[線代] one-to-one and onto

[mqazz1 (無法顯示)]

let D be the differentiation operator on P[3], and let

S = {p屬於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


請問這題該怎麼證呢?

謝謝!!

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→ ricestone   :(a)P[3]裡面每個只差常數的多項式都會映到同一個
→ ricestone   :(b)P[2]中任一元素都可以積分回去唯一的P[3]元素
→ ricestone   :上一行說錯了
→ ricestone   :(b)S中任兩不同元素經過映射後會射到不同兩點
→ ricestone   :(b)可以用兩式相減來證
推 aegius1r    :Kernel, image
→ ricestone   :不過你從以前到現在問了這麼多線代問題，到了現在還
→ ricestone   :問這類題目還蠻怪的...
→ recorriendo :樓上+1 從以前就看到你問類似的題目了