设 A 为元素为 a11 = a12 = a21 = +1 且 a22 = -1 的 2 × 2 矩阵。那么矩阵 A 19的特征值是
(一) A
(乙)乙
(C)丙
(四)丁答案： (D)
解释：
A = 1 1
1 -1
A2 = 2 0
0 2
A4 = A2 X A2
A4 = 4 0
0 4
A8 = 16 0
0 16
A16 = 256 0
0 256
A18 = A16 X A2
A18 = 512 0
0 512
A19 = 512 512
512 -512
Applying Characteristic polynomial
512-lamda 512
512 -(512+lamda) = 0
-(512-lamda)(512+lamda) - 512 x 512 = 0
lamda2 = 2 x 5122
替代解决方案：
det(A) = -2.
det(A^19) = (det(A))^19 = -2^19 = lambda1*lambda2.
The only viable option is D. 感谢 Matan Mandelbrod 提出这个解决方案。
这个问题的测验