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. 

Suppose $V$ is infinite dimensional. Let $\{u_1, \dots, u_m\}$ be a basis for $\ker{(T)}$. Then we can extend $\{u_1,\dots ,u_m\}$ to a linearly independent set $\{u_1,\dots ,u_m,v_1,\dots , v_n\}$, with $n> \dim \text {Im}(T)$. The image of the span of $\{v_1, \dots ,v_n\}$ has dimension $n,$ contradiction.
