Proof: Span of Vectors after a Surjective, Linear Transformation Let $T: \mathbb{R}^m \to \mathbb{R}^n$ be a linear transformation and let $S=\lbrace v_1, v_2,..., v_k\rbrace$ be in $\mathbb{R}^m$ where $S$ spans $\mathbb{R}^m$. If $T$ is surjective, then prove that $\lbrace T(v1), T(v2),..., T(v_k)\rbrace$ spans $\mathbb{R}^n$.
I'm not really sure how to go about proving this.
 A: Choose a vector $v\in\mathbb{R}^{n}$. Since $T$ is surjective, we can find $w\in\mathbb{R}^{m}$ such that $T(w)=v$. Now, since $S$ spans $\mathbb{R}^{m}$, it holds that
$$w=\sum_{i=1}^{k}\lambda_{i}v_{i}$$
for some $\lambda_{1},\ldots,\lambda_{k}\in\mathbb{R}$. Using linearity of $T$, we get
$$v=T(w)=T\left(\sum_{i=1}^{k}\lambda_{i}v_{i}\right)=\sum_{i=1}^{k}\lambda_{i}T(v_{i}).$$
So $v$ can be written as a linear combination of $T(v_{1}),\ldots,T(v_{k})$, which proves the statement.
A: The intuition behind this problem is this:
$T$ is surjective, by definition means $|Im(R^m)|$=$|R^n|$. (The Co-Domain and the Image of $R^m$ by $T$ have the same order). And since $S$ spans $R^m$ we have  $R^m=S$ $\rightarrow$ $|Im(S)|$=$|R^n|$ $\rightarrow$ $Im(S)$ spans $R^n$ $\rightarrow$ $T(S)$ spans $R^n$ $\rightarrow$ $T(v_i)$ $1$ $\le$ i $\le$ $k$ spans $R^n$.
For the proof, you have to show that {T(S)} spans $R^n$. So you have to show for any vector $w$ in $R^n$, you can write it as a linear combination of $\{T(v_1),T(v_2),...,T(v_k)\}$.
You have the following: $S$ contains $k$ vectors (of which $n$ are linearly independent, you don't need to know this, it's just so you can start to smell the Rank-Nullity theorem approaching). Which means any vector $v_i$  $1$ $\le$ i $\le$ $k$ in $S$ can be written as a linear combination of all its vectors. 
$T$ is surjective, that means that every vector in $R^n$ has a pre-image in $R^m$, (these same pre-images can be written as a linear combination of all vectors $v_i$ in $R^m$).
So when you take the transformation of any vector $v$ in $R^m$, you are also taking the linear combination of all vectors $v_i$ that give $v$. So for all $v$ in $R^n$ we have:
$v=x_1*v_1+x_2*v_2+...x_k*v_k$ and let $T(v)=w$ with $w$ in $R^m$
$T(v)=T(x_1*v_1+x_2*v_2+...x_k*v_k)$ by linear combination.
$T(v)=T(x_1*v_1)+T(x_2*v_2)+...+T(x_k*v_k)$ by the fact that $T$ is linear.
$T(v)=(x_1+x_2+...+x_k)*T(v_1)+T(v_2)+...+T(v_k)=w$
We just proved that $w$ can be written as a combination of $\{T(v_1),T(v_2),...,T(v_k)\}$. And as a result, $\{T(v_1),T(v_2),...,T(v_k)\}$ spans $R^n$.
