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?

$$a_1v_1 = 0$$ 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.)

• This is what I want to read! +1 for that! Thanks a lot for that! – ZFR Feb 12 at 17:59

The contradiction is that for $$a_2=\dots=a_m=0$$ and $$a_1\ne0,\ v_1\ne 0$$, we get $$0=a_1v_1+\dots +a_mv_m=a_1v_1\ne 0$$

• Sorry but why $a_1v_1\neq 0$? It is not so obvious for me. – ZFR Feb 9 at 18:14
• @ZFR: It's an elementary consequence of vector space axioms; see my answer – J. W. Tanner Feb 10 at 3:27

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.