If a set of vectors in $\mathbb{R}^n$ is linearly independant, then does that mean the set spans $\mathbb{R}^n$? If we have a set of vectors in $\mathbb{R}^3$ for example, and they are linearly independant that means the set does not lie in a plane nor a line, so it must span $\mathbb{R}^3$. Alternatively if the same set is linearly dependant then at most, the set will span $\mathbb{R}^2$. Is my intuition correct?
I know how to show a weather a set of vectors in $\mathbb{R}^n$ are linearly independant or dependant but I'm not sure how to show that they indeed span $\mathbb{R}^n$. Could I get some help understanding?
Thanks
 A: No, a set of linearly independent vectors of $\mathbb{R}^n$ need not generate all of $\mathbb{R}^n$: consider, as mentioned in the comments, $\operatorname{span}(\{e_1, e_2 \})$: it's clearly $2$-dimensional and doesn't generate all of $\mathbb{R}^3$. On the other hand, a set of $n$ linearly independent vectors $\{v_1, \cdots, v_n\}$ of $\mathbb{R}^n$ must necessarily span all of $\mathbb{R}^n$. Otherwise, there would exist $v \in \mathbb{R}^n$ such that $\{v_1, \cdots, v_n, v\}$ is a set of $n + 1$ linearly independent vectors, which cannot happen. And no, if a set  $\{v_1, \cdots, v_m \}$ is linearly dependent, then it can generate any subspace of $\mathbb{R}^n$ you desire: this can easily be seen by taking the vectors $e_1, \cdots, e_n, 2e_1$, for example - it's a linearly dependent set but it generates all of $\mathbb{R}^n$. What you need to look at in that case is $\dim(\operatorname{span}(\{v_1, \cdots, v_m \}))$.
A: The simplest example I can think of is singleton sets.
Consider a vector $e \ne 0$ in $\mathbb{R}^n$. Then it is easy to see that $\{e\}$ is linearly independent (by definition) but obviously it does not span $\mathbb{R}^n$ for $n >1$
