# 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:

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

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.

### Set

Let $S \subseteq G$.

Then $S$ is a linearly independent set (over $R$) if and only if every finite sequence of distinct terms in $S$ is a linearly independent sequence.

That is, such that:

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

where $a_1, a_2, \ldots, a_k$ are distinct elements of $S$.

### Linearly Independent Set on a Real Vector Space

Let $\tuple {\R^n, +, \cdot}_\R$ be a real vector space.

Let $S \subseteq \R^n$.

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

That is, such that:

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

where $\mathbf v_1, \mathbf v_2, \ldots, \mathbf v_n$ are distinct elements of $S$.