2 主要结论
设$E(H)$为Hillbert空间$H$上的效应代数, $P(H)$为$B(H)$中所有投影的全体, $E(\mathcal{A})$为von-Neumann代数$\mathcal{A}$上的效应代数.用$M_n(\mathbb{C})$表示$n$阶矩阵代数, $E(\mathbb{C}^{n})$表示所有大于等于0小于等于I的半正定矩阵, 则$E(\mathbb{C}^{n})$是矩阵代数$M_n(\mathbb{C})$上的效应代数.
定义2.1[1] 设$E$和$F$为效应代数, 如果映射$\varphi:E\rightarrow F$满足
(1) $a\bot b\Rightarrow \varphi(a)\bot\varphi(b),$
(2) $\varphi(a\oplus b)=\varphi(a)\oplus \varphi(b),$
(3) $\varphi(1_{E})=1_{F},$
则称$\varphi$为态射. (条件 (1) 和条件 ($1^{'}$) $a \leq b\Rightarrow \varphi(a)\leq \varphi(b),$是等价的.)
定义2.2[1] 设$E$和$F$为效应代数, $\varphi:E\rightarrow F$是态射, 并且满足对任意$a,b\in E$, 若$\varphi(a)\bot \varphi(b)$, 则$ a\bot b,$我们就称$\varphi$为单态射.
定义2.3[1] 设$E$和$F$为效应代数, $\varphi:E\rightarrow F$是态射, 如果满足保存在的有限的上确界和下确界, 则称$\varphi$为同态.
定义2.4[1] 设$\varphi:E(H)\rightarrow E(H)$是同态, 如果满足对于$E(H)$中的任意单调递增的序列$(A_{i},i\in\Lambda),$有$\varphi(\bigvee_{i\in\Lambda} (A_{i})=\bigvee_{i\in\Lambda} \varphi(A_{i})$, 则称$\varphi$为$\sigma$-正交完备的同态.
定义2.5 设$\varphi:E(H)\rightarrow E(H)$是态射, 如果满足对于任意投影$P,Q\in P(H)$, $\varphi(P)\bot \varphi(Q)\Rightarrow P\bot Q$, 则称$\varphi$是强态射.
下面引理来自文献[1], 为方便读者我们给出其证明.
引理2.6[1] 设$\varphi:E(H)\rightarrow E(H)$为$\sigma$-正交完备的同态, 则$\varphi(\lambda A)=\lambda \varphi(A),$任意$\lambda\in[0,1],A\in E(H).$
证 因为$\varphi(A)=\varphi(n/nA)=\varphi(1/nA)+\varphi(1/nA)+\cdots+ \varphi(1/nA)=n\varphi(1/nA),\forall n\in \mathbb{N}$, 所以$\varphi(1/nA)=1/n\varphi(A)$.同理可得$\varphi(m/nA)=m/n\varphi(A),$任意$n,m\in\mathbb{N},m\leq n$.故对于任意有理数$\mu \in [0,1],\varphi(\mu A)=\mu \varphi(A)$成立.对于任意无理数$\lambda \in [0,1],$取一列单调递增的有理数列$a_{i},i=1,2\cdots$, 使得$\lim\limits_{i\rightarrow \infty}a_{i}=\lambda$, 即$\sup\limits_{i\in\Lambda} a_{i}=\lambda.$对于任意$A\in E(H)$, $\{a_{i}A\}$是一个单调递增序列, 强算子收敛到它的上确界, 即$\bigvee_{i\in \Lambda}a_{i}A=\lambda A.$又$\varphi$是$\sigma$-正交完备的同态, 所以$\varphi(\bigvee_{i\in \Lambda}a_{i}A)=\bigvee \varphi(a_{i}A)=\bigvee a_{i}\varphi(A),$即$\varphi(\lambda A)=\lambda \varphi(A).$综上知结论成立.
定理2.7 设$\varphi:E(H)\rightarrow E(H)$是满的$\sigma$-正交完备的强同态, 则存在$H$上的酉算子或反酉算子$U$, 使得任意$ A\in E(H),$有$\varphi(A)=UAU^{*}$成立.
