Let $V$ be a vector space having dimension $n$, and let $S$ be a subset of $V$ that generates $V$.
Prove that there is a subset of $S$ that is a basis for $V$.
So if I let $\beta={u_1, u_2,….,u_n}$ be a basis for $V$, then each vector in $V$ can be written as a linear combination of the vectors in $\beta$. Since $S$ generates $V$ and $\beta$ $\subseteq$ V, then $S$ generates $\beta$.
I'm not sure where to go from here. I know I can somehow use the fact that the vectors in $\beta$ are linearly independent, but I still don't fully see how.
Any tips are appreciated.
Thanks
Best Answer
You can proceed with an inductive argument:
Take $\;0\neq s_1\in S\;$ . If $\;$ Span$\{s_1\}=V\;$ then we're done, otherwise there must be $\;s_2\in S\setminus \{s_1\}\;$ (why?).
If $\;$Span$\{s_1,s_2\}=V\;$ we're done, otherwise there's $\;s_3\in S\setminus\{s_1,s_2\}\;$ ...etc.
The above process is finite since $\;\dim V=n\;$ . Now polish, finise and serve hot.