由PAP=A,得AP=PA, 1 即4(n1,p2…,pn)=(n1,P2,…,pn 2 =(巩1D1,2P2,…,,Dn A(m1,D2,…pn)=(41,42…,pn) =(41p1,42,…,4n) 于是有42=4p1(=12…m小 上页( ) ( )               = n n n A p p p p p p       2 1 1 2 1 2 即 , , , , , , ( , , , ). = 1 p1 2 p2   n pn ( ) ( ) A p p pn Ap Ap Apn , , , , , ,  1 2  = 1 2  Ap p (i 1,2, ,n). 于是有 i = i i =  ( )  p p pn , , , = 1 1 2  , , 1 =  =  − 由P AP 得AP P