Question

设f（x1x2x3）=x1²+X2²+X3²+4X1X2+4X1X3+4X2X3

（1）求一正交变换化f为标准型
（2）判定f的正定性

Answer 1

解: f的矩阵A=
1 2 2
2 1 2
2 2 1

|A-λE| = (5-λ)(1+λ)^2.
所以A的特征值为 5, -1, -1

(A-5E)X = 0 的基础解系为: a1 = (1, 1, 1)'
(A+E)X = 0 的基础解系为: a2 = (1, -1, 0)', a3 = (1, 0, -1)'

将 a2,a3 正交化得 b2 = (1,-1,0)', b3 = (1/2,1/2,-1)'

单位化得
c1 = (1/√3, 1/√3, 1/√3)',
c2 = (1/√2, -1/√2, 0)',
c3 = (1/√6,1/√6,-2/√6)'

令矩阵P = (c1,c2,c3), 则P为正交矩阵,
且 P^(-1)AP = diag(5,-1,-1).

正交变换 X=PY, f = 5y1^2-y2^2-y3^2.
f不是正定的, 也不负定的.

Answer 2

=x1²等于X6