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!

  • 4
    $\begingroup$ 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

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.


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.