由P1AP=A,得AP=PA, 1 即4(n1,n2,,pn)=(n1 2 1,2,”5n =(1p 119229 .AnPu). A 1929 ,pn)=(4,42,…,4pn) =(41D1,2…,n) 于是有41=41P1(=1,2,…,n( ) ( )               = 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