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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 $v_2=cw_2$?

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  • $\begingroup$ 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 $0$. $\endgroup$ May 25, 2017 at 11:23
  • $\begingroup$ @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 $\endgroup$
    – Yos
    May 25, 2017 at 11:43
  • $\begingroup$ @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. $\endgroup$ May 25, 2017 at 11:51
  • $\begingroup$ @JoséCarlosSantos so can they be dependent if both a and b are zeroes? $\endgroup$
    – Yos
    May 25, 2017 at 11:55
  • $\begingroup$ @Yoz Of course they can! $0.(1,0)=0.(1,0)$, right?! $\endgroup$ May 25, 2017 at 12:00

2 Answers 2

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$v_1$ 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.

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  • $\begingroup$ Thank your for your answer. Is my attempt to prove correct? $\endgroup$
    – Yos
    May 25, 2017 at 11:53
  • $\begingroup$ I can't see why you think $v_2=c w_2$. $\endgroup$ May 25, 2017 at 12:19
  • $\begingroup$ This is the answer which should be accepted $\endgroup$
    – JJR
    May 25, 2017 at 20:38
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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.

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

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