入-10 0 (2)fB()=2X-52=(X-1)2(X-3),口有A1=1(2重; A2=3(1重 解(A1-A)X=0,即2-42x=0,得到基础解代1 解(2-4)X=0,即2-22x=0,得到基础解代521= 口A可对角化,令P=(5151521),则P-1AP=1 亦可令Q=101,则Q-AQ 111 例2计算A0,其结A= 10 解根上例方法求得存化可逆阵P 使得P-1AP 10 1 0 例3设3阶矩阵A的特征值为1,1,3,对应的特征向量依次为(2,1,0)y,(-1,0.,1)y (0,1,1).求矩阵A 2-10 解令P=101,B 则P1AP=B.从而A 011 100 PBP-I 例4求证: (1)若矩阵A,合A2=Ln,则A必可对角化; (2)若矩阵A,合A2=A,则A必可对角化;(2) fB(λ) =   λ − 1 0 0 2 λ − 5 2 2 −4 λ + 1   = (λ − 1)2 (λ − 3), 23 λ1 = 1 (2 I); λ2 = 3 (1 I). I (λ1 − A)X = 0, <   0 0 0 2 −4 2 2 −4 2   X = 0, ;I ξ11 =   2 1 0  , ξ12 =   −1 0 1  . I (λ2 − A)X = 0, <   2 0 0 2 −2 2 2 −4 4   X = 0, ;I ξ21 =   0 1 1  . 2 A M#E9
Z P = ( ξ11 ξ12 ξ21 ), : P −1AP =   1 1 3  . +MZ Q =   1 −1 0 1 0 1 1 1 1  , : Q−1AQ =   1 1 3  . U 2 > A10 , kH A =  1 0 1 −2  . R 1wS*(n9Mh> P =  1 0 1 1  , | P −1AP =  1 0 0 −2  , 2 A10 =  P  1 0 0 −2  P −1 10 = P  1 0 0 −2 10 P −1 =  1 0 1 − 2 10 2 10  . U 3 y 3 GJ> A ?D 1,1,3, #0? V) (2, 1, 0)′ , (−1, 0, 1)′ , (0, 1, 1)′ . nJ> A. R Z P =   2 −1 0 1 0 1 0 1 1  , B =   1 1 3  . : P −1AP = B. & A = P BP −1 =   1 0 0 −2 5 −2 −2 4 −1  . U 4 n (1) uJ> A ￾7 A 2 = In, : A M#E9 (2) uJ> A ￾7 A2 = A, : A M#E9 5