设A=3-2-23试利用A的正交相似对角化,求￠（A）=A10-5A9.

Question

设A= 3 -2 -2 3 试利用A的正交相似对角化,求￠（A）=A10-5A9.

Answer 1

|A-λE| = (3-λ)^2-(-2)^2 = (1-λ)(5-λ)
A的特征值为1,5.
由(A-E)X=0得 (1,1)',单位化得 a1=(1/√2,1/√2)'
由(A-5E)X=0得 (1,-1)',单位化得 a2=(1/√2,-1/√2)'
令P=(a1,a2)=
1/√2 1/√2
1/√2 -1/√2
则P为正交矩阵,满足 P^-1AP =diag(1,5)
所以 A=Pdiag(1,5)P^-1
所以 A^10-5A^9
= P [diag(1,5)^10 - 5diag(1,5)^9] P^-1
= P diag(-4,0) P^-1
= -2 -2
-2 -2