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Being $V$ a linear space of $\Bbb K$. Show that if $v_1, v_2, v_3$ are vectors of $V$ and $𝛼_1, 𝛼_2, 𝛼_3 ∈ 𝕂$ then the set $${𝛼_2v_3 − 𝛼_3v_2, 𝛼_1v_2 − 𝛼_2v_1, 𝛼_3v_1 − 𝛼_1v_3}$$ is linearly dependent.

My attempt:

If the set is linearly dependent then there are scalars $\beta1$ $\beta2$ and $\beta3$ not all zero that: $$\beta_1(𝛼_2v_3 − 𝛼_3v_2) + \beta_2(𝛼_1v_2 − 𝛼_2v_1) + \beta_3(𝛼_3v_1 − 𝛼_1v_3) = 0 $$

I started solving the equation and I decided to isolate $v_3$, so that I can write that I can express $v_3$ as a linear combination of $v_2$ and $v_1$. But for that $\beta_1\alpha_2 + \beta_3\alpha_1$ might be different that zero $\dots$

But how can I justify that?

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  • $\begingroup$ Let $v_1 = v_2 = v_3 = 0$. $\endgroup$ – Henricus V. Apr 2 '16 at 17:09
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My thought: Can I write the thirs vector as a linear combination of the first two? Where to start: The scalar of $v_3$ in the third vector is $α_3$ but in the first vector it is $α_2$. So let's multiply the first vector by $\frac{α_3}{α_2}$ assuming that $α_2\neq 0$. Then repeat for $v_1$ and the second vector and see if it works (which it does): $$α_3v_1-α_1v_3=-\frac{α_1}{α_2}(α_2v_3-α_3v_2)-\frac{α_3}{α_2}(α_1v_2-α_2v_1)$$ if of course $α_2\neq 0$. Else...

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  • $\begingroup$ Ok I understood your proof but how can I conclude that alfa 2 is zero? $\endgroup$ – Granger Obliviate Apr 2 '16 at 18:16
  • $\begingroup$ Hmm, you cannot conclude that $\alpha_2=0$ and this is not implied nowhere in the answer. So seemingly you did not understand my proof $\endgroup$ – Jimmy R. Apr 3 '16 at 7:00
  • $\begingroup$ *is not zero was what i meant, sorry! $\endgroup$ – Granger Obliviate Apr 3 '16 at 12:46
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Since it's possible that $\vec v_1, \vec v_2, \vec v_3$ are linearly independent vectors, we want to solve for $\beta_1, \beta_2, \beta_3$ in terms of $\alpha_1, \alpha_2, \alpha_3$ subject to the constraints that: $$\begin{cases} \alpha_3\beta_3 - \alpha_2\beta_2 = 0 \\ \alpha_1\beta_2 - \alpha_3\beta_1 = 0 \\ \alpha_2\beta_1 - \alpha_1\beta_3 = 0 \end{cases}$$ By inspection, note that we can choose: $$\begin{cases} \beta_1 = \alpha_1 \\ \beta_2 = \alpha_3 \\ \beta_3 = \alpha_2 \end{cases}$$

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  • $\begingroup$ Right, so if I choose those values for my beta I will get the zero vector and all the betas are not zero so my set in linearly dependent, right? $\endgroup$ – Granger Obliviate Apr 2 '16 at 18:14
  • $\begingroup$ Right. If we assume instead that all the betas are zero, then all the alphas are zero, so the given set contains only the zero vector and so is still linearly dependent. $\endgroup$ – Adriano Apr 2 '16 at 18:19
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Multiply the first by $a_1$ then you get $a_2a_1v_3-a_3a_1v_2$. Multiply the second by $a_3$ then you get $a_1a_3v_2-a_2a_3v_1$. Summing you obtain $a_2(a_1v_3-a_3v_1)$. If $a_2=0$ the set is trivially dependend because the first vector is $a_1v_2$ and the second is $a_3v_2$. If not, then $$\frac{a_1}{a_2}(a_2v_3-a_3v_2)+\frac{a_3}{a_2}(a_1v_2-a_2v_1)=a_1v_3-a_3v_1$$

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