Linear independence

Definition

Given an arbitrary subset, $S$ of a vector space $(V,\mathbb{F})$ (that is to say $S\subseteq V$) if:

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

This can be stated more concretely as:

• $$\forall \{v_\alpha\}_{\alpha\in I}\in\mathcal{P}(S)\left[\left(n:=\big\vert\{v_\alpha\}_{\alpha\in I}\big\vert\right)\in\mathbb{N}\implies\left[\left(\sum^n_{i=1}\lambda_i v_i=0\right)\implies\left(\lambda_i=0\forall i\in I\right)\right]\right]$$ (with the slight abuse of notation that once $\{v_\alpha\}_{\alpha\in I}$ is known to be finite we can index it by integers from $1$ to $n$)

  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 $S$ is a linearly independent set of vectors

Equivalent condition

The statement that:

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

Is the same as:

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

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TODO: Document sums and directness

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For a finite $S$

When $S\subseteq V$ is finite the definition becomes so much nicer it is worth mentioning.

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

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

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

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See also

• Basis
• Span
• Addition of vector spaces