Question

急求解大学数学矩阵，题目如下

Answer 1

解: |A-λE| =
1-λ 0 -2
0 2-λ 0
-2 0 1-λ

= (2-λ)[(1-λ)^2 - 2^2]
= (2-λ)(3-λ)(-1-λ).

所以A的特征值为 2,3,-1.

(A-2E)X=0 的基础解系为: (0,1,0)', 单位化得 a1=(0,1,0)'
(A-3E)X=0 的基础解系为: (1,0,-1)', 单位化得 a2=(1/√2,0,-1/√2)'
(A+E)X=0 的基础解系为: (1,0,1)', 单位化得 a3=(1/√2,0,1/√2)'

令 P=(a1,a2,a3) =
0 1/√2 1/√2
1 0 0
0 -1/√2 1/√2
则P为正交矩阵, P^-1 = P'.
且 P^-1AP = diag(2,3,-1).

所以 A = Pdiag(2,3,-1)P^-1.

记 f(x) = x^6-4x^5+2x^4+3x^3-3x^2+8x+7
则 f(A) = Pdiag(f(2),f(3),f(-1))P^-1
= Pdiag(3,4,0)P^-1
=
2 0 -2
0 3 0
-2 0 2