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Show that vectors $v_1$,...,$v_k \in \Bbb R^n$ are linearly dependent if and only if one of these vectors is a linear combination of the others.

I know that $v_1$,...,$v_k$ being linearly dependent means that there exists $c_1$,...,$c_k \in \Bbb R^n$, not all zero, such that $c_1v_1+\dots+c_1v_k = 0$ (vector). Also a linear combination of $v_1,\dots,v_k$ means that a vector $u = c_1v_1+\cdots+c_kv_k$. I am unsure of where to go with this information.

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Chose an index $j$ such that $c_j\ne 0$ and then divide everything by $c_j$. What can you deduce from there?

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  • $\begingroup$ so then (c1/cj)v1+...+vj+...+(ck/cj)vk = 0, then vj = -((c1/cj)v1+...+(ck/cj)vk), so vj is a linear combination of the others? $\endgroup$ – Tommy Dec 6 '16 at 17:10
  • $\begingroup$ @Tommy yes, exactly. $\endgroup$ – TZakrevskiy Dec 6 '16 at 17:10
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Hints:

  • For the forward direction, since one of the $c_i$'s is nonzero, you can solve for $v_i$ in terms of the other vectors (you can divide by $c_i$ because its nonzero).

  • For the backwards direction, if $v_i$ is a linear combination of the others, then $$ v_i=a_1v_1+a_2v_2+\cdots+a_{i-1}v_{i-1}+a_{i+1}v_{i+1}+\cdots a_nv_n. $$ Observe that $v_i$ is missing from the RHS. Put all your vectors on one side and show that you can write $0$ as a linear combination and some of the coefficients (at least one) is nonzero.

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