# Linear independence

## Contents

## 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^{[1]}:

- 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:

- [math]\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][/math]
^{[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 [math]\sum^n_{i=1}s_i\lambda_i=0[/math] 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

## See also

## Notes

- ↑ Note that the use of [ilmath]\subseteq[/ilmath] just means we don't require a proper subset, in fact as the vector space contains a zero any linearly independent set
*must*be a proper subset, as to have the zero vector allows the scalar coefficient of it to be arbitrary! - ↑ As if you need to find some [ilmath]\lambda_i[/ilmath]s such that the summation is zero, it is trivial to choose [ilmath]\lambda_i=0[/ilmath] for all [ilmath]i[/ilmath]

## References

- ↑ Topology & Manifolds

## Corrections applied to references

- ↑ Corrected from
*Tensors and manifolds*which claimed that:- If for all finite sums [ilmath]\sum_i a_iv_i[/ilmath] with [ilmath]v_i\in S[/ilmath] we have that [ilmath]\sum_i a_iv_i=0\ \implies\ a_i=0\forall i[/ilmath]

- The [ilmath]v_i[/ilmath] in the sum be distinct. Suppose that [ilmath]s\in S[/ilmath], then [ilmath]-3s+3s=0[/ilmath], but [ilmath]3\ne 0[/ilmath] and [ilmath]-3\ne 0[/ilmath] thus with this definition only the empty set is linearly independent. It is clear it wanted to say they were distinct.