证 我们先证$\varphi$保投影.如果$P$是投影, 则由文[3]中推论2.8可知, $P\wedge P^{'}=0$.由$\varphi$保单位可得, $I=\varphi(I)=\varphi(P)+\varphi(P^{'})\Rightarrow \varphi(P^{'})=I-\varphi(P)=\varphi(P)^{'}.$因为$\varphi$为同态, 所以$\varphi(P)\wedge \varphi(P)^{'}=\varphi(P)\wedge \varphi(P')=\varphi(P\wedge P^{'})=0,$故$\varphi$保投影.由文[1]中引理3.3可知, $\varphi$是单的.则易证$\varphi|_{P(H)}:P(H)\rightarrow P(H)$为单的满的$\sigma$-正交完备的强同态, 由强同态的定义可知, $\varphi$对投影是双边保序的.因此$\varphi|_{P(H)}:P(H)\rightarrow P(H)$为正交序自同构, 由Ludwig在文[4]中的结论可知, 存在酉算子或反酉算子$U$, 使得任意$ P\in P(H),$有$\varphi(P)=UPU^{*}$成立.
下证$\varphi(A)=UAU^{*},$任意$A\in E(H)$.由引理2.6可知, 任意$A\in E(H),\lambda \in[0,1],$有$\varphi(\lambda A)=\lambda \varphi(A).$又任意$ A\in E(H)$, 它是所有$\lambda_{i}P_{i}$形式的元的上确界, 其中$\lambda_{i}\in[0,1]$, $P_{i}$为秩一投影.所以存在相互正交的有限秩投影$P_{1},P_{2},\cdots P_{n}\cdots$, 及$\lambda_{1},\lambda_{2},\cdots \lambda_{n}\cdots\in \sigma(A)\subseteq[0,1],$使得
| $A=\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {{\lambda _i}} {P_i}{i=1}=\bigvee \lambda_{i}P_{i},\\ \varphi(A)=\varphi(\bigvee \lambda_{i}P_{i})=\bigvee\varphi(\lambda_{i}P_{i})=\bigvee\lambda_{i}\varphi(P_{i})=\bigvee \lambda _{i}UP_{i}U^{*}=U\bigvee \lambda_{i}P_{i}U^{*}=UAU^{*}.$ |
定理2.8 设$\mathcal{A}$是一个von-Neumann代数.令$\varphi:E(\mathcal{A})\rightarrow E(H)$是同态, 且满足
| $\varphi(A)\leq \varphi(B)\Rightarrow A\leq B,\\ \varphi(\lambda A)=\lambda \varphi(A),\ \forall \lambda \in[0,1],\ A\in E(\mathcal{A}).$ |
则$\varphi$可以延拓到$\psi:\mathcal{A}\rightarrow B(H)$上有界的Jordan $*$同态.
证 我们先证$\varphi$保补元.因为$\varphi$保单位, 所以对于任意元$ A$, $I=\varphi(I)=\varphi(A+A^{'})=\varphi(A)\oplus\varphi(A^{'}),$所以$\varphi(A^{'})=I-\varphi(A)=\varphi(A)^{'}.$即$\varphi$保补元.
下证$\varphi$保投影.如果$P$是投影, 即$P\wedge P^{'}=0.$因为$\varphi$是同态, 所以有$\varphi(P)\wedge \varphi(P)^{'}=\varphi(P)\wedge\varphi(P^{'})=\varphi(P\wedge P^{'})=0,$即$\varphi(P)$是投影.
