12}+E_{21})$是部分等距,因而$\frac{1}{2}(I+E_{12}+a_{21}E_{21})$是部分等距,则有

$$\left[{\begin{array}{*{20}{c}} \frac{1}{2}~~& \frac{1}{2} \overline{a_{21}}\\ \frac{1}{2} ~~& \frac{1}{2} \end{array}} \right] \left[{\begin{array}{*{20}{c}} \frac{1}{2}~~& \frac{1}{2}\\ \frac{1}{2} a_{21}~~& \frac{1}{2}\\ \end{array}} \right]= \left[{\begin{array}{*{20}{c}} \frac{1}{4}+\frac{1}{4}| a_{21}| ^{2}~~& \frac{1}{4}+ \frac{1}{4}\overline{a_{21}}\\ \frac{1}{4}+ \frac{1}{4}a_{21}~~& \frac{1}{2} \end{array}} \right] $$ 是投影,因此有$\frac{1}{16}(1+a_{21})(1+\overline{a_{21}})+\frac{1}{4}=\frac{1}{2}.$ 又因为$a_{21}\in {\Bbb T}$,故$a_{21}=1$,因此,$\Phi_{3}(E_{21})=E_{21}$. 由于$\Phi$在${\cal PI(H)}$上正交可和,由引理2.3知 $\forall \alpha,\ \beta \in {\Bbb T}$,$\Phi_3(\alpha E_{12}+\beta E_{21})= \alpha E_{12}+\beta E_{21}$.

第四步: $\Phi_{3}(\alpha E_{ij}+\beta E_{ik})=\alpha E_{ij}+\beta E_{ik}$,$\Phi_{3}(\alpha E_{ij}+\beta E_{kj})=\alpha E_{ij}+\beta E_{kj}$($i,\ j,\ k=1,2$).

假设$\alpha,\ \beta\neq 0,$ 由于$\Phi_{3}$是齐次的,因而假设$\alpha =1.$ 下面只证明$\Phi_{3}(E_{11}+\beta E_{12})=E_{11}+\beta E_{12}$,其余情形类似可得. 设$\Phi_{3}(E_{11}+\beta E_{12})=\left[{\begin{array}{*{20}{c}} a_{11}~&a_{12}\\ a_{21}~&a_{22}\\ \end{array}} \right]$.

1. 若$a_{11}=0,$ 由于$\Phi_{3}$双边保一秩且$\Phi_{3}$是单射,则$a_{21}$与$a_{12}$有且只有一个为0.

(a) $a_{21}=0$且$a_{12}\neq 0.$

此时,$\Phi_{3}(E_{11}+\beta E_{12})=a_{12}E_{12}+\lambda a_{12} E_{22}$,由于$\frac{1}{\overline{a_{12}}}(a_{12}E_{12}+\lambda a_{12} E_{22})-\frac{\lambda a_{12}}{\overline{a_{12}}}E_{22}$是部分等距,因而 $\frac{1}{\overline{a_{12}}}(E_{11}+\beta E_{12})-\frac{\lambda a_{12}}{\overline{a_{12}}}E_{22}$是部分等距,又因为该部分等距是2秩的,从而是酉算子. 因此,$\beta=0$,与题设矛盾. 故该情况不成立.

(b) $a_{21}\neq0$且$a_{12}= 0.$

此时,$\Phi_{3}(E_{11}+\beta E_{12})=a_{21}E_{21}+\lambda a_{21}E_{22}$,由于$\frac{1}{\overline{a_{21}}}(a_{21}E_{21}+\lambda a_{21}E_{22})-\frac{\lambda a_{21}}{\overline{a_{21}}}E_{22}$是部分等距,因而$\frac{1}{\overline{a_{21}}}(E_{11}+\beta E_{12})-\frac{\lambda a_{21}}{\overline{a_{21}}}E_{22}$是部分等距,又因为该部分等距是2秩的,从而是酉算子,因此,$\beta=0$,与题设矛盾. 故该情况不成立.

综上可知,$a_{11}\neq 0.$

2. 若$a_{12}=0,$ 因为$\Phi_{3}$双边保一秩,则$a_{22}=0$.

