Remark

The definition of linear dependence is equivalent to having a linear combination $$a_1v_1+...+a_kv_k=0$$ where at least one $a_i\not=0$.

Remark

This proposition is useful for two things:

1. Choosing a vector to remove from a linearly dependent set without affecting span
2. If we have a linearly independent set $\{v_1,...,v_k\}\subset V$ such that $span(v_1,..,v_k)\not=V$, then there is some element $v_{k+1}\in V \setminus span(v_1,...,v_k)$ . If the set $\{v_1,...,v_k,v_{k+1}\}$ were linearly dependent, then the dependence lemma would tell us $v_j\in span(v_1,...,v_{j-1})$ for some $j=2,...,k+1$. Because the set $\{v_1,...,v_k\}$ is linearly independent, our $v_j$ cannot be any of the $v_1,...,v_k$. Then $v_j=v_{k+1}$, which is a contradiction because we specifically chose $v_{k+1}\notin span(v_1,...,v_k)$. Therefore, the set $\{v_1,...,v_{k+1}\}$ is linearly independent.

Remark

What this tells us is that the largest linearly independent set has fewer elements than the smallest spanning set.

Remark

Because this theorem will hold true when we compare any linearly independent set to any spanning set in $V$, what this theorem is really telling us is that: $$max\{m|\{v_1,...,v_m\} \text{ is linearly independent}\}\leq min\{n|span(w_1,...,w_n)=V\}$$ So this brings us to bases which will tell us that the largest linearly independent set is a spanning set and also that the smallest spanning set is linearly independent. A set that satisfies both is a basis.