Proof help, span and invertible matrix 

Let $A$ be an $n \times n$ invertible matrix and $v_1,v_2,...,v_m$ an element of $\mathbb{R}^{n}$. Prove if $\{v_1,v_2,...,v_m\}$ spans $\mathbb{R}^{n}$ then $\{Av_1, Av_2,...,Av_m\}$ also spans $\mathbb{R}^{n}$. 


I'm not sure how to use the definition of spanning to prove this statement...do I have to get a homogeneous linear system?
 A: Let $v \in \mathbb{R}^n$ be arbitrary.  You want to show $v$ is contained in the spanning set of the vectors $Av_1, Av_2, \ldots, Av_m$.
Well, that means you want
$$
v = a_1 Av_1 + a_2 Av_2 + \cdots + a_m A v_m \tag{1}
$$
How might you find such $a_1, \ldots, a_m$?  Well, just multiply the above by $A^{-1}$, to get
$$
A^{-1} v = a_1 v_1 + a_2 v_2 + \cdots + a_m v_m \tag{2}
$$
Now, just find scalars $a_1, \ldots , a_m$ which satisfy (2).  Such scalars exist because $v_1, \ldots, v_m$ is a spanning set.  Then, apply the matrix $A$ to (2) to arrive at (1), and the desired statement is proved.
A: Your vectors spanning $\mathbb{R}^n$ means we can find $n$ of these vectors which form a basis for $\mathbb{R}^n$. Then if we note applying an invertible matrix is just a change of basis. So if we perform a change of basis on our basis which is a subset of our original vectors, we still have a basis for $\mathbb{R}^n$.
A: If you have an arbitrary $w\in \mathbf{R}^n$, then you can find $\lambda_1,\ldots,\lambda_n$ so that $\lambda_1v_1 + \cdots +\lambda_nv_n = w$ - that's what spanning means. Now you can apply $A$ to both sides of this equation - then what does the invertibility of $A$ tell you?
