Given vectors $v_1,v_2,w_1,w_2$ are in linear space $V$ such that $Sp\{v_1,v_2\}=Sp\{w_1,w_2\}$ and {$v_1,w_2$} are linearly independent prove that {$v_1,v_2$} are linearly independent.
I thought to prove this by showing a contradiction.
Let $U=Sp\{v_1,v_2\}$, $W=Sp\{w_1,w_2\}$.
Suppose that {$v_1, v_2$} are linearly dependent. Then scalars $a, b \in \mathbb F$ exist such that $av_1=bv_2$ and $a,b$ are not all zeroes.
Because $U=W$ then every vector in {$v_1,v_2$} is a linear combination of {$w_1,w_2$} and vice versa.
Then $c \in \mathbb F$ exists such that $v_2=cw_2$.
Therefore: $$ av_1=bv_2=b\cdot c\cdot w_2 $$
According to our hypothesis {$v_1,v_2$} are linearly dependent then in the above equality $a, bc$ are not all zeroes. But this contradicts the given that {$v_1,w_2$} are linearly independent therefore $a, bc$ must be all zeroes.
EDIT: can I really claim that because $U=W$ then $v_2=cw_2$?