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Say you have three vectors $u,v$, and $w$ in $\mathbb{R}^3 $ that are linearly independent. Prove that the two vectors $u+w$ and $v+w$ have to be linearly independent.

(start by assuming $c_1(u+w)+c_2(v+w)=0$, and use the given information to prove that $c_1$ and $c_2$ must both be $0$.

So I understand that linear dependence means that one vector is a scalar multiple of another, therefore I would want to show that $c_1=c_2=0$. Would I go about this by assuming this is not the case? Step by step explanation please!

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  • $\begingroup$ $c_1\mathbf{u}+(c_1+c_2)\mathbf{v}+c_2\mathbf{w}$... $\endgroup$ – JP McCarthy Feb 9 '16 at 14:00
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$$c_1(u+v)+c_2(v+w)=0$$ So $$c_1u+(c_1+c_2)v+c_2w=0$$ As $u,v,w$ are linearly independent, $$c_1=0, c_2=0$$

If they were not zero, one could be written in terms of the other.

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  • $\begingroup$ My bad it should be c1(u+w) +c2(v+w) so would I distribute and then write it as c1u+(c1+c2)w+c2v=0 but how does this show they are independent? $\endgroup$ – Lil Feb 9 '16 at 14:05
  • $\begingroup$ I'm just confused because c1 and c2 are both being multiplied by w so wouldn't tat mean they are multiples of each other? $\endgroup$ – Lil Feb 9 '16 at 14:06
  • $\begingroup$ @Lil Look at it this way: If exactly one coefficient was zero, then the remaining two vectors would be scalar multiples of each other. If two coefficients were zero, the equation would be of the form $cv=0$, forcing $c$ to be zero. Thus, for linear independence, all of them must be zero. $\endgroup$ – GoodDeeds Feb 9 '16 at 14:09
  • $\begingroup$ sorry I'm still slightly confused. I see if either c1 or c2 were 0 the whole equation would equate to 0 which is why it would be in the form cv=0, but can you demonstrate that if one coefficient was 0 the remaining two would be scalar multiples of each other? $\endgroup$ – Lil Feb 9 '16 at 14:11
  • $\begingroup$ Let's say c1 was 0 then we would have c2v+c2w=0 and since v and w are both being multiplied by c2 they are scalar multiples of each other? $\endgroup$ – Lil Feb 9 '16 at 14:12
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Then $c_1\cdot u+c_2\cdot w+(c_1+c_2)\cdot v=0$.

Thus $c_1=0$, $c_2=0$ and $c_1+c_2=0$ by the independence of $u,v,w$. Doesn't that show what you need?

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