Is the span of any 3 n-dimensional, linearly independent vectors, $\mathbb R^3$? I am not sure if the span of a set of any 3 (n-dimensional) vectors that are linearly independent is  $\mathbb R^3$.
I think, since any two (2 dimensional vectors) that are independent always span $\mathbb R^2$, that if the dimension of those vectors is 3 or more,they should span a space that looks like  $\mathbb R^3 $, since 2 3D vectors that are independent span a plane. 

Any two $n$-dimensional vector spaces (over the same field ) are isomorphic. ..  It would be easy to write down an isomorphism:  just send basis vectors to basis vectors. ..  i.e. $\mathcal i:V\to W$ by $\mathcal i (v_i)=w_i $ where $\{v_1, \dots, v_n\} $ and $\{w_1, \dots,v_n\} $are bases and extend linearly. ..
Secondly,  the span of  $3$ linearly independent vectors is a $3$-dimensional vector space...
Therefore,  any $n $-dimensional vector space (over $\mathbb R  $) is isomorphic to  $\mathbb R^n $.  In particular, the span of  $3$ linearly independent vectors (over $\mathbb R $)is isomorphic to  $\mathbb R^3$.
