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If $v_1,v_2,v_3,v_4$ are linearly dependent, then there exists a,b,c,$\in\Bbb{R}$ such that $v_1=av_2+bv_3+cv_4$.

My attempt:

Assume $v_2=o$, then clearly, $0v_1+1v_2+0v_3+0v_4$=o so still linearly dependent but in this case statement not true since b.o=o for any b$\in\Bbb{R}$. So statement true iff all 4 vectors are different than zero vector.

Is this correct, anything to add?

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  • $\begingroup$ note that we could have $v_1$ linearly independent from $v_2,v_3$ and $v_4$ $\endgroup$ – user Mar 30 '18 at 22:36
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If $v_1,v_2,v_3,v_4$ are linearly dependent then exists $(a_1,a_2,a_3,a_4)\neq(0,0,0,0)$ such that

$$a_1v_1+a_2v_2+a_3v_3+a_4v_4=0$$

then if $a_1 \neq 0$

$$v_1=-\frac{a_2}{a_1}v_2-\frac{a_3}{a_1}v_3-\frac{a_4}{a_1}v_4=0$$

but if $a_1=0$ the statement is not true.

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  • $\begingroup$ yes , I get that, but what if one of the vectors equal zero vector, do we consider that case? $\endgroup$ – Rivaldo Mar 30 '18 at 22:36
  • $\begingroup$ @Rivaldo Are you excluding this case? $\endgroup$ – user Mar 30 '18 at 22:37
  • $\begingroup$ @Rivaldo The statement is not true in the sense that it is not true in general, even if it can be true in some cases but not always. $\endgroup$ – user Mar 30 '18 at 22:39
  • $\begingroup$ ahh makes sense thanks $\endgroup$ – Rivaldo Mar 30 '18 at 22:43
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This is false in general, as $v_1 \neq 0$, $v_2= v_3 = v_4 = 0$ is surely linearly dependent, but no such $a,b,c$ can be found.

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