2
$\begingroup$

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.

$\endgroup$
3
  • 3
    $\begingroup$ 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 $c_1,c_2,c_3$? $\endgroup$
    – Zuriel
    Feb 28, 2018 at 22:04
  • $\begingroup$ @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 $\endgroup$
    – nova_star
    Feb 28, 2018 at 22:21
  • 1
    $\begingroup$ No. Since $M$ is linearly independent, the equation in my first comment implies that $c_1+c_2=c_1+c_3=c_2+c_3=0$. $\endgroup$
    – Zuriel
    Mar 1, 2018 at 6:14

2 Answers 2

Reset to default
2
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ +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 $c_i = 0$) $\endgroup$
    – pjs36
    Mar 1, 2018 at 5:08
  • 1
    $\begingroup$ 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. $\endgroup$
    – yifan dai
    Mar 1, 2018 at 7:15
  • $\begingroup$ Thanks for the contra-positive proof. Based on other inputs, I also posted the direct proof. Appreciate your feedback. $\endgroup$
    – nova_star
    Mar 1, 2018 at 17:21
0
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ 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. $\endgroup$
    – pjs36
    Mar 1, 2018 at 19:20
  • $\begingroup$ 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. $\endgroup$
    – nova_star
    Mar 1, 2018 at 19:52
  • 1
    $\begingroup$ 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 $x_i \text{ independent} \implies y_i \text{ independent}$) $\endgroup$
    – pjs36
    Mar 1, 2018 at 19:57
  • 1
    $\begingroup$ 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. $\endgroup$
    – yifan dai
    Mar 2, 2018 at 2:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .