If $T:V \to W$ is such that both $\ker(T)$ and $\operatorname{Im}(T)$ are finite-dimensional, then $V$ is finite-dimensional Let $V,W$ be vector spaces. Prove that if there exists a linear transformation $T:V \to W$ such that both $\ker(T)$ and $\operatorname{Im}(T)$ are finite-dimensional then $V$ is finite-dimensional as well.
I'm not sure how to prove this. My first intuition was to use the dimension theorem, but I can't because it requires that the domain is finite dimensional, and that's what I want to prove. 

Pick $\{v_1,\dots,v_n\}$ such that $\{T(v_1),\dots,T(v_n)\}$ is a spanning set for $\operatorname{Im}(T)$ and prove that, if $\{u_1,\dots,u_m\}$ is a spanning set for $\ker(T)$, then
$$
\{u_1,\dots,u_m,v_1,\dots,v_n\}
$$
is a spanning set for $V$.
Hint: take $v\in V$; then $T(v)=\sum_{i=1}^n\beta_iT(v_i)$ and, if we set $v'=\sum_{i=1}^n\beta_iv_i$, we have $T(v)=T(v')$, so $v-v'\in\ker(T)$.
