TITLE: Is correct this way to prove that $\dim(V)<\infty$?
QUESTION [1 upvotes]: Let be $F$ a field, $V$ and $W$ are vector spaces over $F$. Prove that
if it exist a linear tranformation $T:V\rightarrow W$ with
$\dim(\operatorname{Im}(T))$ and $\dim(\operatorname{Ker}(T))$ both finite, then $\dim(V)$ is finite.

My attempt:
Since we know that $\dim(\operatorname{Ker}(T))<\infty$, there exist a finite basis $B_0 \subset V$ of $\operatorname{Ker}(T)$
Also, there exists $B$ a basis of $V$.
On  another side, let be $x \in \operatorname{Im}(T)$, then $x=T(y)$ for some $y \in V$. We have to note that $y$ is not a linear combination of elements from $B_{0}$, since $T(B_{0})=0$.
Then, $y$ is a linear combination of elements of $B \setminus B_{0}$.
\begin{align}
&\Rightarrow x \text{ is a linear combination of elements from } T(B \setminus B_{0}) \text{ for every } x \in Im(T) \\
&\Rightarrow T(B \setminus B_{0}) \text{ is a basis of } Im(T)\\
&\Rightarrow  T(B \setminus B_{0}) \text{ is finite }\\
&\Rightarrow B \setminus B_{0} \text{ is finite } \\
&\Rightarrow  B \text{ is finite}\\
&\Rightarrow  \dim(V)<\infty
\end{align}
Is my proof correct? In case that it is not, please help me to write it fine. I really appreciate your help.

REPLY [2 votes]: Hint Take a finite basis $B_0$ of $\text{Ker} T$ and a finite basis $Y$ of $\text{Im} T$. Use this  to create a basis $B_0 \cup X$ of $V$.
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