$\varphi$对投影保正交性.任意$P,Q\in E(\mathcal{A})$是投影, 如果$PQ=0$, 则$P\leq Q^{'}$, 又$\varphi$保序, 所以$\varphi(P)\leq\varphi(Q^{'}).$ $\varphi(P)\varphi(Q)=\varphi(P)(I-\varphi(Q)^{'})=\varphi(P)(I-\varphi(Q^{'})=\varphi(P)-\varphi(P)\varphi(Q^{'})=\varphi(P)-\varphi(P)=0.$
先将$\varphi$延拓到$\mathcal{A}$的所有正元$\mathcal{A}^{+}$上.令$\varphi_{1}: \mathcal{A}^{+}\rightarrow B(H)^{+},$定义$\varphi_{1}(A)=\|A\|\varphi(A/\|A\|),\ 0\neq A\in \mathcal{A}^{+}.$若$ A=0$, 定义$\varphi_{1}(A)=0$.则$\varphi_{1}$保单位, 正的可加的.且$\varphi_{1}(\lambda A)=\lambda \varphi_{1}(A),$任意$ A,B\in \mathcal{A}^{+}.$ $\varphi_{1}(A)+\varphi_{1}(B)=\|A\|\varphi(A/\|A\|)+\|B\|\varphi(B/\|B\|).$设$\|A\|\ \geq \|B\|$, 则
| $\varphi_{1}(A)+\varphi_{1}(B)=\|A\|(\varphi(A/\|A\|)+\|B\|/\|A\|\varphi(B/\|B\|))\\ =\|A\|(\varphi(A/\|A\|)+\varphi(B/\|A\|))\\=\|A\|\varphi((A+B)/\|A\|)\\ =(\|A\|\|A+B\|/\|A+B\|)\varphi(A+B)/\|A\|\\=\|A+B\|\varphi(A+B/\|A+B\|)=\varphi_{1}(A+B).$ |
因此$\varphi_{1}$是正的可加的.故$\varphi_{1}(\lambda A)=\|\lambda A\|\varphi(\lambda A/\|\lambda A\|)=\lambda \varphi_{1}(A)),$任意$ \lambda\geq 0.$且$\varphi_{1}$保投影.如果$P$是投影, 即$P\wedge P^{'}=0.$因为$\varphi$是同态, 且保正交补, 则
| $\varphi(P)\wedge \varphi(P)^{'}=\varphi(P)\wedge\varphi(P^{'})=\varphi(P\wedge P^{'})=0.$ |
所以$\varphi$保投影, 且对投影保正交性.如果$P, Q$是投影, 且$PQ=0$, 则$\varphi_{1}(P)\varphi_{1}(Q)=\varphi(P)\varphi(Q)=0.$ $\varphi_{1}|_{E(\mathcal{A})}=\varphi$, $\|\varphi_{1}(A)\|=\| (\|A\|\varphi(A/\|A\|))\|\leq \|A\|$, 且$\varphi_{1}(I)=I .$
进一步将$\varphi_{1}$延拓到$\mathcal{A}$的所有自伴元上. $\varphi_{2}:\mathcal{A}_{s}\rightarrow B(H)_{s},$定义$\varphi_{2}(A)=\varphi_{1}(A^{+})-\varphi_{1}(A^{-})$, $A^{+}$和$A^{-}$分别为$A$的正的部分和负的部分.则可证$\varphi_{2}$是实线性的, 保投影及投影的正交性.任意$A,B\in \mathcal{A}_{s}\Rightarrow A+B\in \mathcal{A}_{s}$, 且$A=A^{+}-A^{-},B=B^{+}-B^{-}.$
| $\varphi_{2}(A+B)=\varphi_{1}(A^{+}+B^{+})-\varphi_{1}(A^{-}+B^{-})=\varphi_{2}(A)+\varphi_{2}(B),$ |
易证任意$ \lambda \in \mathbb{R}, A\in \mathcal{A}_{s}, \varphi_{2}(\lambda A)=\lambda \varphi_{2}(A).$设
| $A=\|A\|I+A/2-(\|A\|I-A/2),A^{+}=\|A\|I+A/2,A^{-}=\|A\|I-A/2,$ |
$(\lambda A)^{+}=\|\lambda A\|I+\lambda A/2$, $(\lambda A)^{-}=\|\lambda A\|I-\lambda A/2$, 如果$\lambda \geq 0,(\lambda A)^{+}=\lambda A^{+}, (\lambda A)^{-}=\lambda A^{-},$显然$\varphi_{2}(\lambda A)=\lambda \varphi_{2}(A).$如果
| $\lambda\leq 0,(\lambda A)^{+}=-\lambda A^{+}, (\lambda A)^{-}=-\lambda A^{-}.$ |
则可得
| $\varphi_{2}(\lambda A)=\lambda \varphi_{2}(A).$ |
所以结论成立.
