线性代数,求详细过程,谢谢
展开全部
|λE-A| =
| λ-1 2 0|
| 2 λ-2 2|
| 0 2 λ-3|
= (λ-1)(λ-2)(λ-3) - 4(λ-1) - 4(λ-3)
= (λ-1)(λ-2)(λ-3) - 8(λ-2)
= (λ+1)(λ-2)(λ-5)
特征值 λ = -1, 2, 5
对于特征值 λ = -1,λE-A =
[-2 2 0]
[ 2 -3 2]
[ 0 2 -4]
初等变换为
[-1 1 0]
[ 0 -1 2]
[ 0 2 -4]
初等变换为
[-1 0 2]
[ 0 -1 2]
[ 0 0 0]
得特征向量 (2, 2, 1)^T,
单位化为 (2/3, 2/3, 1/3)^T.
对于特征值 λ = 2,λE-A =
[ 1 2 0]
[ 2 0 2]
[ 0 2 -1]
初等变换为
[ 1 2 0]
[ 0 -4 2]
[ 0 2 -1]
初等变换为
[ 1 0 1]
[ 0 -2 1]
[ 0 0 0]
得特征向量 (-2, 1, 2)^T,
单位化为 (-2/3, 1/3, 2/3)^T.
对于特征值 λ = 5,λE-A =
[ 4 2 0]
[ 2 3 2]
[ 0 2 2]
初等变换为
[ 2 1 0]
[ 0 2 2]
[ 0 2 2]
初等变换为
[ 2 0 -1]
[ 0 1 1]
[ 0 0 0]
得特征向量 (1, -2, 2)^T,
单位化为 (1/3, -2/3, 2/3)^T.
则正交矩阵 P =
[2/3 -2/3 1/3]
[2/3 1/3 -2/3]
[1/3 2/3 2/3]
使得 P^(-1)AP = P^TAP = diag(-1, 2, 5)
| λ-1 2 0|
| 2 λ-2 2|
| 0 2 λ-3|
= (λ-1)(λ-2)(λ-3) - 4(λ-1) - 4(λ-3)
= (λ-1)(λ-2)(λ-3) - 8(λ-2)
= (λ+1)(λ-2)(λ-5)
特征值 λ = -1, 2, 5
对于特征值 λ = -1,λE-A =
[-2 2 0]
[ 2 -3 2]
[ 0 2 -4]
初等变换为
[-1 1 0]
[ 0 -1 2]
[ 0 2 -4]
初等变换为
[-1 0 2]
[ 0 -1 2]
[ 0 0 0]
得特征向量 (2, 2, 1)^T,
单位化为 (2/3, 2/3, 1/3)^T.
对于特征值 λ = 2,λE-A =
[ 1 2 0]
[ 2 0 2]
[ 0 2 -1]
初等变换为
[ 1 2 0]
[ 0 -4 2]
[ 0 2 -1]
初等变换为
[ 1 0 1]
[ 0 -2 1]
[ 0 0 0]
得特征向量 (-2, 1, 2)^T,
单位化为 (-2/3, 1/3, 2/3)^T.
对于特征值 λ = 5,λE-A =
[ 4 2 0]
[ 2 3 2]
[ 0 2 2]
初等变换为
[ 2 1 0]
[ 0 2 2]
[ 0 2 2]
初等变换为
[ 2 0 -1]
[ 0 1 1]
[ 0 0 0]
得特征向量 (1, -2, 2)^T,
单位化为 (1/3, -2/3, 2/3)^T.
则正交矩阵 P =
[2/3 -2/3 1/3]
[2/3 1/3 -2/3]
[1/3 2/3 2/3]
使得 P^(-1)AP = P^TAP = diag(-1, 2, 5)
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询