Definition:Linearly Independent/Sequence

From ProofWiki
Jump to navigation Jump to search


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 $\struct {G, +_G, \circ}_R$ be a unitary $R$-module.

Let $\sequence {a_n}$ be a sequence of elements of $G$ such that:

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

That is, the only way to make $e$ with a linear combination of $\sequence {a_n}$ is by making all the terms of $\sequence {\lambda_n}$ equal to $0_R$.

Such a sequence is linearly independent.

Linearly Independent Sequence on a Real Vector Space

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

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

Then $\sequence {\mathbf v_n}$ is linearly independent if and only if:

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

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

Also see