$T$ is linear. Show: $T$ is onto if and only if that $T$ maps spanning sets to spanning sets Prove that $T$ is surjective if and only if for every set of vectors $\{v_1,...v_k\}$ that span $\mathbb{R}^m$, the set $\{T(v_1)...T(v_k)\}$ spans $\mathbb{R}^n$.
How would I prove this?
a surjective function is a function whose codomain and image are equal. A span of a trasformation is the combination of subspaces.
Should I find the span of the first set, then the second set to prove the first part of the statement? 
I don't know how to do the second part.

Let $T:\Bbb R^m\to\Bbb R^n$ be a surjective linear map and let $v_1,\dotsc,v_k\in\Bbb R^m$ span $\Bbb R^m$. 
To show that $\{T(v_1),\dotsc,T(v_k)\}$ spans $\Bbb R^n$, let $w\in\Bbb R^n$. Since $T$ is surjective there exists a $u\in\Bbb R^m$ such that $T(u)=w$. Since $\{v_1,\dotsc,v_k\}$ spans $\Bbb R^m$ there exist $\lambda_1,\dotsc,\lambda_k\in\Bbb R$ such that
$$
\lambda_1\, v_1+\dotsb+\lambda_k\,v_k=u
$$
It follows that
$$
\lambda_1\, T(v_1)+\dotsb+\lambda_k\, T(v_k)= w
$$
Hence $\{T(v_1),\dotsc,T(v_k)\}$ spans $\Bbb R^n$.
