Question

设3阶矩阵A=| 1,2,32,1,33,3,6 |,求可逆矩阵P,使得P＾（-1）AP=A

Answer 1

解: |A-λE| = λ(9-λ)(1+λ).

所以A的特征值为 0, 9, -1

AX = 0 的基础解系为: a1 = (1,1,-1)'

(A-9E)X = 0 的基础解系为: a2 = (1,1,2)'

(A+E)X = 0 的基础解系为: a3 = (1,-1,0)'

令矩阵P = (a1,a2,a3), 则 P^(-1)AP = ∧=diag(0,9,-1).