设A是n(n geq 2)阶矩阵，alpha_1, alpha_2, ..., alpha_n是n维列向量，且Aalpha_1 = alpha_2, Aalpha_2 = alpha_3, ..., Aalpha_(n-1) = alpha_n, Aalpha_n = 0, alpha_n neq 0。(1) 证明：alpha_1, alpha_2, ..., alpha_n线性无关；(2) 令P = (alpha_1, alpha_2, ..., alpha_n)，若P^-1AP = B，求B.
设$A$是$n(n \geq 2)$阶矩阵，$\alpha_1, \alpha_2, \cdots, \alpha_n$是$n$维列向量，且$A\alpha_1 = \alpha_2, A\alpha_2 = \alpha_3, \cdots, A\alpha_{n-1} = \alpha_n, A\alpha_n = 0, \alpha_n \neq 0$。 (1) 证明：$\alpha_1, \alpha_2, \cdots, \alpha_n$线性无关； (2) 令$P = (\alpha_1, \alpha_2, \cdots, \alpha_n)$，若$P^{-1}AP = B$，求$B$.
题目解答
答案
(1) 证明线性无关
假设 $k_1\alpha_1 + k_2\alpha_2 + \cdots + k_n\alpha_n = 0$，两边左乘 $A$ 得 $k_1\alpha_2 + k_2\alpha_3 + \cdots + k_{n-1}\alpha_n = 0$。重复此过程，最终得到 $k_1 = k_2 = \cdots = k_n = 0$，故 $\alpha_1, \alpha_2, \cdots, \alpha_n$ 线性无关。
(2) 求 $B$
令 $P = (\alpha_1, \alpha_2, \cdots, \alpha_n)$，则 $AP = (\alpha_2, \alpha_3, \cdots, \alpha_n, 0)$。
由 $P^{-1}AP = B$，得 $AP = PB$，其中 $B$ 为 $n \times n$ 矩阵。
比较列向量，得 $B = \begin{pmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ 0 & 0 & 0 & \cdots & 0 \end{pmatrix}$。
答案：
(1) $\alpha_1, \alpha_2, \cdots, \alpha_n$ 线性无关。
(2) $B = \boxed{\begin{pmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ 0 & 0 & 0 & \cdots & 0 \end{pmatrix}}$。