How to prove that {$v_1,v_2$} are linearly independent if $Sp\{v_1,v_2\}=Sp\{w_1,w_2\}$ and {$v_1,w_2$} are linearly independent?

Question

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 $v_2=cw_2$?

Answer 1

$v_1$ and $w_2$ both lie in $U:=\text{Span}\{v_1,v_2\}=\text{Span}\{w_1,w_2\}$ which is a subspace of dimension at most $2$. As they are given to be LI they form a basis of $U$ which is therefore of dimension exactly $2$. Were $\{v_1,v_2\}$ not LI we would have that $U$ is at most one-dimensional.

Answer 2

The last two steps of this proof are dubious at best.

Here is my proof.

Suppose that $\{v_1,v_2\}$ are linearly dependent. Then there exist $a$ and $b$ not both zero such that $av_1+bv_2=0$. Suppose that $b=0$, then $a\neq0$ and $v_1=0$. Therefore $\{v_1,w_2\}$ are linearly dependent, which is a contradiction. Hence $b\neq0$ and $v_2=-(a/b)v_1$. Since $w_2\in{}Span(\{v_1,v_2\})$ there exist $c$ and $d$ such that $w_2=cv_1+dv_2$. Hence $bw_2=bcv_1+bdv_2=-(ad-bc)v_1$ and $bw_2+(ad-bc)v_1=0$. Since $b\neq0$ $\{v_1,w_2\}$ are linearly dependent, which is a contradiction. Hence our initial hypothesis was false and $\{v_1,v_2\}$ are linearly independent.