Re: [線代] 維度定理的證明

看板Math作者 (龐加萊)時間8年前 (2018/01/09 08:40), 編輯推噓0(002)
留言2則, 1人參與, 8年前最新討論串2/2 (看更多)
整理大家說的: Let {v_i} (i = 1,...,m) be a basis of ImT, and say v_i = T(u_i) for some u_i in U, i = 1,...,m. Clearly {u_i} (i = 1,...,m) is linearly independent, so we may extend it to a basis {u_1,...,u_m, w_1,...,w_n} of U. Now for i = 1,...,n, we write T(w_i) = Σa_ij v_j since T(w_i) is in ImT. Then T(w_i-Σa_ij u_j) = 0 for i = 1,...,n. It's obvious that {w_i-Σa_ij u_j} (i = 1,...,n) is linearly independent because {u_1,...,u_m, w_1,...,w_n} forms a basis of U. Suppose x is in KerT, and write x = Σb_j u_j + Σc_k w_k. Then 0 = T(x - Σc_k (w_k - Σa_kj uj)) = T(Σ(...)u_j) = Σ(...)v_j, where the first equality is because the both are in KerT. By the linear independence, we have the coefficients of all v_j, explicitly Σc_k a_kj, are 0. Hence x = Σc_k (w_k - Σa_kj uj), showing that {w_i-Σa_ij u_j} (i = 1,...,n) forms a basis of KerT. -- 名字不會定義一個人,名字只會留在一個人所踏過的足跡裡。 --殺老師 -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 140.112.51.108 ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1515458449.A.8CC.html

01/11 11:27, 8年前 , 1F
Σc_k a_kj應該不會是0吧 你少含了b_j
01/11 11:27, 1F

01/11 11:36, 8年前 , 2F
可是接下來似乎就沒辦法得到你的結論使用linear ind
01/11 11:36, 2F
文章代碼(AID): #1QL0-HZC (Math)
文章代碼(AID): #1QL0-HZC (Math)