作者mqazz1 (无法显示)
看板Math
标题[线代] one-to-one and onto
时间Sat May 14 22:25:30 2011
let D be the differentiation operator on P[3], and let
S = {p属於P[3] | p(0)=0}
show that
(a) D maps P[3] onto the subspace P[2], but
D: P[3] -> P[2] is not one-to-one
(b) D: S -> P[3] is one-to-one but not onto
请问这题该怎麽证呢?
谢谢!!
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1F：→ ricestone :(a)P[3]里面每个只差常数的多项式都会映到同一个 05/14 22:35
2F：→ ricestone :(b)P[2]中任一元素都可以积分回去唯一的P[3]元素 05/14 22:37
3F：→ ricestone :上一行说错了 05/14 22:39
4F：→ ricestone :(b)S中任两不同元素经过映射後会射到不同两点 05/14 22:41
5F：→ ricestone :(b)可以用两式相减来证 05/14 22:42
6F：推 aegius1r :Kernel, image 05/14 22:43
7F：→ ricestone :不过你从以前到现在问了这麽多线代问题，到了现在还 05/14 22:45
8F：→ ricestone :问这类题目还蛮怪的... 05/14 22:45
9F：→ recorriendo :楼上+1 从以前就看到你问类似的题目了 05/15 00:21