Question

求下列实对称矩阵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

|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^-1AP = diag(5,-1,-1).

Answer 2

P为
1/√2 1/√6 1/√3
-1/√2 1/√6 1/ √3
0 -2/√6 1/√3