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There is a lemma saying that if $v_1, ..., v_n$ are linearly dependent in a vector space $V$ and $v_1 \neq 0$ then there exists $j \in \{2, ..., n\}$ such that $v_j \in span(v_1,...,v_{j-1})$. The proof uses that in $a_1 v_1 + ....+a_nv_n = 0$ not all $a_i$ with $i>1$ can equal $0$ and then goes on by letting $j$ be the largest index in $2,....,n$ with $a_j \neq 0$. Then $v_j = -{a_1 \over a_j}v_1 - .... -{a_{j-1} \over a_j}v_{j-1}$.

Why does $j$ have to be the largest index? Wouldn't any $j$ with $a_j$ non-zero do the trick?

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If $j$ weren't the largest index then there would be an $a_i \neq 0$ with $i > j$. The result would then not follow as $v_j$ would be a linear combination of a set of vectors that includes $v_i$.

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