✔ 最佳答案
(a) 令 A 為 m×n, 元素為 a(i,j) 的矩陣, x=[x_1,...,x_n]'.
則 l_i 即是 Ax 的第 i 個元素.
dim(span(l_1,...,l_m)) = rank(A).
w in W iff. wAx = 0 for all x iff wA=0, 故
W=(RS(A))^⊥ (RS(A) 的 orthogonal complement).
dim(W)+rank(A) = m, 即 rank(A) = m - dim(W).
N={x in R^n | Ax'=0}, 即 x in N iff. x' in Ker(A),
故 dim(N)+rank(A) = n.
所以,
dim(N) = n - rank(A) = n - (m - dim(W)) = dim(W) + n - m.
(b) 取 V 的一 subdpace Z 使 V 為 Z 與 span(S) 的 direct sum.
所有 v in V 有唯一方式表示為 v = z + x where z in Z and x in span(S).
故,
U_S = {T in L(V,W) | T(z+x) = T(z) for all z in Z, x in span(S)}
則
dim(U_S) = dim(Z)．dim(W) = (dim(V)-dim(span(S)))．dim(W)
= (dim(V) - d)．dim(W).
2011-05-24 01:17:03 補充：
想問為什麼
U_S = {T in L(V,W) | T(z+x) = T(z) for all z in Z, x in span(S)}之後
為什麼dim(U_S)會等於dim(Z)．dim(W)？
因為 T 把 Z 的元素映至 W 並無額外限制.
就像 L(V,W) 的 dimension 是 dim(V)．dim(W),
所以 dim(U_S) = dim(Z)．dim(W).
2011-05-28 03:15:48 補充：
舉個例子: F=R, V=R^3, W=R^1, S={(0,1,0),(0,0,1)}, 故 d=2.
T(v)=0 for all v in S, 則 T((x,y,z)) = kx, for any (x,y,z) in V, for some k in R.
所以
Us = { T | T(x,y,z) = kx for all (x,y,z) in V }
的 dimension 是 1 = (3-2)*1 而不是 3*1-2^2=-1.
