Question

线性代数 设A=(1,1,1;2,2,2;3,3,3 )则A^2005=A^2004*A

Answer 1

解题过程如图

Answer 2

5. r(A) = 1, A 有重特征值 λ = 0， 非零特征值 λ = tr(A) = 6.
对于 λ = 6， λE-A =
[ 5 -1 -1]
[-2 4 -2]
[-3 -3 3]
行初等变换为
[1 1 -1]
[0 -6 4]
[0 6 -4]
行初等变换为
[1 0 -1/3]
[0 1 -2/3]
[0 0 0]
得特征向量 (1 2 3)^T;
对于 λ = 0， λE-A = -A =
[-1 -1 -1]
[-2 -2 -2]
[-3 -3 -3]
行初等变换为
[1 1 1]
[0 0 0]
[0 0 0]
得特征向量 (1 -1 0)^T, (1 0 -1)^T
取 P =
[1 1 1]
[2 -1 0]
[3 0 -1]
得 P^(-1) = (1/6)*
[1 1 1]
[2 -4 2]
[3 3 -3]
P^(-1)AP = ∧ = diag(6, 0, 0),
A = P∧P^(-1)
A^2005 = P∧P^(-1)P∧P^(-1)P∧P^(-1) ...... P∧P^(-1)P∧P^(-1)
= P∧^2005P^(-1)
= (1/6)6^2005*
[1 1 1]
[2 2 2]
[3 3 3]
= 6^2004 A