Question

对下列实对称矩阵A,求一个正交矩阵P,使P^-1AP=P^TAP=D为对角矩阵 【1,2,2；2,1,2；2,2,1】

Answer 1

做好了 刚才那个不见了

|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 = P^TAP=diag(5,-1,-1).