Prove the generalization of the Elimination Theorem: Let S={v1, v2,...,vn} be a linearly dependent subset of a vector space v, and let a vector w in V, with the dependence equation: c1v1 + c2v2+...+cnvn=0 where cn is not equal to 0. Then Span({v1, v2, ..., vn})=Span(v1, v2,...,v(n-1)}) that is, we can reduce the size of S by one vector, vn, and the smaller set still has the same Span as the original set S.
if \(c_1v_1+c_2v_2+\dot+c_nv_n=0\) with \(c_n\neq0\) prove that \[ \large K:=\langle v_1,v_2,\dots,v_n\rangle= \langle v_1,v_2,\dots,v_{n-1}\rangle=:M \]
by definition, let V is a vector space over a field \(\mathbb{F}\), if \(w_1,\dots,w_n\in V\) then \[ \large \langle w_1,w_2,\dots,w_r\rangle= \left\{\sum_{i=1}^r\alpha_iw_i:\alpha_i\in\mathbb{F}\right\} \]
this means that \(M\subseteq K\) is trivial.
for the converse, from the hypothesis \[ \large v_n=-\frac{c_1}{c_n}v_1-\frac{c_2}{c_n}v_2-\dots- \frac{c_{n-1}}{c_n}v_{n-1} \]
let \(w\in K\) so there are \(\beta_1,\dots,\beta_n\in\mathbb{F}\) such that \[ \large w=\sum_{i=1}^n\beta_iv_i=\sum_{i=1}^{n-1}\beta_iv_i+\beta_nv_n \] \[ \large =\sum_{i=1}^{n-1}\left(\beta_i-\frac{c_i\beta_n}{c_n}\right)v_i\in M \] that is \(K\subseteq M\)
Umm, I think you may have overdone it...however I do appreciate the help. I simply need to prove that \[Span(S)= Span(S-v_n)\]
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