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

解: |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).