Prove: If $T:V \rightarrow W$ is a linear map, and $v_1, v_2, …, v_n$ span $V$, then $T(v_1), T(v_2), …, T(v_n)$ span $\operatorname{range}(T)$.
This is what I have so far…
Proof: For any Linear map $T:V \rightarrow W$, then $\dim(V)=dim \operatorname{null}(T) + \dim \operatorname{range}(T)$ by definition of the fundamental theorem of linear maps.
So if $v_1, v_2, …, v_n$ span $V$, then $\dim(V)=n$.
Furthermore, if $T(v_1), T(v_2), …, T(v_n)$ span $\operatorname{range}(T)$, then $\dim \operatorname{range}(T) = n$.
Therefore it follows that $\dim \operatorname{null}(T) =0$, for T: to be a linear map.
Am I on the right track?
Best Answer
I would make it simpler.
Supposing that $T(v_1), T(v_2), ..., T(v_n)$ span $\operatorname{range}(T)$, take $y \in \operatorname{range}(T)$. By definition, there exists $x\in V$ such that $T(x)=y$.
As $v_1,v_2,\dots, v_n$ span $V$, there exists $\alpha_1, \dots, \alpha_n$ such that $$x = \alpha_1 v_1+ \dots + \alpha_n v_n.$$
Then
$$T(x)=y=\alpha_1 T(v_1)+ \dots \alpha_nT(v_n),$$
allowing us to conclude.