Linear independence of a set of vectors in $\mathbb{R}^n$

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There is a theorem that states: $$Let \space S=\{v_1,v_2,...v_r\}\space be \space a \space set \space of\space vectors \space in\space \mathbb{R}^n. \space If\space r>n,\space then\space S\space is\space linearly\space dependent.$$ My question is, does the 'other' case apply as well? By other case, I mean if $r<n$, does this mean the set $S$ is linearly independent or linearly dependent? No proofs please, just some intuition behind the answer if possible!

Thanks in advance!

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If $r\leqslant n$ then anything could happen.

For instance, if one of the vectors is $0$, or if two of the vectors are the same, then the set is linearly dependent no matter how small $r$ is.

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Not without more information. We can choose $r<n$ linearly independent vectors. But we can also choose them to be linearly dependent: we can choose them all from a subspace of dimension less than $r$, in which case they must be dependent.

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There's nothing to be said (without further information) about a set with less elements than the dimension of the space, except when the size of the set is exactly one (in which case it's trivially linearly independent iff it's not the zero subspace). Consider, for example, $v \in \mathbb{R}^k \setminus\{0\}$ with $k \geq 3$ and the set

$$ \{v,2v,3v,...,(k-1)v\} $$

which is clearly linearly dependent, but has less than $k$ elements. The intuition here is that you could easily choose vectors that belong to the subspace generated by the previously chosen, and if you choose more vectors than the dimension of the space, this property is guaranteed.

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Just from $r<n$, you cannot say whether the set S is linearly independent or linearly dependent. E.g., in $R^3$, two parallel vectors are linearly dependent, while two non-parallel vectors are linearly independent.