Let $\{v_1, v_2, v_3\}$ be a linearly independent set of vectors in a vector space $V$. Show that the set $\{v_1 + v_2, v_2 + v_3, v_1 + v_3\}$ is also a linearly independent set.

I put the equations in the form: $c_1(v_1 + v_2) + c_2(v_2 + v_3) + c_3(v_1 + v_3) = 0$, to see if it implies that $c_1 = c_2 = c_3 = 0$.

Then I distributed to change it to the form: $v_1(c_1 + c_3) + v_2(c_1 + c_2) + v_3(c_2 + c_3) = 0$

I'm not sure what to do from here. I need to show the linear independence of: $c1 + c3 = 0$, $c1 + c2 = 0$, and $c2 + c3 = 0$

Do I just put this in reduced row echelon form?


The set $\{v_1,v_2,v_3\}$ is linearly independent.

Since you've reduced it to the form $$(c_1+c_3)v_1+(c_1+c_2)v_2+(c_2+c_3) = 0,$$ you know that this is only true if all $c_i = 0$, since we assumed linear independence i.e. let $d_1 = (c_1+c_3)$, $d_2 = (c_1+c_2)$, $d_3=(c_2+c_3)$. Then we have $$d_1v_1 + d_2v_2 + d_3v_3$$ which is only possible if the $d_i =0$. Ergo, your second set is linearly independent, since they may only sum to $0$ if the coefficients are all $0$. Is this clear?

  • $\begingroup$ To solve d1=(c1+c3), d2=(c1+c2), d3=(c2+c3), do I put this in reduced row echelon form, where the columns are c1, c2, and c3? $\endgroup$ – Neel Oct 8 '15 at 8:58
  • $\begingroup$ It's not clear to me what you're solving. You know the $d$'s have to be 0. So you can substitute variables around or just look at the signs of the coefficients to deduce they must all be 0 $\endgroup$ – Anthony Peter Oct 8 '15 at 9:01

So once you're at the step $$(c_1+c_3)v_1+(c_1+c_2)v_2+(c_2+c_3)v_3=0$$ because you know the set $\{v_1,v_2,v_3\}$ is linearly independent, we must have all these coefficients equal to zero, i.e. $$c_1+c_3=c_1+c_2=c_2+c_3=0$$

From the first equality we can say we must have $c_2=c_3$ (by subtracting $c_1$ from both). Then $$0=c_2+c_3=c_3+c_3=2c_3\implies c_3=0$$ and since $c_2=c_3$, we must have $c_2=0$. Then $0=c_1+c_3=c_1$, so all the $c_i$ are zero and the set is linearly independent.

  • $\begingroup$ @AnthonyPeter I thought you had made a jump you shouldn't have, once I saw you edited your answer I fixed my post. Sorry about that $\endgroup$ – Alex Mathers Oct 8 '15 at 7:33
  • $\begingroup$ it was the same idea without the added line. $\endgroup$ – Anthony Peter Oct 8 '15 at 7:35
  • $\begingroup$ You're basically just saying "because $d_i=0$, $c_i=0$". You're not giving any explanation as to why. $\endgroup$ – Alex Mathers Oct 8 '15 at 7:36
  • $\begingroup$ mine was on the basis that your answer adds absolutely nothing to this discussion. yours is childish. $\endgroup$ – Anthony Peter Oct 8 '15 at 7:42

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