Linear Dependence lemma - an unclear moment from the proof 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?

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.
