1234 例1(教材P176例3-8).A 12求 [解]1°求出J及P 1100 8400 2-2-10 420 P P 16 816 16 2°求出f(A),f(入),…,升(入)并构成f() =1,m1=4f(z) f(1)=1, Z 82z=8 168-21 168 f(j) 16816 16 3°合成f(U)=f( 40 f(A)=Pf(])P-1, f(A= 说明 123 (1)o=例 1 （教材 P176 例 3-8）.             1 2 3 4 1 2 3 A = 1 2 1 ,求 A [解] 1 o 求出 J 及 P                                     -1 1 1 0 0 8 4 0 0 2 -2 -1 0 1 1 0 4 -1 1 4 2 0 1 J = ,P = ,P = 1 1 2 -2 8 16 16 1 1 16 2 o 求出 ( )  i m -1 i i i f(λ),f (λ), ,f (λ) 并构成 i f(J ) ： λ1 1 = 1,m = 4,f(z)= z f(1)=1,    1 3 5 - - - 2 2 2 1 1 1 1 3 3 f (1)= z | = ,f (1)= - z | = - ,f (1)= z | = 2 2 4 4 8 8 z=1 z=1 z=1             1 16 8 -2 1 16 8 -2 1 f(J )= 16 8 16 16 3 o 合成 1 f(J)= f(J ) 4 o 求 -1 f(A)= Pf(J)P ，             1 1 1 1 1 1 1 f(A)= 1 1 1 说明： （1）                         2 2 1 1 1 1 1 2 3 4 1 1 1 1 2 3 [f(A)] = = = A 1 1 1 2 1 1