Let $V$ be a vector space and $S\subseteq V$ be a subset.
$S$ is linearly independent if there is no $T\subset S$ such that
\[\langle T\rangle = \langle S\rangle.\]$S$ is linearly dependent if it is not linearly independent.
Let $V$ be a vector space over $F$ and
\[S = \{v_1,\cdots,v_n\} \subseteq V\]be a subset $(n\geq 1)$. $S$ is linearly independent if and only if for every $x\in V$ there are unique $\lambda_1$, $\cdots$, $\lambda_n\in F$ such that
\[\sum_{i=1}^n \lambda_i v_i = x.\](In other words,
\[f : F^n \to V, \,% (\lambda_1,\cdots,\lambda_n) \mapsto \sum_{i=1}^n \lambda_i v_i\]is a monomorphism.)
In particular, since
\[\sum_{i=1}^n 0\cdot v_i = 0,\]we have:
Let $V$ be a vector space over $F$ and
\[S = \{v_1,\cdots,v_n\} \subseteq V\]be a subset $(n\geq 1)$. If for all $\lambda_1$, $\cdots$, $\lambda_n\in F$,
\[\sum_{i=1}^n \lambda_i v_i = 0 \enspace\rimp\enspace \forall i\in\llbra n\rrbra : \lambda_i = 0,\]then $S$ is linearly independent.
Let $V$ be a vector space over $F$.
For any $v\in V$, $\{v\}$ is linearly independent if and only if $v\neq 0$.
For any finite $S\subseteq V$, if $0\in S$, then $S$ is linearly dependent.
For any finite $S\subseteq V$, $S$ is linearly dependent if and only if some vector from $S$ is a linearly combination of others:
\[\exists v\in S : v\in\langle S\setminus\{v\}\rangle.\]
Notation
- $\langle S\rangle$
- The linear span of $S$.
- $\llbra n\rrbra$
- $\{1,2,\cdots,n\}$.