线性代数二次型化为标准型

Question

如题所述

Answer 1

二次型矩阵 A =
[ 2 -2 0]
[-2 1 -2]
[ 0 -2 0]
|λE-A| =
|λ-2 2 0|
| 2 λ-1 2|
| 0 2 λ|
= λ(λ-1)(λ-2) - 4(λ-2) - 4λ
= λ(λ-1)(λ-2) - 8(λ-1)
= (λ-1)(λ^2-2λ-8) = (λ-1)(λ-4)(λ+2)
特征值λ = 4，1, -2.
对于特征值 λ = 4，λE-A =
[ 2 2 0]
[ 2 3 2]
[ 0 2 4]
初等行变换为
[ 1 1 0]
[ 0 1 2]
[ 0 2 4]
初等行变换为
[ 1 0 -2]
[ 0 1 2]
[ 0 0 0]
得特征向量(2 -2 1)^T,
单位化是(2/3 -2/3 1/3)^T;
对于特征值 λ = 1，λE-A =
[-1 2 0]
[ 2 0 2]
[ 0 2 1]
初等行变换为
[ 1 -2 0]
[ 0 4 2]
[ 0 2 1]
初等行变换为
[ 1 0 1]
[ 0 2 1]
[ 0 0 0]
得特征向量(2 1 -2)^T,
单位化是(2/3 1/3 -2/3)^T;
对于特征值 λ = -2，λE-A =
[-4 2 0]
[ 2 -3 2]
[ 0 2 -2]
初等行变换为
[ 2 -1 0]
[ 0 -2 2]
[ 0 2 -2]
初等行变换为
[ 2 0 -1]
[ 0 1 -1]
[ 0 0 0]
得特征向量(1 2 2)^T,
单位化是(1/3 2/3 2/3)^T.
得正交矩阵 P =
[ 2/3 2/3 1/3]
[-2/3 1/3 2/3]
[ 1/3 -2/3 2/3]
作正交变换 x = Py
使得 f = x^TAx = y^T(P^TAP)y = 4(y1)^2 + (y2)^2 - 2(y3)^2本回答被提问者和网友采纳