→ musicbox810 : Σc_k a_kj應該不會是0吧 你少含了b_j 01/11 11:27
→ musicbox810 : 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
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.
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