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 $v_2=cw_2$?

  • $\begingroup$ 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


$v_1$ 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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.