此时,$\Phi_{3}(E_{11}+\beta E_{12})=a_{11}E_{11}+\lambda a_{11}E_{21}$,由于$(E_{11}+\beta E_{12})-\beta E_{12}$是部分等距,因而$(a_{11}E_{11}+\lambda a_{11}E_{21})-\beta E_{12}$是部分等距,又因为该部分等距是2秩的,从而是酉算子,于是有 $a_{11}=0.$ 矛盾.故$a_{12}\neq 0.$

由于$\Phi_{3}$双边保一秩,因而可设$\Phi_{3}(E_{11}+\beta E_{12})=\left[{\begin{array}{*{20}{c}} a_{11}~~&a_{12}\\ \lambda a_{11}~~&\lambda a_{12} \end{array}} \right].$ 假设$\lambda \neq 0$,由于$\Phi_{3}(E_{11}+\beta E_{12})-\beta E_{12}$ 以及$\frac{1}{\beta}\Phi_{3}(E_{11}+\beta E_{12})-\frac{1}{\beta} E_{11}$均是秩为2的部分等距,从而是酉算子,则有

\begin{eqnarray*} &&[\Phi_{3}(E_{11}+\beta E_{12})-\beta E_{12}][\Phi_{3}(E_{11}+\beta E_{12})-\beta E_{12}]^{*}\\&=&\left[{\begin{array}{*{20}{c}} | a_{11}| ^{2}+| a_{12}-\beta| ^{2}~~&\overline{\lambda}| a_{11}| ^{2}+\overline{\lambda a_{12}}(a_{12}-\beta)\\ \lambda | a_{11}| ^{2}+\lambda a_{12}\overline{(a_{12}-\beta)}~~&| \lambda| ^{2}| a_{11}| ^{2}+| \lambda| ^{2}| a_{12}| ^{2}\\ \end{array}} \right]\\ &=&I \end{eqnarray*} 以及 \begin{eqnarray*} &&\bigg[\frac{1}{\beta}\Phi_{3}(E_{11}+\beta E_{12})-\frac{1}{\beta} E_{11}\bigg]\bigg[\frac{1}{\beta}\Phi_{3}(E_{11}+\beta E_{12})-\frac{1}{\beta} E_{11}\bigg]^{*}\\&=&\frac{1}{| \beta| ^{2}}\left[{\begin{array}{*{20}{c}} | a_{11}-1| ^{2}+| a_{12}| ^{2}~&\overline{\lambda a_{11}}( a_{11}-1)+\overline{\lambda }| a_{12}| ^{2}\\ \lambda a_{11}\overline{( a_{11}-1)}+\lambda | a_{12}| ^{2}~&| \lambda| ^{2}| a_{11}| ^{2}+| \lambda| ^{2}| a_{12}| ^{2}\\ \end{array}} \right]\\ &=&I. \end{eqnarray*} 由上述两式可得

\[\overline {\lambda {a_{11}}} ({a_{11}} - 1) + \bar \lambda |{a_{12}}{|^2} = 0.\]

(2.13)

\[\frac{1}{{|\beta {|^2}}}(|{a_{11}} - 1{|^2} + |{a_{12}}{|^2}) = 1.\]

(2.14)

\[\frac{1}{{|\beta {|^2}}}(|\lambda {|^2}|{a_{11}}{|^2} + |\lambda {|^2}|{a_{12}}{|^2}) = 1.\]

(2.15)

\[|\lambda {|^2}|{a_{11}}{|^2} + |\lambda {|^2}|{a_{12}}{|^2} = 1.\]

(2.16)

由(2.13)式可得 $\overline{ a_{11}}=| a_{11}| ^{2}+| a_{12}| ^{2}$,将其代入(2.14)式可得 $\frac{1}{| \beta| ^{2}}(1-a_{11})=1.$ 同样的,代入(2.15)式可得 $\frac{| \lambda| ^{2}}{| \beta| ^{2}}a_{11}=1$. 再将其代入(2.16)式可得 $| \lambda| ^{2}a_{11}=1$,计算可得$| \beta| =1$,于是,$a_{11}=0.$ 矛盾. 故$\lambda = 0$.

