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Let $v_1$, $v_2$, $v_3$, be linearly independent vectors in $\mathbb{R}^n$. Let $y_1$ = $v_2$-$v_1$, $y_2$ = $v_3$-$v_2$, $y_3$ = $v_3$-$v_1$. Are $y_1$, $y_2$, $y_3$ linearly dependent or independent? Prove the answer.

Could someone point me in the right direction? I am having a lot of difficulties in my class, and I'm currently stuck on this problem.

I saw a similar example, in which the coefficients were zero, which showed linear independence. I understand a little bit of the intuition, but I am still confused on how to prove this problem.

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2 Answers 2

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Hints:

Since the list $(v_1,v_2,v_3)$ is linearly independent, then for constants $\alpha, \beta, \gamma$ we have that

$$ \alpha v_1 + \beta v_2 + \gamma v_3 = \Theta $$

implies that $\alpha = \beta = \gamma = 0 $. Now, you want to show that the list $(v_2 - v_1, v_3 - v_2, v_3 - v_1 ) $ is a linearly independent list. Again, using the definition and what you have: Say $\alpha', \beta', \gamma' $ are constants and say

$$ \alpha' ( v_2 - v_1) + \beta' ( v_3 - v_2) + \gamma' (v_3 - v_1 ) = \Theta $$

Now, it is easy to see that $\alpha' = \beta' = \gamma' = 0 $. $\mathbf{Show} $ $\mathbf{this}$.

PS: $\Theta$ denotes the zero vector.

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    $\begingroup$ But $\alpha^\prime =1, \beta^\prime=1, \gamma^\prime=-1$ gives: $$v_2 - v_1 + v_3 - v_2 + v_1 - v_3 = 0$$ (so infact they're linearly dependent) $\endgroup$ Nov 21, 2022 at 4:48
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From the definition of $\newcommand{\vect}[1]{\boldsymbol{#1}}\vect{y_1},\:\vect{y_2},\:\vect{y_3}$ above we can see that: $$ \vect{y_1} + \vect{y_2} = \vect{v_3} - \vect{v_1} = \vect{y_3}$$ $$\implies \vect{y_1} + \vect{y_2} - \vect{y_3} = \vect{0} $$ Three vectors $\vect{y_1},\:\vect{y_2},\:\vect{y_3}$ are linearly independent if: $$\lambda\vect{y_1} + \mu\vect{y_2} + \zeta\vect{y_3} = \vect{0} \iff \lambda, \mu, \zeta = 0$$ But above we have found a solution to $\lambda\vect{y_1} + \mu\vect{y_2} + \zeta\vect{y_3} = \vect{0}$ with $\lambda,\: \mu =1 $ and $\zeta = -1$, so we can conclude that these vectors are not linearly independent.

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