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 $\ds \sum_{j \mathop = 1}^m \mu_j b_j = 0$.
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:
- $\ds 0 = \sum_{j \mathop = 1}^m \mu_j b_j = \sum_{k \mathop = 1}^n \lambda_k a_k$
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.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Vector Spaces: $\S 33$. Definition of a Basis