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Sorry I'm asking such a stupid question I'm a newbie student and we just started the span material.

Let's say we have $3$ linearly dependent vectors in some vector space.

In my opinion it's true. It's like taking the vectors: $u=\{1,2,3\},v=\{2,4,6\},w=\{3,6,9\}.$

$\mathrm{span}\{u,v\}=\mathrm{span}\{u,w\}$ because its mapping the $\mathbb{R}^3$ for this example.

Am I right? How can I better prove it for each vector space? use the rule thats say $xu + yv + zw =0$ where $x$ or $y$ or $z$ is not $0$?

Thank you very much and have a nice day!

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    $\begingroup$ Consider the case $\vec u=\vec 0$ and $\vec v, \vec w$ linearly independent. $\endgroup$ – lulu Aug 30 '20 at 13:15
  • $\begingroup$ so it means the span{v} is not equal to the span{w} because its they are linear independent - we can't reach w from v or v from w. right? $\endgroup$ – Roach87 Aug 30 '20 at 13:50
  • $\begingroup$ Yes, that is correct. You could also take, e.g., $\vec u=\vec v$, so long as $\vec w$ is independent of them. $\endgroup$ – lulu Aug 30 '20 at 13:52
  • $\begingroup$ if the question was that all the scalars are not 0 then it would be true because even if u=0 then v and w are still dependent so the span is equal ? So basically , every time I see 2 independent vectors I should know that their span can't equal each other? $\endgroup$ – Roach87 Aug 30 '20 at 14:01
  • $\begingroup$ Yes. If $a\vec u +b\vec v +c\vec w=\vec 0$ with $a,b,c$ all non-zero then $\vec w=-\frac ac\vec u -\frac bc\vec v\in \text {Span}\{\vec u, \vec v\}$ and similarly $\vec v\in \text {Span}\{\vec u, \vec w\}$ As you can see, you only need $b,c$ to be non-zero...$a$ doesn't matter. $\endgroup$ – lulu Aug 30 '20 at 14:04
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False take 2 standard basis vactor e1 ,e2 and 0 also these three are linear dependent so span of e1,0 is not equal to span of e2,0.

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