Rank Nullity "Converse"? The rank nullity theorem requires a linear map $T:V \longrightarrow W$ between a finite dimensional domain VS and some VS W.
The conclusion of the theorem is that $\text{Dim}(V) = \text{Dim}(\text{Ker}(T))+\text{Dim}(\text{Ran}(T))$.
My question is, if I have a linear map $T: V \longrightarrow W$ between any two vector spaces and $\text{Dim}(\text{Ker}(T))+\text{Dim}(\text{Ran}(T)) < \infty$, then is $\text{Dim}(V) < \infty$ with $\text{Dim}(V) = \text{Dim}(\text{Ker}(T))+\text{Dim}(\text{Ran}(T))$?
Apologies if this is blatantly true or false.
Thank you.

Suppose $\dim{V} = \infty$, then we can write $V = \ker{T} \oplus U$, where $U$ is an infinite dimensional subspace of $V$. Let $\{u_i\}_{i \in I}$ be a Hamel basis for $U$, then $\{T(u_i)\}$ would be a Hamel basis for $\operatorname{Ran}{T}$. This means $\operatorname{Ran}{T}$ would need to be infinite dimensional, which it isn't. Hence, we have a contradiction, and $\dim V < \infty$.
