※ 引述《PttFund (批踢踢基金只進不出)》之銘言：
: Let P_2 = { ax^2 + bx + c : a,b,c in R } and let T: P_2 -> R
︿︿修正一下.
: be the linear transformation defined by
: 1
: T(P(x)) = ∫ P(x) dx
: 0
: (a) Find a basis for ker(T).
: (b) Find a basis for range(T).
這一題應該不難吧, 我們先看: 取 P(x) = ax^2 + bx + c in P_2,
則
T(P(x)) = T(ax^2 + bx + c)
1
= ∫ (ax^2 + bx + c) dx
0
= a/3 + b/2 + c.
所以當 P(x) in ker(T) <=> T(P(x)) = 0.
<=> a/3 + b/2 + c = 0.
(Let a = t, b = s, then c = -t/3 - s/2)
<=> P(x) = tx^2 + sx + (-t/3 - s/2)
= t(x^2 - 1/3) + s(x - 1/2)
所以 ker(T) 的基底可以是 { x^2 - 1/3, x - 1/2 }.
range(T) 的基底則是 {1}.
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