Let $V$ be a vector space over $K$ and $v_1, ... v_n$ linearly independent, and $v_{n+1} \in V$ \ Span{$v_1, ... v_n$}. Show that $v_1, ... v_n, v_{n+1}$ are linearly independent.
Any idea how to prove that ?
My idea is that if $v_{n+1} \notin$ Span{$v_1, ... v_n$}, then it cannot be expressed as linear combination of those vectors. Also $v_1, ... v_n$ linearly independent means $$a_1v_1 + .... a_nv_n = 0 \implies a_1 = ... = a_n = 0$$
But if we write $$ a_1v_1 + .... a_nv_n + a_{n+1}v_{n+1} = 0$$ then we have $$ a_1v_1 + .... a_nv_n = - a_{n+1}v_{n+1} $$
Which isn't necessarily zero, isn't it ? Or is what I did until now wrong ?
Suppose $v_1,...,v_{n+1}$ are linearly dependent. Then there exist $a_i$, not all zero such that $$a_1v_1+...+a_nv_n+a_{n+1}v_{n+1} = 0.$$
If $a_{n+1}=0$ then the linear independence of $v_1,...,v_n$ implies that $a_i=0$ for $1 \leq i \leq n$, which contradicts our choice of $a_i$ (not all zero).
If $a_{n+1} \neq 0$ then $$ v_{n+1} = -\frac{a_1}{a_{n+1}}v_1-...-\frac{a_n}{a_{n+1}}v_n$$ showing that $v_{n+1}$ is in the subspace generated by $v_1,...v_n$.