# 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.

• Write your equation as $(c_1+c_2)x_1+(c_1+c_3)x_2+(c_2+c_3)x_3=0$ and use the fact that $M$ is linearly independent. What do you conclude about $c_1,c_2,c_3$? Feb 28, 2018 at 22:04
• @Zuriel, can we say that $c_1+c_2=k_1, c_1+c_3=k_2, c_2+c_3=k_3$? And thus it is now expressed as a linear combinations of the vectors in M. Since M is linearly independent, so is N Feb 28, 2018 at 22:21
• No. Since $M$ is linearly independent, the equation in my first comment implies that $c_1+c_2=c_1+c_3=c_2+c_3=0$. Mar 1, 2018 at 6:14

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.

• +1, hopefully this helps OP formalize their thoughts. I also want to note that the same algebra can be used for a direct proof, rather than contrapositive (so one need not assume anything, instead using linear independence of the $x_i$ to show that each $c_i = 0$) Mar 1, 2018 at 5:08
• Yes, this statement can be proved directly, and I think they have been discussing on this. I gave a contrapositive proof to provide a different view on this question. Mar 1, 2018 at 7:15
• Thanks for the contra-positive proof. Based on other inputs, I also posted the direct proof. Appreciate your feedback. Mar 1, 2018 at 17:21

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.

• But those last equations do only have the trivial solution $c_1 = c_2 = c_3 = 0$. In the contrapositive proof, we prove “$P \implies Q$” by proving “$\text{not }Q \implies \text{not } P$”, since the two are logically equivalent. The negations here are “not linearly independent,” that is, “linearly dependent,” that’s why there were statements that certain sets were dependent. Mar 1, 2018 at 19:20
• The answer I posted was my attempt at trying to prove using direct proof. So my last statement is wrong then? So, the set N is linearly independent?Sorry, I’m a tad bit confused. Mar 1, 2018 at 19:52
• Yeah, if you actually try to solve that system of equations with the $c_i$, you should find that they all have to be $0$. I brought up the contrapositive because I assumed that it was the source of the confusion here -- that you saw things being shown to be linearly dependent in the other answer, but didn't realize that this came from having different assumptions (namely, that if the $y_i$ are dependent, then the $x_i$ are too; that's the structure of the contrapositive statement, when proving that $x_i \text{ independent} \implies y_i \text{ independent}$) Mar 1, 2018 at 19:57
• When using the direct proof to prove this statement, you have to show that M is a linearly independent set $\implies$ N is a linearly independent set. In other words, $c_1x_1+c_2x_2+c_3x_3=0$ only has the trivial solution $\implies$ $d_1(x_1+x_2)+d_2(x_2+x_3)+d_3(x_3+x_1)=0$ only has the trivial solution. Mar 2, 2018 at 2:46