此时,$\Phi_{3}(E_{11}+\beta E_{12})=a_{11}E_{11}+a_{12}E_{12}.$由于$\frac{1}{a_{11}}(a_{11}E_{11}+a_{12}E_{12})-\frac{a_{12}}{a_{11}}E_{12}$是部分等距,因而 $\frac{1}{a_{11}}(E_{11}+(\beta-a_{12})E_{12})$是部分等距,从而 $[\frac{1}{a_{11}}(E_{11}+(\beta-a_{12})E_{12})][\frac{1}{a_{11}}(E_{11}+(\beta-a_{12})E_{12})]^{*}=\frac{1+| \beta-a_{12}| ^{2}}{| a_{11}| ^{2}}E_{11}$ 是投影,所以

\[\frac{{1 + |\beta - {a_{12}}{|^2}}}{{|{a_{11}}{|^2}}} = 1.\]

(2.17)

又因为$\frac{1}{a_{12}}(a_{11}E_{11}+a_{12}E_{12})-\frac{a_{11}}{a_{12}}E_{11}$是部分等距,因而 $\frac{1}{a_{12}}((1-a_{11})E_{11}+\beta E_{12})$是部分等距.类似地,可得

\[\frac{{|1 - {a_{11}}{|^2} + |\beta {|^2}}}{{|{a_{12}}{|^2}}} = 1.\]

(2.18)

又因为$\frac{1}{\sqrt{1+| \beta| ^{2}}}(a_{11}E_{11}+a_{12}E_{12})$是部分等距,计算可得

\[\frac{{|{a_{11}}{|^2} + |{a_{12}}{|^2}}}{{1 + |\beta {|^2}}} = 1.\]

(2.19)

结合(2.18),(2.19)式可得

\[|{a_{11}}{|^2} + |1 - {a_{11}}{|^2} = 1.\]

(2.20)

故$| a_{11}| \leq1.$ 由(2.17)式可得 $| a_{11}| \geq1$,因而$| a_{11}| =1$. 将其分别代入(2.17),(2.20)式可得 $a_{11}=1,\ a_{12}=\beta.$ 因此,$\Phi_{3}(E_{11}+\beta E_{12})=E_{11}+\beta E_{12}.$

第五步: $\Phi_{3}\left(\left[{\begin{array}{*{20}{c}} \alpha~&\beta\\ 0~&\delta\\ \end{array}} \right]\right)=\left[{\begin{array}{*{20}{c}} \alpha~&\beta\\ 0~&\delta\\ \end{array}} \right]$,$\Phi_{3}\left(\left[{\begin{array}{*{20}{c}} 0~&\alpha\\ \delta~&\beta\\ \end{array}} \right]\right)=\left[{\begin{array}{*{20}{c}} 0~&\alpha\\ \delta~&\beta\\ \end{array}} \right]$.

假设$\alpha,\ \delta\neq 0$,且$\Phi_{3}\left(\left[{\begin{array}{*{20}{c}} \alpha~&\beta\\ 0~&\delta\\ \end{array}} \right]\right)=\left[{\begin{array}{*{20}{c}} a_{11}~&a_{12}\\ a_{21}~&a_{22}\\ \end{array}} \right]$. 由于$\Phi_{3}$双边保持算子束部分等距,因此

\[\left[ {\begin{array}{*{20}{c}} 0&{ - \frac{\beta }{\alpha }}\\ 0&{ - \frac{\delta }{\alpha }} \end{array}} \right] + \frac{1}{\alpha }\Phi \left( {\left[ {\begin{array}{*{20}{c}} \alpha &\beta \\ 0&\delta \end{array}} \right]} \right),\]

(2.21)

\[\left[ {\begin{array}{*{20}{c}} {0}&{ - \frac{\beta }{\alpha }}\\ { 0}&{1 - \frac{\delta }{\alpha }} \end{array}} \right] + \frac{1}{\alpha }{\Phi _3}\left( {\left[ {\begin{array}{*{20}{c}} {\alpha }&\beta \\ {0}&\delta \end{array}} \right]} \right),\]

(2.22)

\[\frac{1}{\delta }\left[ {\begin{array}{*{20}{c}} {\alpha }&\beta \\ {0}&0 \end{array}} \right] - \frac{1}{\delta }{\Phi _3}\left( {\left[ {\begin{array}{*{20}{c}} {\alpha }&\beta \\ {0}&\delta \end{array}} \right]} \right),\]

