# Linear independence

## Definition

Given an arbitrary subset, [ilmath]S[/ilmath] of a vector space [ilmath](V,\mathbb{F})[/ilmath] (that is to say [ilmath]S\subseteq V[/ilmath][Note 1]) if:

• For all finite sums [ilmath]\sum_i a_iv_i[/ilmath] with distinct [ilmath]v_i\in S[/ilmath] we have that [ilmath]\sum_i a_iv_i=0\ \implies\ \forall i[a_i=0][/ilmath]

This can be stated more concretely as:

• $\forall \{v_\alpha\}_{\alpha\in I}\in\mathcal{P}(S)\left[\left(n:=\big\vert\{v_\alpha\}_{\alpha\in I}\big\vert\right)\in\mathbb{N}\implies\left[\left(\sum^n_{i=1}\lambda_i v_i=0\right)\implies\left(\lambda_i=0\forall i\in I\right)\right]\right]$[CorrectedRef 1] (with the slight abuse of notation that once [ilmath]\{v_\alpha\}_{\alpha\in I}[/ilmath] is known to be finite we can index it by integers from [ilmath]1[/ilmath] to [ilmath]n[/ilmath])
Notice that we distill the distinct part by using the power set, and then we require finite-ness before we consider solutions to the summation of the subset being zero.

Then we say that [ilmath]S[/ilmath] is a linearly independent set of vectors

### Equivalent condition

The statement that:

• [ilmath]\forall \{v_\alpha\}_{\alpha\in I}\in\mathcal{P}(S)\left[\left(n:=\big\vert\{v_\alpha\}_{\alpha\in I}\big\vert\right)\in\mathbb{N}\implies\left[\left(\sum^n_{i=1}\lambda_i v_i=0\right)\implies\left(\lambda_i=0\forall i\in I\right)\right]\right][/ilmath] (given above)

Is the same as:

• [ilmath]0\notin S[/ilmath] and:
• For every finite sum of the subspaces of the form [ilmath]\ell_v=\{av\vert a\in\mathbb{F}\}[/ilmath] is direct.

TODO: Document sums and directness

## For a finite [ilmath]S[/ilmath]

When [ilmath]S\subseteq V[/ilmath] is finite the definition becomes so much nicer it is worth mentioning.

Given a finite subset, [ilmath]S\subseteq V[/ilmath] of a vector space [ilmath](V,\mathbb{F})[/ilmath], where [ilmath]S=\{s_1,s_2,\ldots,s_n\}[/ilmath], we say the vectors in [ilmath]S[/ilmath] are linearly independent if:

• The only solution (of [ilmath]\lambda_1,\ldots,\lambda_n[/ilmath]) to $\sum^n_{i=1}s_i\lambda_i=0$ is [ilmath]\lambda_i = 0[/ilmath] [ilmath]\forall i\in\{1,2,\ldots,n\} [/ilmath] - this is sometimes described as the trivial solution[Note 2]

TODO: Find reference, I didn't bring any undergraduate Lin. Algebra books with me