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.


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.

  • $\begingroup$ 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

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.