Proof of Linearly Independent system by deleting a vector Question: 
Let $\{v_1, v_2, . . . , v_k\}$ be a linearly independent
set of vectors in a vector space $V$. Delete the vector $v_k$
from this set and prove that the set $\{v_1, v_2, . . . , v_{k−1}\}$
cannot span $V$.
What I Know:
so if we set $S = \{v_1, v_2,...,v_k\}=0$, we can do this since we know it is linearly independent. Then the combination would be as follows:
$c_1v_1+c_2v_2+c_3v_3+....+c_kv_k=0$. Since this set is linearly independent it contains the trivial solution and $c_1=c_2=c_k=0$. 
But when we delete the vector $v_k$ we cannot assume that the set is linearly independent. Now I am stuck :/

If the vectors $$ v_1, v_2,..., v_{k-1} $$ span $v_k$ then you have $$c_1v_1 + c_2 v_2 +...+c_{k-1}v_{k-1} -v_k =0$$ but that is not possible because in this case $c_k=-1$ which contradicts the linear independence of your vectors.
