Showing V is finite dimensional if there is a linear map $T$ on $V$ such that $\ker T$ and range $T$ are finite dimensional. There are a couple other answers for this question here, but I'm actually looking for a proof verification (or really a foot foward for my own take in a different direction as the others).
My proof (so far):
Choose a basis $u_1,\dots,u_n$ of range $T$. Then for $v\in V$, we have $Tv=\lambda_1u_1+\dots+\lambda_n u_n$. Since $Tv\in$ range $T\subseteq V$, we have that $Tv$, an arbitrary element in $V$, is written in terms of a basis of range $T$, which is finite.
Assuming this is correct, what does it tell us about $\dim V$? A better question might be: is it possible to move on in the right direction from here?
My tired self wants to say that we would have $V=$ range $T$ (implying $\dim \ker T=0$ i.e. $T$ injective?), but I also feel like that's... wrong, to say the least.
 A: It is possible to move on in the correct direction, if you adjust things slightly.
Choose a basis $Tu_1,\ldots, Tu_n$ of the range, $T$. Then for $v\in V$ we have $Tv=\lambda_1Tu_1+\cdots + \lambda_nTu_n=T(\lambda_1u_1+\cdots+\lambda_nu_n)$.
Therefore $Tv-T(\lambda_1u_1+\cdots+\lambda_nu_n) = T(v-\sum_i \lambda_iu_i)= 0$.
And so $(v-\sum_i\lambda_iu_i)\in\ker T$. Can you finish it from here?
A: Let $\beta=\{u_1,u_2,\ldots,u_m\}$ be a basis for ${\rm range}(T)$ and 
$\gamma=\{v_1,v_2,\ldots,v_n\}$ be a basis for ${\rm ker}(T)$. Then we may write
$u_i=Tw_i$ for some $w_i\in V$, where $1\le i\le m$. Now, it suffices to claim that
\begin{align}
\alpha=\{v_1,v_2,\ldots,v_n,w_1,w_2,\ldots,w_m\}
\end{align}
is a basis for $V$.


*

*Given $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_m$ be scalars such that
\begin{align}
\sum_{i=1}^na_iv_i+\sum_{j=1}^mb_jw_j={\it 0}.\tag{1}
\end{align}
Then
\begin{align}
{\it 0}
&=T\left(\sum_{i=1}^na_iv_i+\sum_{j=1}^mb_jw_j\right)\\
&=\sum_{i=1}^na_iTv_i+\sum_{j=1}^mb_jTw_j\\
&=\sum_{j=1}^mb_ju_j,
\end{align}
which follows that $b_j=0$ for each $j$ because $\beta$ is a basis for ${\rm range}(T)$. Thus equation $(1)$ becomes $\sum_{i=1}^na_iv_i={\it 0}$,
but since $\gamma$ is a basis for ${\rm ker}(T)$, we must have $a_i=0$ for each $i$. Hence $\alpha$ is linearly independent.

*Given $v\in V$, then $Tv\in{\rm range}(T)$ and we can write 
$Tv=\sum_{i=1}^m\lambda_iTw_i=T\left(\sum_{i=1}^m\lambda_iw_i\right)$
for some scalars $\lambda_1,\lambda_2,\ldots,\lambda_m$. It follows that
$$T\left(v-\sum_{i=1}^m\lambda_iw_i\right)=Tv-T\left(\sum_{i=1}^m\lambda_iw_i\right)={\it 0},$$
that is, $v-\sum_{i=1}^m\lambda_iw_i\in{\rm ker}(T)$. Thus we can write 
$v-\sum_{i=1}^m\lambda_iw_i=\sum_{j=1}^n\mu_jv_j$ for some scalars
$\mu_1,\mu_2,\ldots,\mu_n$. That is, 
$$v=\sum_{i=1}^m\lambda_iw_i+\sum_{j=1}^n\mu_jv_j.$$
Hence $\alpha$ generates $V$.


By checking the above two steps, we conclude that $\alpha$ is a basis for $V$, and therefore $V$ is finite-dimensional.
