In Linear Algebra Done Right, it presents the Linear Dependence Lemma which states that :

Suppose $v_1,...,v_m$ is a linearly dependent list in $V$. Then there exists $j \in {1,2,...,m}$ such that the following hold: $v_j \in span(v_1,...,v_{j-1})$. I follow the proof, but get confused on a special case where $j=1$. The book said choosing $j=1$ means that $v_1=0$, because if $j=1$ then the condition above is interpreted to mean that $v_1 \in span()$.

I tried to follow the example in the proof:

Because the list $v_1,...,v_m$ is linearly dependent, there exist numbers $a_1,...,a_m \in \mathbb{F}$ , not all $0$ such that $$a_1v_1+...+a_mv_m = 0$$

Let $j$ be the first element of {1,...,m}, such that $a_j \neq 0$. Then $$a_1v_1 = -a_2v_2 -...-a_jv_j$$ $$v_1 = \frac{-a_2}{a_1}v_2-...-\frac{a_j}{a_1}v_j$$

Then does it mean $v_1$ is the span of $v_2,....v_j$?


You've a mistake in the line after "Let $\;j\;$ be the first ..." . It must be;

$$a_jv_j=-a_1v_1-\ldots-a_{j-1}v_{j-1}\implies v_j=-\frac{a_1}{a_j}v_1-\ldots-\frac{a_{j-1}}{a_j}v_{j-1}$$

and thus



I have misunderstood the proof. It should be :

Let $j$ be the largest element of {1,...,m} such that $a_j \neq 0$. It means that $a_{j+1} = 0, a_{j+2} = 0$, and so on.

Therefore, $v_j \in span(v1,...,v_{j-1})$ actually means if the list is linearly independent, one of the vectors is in the span of the PREVIOUS ones.

So, in the case of $j=1$, $v_1 \in span()$

More reference here


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