(2.23)

\[\frac{1}{{\sqrt 2 }}\left( {\frac{1}{\delta }{\Phi _3}\left( {\left[ {\begin{array}{*{20}{c}} {\alpha }&\beta \\ {0}&\delta \end{array}} \right]} \right) - \left[ {\begin{array}{*{20}{c}} {\frac{\alpha }{\delta }}&{1 + \frac{\beta }{\delta }}\\ {0}&0 \end{array}} \right]} \right),\]

(2.24)

\[\frac{1}{\delta }{\Phi _3}\left( {\left[ {\begin{array}{*{20}{c}} {\alpha }&\beta \\ {0}&\delta \end{array}} \right]} \right) - \left[ {\begin{array}{*{20}{c}} {1 + \frac{\alpha }{\delta }}&{\frac{\beta }{\delta }}\\ {0}&0 \end{array}} \right],\]

(2.25)

\[\frac{1}{{\sqrt 2 }}\left( {\frac{1}{\alpha }{\Phi _3}\left( {\left[ {\begin{array}{*{20}{c}} {\alpha }&\beta \\ {0}&\delta \end{array}} \right]} \right) + \left[ {\begin{array}{*{20}{c}} {0}&{1 - \frac{\beta }{\alpha }}\\ { 0}&{ - \frac{\delta }{\alpha }} \end{array}} \right]} \right)\]

(2.26)

均为部分等距,计算可得

\[\frac{1}{{|\alpha {|^2}}}\left[ {\begin{array}{*{20}{c}} {|{a_{11}}{|^2} + |{a_{21}}{|^2}}&{\overline {{a_{11}}} ({a_{12}} - \beta ) + \overline {{a_{21}}} ({a_{22}} - \delta )}\\ {{a_{11}}\overline {({a_{12}} - \beta )} + {a_{21}}\overline {({a_{22}} - \delta )} }&{|{a_{12}} - \beta {|^2} + |{a_{22}} - \delta {|^2}} \end{array}} \right],\]

(2.27)

\[\frac{1}{{|\alpha {|^2}}}\left[ {\begin{array}{*{20}{c}} {|{a_{11}}{|^2} + |{a_{21}}{|^2}}&{\overline {{a_{11}}} ({a_{12}} - \beta ) + \overline {{a_{21}}} ({a_{22}} + \alpha - \delta )}\\ {{a_{11}}\overline {({a_{12}} - \beta )} + {a_{21}}\overline {({a_{22}} + \alpha - \delta )} }&{|{a_{12}} - \beta {|^2} + |{a_{22}} + \alpha - \delta {|^2}} \end{array}} \right],\]

(2.28)

\[\frac{1}{{|\delta {|^2}}}\left[ {\begin{array}{*{20}{c}} {|\alpha - {a_{11}}{|^2} + |{a_{21}}{|^2}}&{\overline {(\alpha - {a_{11}})} (\beta - {a_{12}}) + \overline {{a_{21}}} {a_{22}}}\\ {(\alpha - {a_{11}})\overline {(\beta - {a_{12}})} + {a_{21}}\overline {{a_{22}}} }&{|\beta - {a_{12}}{|^2} + |{a_{22}}{|^2}} \end{array}} \right],\]

(2.29)

\[\frac{1}{{2|\delta {|^2}}}\left[ {\begin{array}{*{20}{c}} {|{a_{11}} - \alpha {|^2} + |{a_{21}}{|^2}}&{\overline {({a_{11}} - \alpha )} ({a_{12}} - \beta - \delta ) + \overline {{a_{21}}} {a_{22}}}\\ {({a_{11}} - \alpha )\overline {({a_{12}} - \beta - \delta )} + {a_{21}}\overline {{a_{22}}} }&{|{a_{12}} - \beta - \delta {|^2} + |{a_{22}}{|^2}} \end{array}} \right],\]

(2.30)

\[\frac{1}{{|\delta {|^2}}}\left[ {\begin{array}{*{20}{c}} {|{a_{11}} - \alpha - \delta {|^2} + |{a_{21}}{|^2}}&{\overline {({a_{11}} - \alpha - \delta )} ({a_{12}} - \beta ) + \overline {{a_{21}}} {a_{22}}}\\ {({a_{11}} - \alpha - \delta )\overline {({a_{12}} - \beta )} + {a_{21}}\overline {{a_{22}}} }&{|{a_{12}} - \beta {|^2} + |{a_{22}}{|^2}} \end{array}} \right],\]

