A set is linearly independent iff all its finite subsets are also L.I.

up:: 021a MOC Linear Algebra

Let $(V,+,\cdot )$ be a Vector Space, and $S\subset V$. Then we have that $S$ is Linearly Independent iff all of its subsets are linearly independent.

This follows from the contrapositive of A set is linearly dependent iff there is a vector which is a linear combination of previous vectors (provided adequate rearrangements of the set $S$).

Constructive proof

Entire set is L.I. $⟹$ All finite subsets are L.I.

Let $S$ be linearly independent. Then, for all finite combinations of $\{v_1,\dots ,v_k\}\subseteq S$, we’ll have that their Linear Combination will only be zero when all of them are multiplied by $0$ ─ which makes all of these finite subsets L.I. in their own right.

All finite subsets are L.I. $⟹$ Entire set is L.I.

Proof by contrapositive: assume the entire set is Linearly Dependent. Then there is some finite combination of vectors in $\{v_1,\dots ,v_k\}\subseteq S$ such that, for at least some $j$, ${\alpha }_j\ne 0$, for which we have

$$v_j=-\sum _{i\ne j}\frac{{\alpha }_i}{{\alpha }_j}{v}_i$$

Thus, there is some subset $\{v_1,\dots ,v_k\}\subset S$ which is L.D..