Re: [線代] 106年台大線代(A)考古題

[XII (Mathkid)]

※ 引述《Desperato (Farewell)》之銘言：
: 1. Show that if R^n = Union W_i, i = 1, 2, 3, ...
: W_i are subspaces of R^n
: Then R^n = W_i for some i

重新整理一下好了..

Fact.
If V is a vector space over field F,
I is an index set with Card(I)<Card(F),
and {W_i}_{i in I} is a set of proper subspaces of V,
then V≠∪_{i in I} W_i

Proof.
  Suppose V=∪_{i in I} W_i.
  W.O.L.G. we may assume W_i is not contained in ∪_{j≠i} W_j for all i in I.
  (If I is an infinite set,
  we use Zorn's Lemma to find a minimal index set I' contained in I s.t.
  V=∪_{i in I'} W_i, then replace I by I')
  Hence there are w_i in W_i-∪_{j≠i} W_j for all i in I.
  Since all W_i are proper subspaces, we have Card(I)≧2.
  Let p,q in I. Consider S={a*w_p+w_q: a in F}.
  Since Card(S)=Card(F)>Card(I),
  there is no injection from F to I (Pigeonhole Principle or definition of >).
  Hence are distinct a,b in F s.t.
  u:=a*w_p+w_q and v:=b*w_p+w_q are in W_n for some n in I.
  Since (a-b)^(-1)(u-v)=w_p in W_p-∪{j≠p} W_j and W_n, so n=p.
  Thus w_q=u-a*w_p in W_p and W_q with p≠q, a contradiction (w_q not in W_p)
  Therefore V≠∪_{i in I} W_i. □

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※ 編輯: XII (110.50.138.101), 01/16/2019 17:46:40
推 Desperato   : 沒錯就是這樣XD
推 panda0315   : 推
推 Sfly        : 有個地方錯了
→ XII         : 哪裡?