1
$\begingroup$

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?

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ Let $v_1 = v_2 = v_3 = 0$. $\endgroup$ – Henricus V. Apr 2 '16 at 17:09
2
$\begingroup$

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...

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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: $$\begin{cases} \beta_1 = \alpha_1 \\ \beta_2 = \alpha_3 \\ \beta_3 = \alpha_2 \end{cases}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $$\frac{a_1}{a_2}(a_2v_3-a_3v_2)+\frac{a_3}{a_2}(a_1v_2-a_2v_1)=a_1v_3-a_3v_1$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.