I'm trying to prove the following question with a different path,
Suppose there exists a linear map on V whose null space and range are both finite-dimensional, prove that V is finite-dimensional.
Now, instead of trying to build a basis from null space and rang T, I want to say that since V is infinite-dimensional, I can construct a list of linearity independent vectors of any length k>0. Specifically, $k>\dim $ range$ T =n$, now suppose $v_1,…,v_k$ are such linearity independent vectors, then $Tv_1,\dots, Tv_k$ are linearly independent in the range, but this is a contradiction since we can't have a length of independent list bigger than the dimension of the space.
Best Answer
I found the problem interesting and I think I found a proof.
Firstly, I have been told that this theorem only holds for vector spaces over the same field. So the question becomes the following.
Let $V$ and $W$ be vector spaces defined over the same field $F$. Let $T : V\to W$ be a linear map. Suppose that range$T$ and null$T$ are finite dimensional. Prove that $V$ is finite dimensional.
Proof: Given that range$T$ is finite dimensional, there exists a basis of range$T$. Let $Tv_1,. . .,Tv_n$ be the basis of range$T$. Assuming that the function $T$ is not a partial function. We see that for every $v\in V$, $Tv\in$ range$T$. Because $Tv_1,. . .,Tv_n$ is a basis, there exists scalars $a_1,. . .,a_n$ such that $Tv=a_1Tv_1+. . .+a_nTv_n$. Subtracting $Tv$ from both sides and using the linearity of $T$ we get $T(a_1v_1+. . .+a_nv_n-v)=0$.
Given that null$T$ is finite dimensional. Let $w_1,. . .,w_m$ be a basis of null$T$. Then, $a_1v_a+. . .+a_nv_n-v=c_1w_1+. . .+c_mw_m$ $\implies$ $a_1v_a+. . .+a_nv_n-c_1w_1+. . .+c_mw_m=v$.
The list $v_1,. . .,v_n,w_1,. . .,w_m$ spans $V$ because $v$ was an arbitrary vector. Hence, $V$ is finite dimensional.
Note that this result is also consistent with the rank-nullity theorem. Note that you cannot use the rank nullity theorem to prove this as the rank nullity theorem already assumes that $V$ is finite dimensional.
I did post this proof as another question. Although I think the proof is correct. If someone finds a mistake, I will update you on it.