又易证$\varphi_{2}|_{\mathcal{A}^{+}}=\varphi_{1},$因此$\varphi_{2}$保投影及投影的正交性, $\varphi_{2}(I)=I$.
| $\|\varphi_{2}(A)\|=\|\varphi_{1}(A^{+})-\varphi_{1}(A^{-})\|\leq \|\varphi_{1}(A^{+})\|+\|\varphi_{1}(A^{-})\|\leq 2\|A\|,$ |
将$\varphi_{2}$延拓到整个von-Neumann代数$\mathcal{A}$上.定义$\psi(A)=\varphi_{2}({\rm Re}A)+i\varphi_{2}({\rm Im}A),$任意$ A\in \mathcal{A}.$其中 (Re$A$) 和 (Im$A$) 分别是$\mathcal{A}$的实部和虚部.则可证$\psi$为线性的, 保投影及保投影的正交性, 且$\psi|\mathcal{A}_{s}=\varphi_{2}$.任意$\lambda\in\mathbb{C}$, 设$\lambda=a+ib,a,b\in\mathbb{R}$, 任意$ A\in \mathcal{A}.$
| $\psi(\lambda A)=\varphi_{2}({\rm Re}\lambda A)+i\varphi_{2}({\rm Im}\lambda A)\\ =\varphi_{2}(\lambda A+\lambda A^{*})/2+i\varphi_{2}(i\overline{\lambda}A^{*}-i\lambda A)/2\\ =\varphi_{2}((aA+aA^{*})/2+(ibA-ibA^{*})/2)+i\varphi_{2}((iaA^{*}-iaA)/2+(bA+bA^{*})/2)\\ =a\varphi_{2}(A+A^{*})/2+b\varphi_{2}(iA-iA^{*})/2+ia\varphi_{2}(iA^{*}-iA)/2+ib\varphi_{2}(A+A^{*})/2\\ =(a+ib)\varphi_{2}(A+A^{*})/2+i(a+ib)\varphi_{2}(iA^{*}-iA)/2\\ =\lambda\varphi_{2}({\rm Re}A)+i\lambda \varphi_{2}({\rm Im}A)\\ =\lambda \psi(A).$ |
任意$A, B\in \mathcal{A}$,
| $\psi(A+B)=\varphi_{2}({\rm Re}(A+B))+i\varphi_{2}({\rm Im}(A+B))\\ =\varphi_{2}((A+B)/2+(A^{*}+B^{*})/2)+i\varphi_{2}(i(A^{*}+B^{*})/2-i(A+B)/2))\\ =\varphi_{2}(A+A^{*})/2+\varphi_{2}(B+B^{*})/2+i\varphi_{2}(iA^{*}-iA)/2+i\varphi_{2}(iB^{*}-iB)/2\\ =\psi(A)+\psi(B),$ |
综上$\psi$是线性的, 因为易证$\psi|_{E(\mathcal{A})}=\varphi.$所以$\psi$保投影及其正交性.
