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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!

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  • $\begingroup$ What do you mean by list? Is it a set of two vectors $s, t$? $\endgroup$
    – Hendrra
    Commented Mar 12, 2018 at 22:59
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    $\begingroup$ Yes it's the set of the two vectors s and t, but the order of the vectors matter. $\endgroup$
    – user508748
    Commented Mar 12, 2018 at 23:03

1 Answer 1

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$(\rightarrow)$ 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.

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