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$\textbf{Problem:}$ Show that if Span$\{v_1,v_2,v_3\}$=$\mathbb{R}^3$ then $\{v_1,v_2,v_3\}$ is linearly independent.

$\textbf{Proof:}$ Assume that Span$\{v_1,v_2,v_3\}$=$\mathbb{R}^3$ and assume to the contrary that the $\{v_1,v_2,v_3\}$ is linearly dependent. It follows $$ c_1v_1+c_2v_2+c_3v_3=\vec{0} \text{ if at least one $c_i\neq0$ for $i=1,2, \text{ or } 3$}$$ It follows that at least one vector is a linear combination of the other two vectors. This means that the span reduces to a plane in $\mathbb{R^3}$ which is a contradiction.

I feel that this proof lacks some generalization or I could add more detailed steps where I have "it follows that atleast one vector..." making the proof more formal. Thanks!

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  • $\begingroup$ I think you mean "$c_1v_1+c_2v_2+c_3v_3=\vec{0} \text{ where at least one }c_i\neq 0\text{ for } i=1,2, \text{ or } 3$" $\endgroup$ – Arthur Sep 20 '17 at 6:08
  • $\begingroup$ All this uses the fact that $\mathbb{R}^3$ cannot be spanned by $2$ vectors. $\endgroup$ – Orest Bucicovschi Sep 20 '17 at 6:23
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Because of symmetry it is enough to consider the case when $c_1\ne 0$ (in the definition of linear independence it is assumed that not all scalars are zero). Then $v_1=-\dfrac{c_2}{c_1}v_2-\dfrac{c_3}{c_1}v_3$ which proves the claim you are asking for.

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I don't know if you study this already but if this question was given to me I would say as you said with a little note to make it clearer: if, let's say, $v_1$ could be created by the other 2, let's say, $v_1=k_1v_2+k_2v_2$ than the span of $v_1,v_2,v_3$ was equal to the span of $v_2,v_3$ and then I would add the proof that if the span of $v_1,v_2,\cdots ,v_n$ is $ℝ^k$ then $n\ge k$.

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