已知2阶矩阵A=(第一行3,-2,第二行-2,3),求f(A)=A^10-5A^9已知A=3,-2-2,3求f(A)=A^10-5A^9

Question

已知2阶矩阵A=(第一行3,-2,第二行-2,3),求f(A)=A^10-5A^9
已知A=
3,-2
-2,3
求f(A)=A^10-5A^9

Answer 1

求得特征值为:5,1,对应的特征i向量为(1,-1)转置,(1,1)转置.
它们正交,再单位化为:(根号2)/2 *(1,-1)转置,(根号2)/2 *(1,1)转置
构成正交阵:p=(根号2)/2* (1 1 )
(-1 1 )
有p逆 A p= diag (5,1)
故A=p diag(5 ,1) p逆.
故:f(A)=A^10-5A^9 =A^9 (A-5E)= (p [diag(5,1)]^9 p逆)(A-5E)=p[diag(5^9,1)p逆](A-5E)
= (1/2)(5^9 +1 - 5^9+1 )
(-5^9 +1 5^9 +1 ) (A-5E)
求得A-5E = - (2 2 )
(2 2 )
最后得f(A)=-2( 1 1)
(1 1)