(2.31)

\[\frac{1}{{2|\alpha {|^2}}}\left[ {\begin{array}{*{20}{c}} {|{a_{11}}{|^2} + |{a_{21}}{|^2}}&{\overline {{a_{11}}} ({a_{12}} + \alpha - \beta ) + \overline {{a_{21}}} ({a_{22}} - \delta )}\\ {{a_{11}}\overline {({a_{12}} + \alpha - \beta )} + {a_{21}}\overline {({a_{22}} - \delta )} }&{|{a_{12}} + \alpha - \beta {|^2} + |{a_{22}} - \delta {|^2}} \end{array}} \right]\]

(2.32)

均为投影,又因为$M_{2}({\Bbb C})$中的非零投影为$E_{11},\ E_{22},\ I$或 $\frac{1}{1+| \lambda| ^{2}}\left[{\begin{array}{*{20}{c}} 1~&\lambda\\ \overline{\lambda}~&| \lambda| ^{2} \end{array}} \right]$ ($\forall \lambda \in {\Bbb C}$且$\lambda \neq 0$),因而比较(2.27),(2.28) 式可得

\[|{a_{12}} - \beta {|^2} + |{a_{22}} - \delta {|^2} = |\alpha {|^2},\;|{a_{12}} - \beta {|^2} + |{a_{22}} + \alpha - \delta {|^2} = 0\]

(2.33)

或

\[|{a_{12}} - \beta {|^2} + |{a_{22}} - \delta {|^2} = 0,\;|{a_{12}} - \beta {|^2} + |{a_{22}} + \alpha - \delta {|^2} = |\alpha {|^2}.\]

(2.34)

假设(2.33)式成立,则有 $$a_{12}=\beta,\ a_{22}=\delta-\alpha,\ | a_{11}| ^{2}+| a_{21}| ^{2}=| \alpha| ^{2},\ a_{21}=0,\ | a_{11}| =| \alpha| . $$ 将其代入(2.32)式可得$\left[{\begin{array}{*{20}{c}} \frac{1}{2}&\alpha \overline{ a_{11}}\\ \overline{\alpha}a_{11}&1\\ \end{array}} \right]$,易知其不是投影,矛盾. 因而 $$a_{12}=\beta,\ a_{22}=\delta,\ | a_{11}| ^{2}+| a_{21}| ^{2}=| \alpha| ^{2},\ a_{21}=0,\ | a_{11}| =| \alpha| . $$ 比较(2.29),(2.31)式可得

\[|\alpha - {a_{11}}{|^2} + |{a_{21}}{|^2} = 0,\;|{a_{11}} - \alpha - \delta {|^2} + |{a_{21}}{|^2} = |\delta {|^2},\]

(2.35)

或

\[|\alpha - {a_{11}}{|^2} + |{a_{21}}{|^2} = |\delta {|^2},\;|{a_{11}} - \alpha - \delta {|^2} + |{a_{21}}{|^2} = 0.\]

(2.36)

假设(2.36)式成立,则有 $a_{11}=\delta+\alpha,$ 将其代入(2.30)式可得$\left[{\begin{array}{*{20}{c}} \frac{1}{2}&-\frac{1}{2}\\ -\frac{1}{2}&1\\ \end{array}} \right]$,易知其不是投影,矛盾. 因而$a_{11}=\alpha$. 综上可知,$ a_{11}=\alpha,\ a_{12}=\beta,\ a_{22}=\delta,\ a_{21}=0.$ 另一个等式类似可得.

第六步: $\Phi_{3}\left(\left[{\begin{array}{*{20}{c}} \alpha~&\beta\\ \gamma~&\delta\\ \end{array}} \right]\right)=\left[{\begin{array}{*{20}{c}} \alpha~&\beta\\ \gamma~&\delta\\ \end{array}} \right]$.

