A set is linearly dependent iff there is a vector which is a linear combination of previous vectors

up:: Linear Dependence

Let $(V,+,\cdot )$ be a Vector Space, and $S=\{v_1,\dots ,v_n\}\subset V$ (out of which no vector is $0$).

We seek to prove that

$S$ is Linearly Dependent $⟺$ there is some index $k$ in $\{2,\dots ,n\}$ for which

$$v_k=\sum _{i=1}^{k-1}{\alpha }_i{v}_i$$

Linear dependence implies there is a vector dependent on previous ones

We prove by contrapositive: if for all indexes, $v_k\notin [v_i{]}_{i=1}^{k-1}$, we seek to prove that $S$ is Linearly Independent.

Since all vectors are non-null, $\{v_1\}$ will be L.I.. By hypothesis, all subsequent vectors will not be in the Span of the previous ones, for which The union of a linearly independent set with a vector outside of its span is also linearly independent. Thus, we can construct

$$\{v_1\}L.I.⟹\{v_1,v_2\}L.I.⟹\cdots ⟹S=\{v_1{\}}_{i=1}^{n}L.I.$$

If there is a vector dependent on previous ones, the set is linearly dependent

If there is at least one index $k\in \{2,\dots ,n\}$ such that $v_k\in [v_i{]}_{i=1}^{k-1}$, then the set $\{v_1,\dots ,v_k\}$ will be Linearly Dependent, which means that adding the rest of the vectors will still keep it L.D..¹

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References

• Notas sobre Álgebra Linear

Footnotes

1. Because if $\{v_1,\dots ,v_k\}$ is already L.D., we have that $\{v_1,\dots ,v_k,\dots \}$ is L.D., since $$
   v_k = \sum\limits_{i=1}^{k-1} \alpha_i v_i + \sum\limits_{i\geq k} 0 v_i ↩