# 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?

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 $v_2=cw_2$?

• If two vectors $v$ and $w$ are linearly dependent and if $a$ and $b$ are numbers such that $av=bw$, it is not necessarily true that none of the numbers $a$ and $b$ is $0$. May 25, 2017 at 11:23
• @JoséCarlosSantos I said that if 2 vectors are linearly dependent then it can't be that both $a,b$ their coefficients are $0$ else they'd be linearly independent
– Yos
May 25, 2017 at 11:43
• @Yoz You wrote "According to our hypothesis $\{v_1,v_2\}$ are linearly dependent then in the above equality $a,bc$ are not all zeroes". This is plain wrong. May 25, 2017 at 11:51
• @JoséCarlosSantos so can they be dependent if both a and b are zeroes?
– Yos
May 25, 2017 at 11:55
• @Yoz Of course they can! $0.(1,0)=0.(1,0)$, right?! May 25, 2017 at 12:00

$v_1$ 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.

– Yos
May 25, 2017 at 11:53
• I can't see why you think $v_2=c w_2$. May 25, 2017 at 12:19
• This is the answer which should be accepted
– JJR
May 25, 2017 at 20:38

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.

• I have $bw_2=bcv_1+bd\cdot (-\frac{a}{b})v_1=v_1(bc-da)$ how did you get $(ad-bc)v_1$?
– Yos
May 25, 2017 at 13:31