How to proof $\operatorname{span} (v_1...v_n)$ not equal to $\operatorname{span }(v_2...v_n)$ Suppose that $S = \{ {v}_1 , \ldots , {v}_n \}$ is a linearly independent set of vectors in a vector space $V$.  Show that $\operatorname{span}(S) \neq \operatorname{span}({v}_2 , \ldots , {v}_n)$
I set up two groups of coefficient $a_1 \ldots a_n$ and $b_2 \ldots b_n$ and assumed the ${v}_2 , \ldots , {v}_n$ is linearly dependent.
Then I assumed S = ${v}_2 , \ldots , {v}_n$ .. and found that it would make $S$ linearly dependent, which is contradiction, does it make sense?
Thanks!
 A: HINT
For the proof simply show that


*

*$v_1 \in \operatorname{span}(S) $ that's true by definition

*$v_1 \not \in \operatorname{span}({v}_2 , \ldots , {v}_n )$ by contradiction assuming $v_1=\sum_{i=2}^n c_i v_i$

A: No, it does not make sense to assume $S=\{v_2,\cdots, v_n\}$, because $S$ is in fact not that set. Since you want to show that the assumption that for all $a_1,\cdots, a_n$ there are $b_2,\cdots, b_n$ such that $\sum_{i=1}^na_iv_i=\sum_{j=2}^n b_jv_j$ yields a contradiction, you only need to show that there are some $a_1,\cdots,a_n$ such that any choice of $b_2,\cdots, b_n$ such that the identity holds would produce a non-trivial linear correlation between the $v_1,\cdots, v_n$. I suggest you choose $a_1=1$ and $a_2=\cdots=a_n=0$.
A: By contraposition: 
if span$\{v_1,\cdots, v_n\}=$span$\{v_2,\cdots, v_n\}$ than, $\forall a_i$ scalars, we have:
$$
a_1v_1+a_2v_2+\cdots a_nv_n=b_2v_2+\cdots+b_nv_n
$$
for somes $b_i$. But this gives:
$$
a_1v_1+(a_2-b_2)v_2+ \cdots(a_n-b_n)v_n=0
$$
and this means that $\{v_1,v_2,\cdots, v_n\}$ are not linearly independent.
