设A是n阶方阵，X1是A的属于k的特征向量，又 (A-kI ) Xj+1=Xj, j=1,2,... ，s-1） 证明；X1，X2，Xs 线性无

由已知得
(A-kE)x1 = 0,
(A-kE)x2 = x1,
(A-kE)x3 = x2,
....
(A-kE)xs = x(s-1).

叠代一下就得到:
当 m≥j 时 (A-kE)^m xj = 0
且 (A-kE)^(j-1) xj = x1.

设 k1x1+k2x2+...k(s-1)x(s-1)+ksxs = 0 (*)
等式两边左乘 (A-kE)^(s-1) 得
k1(A-kE)^(s-1)x1+k2(A-kE)^(s-1)x2+...k(s-1)(A-kE)^(s-1)x(s-1)+ks(A-kE)^(s-1)xs = 0
则有 0+0+...+ksx1 = 0
即 ksx1 = 0.
而 x1 是A的特征向量, 所以 x1≠0, 故 ks = 0.
代入(*)式得 k1x1+k2x2+...k(s-1)x(s-1) = 0.

同理, 等式两边左乘 (A-kE)^(s-2) 得 k(s-1)=0.
如此下去, 得 ks = k(s-1)= ...= k2 = 0.
代入(*)式得 k1x1 = 0, 自然有 x1 = 0.

即 ks = k(s-1)= ...= k2 = k1 = 0
所以 x1,x2,...,xs 线性无关.
证毕#

呵呵,这题目有点难度,加点分哈!

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