Statement:

  Let \(\pmb{(V,+,\cdot)} \) be a non null vector space over a field \(\pmb{(F,+,\cdot)} \) and \(\pmb{S} \) be a finite set. If \(\pmb{S} \) be a linearly dependent set and \(\pmb{L(S)=V} \) then there exists proper subset \(\pmb{T} \) of \(\pmb{S} \) such that \(\pmb{L(T)=V} \).

  Proof:
  Given that \(\pmb{(V,+,\cdot)} \) is a vector space over a field \(\pmb{(F,+,\cdot)} \).
  Let \(\pmb{ S=\{\alpha_{1},\alpha_{2},…,\alpha_{n} \} } \) be linearly dependent set and \(\pmb{L(S)=V} \).
  Then there exists a vector of \(\pmb{\alpha_{j}} \) in \(\pmb{S} \) such that
\begin{align} &\pmb{\alpha_{j} =\displaystyle\sum_{\substack{i=1 \\ i\ne j}}^{n}d_{i}\cdot\alpha_{i} }\nonumber\\ \implies &\pmb{\alpha_{j} =\displaystyle\sum_{i=1 }^{j-1}d_{i}\cdot\alpha_{i}+\displaystyle\sum_{i=j+1 }^{n}d_{i}\cdot\alpha_{i} } \end{align} for some scalars \(\pmb{d_{1},d_{2},…,d_{j-1},d_{j+1},..,d_{n}} \) in \(\pmb{F} \)
  Let \(\pmb{ T=\{\alpha_{1},\alpha_{2},…,\alpha_{j-1},\alpha_{j+1},…,\alpha_{n} \} } \) implies \(\pmb{T\subset S} \).
  To prove \(\pmb{L(T)=V} \).
  That is to prove \(\pmb{L(T)=L(S)} \) since \(\pmb{L(S)=V} \).

• To prove \(\pmb{L(T)\subseteq L(S)} \)
  Let \(\pmb{\beta\in L(T) } \) then for some scalars \(\pmb{e_{1},e_{2},…,e_{j-1},e_{j+1},..,e_{n}} \) in \(\pmb{F} \)
  \begin{align*} &\pmb{\beta = \displaystyle\sum_{\substack{i=1 \\ i\ne j}}^{n}e_{i}\cdot\alpha_{i}}\nonumber\\ \implies&\pmb{\beta =\displaystyle\sum_{i=1 }^{j-1}e_{i}\cdot\alpha_{i}+\displaystyle\sum_{i=j+1 }^{n}e_{i}\cdot\alpha_{i} }\nonumber\\ \implies &\pmb{\beta =\displaystyle\sum_{i=1 }^{j-1}e_{i}\cdot\alpha_{i}+0\cdot\alpha_{j}+\displaystyle\sum_{i=j+1 }^{n}e_{i}\cdot\alpha_{i} }\nonumber\\ \implies &\pmb{\beta\in L(S) } \end{align*} Therefore \(\pmb{L(T)\subseteq L(S)} \).
• To prove \(\pmb{L(S)\subseteq L(T)} \)
  Let \(\pmb{\gamma\in L(S) } \) then for some scalars \(\pmb{f_{1},f_{2},…,f_{j-1},f_{j},f_{j+1},..,f_{n}} \) in \(\pmb{F} \)
  \begin{align*} &\pmb{\gamma =\displaystyle\sum_{i=1 }^{n}f_{i}\cdot\alpha_{i} }\nonumber\\ \implies&\pmb{\gamma =\displaystyle\sum_{i=1 }^{j-1}f_{i}\cdot\alpha_{i}+f_{j}\cdot\alpha_{j}+\displaystyle\sum_{i=j+1 }^{n}f_{i}\cdot\alpha_{i} }\nonumber\\ \implies&\pmb{\gamma =\displaystyle\sum_{i=1 }^{j-1}f_{i}\cdot\alpha_{i}+f_{j}\cdot\left[\displaystyle\sum_{i=1 }^{j-1}d_{i}\cdot\alpha_{i}+\displaystyle\sum_{i=j+1 }^{n}d_{i}\cdot\alpha_{i} \right]+\displaystyle\sum_{i=j+1 }^{n}f_{i}\cdot\alpha_{i} }\text{ using (1)}\nonumber\\ \implies&\pmb{\gamma =\displaystyle\sum_{i=1 }^{j-1}f_{i}\cdot\alpha_{i}+\displaystyle\sum_{i=1 }^{j-1}\left(f_{j}.d_{i}\right)\cdot\alpha_{i}+\displaystyle\sum_{i=j+1 }^{n}\left(f_{j}.d_{i}\right)\cdot\alpha_{i} +\displaystyle\sum_{i=j+1 }^{n}f_{i}\cdot\alpha_{i} }\nonumber\\ \implies&\pmb{\gamma =\displaystyle\sum_{i=1 }^{j-1}\left(f_{i}+f_{j}.d_{i}\right)\cdot\alpha_{i}+\displaystyle\sum_{i=j+1 }^{n}\left(f_{i}+f_{j}.d_{i}\right)\cdot\alpha_{i} }\nonumber\\ \implies&\pmb{\gamma = \displaystyle\sum_{\substack{i=1 \\ i\ne j} }^{n}\left(f_{i}+f_{j}.d_{i}\right)\cdot\alpha_{i} } \end{align*} Therefore \(\pmb{L(S)\subseteq L(T)} \).

  Hence \(\pmb{L(T)=L(S)} \) implies \(\pmb{L(T)=V} \) since \(\pmb{L(S)=V} \).