Question

一道高等代数题，希望大家帮帮忙，谢谢！

在空间P[x]n 中，设变换 τ 为f(x)→f(x+1)-f(x) ， τ 在基ε(0)=1，ε(i)=[x(x-1)……（x-i+1)]/ i! (i=1，2，……，n-1)下的矩阵。

Answer 1

求τ的矩阵就是求τ在基上的作用结果, 再表示为基的线性组合.

想不清楚的话就从简单的例子入手:
ε(0)为常值多项式, τ(ε(0)) = 1-1 = 0;

ε(1) = x, τ(ε(1)) = (x+1)-x = 1;

ε(2) = x(x-1)/2, τ(ε(2)) = (x+1)x/2-x(x-1)/2 = x.

一般的, ε(i) = (x(x-1)...(x-i+1))/i!,
τ(ε(i)) = ((x+1)x...(x-i+2))/i!-(x(x-1)...(x-i+1))/i!
= ((x+1)-(x-i+1))x(x-1)...(x-i+2)/i!
= i·x(x-1)...(x-i+2)/i!
= x(x-1)...(x-i+2)/(i-1)!
= ε(i-1).

因此τ(ε(0), ε(1), ε(2),..., ε(n-1)) = (ε(0), ε(1), ε(2),..., ε(n-1))·A.
其中A =
0 1 0 ... 0
0 0 1 ... 0
......
0 0 0 ... 1
0 0 0 ... 0.