Is showing that if two vectors in $\mathbb{R}^n$ are linearly independent, then they span $\mathbb{R}^n$? For example, if you show that 
$c_1 = c_2 = 0$
is the only solution to
$c_1\textbf{v}_1 + c_2\textbf{v}_2 = \textbf{0}$ 
Then you've shown that $v_1$ and $v_2$ are linearly independent and span $\mathbb{R}^n$ right?
 A: No. Two vectors linearly independents only span $\mathbb R^{2}$. But, if $n>2$, two vectors linearly independents can only span a subset of $\mathbb R^{n}$ because its basis has the same number of elements than its dimension.
A: No, all that shows is that $\{v_1, v_2\}$ is linearly independent. In order to show that a set $S$ spans a  vector space $V$, you need to show that any vector $v\in V$ can be written as a linear combination of elements of $S$: $$v=a_1s_1+a_2s_2+...+a_ns_n, \quad s_1,s_2, . . . , s_n\in S.$$ There are linearly independent sets that don't span (like $\{(0,0, 1), (0, 1, 0)\}$ in $\mathbb{R}^3$) and spanning sets that aren't linearly independent (like $\mathbb{R}^3$ in $\mathbb{R}^3$).
A: With only two vectors in $$\mathbb{R}^3$$ you can spann a plane if the vectors are linearly independent  or a line if they are parallel or a point if both of them are the zero vector.
You need n linearly independent vectors to spann $$\mathbb{R}^n$$
A: No. The reason is that no set of fewer vectors than the dimension of the vector space can be a spanning set for the vector space, just like no spanning set of vectors more than the dimension of the vector space can be linearly independent. 
