There are several questions posted which are looking for solutions and hints. I was hoping that someone could verify my proof.
Let $V,W$ be vector spaces and $A:V\to W$ be a linear mapping. Suppose that $\dim \ker A = s$ and $\dim A(V) = t$. We need to show that $A$ is finite dimensional.
We assume the contrary that $V$ is not finite dimensional, hence we can find an infinite list of linearly independent vectors.
We have been given that $\dim \ker A = s$, thus, we can find a basis $\{ \alpha_1 , \ldots , \alpha_s \}$ for $\ker A$. Now, here's my proof strategy: I will extend the previous list to $\{ \alpha_1 , \ldots , \alpha_s , \alpha_{s+1}, \ldots , \alpha_{s+t} \}$ such that $\{ A(\alpha_{s+1}) , \ldots , A(\alpha_{s+t}) \}$ is a basis for $A(V)$ and this list $\{ \alpha_1 , \ldots , \alpha_s , \ldots , \alpha_{s+t} \}$ is linearly independent.
Let $\alpha_{s+1} \in V\setminus \text{Sp} \{ \alpha_1 , \ldots , \alpha_s \}$ be arbitrary. Now, we can pick $\alpha_{s+2} \in V\setminus \text{Sp} \{ \alpha_1 , \ldots , \alpha_s , \alpha_{s+1} \}$ such that $A(\alpha_{s+2}) \not\in \text{Sp} \{ A(\alpha_{s+1})\}$. For otherwise we would have that for all $\alpha_{s+2} \in V\setminus \text{Sp} \{ \alpha_1 , \ldots , \alpha_s , \alpha_{s+1} \}$, $A(\alpha_{s+2}) \in \text{Sp} \{ A(\alpha_{s+1})\}$ which implies that $A(V)= \text{Sp} \{ A(\alpha_{s+1})\}$ and which further would contradict our assumption that $\dim A(V) = t$. Again, we can pick $\alpha_{s+3} \in V\setminus \text{Sp} \{ \alpha_1 , \ldots , \alpha_s , \alpha_{s+1}, \alpha_{s+2} \}$ such that $A(\alpha_{s+2}) \not\in \text{Sp} \{ A(\alpha_{s+1}) ,A(\alpha_{s+2})\}$ (The argument for it is same as the previous one!). If we repeat this argument for $t$ times, we would have found vectors such that $\{ \alpha_1 , \ldots , \alpha_s , \alpha_{s+1}, \ldots , \alpha_{s+t} \}$ such that the vectors $\{ A(\alpha_{s+1}) , \ldots , A(\alpha_{s+t}) \}$ are linear independent and this list $\{ \alpha_1 , \ldots , \alpha_s , \ldots , \alpha_{s+t} \}$ is linearly independent. Since, we know that linearly independent vectors of right length is the basis, it must be that $\{ A(\alpha_{s+1}) , \ldots , A(\alpha_{s+t}) \}$ form a basis for $A(V)$.
Now, we bring about a contradiction. Pick a vector $v \in V \setminus \text{Sp} \{ \alpha_1 , \ldots , \alpha_s , \alpha_{s+1}, \ldots , \alpha_{s+t} \}$. $A(v) \in A(V)$, thus, $A(v)=c_1 A(\alpha_{s+1}) + \ldots + c_t A(\alpha_{s+t})$. By definition of linear mappings, we have $A(v-c_1 \alpha_{s+1}- \ldots - c_t \alpha_{s+t})=0$. It follows that $v-c_1 \alpha_{s+1}- \ldots - c_t \alpha_{s+t} \in \ker A$ thus $v-c_1 \alpha_{s+1}- \ldots - c_t \alpha_{s+t}=d_1 \alpha_1 + \ldots + d_s \alpha_s$ which is contradicts our assumption that $v \not \in \text{Sp} \{ \alpha_1 , \ldots , \alpha_s , \alpha_{s+1}, \ldots , \alpha_{s+t} \}$. Thus, $V$ must be finite dimensional.
Is this proof correct? I know it's a long proof. Thanks in advance!