Definition:Linearly Independent

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.

Sequence
Let $\left \langle {a_n} \right \rangle$ be a sequence of elements of $G$ such that:
 * $\displaystyle \forall \left \langle {\lambda_n} \right \rangle \subseteq R: \sum_{k=1}^n \lambda_k \circ a_k = e \implies \lambda_1 = \lambda_2 = \cdots = \lambda_n = 0_R$

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.

A sequence $\left \langle {a_n} \right \rangle \subseteq G$ which is not linearly independent is linearly dependent.

Real Vector Spaces
Suppose the space $\left({G, +_G, \circ}\right)_R$ in question is the real vector space $\left({\R^n,+,\cdot}\right)_{\R}$.

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 iff:


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

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

Set
Let $S \subseteq G$.

Then $S$ is a linearly independent set if every sequence of distinct terms in $S$ is a linearly independent sequence.

Otherwise $S$ is a linearly dependent set.

Also see

 * Linearly Dependent Sequence of Vector Space