# Definition:Linearly Independent/Sequence

## Definition

Let $G$ be an abelian group whose identity is $e$.

Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $\left({G, +_G, \circ}\right)_R$ be a unitary $R$-module.

Let $\left \langle {a_n} \right \rangle$ be a sequence of elements of $G$ such that:

$\displaystyle \forall \left \langle {\lambda_n} \right \rangle \subseteq R: \sum_{k \mathop = 1}^n \lambda_k \circ a_k = e \implies \lambda_1 = \lambda_2 = \cdots = \lambda_n = 0_R$

That is, the only way to make $e$ with a linear combination of $\left \langle {a_n} \right \rangle$ is by making all the elements of $\left \langle {\lambda_n} \right \rangle$ equal to $0_R$.

Such a sequence is linearly independent.

### Linearly Independent Sequence on a Real Vector Space

Let $\left({\R^n, +, \cdot}\right)_\R$ be a real vector space.

Let $\left\langle{\mathbf v_n}\right\rangle$ be a sequence of vectors in $\R^n$.

Then $\left\langle{\mathbf v_n}\right\rangle$ is linearly independent if and only if:

$\displaystyle \forall \left\langle{\lambda_n}\right\rangle \subseteq \R: \sum_{k \mathop = 1}^n \lambda_k \mathbf v_k = \mathbf 0 \implies \lambda_1 = \lambda_2 = \cdots = \lambda_n = 0$

where $\mathbf 0 \in \R^n$ is the zero vector and $0 \in \R$ is the zero scalar.