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I know that $S(v_1,v_2,v_3)$ is the span of the vector space $V$.

Now I need to show that $S'(v_1,v_2,v_1+v_3)$ is also the span of vector space $V$.

**It's actually prove or disprove but I'm pretty sure it's true so I need to prove.

I'm not sure what to do, how do I prove it's a span of $V$ ? If the first statement only means that $a_1\cdot v_1 + a_2\cdot v_2 + a_3\cdot v_3$ is all linear combinations of $V$.

If I look at all linear combinations of $S'$, I see $(4b_1+b_3)\cdot v_1 + b_\cdot v_2 + b_3\cdot v_3$. Now I see it looks the same, I don't know if it's a proof or not, but it looks the same to me, scalar multiplied by the vector, how do I put the proof if it's a proof to words? I'm not sure what I need to do in this exercise.

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Probably you mean V is the span of $\{v_1, v_2,v_3\}$.

This is so because $v_1+v_3\in S(v_1, v_2,v_3)$. Conversely $v_3\in S(v_1, v_2,v_1+v_3)$ since we can write $$v_3=(v_1+v_3)-v_1.$$ Thus the generators of each span belong to the other span. Hence they're equal.

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  • $\begingroup$ I meant V is a vector space, and the group s(v1,v2,v3) spans V. $\endgroup$ – user3575645 Dec 25 '16 at 22:12
  • $\begingroup$ Now I understand v1+v3∈S(v1,v2,v3) and that v3∈S(v1,v2,v1+v3) but didn't quite understand how you got to v3=(v1+v3)−v1. $\endgroup$ – user3575645 Dec 25 '16 at 22:13
  • $\begingroup$ ?? Remove the parentheses and simplify, that's all. $\endgroup$ – Bernard Dec 25 '16 at 22:22
  • $\begingroup$ Got it thanks.' $\endgroup$ – user3575645 Dec 25 '16 at 22:30
  • $\begingroup$ Actually I'm not sure I got it, if the generators of each span belongs to the other span, what is equal, the generators or the spans ? And in any case, why does it make them even ? maybe it means one contains or contained by another ? $\endgroup$ – user3575645 Dec 25 '16 at 23:05

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