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I am reading linear dependence lemma, namely:

If $(v_1,v_2,\dots,v_m)$ is linearly dependent and $v_1\neq 0$, there exists an index $j\in \{2,\dots,m\}$ such that:

$v_j\in \text{span} (v_1,\dots,v_{j-1}).$

Proof: Since $(v_1,\dots,v_m)$ is linearly dependent there exist $a_1,\dots,a_m\in \mathbb{F}$ not all zero such that $a_1v_1+\dots+a_mv_m=0$. Since by assumption $v_1\neq 0$, not all of $a_2,\dots,a_m$ can be zero (why?). Let $j\in \{2,\dots,m\}$ be largest such that $a_j\neq0$. Then we have $$v_j=-\dfrac{a_1}{a_j}v_1-\dots-\dfrac{a_{j-1}}{a_j}v_{j-1}.$$ From here we get our desired result.

Let me ask you question: If $a_2=\dots=a_m=0$ then we have $a_1v_1=0$ where $a_1\neq 0$ and $v_1\neq 0$. And where is the contradiction?

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$a_1v_1 = 0$ with $a_1 \ne 0$ and $v_1 \ne 0$ is a contradiction.

If $a_1\ne0$, then, since $a_1$ is in a field $\mathbf F,$ there exists $a_1^{-1}$ in $ \mathbf F$, so $a_1v_1 = 0$ implies $v_1 =0.^*$

($^*$Multiply both sides by $a_1^{-1}$ to see this.)

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  • $\begingroup$ This is what I want to read! +1 for that! Thanks a lot for that! $\endgroup$ – ZFR Feb 12 at 17:59
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The contradiction is that for $a_2=\dots=a_m=0$ and $a_1\ne0,\ v_1\ne 0$, we get $$0=a_1v_1+\dots +a_mv_m=a_1v_1\ne 0$$

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  • $\begingroup$ Sorry but why $a_1v_1\neq 0$? It is not so obvious for me. $\endgroup$ – ZFR Feb 9 at 18:14
  • $\begingroup$ @ZFR: It's an elementary consequence of vector space axioms; see my answer $\endgroup$ – J. W. Tanner Feb 10 at 3:27
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v1 is not zero. So if a1*v1+....+an*vn=0 where not all coefficients are 0, then certainly a1 cannot be the only nonzer coefficient. So we have at not equal to zero where t is not equal to one. Now we can rearrange to express vt.

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