Question

简单实对称矩阵的对角化如：0 11 0 对角化

Answer 1

|A-λE| =

-λ 1

1 -λ

= λ^2-1

= (λ+1)(λ-1)

A的特征值为1,-1

A-E=

-1 1

1 -1

-->

1 -1

0 0

(A-E)x=0的基础解系为 (1,1)^T

A+E=

1 1

1 1

(A+E)x=0的基础解系为 (1,-1)^T

令 P=

1 1

1 -1

则P可逆,且 P^-1AP =

1 0

0 -1