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Suppose that $v_1,v_2,v_3,v_4$ spans $V$. Prove that the list $v_1 - v_2, v_2 - v_3, v_3-v_4,v_4$ also spans $V$.

attempt: Suppose $v_1,v_2,v_3,v_4$ spans $V$,

then let $v \in V$.

So $v = a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4$ for some $a_1,a_2,a_3,a_4 \in F$.

Then we will show $v \in span(v_1 - v_2, v_2 - v_3, v_3-v_4,v_4) $.

So $v = c_1(v_1 -v_2) + c_2(v_2-v_3) + c_3(v_3 - v_4) + c_4v_4$ for some $c_1,..,c_4 \in F.$.

Then setting the two equations equal we have $a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 = c_1(v_1 -v_2) + c_2(v_2-v_3) + c_3(v_3 - v_4) + c_4v_4$.

I am stuck and don't really know how to conclude. Can someone please help me? Thank you

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$\begin{eqnarray*}a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 &=& c_1(v_1 -v_2) + c_2(v_2-v_3) + c_3(v_3 - v_4) + c_4v_4\\&=&c_1v_1+(c_2-c_1)v_2+(c_3-c_2)v_3+c_4v_4\end{eqnarray*}$

Take $a_1=c_1, a_4=c_4, c_2=a_2+a_1$ and $c_3=a_3+a_2+a_1$. Then $$c_1(v_1 -v_2) + c_2(v_2-v_3) + c_3(v_3 - v_4) + c_4v_4=v.$$

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Set $c_1=a_1$. This is going to produce the same coefficient accompanying $v_1$. Then we have to mend $v_2$. For that, set $c_2=a_1+a_2$. Then we have to mend $v_3$, and so on... The idea is that each vector of $v_1 - v_2, v_2 - v_3, v_3-v_4,v_4$ is going to mend the previous one, until the last one, $v_4$, fixes itself. It is easily generalizable for n vectors , and this is a way( my way) to prove that if a vector space has a basis, then it has infinite of them.

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Just prove that $v_1$, $v_2$, $v_3$, and $v_4$ are in the span of $v_1-v_2$, $v_2-v_3$, $v_3-v_4$, and $v_4$:

$$v_1=(v_1-v_2)+(v_2-v_3)+(v_3-v_4)+(v_4)$$ $$v_2=(v_2-v_3)+(v_3-v_4)+(v_4)$$ $$v_3=(v_3-v_4)+(v_4)$$ $$v_4=v_4$$

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