| $\|\psi(A)\|=\|\varphi_{2}({\rm Re} A)+i\varphi_{2}({\rm Im} A)\|\leq \|2\varphi_{2}(A)\|\leq4\|A\|,$ |
因此$\psi$是有界的, 且$\psi(I)=I.$
下证$\psi$是Jordan同态.假设$P$是任意一个投影, 因为$\psi$保投影, 所以$\psi(P^{2})=\psi(P)^{2}.$对任意自伴元$A\in\mathcal{A}$, 由文[5]中定理5.2.2知, 存在相互正交的投影$P_{1}, P_{2}, \cdots P_{n}\cdots$及$\lambda_{i}\in \mathbb{R}$, 使得$A=\lim\limits_{n\rightarrow\infty}\sum\limits^{n}_{i=1}\lambda_{i}P_{i}$.又因为$\psi$是保正交的, 因此$\psi(P_{1}), \psi(P_{2}),\cdots \psi(P_{n})\cdots$也是相互正交的, 又$\psi$是有界的, 故有
| $\begin{array}{l}
\psi ({A^2}) = \psi (\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\lambda _i^2} P_i^2) = \mathop {\lim }\limits_{n \to \infty } \psi (\sum\limits_{i = 1}^n {\lambda _i^2} P_i^2) = \mathop {\lim }\limits_{n \to \infty } (\sum\limits_{i = 1}^n {\lambda _i^2} \psi (P_i^2))\\
= \psi {(\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {{\lambda _i}} {P_i})^2} = \psi {(A)^2}.
\end{array}$ |
所以对于任意自伴元$C, D\in \mathcal{A}$, $C+D$也是自伴的, 把$A$换成$C+D$, 则可有$\psi((C+D)^{2})=\psi(C+D)^{2}$, 进而可得$\psi(CD+DC)=\psi(C)\psi(D)+\psi(D)\psi(C),$任意$T \in \mathcal{A}, T=H+iK$, $H$和$K$是自伴的.因此
| $\psi(T^{2})=\psi(H^{2}-K^{2}+iHK+iKH)=\psi(H)^{2}-\psi(K)^{2}+i\psi(HK+KH)\\ =(\psi(H)+\psi(K))^{2}=(\psi(H+iK))^{2}=\psi(T)^{2}.$ |
又$\psi$将自伴元映为自伴元, 所以综上可知, $\psi$是有界的Jordan $*$同态.
推论2.8.1 设$\varphi:E(\mathcal{A})\rightarrow E(H)$是同态, 且满足
| $\varphi(A)\leq \varphi(B)\Rightarrow A\leq B,\\ \varphi(\lambda A)=\lambda \varphi(A),\forall \lambda \in[0,1],A\in E(\mathcal{A}),$ |
如果$\mathcal{A}$是交换的von-Neumman代数, 则$\varphi$可以延拓到$\psi:\mathcal{A}\rightarrow B(H)$上的有界$*$同态.
证 由定理2.8可知, $\varphi$可以延拓到$\psi:\mathcal{A}\rightarrow B(H)$上有界的Jordan$*$同态.对任意投影$ P,Q\in \mathcal{A},$因为$\mathcal{A}$是交换的，所以$PQ=QP.$因此$PQ$是投影, 则存在相互正交的投影$p_{1}, q_{1}, r,$使得$P=p_{1}+r,Q=q_{1}+r.$故$\psi(P)\psi(Q)=(\psi(p_{1})+\psi(r))(\psi(q_{1})+\psi(r))=\psi(r)=\psi(PQ).$又任意$A, B\in E(\mathcal{A})$, 存在相互正交的投影$P_{i}\in E(\mathcal{A}),i=1,2,\cdots n\cdots.$及$Q_{j}\in E(\mathcal{A}),j=1,2,\cdots m\cdots.$使得
| $A = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {{\lambda _i}} {P_i},B = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{j = 1}^m {{\mu _j}} {Q_j},$ |
则
| $\begin{array}{l}
\psi (AB) = \mathop {\lim }\limits_{n \to \infty } \mathop {\lim }\limits_{m \to \infty } \psi (\sum\limits_{i = 1}^n {{\lambda _i}} {P_i}\sum\limits_{j = 1}^m {{\mu _j}} {Q_j}) = \mathop {\lim }\limits_{n \to \infty } \mathop {\lim }\limits_{m \to \infty } \psi (\sum\limits_{i = 1}^n {{\lambda _i}} \sum\limits_{j = 1}^m {{\mu _j}} {P_i}{Q_j})\\
= \mathop {\lim }\limits_{n \to \infty } \mathop {\lim }\limits_{m \to \infty } (\sum\limits_{i = 1}^n {{\lambda _i}} \sum\limits_{j = 1}^m {{\mu _j}} \psi ({P_i}{Q_j})) = \mathop {\lim }\limits_{n \to \infty } \mathop {\lim }\limits_{m \to \infty } (\sum\limits_{i = 1}^n {{\lambda _i}} \sum\limits_{j = 1}^m {{\mu _j}} \psi ({P_i})\psi ({Q_j}))\\
= \mathop {\lim }\limits_{n \to \infty } \mathop {\lim }\limits_{m \to \infty } (\sum\limits_{i = 1}^n {{\lambda _i}} \psi ({P_i})\sum\limits_{j = 1}^m {{\mu _j}} \psi ({Q_j})) = \mathop {\lim }\limits_{n \to \infty } \psi (\sum\limits_{i = 1}^n {{\lambda _i}} {P_i})\mathop {\lim }\limits_{m \to \infty } \psi (\sum\limits_{j = 1}^m {{\mu _j}} {Q_j})\\
= \psi (\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {{\lambda _i}} {P_i})\psi (\mathop {\lim }\limits_{m \to \infty } \sum\limits_{j = 1}^m {{\mu _j}} {Q_j}) = \psi (A)\psi (B).