假设$\alpha,\gamma \neq 0,$ 且$\Phi_{3}\left(\left[{\begin{array}{*{20}{c}} \alpha~&\beta\\ \gamma~&\delta\\ \end{array}} \right]\right)=\left[{\begin{array}{*{20}{c}} a_{11}~&a_{12}\\ a_{21}~&a_{22}\\ \end{array}} \right]$. 由于$\Phi_{3}$双边保持算子束部分等距,因此

\[\frac{1}{\alpha }\left( {{\Phi _3}\left( {\left[ {\begin{array}{*{20}{c}} {\alpha }&\beta \\ {\gamma }&\delta \end{array}} \right]} \right) + \left[ {\begin{array}{*{20}{c}} {0}&{ - \beta }\\ { - \gamma }&{ - \delta } \end{array}} \right]} \right),\]

(2.37)

\[\frac{1}{\alpha }\left( {{\Phi _3}\left( {\left[ {\begin{array}{*{20}{c}} {\alpha }&\beta \\ {\gamma }&\delta \end{array}} \right]} \right) + \left[ {\begin{array}{*{20}{c}} {0}&{ - \beta }\\ { - \gamma }&{\alpha - \delta } \end{array}} \right]} \right),\]

(2.38)

\[\frac{1}{\gamma }\left( {{\Phi _3}\left( {\left[ {\begin{array}{*{20}{c}} {\alpha }&\beta \\ {\gamma }&\delta \end{array}} \right]} \right) - \left[ {\begin{array}{*{20}{c}} {\alpha }&\beta \\ {0}&\delta \end{array}} \right]} \right),\]

(2.39)

\[\frac{1}{{\sqrt 2 \alpha }}\left( {{\Phi _3}\left( {\left[ {\begin{array}{*{20}{c}} {\alpha }&\beta \\ {\gamma }&\delta \end{array}} \right]} \right) - \left[ {\begin{array}{*{20}{c}} {0}&{\beta - \alpha }\\ {\gamma }&\delta \end{array}} \right]} \right)\]

(2.40)

均为部分等距,类似于第五步中计算,比较(2.37),(2.38)两式的计算结果可得

\[|{a_{12}} - \beta {|^2} + |{a_{22}} - \delta {|^2} = |\alpha {|^2},\;|{a_{12}} - \beta {|^2} + |{a_{22}} + \alpha - \delta {|^2} = 0,\]

(2.41)

或

\[|{a_{12}} - \beta {|^2} + |{a_{22}} - \delta {|^2} = 0,\;|{a_{12}} - \beta {|^2} + |{a_{22}} + \alpha - \delta {|^2} = |\alpha {|^2}.\]

(2.42)

假设(2.41)式成立,则有 $$a_{12}=\beta,\ a_{22}=\delta-\alpha,\ | a_{11}| ^{2}+| a_{21}-\gamma| ^{2}=| \alpha| ^{2},\ a_{21}=\gamma,\ | a_{11}| =| \alpha| . $$ 将其代入(2.40)式的计算结果可得$\left[{\begin{array}{*{20}{c}} \frac{1}{2}&\frac{\overline{a_{11}}}{2\overline{\alpha}}\\ \frac{a_{11}}{2\alpha}&1\\ \end{array}} \right]$,易知其不是投影,矛盾. 因而 $$a_{12}=\beta,\ a_{22}=\delta,\ | a_{11}| ^{2}+| a_{21}-\gamma| ^{2}=| \alpha| ^{2},\ a_{21}=\gamma,\ | a_{11}| =| \alpha| . $$ 将其代入(2.39)式的计算结果可得 $| a_{11}-\alpha| ^{2}+| a_{21}| ^{2}=| \gamma| ^{2}$. 因而$a_{11}=\alpha$. 综上可知,$ a_{11}=\alpha,\ a_{12}=\beta,\ a_{22}=\delta,\ a_{21}=\gamma.$

综上所述,$\Phi_{3}(A)=A$. 令$U=U_{1}U_{2}U_{3},\ V=V_{3}V_{2}V_{1}$. 则有

$$\Phi(A)=UAV,~~ \forall A\in M_2({\Bbb C}). $$ 类似地,假设第二种情形成立,则存在酉算子$U,V\in M_2({\Bbb C})$,使得 $$\Phi(A)=UA^{t}V,~~ \forall A\in M_2({\Bbb C}). $$ 证毕.