Find if $N=\langle y_1,y_2,y_3\rangle$ is linearly dependent/independent, given $y_1=x_1+x_2, y_2=x_1+x_3, y_3=x_2+x_3$ Let $M=\langle x_1,x_2,x_3\rangle$ be a set of linearly independent set of vectors in vector space V. Let $N=\langle y_1,y_2,y_3\rangle$, where $y_1=x_1+x_2, y_2=x_1+x_3, y_3=x_2+x_3$. Find if N is linearly dependent or independent.
I figured that I will have to solve this equation and see if $c_n$ have non trivial solutions.
$c_1y_1+c_2y_2+c_3y_3=0 \rightarrow c_1(x_1+x_2)+c_2(x_1+x_3)+c_3(x_2+x_3)=0$
However, I do not how to proceed from here. Any help is much appreciated.
 A: You may try to prove it by contrapositive. Assume that $y_1,y_2,y_3$ are linearly dependent, then there exists scalars $c_1,c_2,c_3$, not all zero, such that $c_1y_1+c_2y_2+c_3y_3=0$. Therefore, we have $(c_1+c_2)x_1+(c_1+c_3)x_2+(c_2+c_3)x_3=0$, where $(c_1+c_2),(c_1+c_3),(c_2+c_3)$ cannot be all zero. (Otherwise it will lead to $c_1=c_2=c_3=0$, which contradicts with our assumption). We have shown that $x_1,x_2,x_3$ are linearly dependent, and this finishes the proof.
A: To find out if N is a linearly dependent/independent set, we need to set the linear combination of the vectors to zero and solve for $c_i$.
$c_1y_1+c_2y_2+c_3y_3=0\rightarrow(c_1+c_2)x_1+(c_1+c_3)x_2+(c_2+c_3)x_3=0$
We know that $M=(x_1,x_2,x_3)$ is a linearly dependent set, therefore,
$c_1+c_2=c_1+c_3=c_2+c_3=0\rightarrow c_1=-c_2,c_1=-c_3, c_2=-c_3$
So,$c_i$'s do not have have to be all zero and can take other non-zero value.Therefore, Y is a linearly dependent set.
Does this direct proof seem okay to you guys? Thanks for your input.
