So the question goes like this:
Let $S$ be a vector space over a field $F$.
Let $s,t \in S$. Prove that only when $s$ is non-zero and $t$ is not a scalar multiple of $s$ that the list $[s,t]$ is linearly independent.
Would I be right in thinking that I need to reword the list $[s,t]$ in accordance with the $2$ requirements stated at the end of the question and from there go and show that that is linearly independent? If so what I'm struggling with is understanding what to do with the piece of information "$t$ is not a scalar multiple of $s$".
Sorry I'm quite new to Linear Algebra, any help would be much appreciated!
$(\rightarrow)$ that implication is trivial. If $s =0$ then of course $s = t \cdot0 = 0 = s$. The same with a scalar multiplication; if $s = \alpha t$ for $\alpha \in F$ we would have $s = \alpha t \rightarrow s-\alpha t = 0$ so $s, t$ would be linearly dependent.
$(\leftarrow)$ suppose that $s \neq 0$ and there is not such $\alpha \in F: s = \alpha t$. Then we won't find any $\alpha_1, \alpha_2 \in F$ such that: $\alpha_1 s + \alpha_2 t = 0 \rightarrow s, t$ are linearly independent.