\end{array}$ |
因此综上可知$\psi$是$\mathcal{A}\rightarrow B(H)$上的有界$*$同态.
引理2.9 设$\mathcal{M}=E(\mathbb{C}^{n}),E(H),$或$E(\mathcal{A})$, $\mathcal{A}$是von-Neumman代数.若$\varphi:\mathcal{M}\rightarrow \mathcal{M}$是单态射, 则$\varphi(\lambda A)=\lambda \varphi(A),$任意$ A\in\mathcal{M}, \lambda\in[0,1].$
证 由文[1]中单态射的定义可知, $\varphi$是双边保序的, 且保单位元.
| $\varphi(A)=\varphi((n/n)A)=\varphi(A/n)+\varphi(A/n)+\cdots \varphi(A/n)=n\varphi(A/n),$ |
任意$ n\in\mathbb{N}$.所以$\varphi(A/n)=1/n\varphi(A).$同理可证$\varphi(nA/m)=n/m\varphi(A),n,m\in\mathbb{N},n\leq m.$则对于任意有理数$\mu\in[0,1]$, 有$\varphi(\mu A)=\mu\varphi(A)$成立.对于无理数$\lambda\in[0,1]$, 取一列单调递增的有理数列$\{a_{i}\}$使得$\lim\limits_{i\rightarrow\infty} a_{i}=\lambda$, 即$\sup_{i} a_{i}=\lambda$. $\forall A\in\mathcal{M}, \bigvee_{i} a_{i}A=\lambda A$. $a_{i}A\leq\lambda A.$因为$\varphi$是保序的, 所以$\varphi(a_{i}A)\leq \varphi(\lambda A).$故$\bigvee\varphi(a_{i}A)\leq\varphi(\lambda A)$.即
| $\lambda\varphi(A)\leq \varphi(\lambda A).$ |
又因为$\bigvee\varphi(a_{i}A)\geq \varphi(a_{i}A)$, $\varphi^{-1}$也是保序的, 所以
| $\varphi^{-1}(\bigvee\varphi(a_{i}A))\geq\varphi^{-1}( \varphi(a_{i}A)),$ |
即$\varphi^{-1}(\bigvee\varphi(a_{i}))\geq (a_{i}A)$.因此$\varphi^{-1}(\bigvee\varphi(a_{i}A))\geq \bigvee(a_{i}A)$, 即$(\bigvee\varphi(a_{i}A))\geq \varphi(\bigvee(a_{i}A))$.即$\lambda\varphi(A)\geq\varphi(\lambda A)$.综上可知$\varphi(\lambda A)=\lambda\varphi(A).$所以对于任意$\lambda\in[0,1],\varphi(\lambda A)=\lambda\varphi(A).$
推论2.9.1 设$\varphi:E(\mathbb{C}^{n})\rightarrow E(\mathbb{C}^{n})$单态射, 则存在$H$上的酉算子或反酉算子$U$, 使得$\varphi(A)=UAU^{*}, A\in E(\mathbb{C}^{n}).$
证 因为$\varphi$是单态射, 由文献[1]中的证明可知, $\varphi$是单射.由单态射的定义知$\varphi$是双边保序的, 且保单位元.所以$\varphi$是保正交补的.任意$P, P^{'}\in E(\mathbb{C}^{n}), P+P^{'}=I,$
| $I=\varphi(P+P^{'})=\varphi(P)+\varphi(P^{'}),$ |
