Span of vector space

STATS·2023년 7월 24일

선형대수학

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Span

벡터공간 VV의 공집합이 아닌 부분집합 SS에 대해, Span(S)Span(S)는 SS의 벡터를 이용해 만들 수 있는 모든 선형 결합의 집합이다.

Property on Span

  1. Span(S)Span(S)는 VV의 부분공간이다.

    Proof)Let S={v1,v2,...,vn}⊂VThen ∀(a1,a2,...,an)∈Rn,∑i=1naivi∈V⇒Span(S)⊂VNow we will show Span(S) is subspace of V.I)∃0∈Span(S) s.t 0=0v1+...+0vnII)∀ s1,s2∈Span(S) s.t s1=∑i=1naivi,s2=∑i=1nbivi,s1+s2=∑i=1n(ai+bi)vi∈Span(S)III)∀c∈F,∀s∈Span(S) s.t s=∑i=1naivi,cs=∑i=1n(cai)vi∈Span(S)∴Span(S) is subspace of VProof) \\ Let \ S = \{v_1, v_2, ..., v_n\} \subset V \\ Then \ \forall(a_1, a_2, ..., a_n) \in \R^n, \sum_{i=1}^n a_iv_i\in V \\ \Rightarrow Span(S) \subset V \\ {} \\ \text{Now we will show Span(S) is subspace of V.} \\ {} \\ I) \exist0 \in Span(S) \ s.t \ 0 = 0v_1 + ... + 0v_n \\ {} \\ II) \forall \ s_1, s_2 \in Span(S) \ s. t\ s_1 = \sum_{i=1}^n a_iv_i, s_2= \sum_{i=1}^n b_iv_i, \\ s_1 + s_2 = \sum_{i=1}^n (a_i + b_i)v_i \in Span(S) \\ {} \\ III) \forall c \in F, \forall s \in Span(S) \ s.t \ s = \sum_{i=1}^n a_iv_i, \\ cs = \sum_{i=1}^n (ca_i)v_i \in Span(S) \\ {} \\ \therefore \text{Span(S) is subspace of V}
  2. Span(S)Span(S)는 SS를 포함한다.

    Proof)Let S={v1,v2,...,vn}⊂VThen ∀(a1,a2,...,an)∈Rn, ∑i=1naivi∈Span(S)The for every vi∈S,there exist Rn vector ai such thata1=(1,0,...,0), a2=(0,1,0,...,0)...,an=(0,0,...,1).Then vi=∑i=1naivi∈Span(S). ∴S⊂Span(S)Proof) \\ Let \ S = \{v_1, v_2, ..., v_n\} \subset V \\ Then \ \forall(a_1, a_2, ..., a_n) \in \R^n, \ \sum_{i=1}^n a_iv_i \in Span(S) \\ {} \\ \text{The for every } v_i \in S, \text{there exist } \R^n \ vector\ a_i\ such \ that \\ a_1 = (1, 0, ...,0), \ a_2 = (0, 1, 0, ..., 0)..., a_n = (0, 0, ..., 1). \\ {} \\ Then \ v_i = \sum_{i=1}^n a_iv_i \in Span(S). \ \therefore S \sub Span(S)
  3. SS를 포함하는 VV의 부분공간은 반드시 Span(S)Span(S)를 포함한다.

Proof)Let W be subspace of V such that S⊂W.Then each vector of S is also in W, we get ∀s1,s2∈S,s1+s2∈W∀c∈F,∀s∈S,cs∈WThen All possible Linear combination of vectors in S is also in W.∴span(S)⊂WProof) \\ \text{Let W be subspace of V such that } S \sub W. \\ \text{Then each vector of S is also in W, we get } \\ {} \\ \forall s_1, s_2 \in S, s_1+s_2 \in W \\ \forall c \in F, \forall s \in S, cs \in W \\ {} \\ \text{Then All possible Linear combination of vectors in S is also in W.}\\{} \\ \therefore span(S) \sub W

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