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 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:
\[\exists v \in S : v \in \langle S \setdif \{v\} \rangle.\]