# If $3$ vectors are linearly dependent then the $\mathrm{span}\{u,v\}=\mathrm{span}\{u,w\}$ , true or false?

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!

• Consider the case $\vec u=\vec 0$ and $\vec v, \vec w$ linearly independent. – lulu Aug 30 '20 at 13:15
• 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? – Roach87 Aug 30 '20 at 13:50
• Yes, that is correct. You could also take, e.g., $\vec u=\vec v$, so long as $\vec w$ is independent of them. – lulu Aug 30 '20 at 13:52
• 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? – Roach87 Aug 30 '20 at 14:01
• 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. – lulu Aug 30 '20 at 14:04