# Subset of Linearly Independent Set is Linearly Independent

## Theorem

A subset of a linearly independent set is also linearly independent.

## Proof

Let $G$ be an unitary $R$-module.

Then $\sequence {a_n}$ is a linearly independent sequence if and only if $\set {a_1, a_2, \ldots, a_n}$ is a linearly independent set of $G$.

So suppose that $\set {a_1, a_2, \ldots, a_n}$ is a linearly independent set of $G$.

Then clearly $\sequence {a_n}$ is a linearly independent sequence of $G$.

Conversely, let $\sequence {a_n}$ be a linearly independent sequence of $G$.

Let $\sequence {b_m}$ be a sequence of distinct terms of $\set {a_1, a_2, \ldots, a_n}$.

Let $\sequence {\mu_m}$ be a sequence of scalars such that $\displaystyle \sum_{j \mathop = 1}^m \mu_j b_j = 0$.

For each $k \in \closedint 1 n$, let:

$\lambda_k = \begin{cases} \mu_j & : j \text { is the unique index such that } a_k = b_j \\ 0 & : a_k \notin \set {b_1, b_2, \ldots, b_m} \end{cases}$

Then:

$\displaystyle 0 = \sum_{j \mathop = 1}^m \mu_j b_j = \sum_{k \mathop = 1}^n \lambda_k a_k$

Thus:

$\forall k \in \closedint 1 n: \lambda_k = 0$

As $\set {\mu_1, \ldots, \mu_m} \subseteq \set {\lambda_1, \ldots, \lambda_n}$, it follows that:

$\forall j \in \closedint 1 m: \mu_j = 0$

and so $\sequence {b_m}$ has been shown to be a linearly independent sequence.

Hence the result.

$\blacksquare$