psie
- 315
- 40
- TL;DR Summary
- I'm reading Linear Algebra by Friedberg, Insel and Spence. Prior to a theorem, they make a statement about linear independence and they claim this can be deduced from the theorem.
Throughout, let ##\mathsf V## be a vector space (the concept of dimension has not been introduced yet). The statement that precedes the theorem below is that if no proper subset of ##T\subset \mathsf V## generates the span of ##T## (where, if I'm not mistaken, ##T## consists of two or more vectors) then ##T## must be linearly independent. Taking the contrapositive,
I'm trying to prove the claim from the theorem. What I struggle with is that I only seem to be able to prove the claim when ##T=S\cup\{v\}##, where ##S## is linearly independent and ##v\notin S##. Then the theorem tells us that ##v\in\operatorname{span}(S)##. This in turn implies ##\operatorname{span}(S)=\operatorname{span}(S\cup\{v\})## (see below for a proof of why this is implied by ##v\in\operatorname{span}(S)##). Hence we can take the proper subset ##S\subset S\cup\{v\}=T## as the set in the claim preceding the theorem. But what if ##T## is not of the form ##S\cup\{v\}##? I feel like I've only proved a very special case.
Regarding ##v\in\operatorname{span}(S)\implies\operatorname{span}(S)=\operatorname{span}(S\cup\{v\})##, the inclusion ##\operatorname{span}(S)\subset\operatorname{span}(S\cup\{v\})## always holds. The reverse inclusion follows since any ##w\in \operatorname{span}(S\cup\{v\})## can be written as ##w=a_1u_1+\cdots +a_nu_n+bv##, where ##u_1,\ldots,u_n\in S##. Since ##v\in\operatorname{span}(S)##, ##w## is actually a linear combination of vectors in ##S##.
Claim: If ##T\subset\mathsf V## is linearly dependent, then there is some proper subset ##S\subset T## such that ##\operatorname{span}(S)=\operatorname{span}(T)##.
Theorem: Let ##S\subset \mathsf V## be linearly independent and let ##v\notin S##. Then ##S\cup\{v\}## is linearly dependent if and only if ##v\in\operatorname{span}(S)##.
I'm trying to prove the claim from the theorem. What I struggle with is that I only seem to be able to prove the claim when ##T=S\cup\{v\}##, where ##S## is linearly independent and ##v\notin S##. Then the theorem tells us that ##v\in\operatorname{span}(S)##. This in turn implies ##\operatorname{span}(S)=\operatorname{span}(S\cup\{v\})## (see below for a proof of why this is implied by ##v\in\operatorname{span}(S)##). Hence we can take the proper subset ##S\subset S\cup\{v\}=T## as the set in the claim preceding the theorem. But what if ##T## is not of the form ##S\cup\{v\}##? I feel like I've only proved a very special case.
Regarding ##v\in\operatorname{span}(S)\implies\operatorname{span}(S)=\operatorname{span}(S\cup\{v\})##, the inclusion ##\operatorname{span}(S)\subset\operatorname{span}(S\cup\{v\})## always holds. The reverse inclusion follows since any ##w\in \operatorname{span}(S\cup\{v\})## can be written as ##w=a_1u_1+\cdots +a_nu_n+bv##, where ##u_1,\ldots,u_n\in S##. Since ##v\in\operatorname{span}(S)##, ##w## is actually a linear combination of vectors in ##S##.