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To show linear independence of 3 vectors is it enough to show that the first two are independent and the third is not a linear combination of the previous two?

The given is $c_1v_1+c_2v_2=0$ has no solution besides the trivial solution and $v_3=k v_2+s v_1$ has no solution. We want to show $c_1v_1+c_2v_2+c_3v_3=0$ has no solution besides the trivial solution. I tried proof by contradiction.

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    $\begingroup$ Suppose there was a set of scalars $c_1,c_2,c_3$ such that $c_1v_1+c_2v_2+c_3v_3=0$. Now... there are two possibilities, either $c_3=0$ or $c_3\neq 0$. Examine each independently in more detail and how that relates to your original hypotheses. $\endgroup$ – JMoravitz Jul 21 '17 at 21:35
  • $\begingroup$ Thanks! ${}{}{}{}{}$. $\endgroup$ – Ahmed S. Attaalla Jul 21 '17 at 21:41
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    $\begingroup$ More generally, if the set $\{v_1,\dots,v_n\}$ is linearly independent and $v$ is not a linear combination thereof, then also $\{v_1,\dots,v_n,v\}$ is linearly independent. $\endgroup$ – egreg Jul 21 '17 at 21:50
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Yes it is enough. To show that, it suffices to show that not being able to write $v_3$ as a combination of $v_1$ and $v_2$ is equivalent to the statement that the only solution for $c_1v_1+c_2v_2+c_3v_3=0$ is the trivial one.

Suppose that $v_3=c_1v_1+c_2v_2$ has no solutions, in particular, that implies that $$0=c_1v_1+c_2v_2+v_3$$ has no solutions. Therefore, the only solution for $$0=c_3c_1v_1+c_3c_2v_2+c_3v_3$$ is if $c_3$ is equal to zero. This implies that a linear combination of the form $$0=c_1v_1+c_2v_2+c_3v_3$$ implies that $c_3$ is $0$ and you can use the independence of $v_1$ and $v_2$ to conclude.

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