则$\varphi(P')=I-\varphi(P)=\varphi(P)^{'}.$如果$\varphi$是满射, 则$\varphi$是正交序自同构.下证$\varphi$是满射.首先将$\varphi$延拓到正矩阵$ M_n(\mathbb{C})^{+}$上.令$\varphi_{1}:M_n(\mathbb{C})^{+}\rightarrow M_n(\mathbb{C})^{+}$.若$A=0$, 定义$ \varphi_{1}(A)=0.$若$ 0\neq A\in M_n(\mathbb{C})^{+},$定义$\varphi_{1}(A)=\|A\|\varphi(A/\|A\|).$则由定理2.7的证明可知, $\varphi_{1}$对正元可加且满足$\varphi_{1}(\lambda A)=\lambda\varphi_{1}(A),$任意$\lambda\geq 0,A\in M_n(\mathbb{C})^{+}$.如果$\varphi_{1}(A)=\varphi_{1}(B)$, 则
| $\|A\|\varphi(A/\|A\|)=\|B\|\varphi(B/\|B\|).$ |
设$\|A\|\geq \|B\|$, 则由引理2.9可得
| $\varphi(A/\|A\|)=\|B\|/\|A\|\varphi(B/\|B\|)=\varphi(B/\|A\|).$ |
因为$\varphi$是单射, 所以可得$A=B$, 即$\varphi_{1}$是单射.
进一步将$\varphi_{1}$延拓到所有实矩阵$M_{s}(\mathbb{C})$上.定义$\varphi_{2}:M_{s}(\mathbb{C})\rightarrow M_{s}(\mathbb{C}),$ $\varphi_{2}(A)=\varphi_{1}(A^{+})-\varphi_{1}(A^{-})$, $A^{+}$和$A^{-}$分别是$A$的正的部分和负的部分.则类似定理2.8可证$\varphi_{2}$是实线性的, 且$\varphi_{2}$也是单射.若$\varphi_{2}(A)=\varphi_{2}(B)$, 即
| $\varphi_{1}(A^{+})-\varphi_{1}(A^{-})=\varphi_{1}(B^{+})-\varphi_{1}(B^{-}).$ |
进一步可得
| $\varphi_{1}(A^{+})+\varphi_{1}(B^{-})=\varphi_{1}(A^{-})+\varphi_{1}(B^{+}).$ |
又$\varphi_{1}$对正元可加, 因此可得$\varphi_{1}(A^{+}+B^{-})=\varphi_{1}(B^{+}+A^{-})$.又因为$\varphi_{1}$是单射, 所以$A^{+}+B^{-}=B^{+}+A^{-}$, 即$A=B.$将$\varphi_{2}$延拓到整个矩阵代数上.定义$\psi:M_{n}(\mathbb{C})\rightarrow M_{n}(\mathbb{C}),$ $\psi(A)=\varphi_{2}({\rm Re}A)+i\psi({\rm Im}A)$, $({\rm Re}A)$和${\rm Im}A$分别是$A$的实部和虚部.则可证$\psi$是线性的, 因为$\varphi_{2}$是单射, 所以可证$\psi$是单射.又因为$\psi$是有限维到有限维的线性映射, 所以$\psi$也是满的.因此$\psi|_{E(\mathbb{C}^{n})}$也是满的, 故$\varphi$是双边保序及保正交补的双射.则由Ludwig的结论[4]可得, 存在酉算子或反酉算子$U$, 使得任意$ A\in E(\mathbb{C}^{n})$, 有$\varphi(A)=UAU^{*}$成立.
2015, Vol. 35 
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