My question is not a duplicate. It is directly related to one answer to this question Category-Theoretic relation between Orbit-Stabilizer and Rank-Nullity Theorems but since the question was asked several years ago and an answer was accepted, I thought it was best to post a separate question requesting further clarification.
Here's the relevant quote from the answer in question:
The first thing to note is that a linear map $A:V\rightarrow V$ also gives a genuine group action: it is the additive group of $V$ acting on the set $V$ by addition. That is, any $v \in V$ acts on $x \in V$ as $v:x↦x+Av$.
Now we see that given any $x$ in $V$ the stabilizer subgroup $\text{stab}(x)$ of this action is precisely the kernel of $A$. The orbit of $x$ is $x$ plus the image of $A$.
If we are working with a vector space over a finite field, we can take the cardinality of these sets as in the formula $|\text{orb}(x)||\text{stab}(x)|=|G|$ and as @Ravi suggests, take the logarithm of this where the base is the size of the field and we get exactly the rank-nullity equation.
The last statement is what has left me puzzled. How do we prove that taking the said logarithm actually gives us the rank-nullity equation? I tried asking my professor about it but he couldn't work it out either. Any hints or clue will also be a great help.
$\newcommand{\im}{\operatorname{im}}$Fixing a finite field $F$ is crucial here. I am skeptical about representing the general rank-nullity theorem over, say, $\Bbb R$, via the orbit-stabiliser theorem.
A finite $F$-vector space has cardinality equal to $|F|^n$ where $n$ is the dimension of that space. To see this, just note any pair $(a_1,a_2,\cdots,a_n)$ uniquely determines an element of the vector space - and visa versa - once a basis is chosen by mapping to $a_1e_1+\cdots+a_ne_n$. This is basically a coordinates theorem.
With this in mind, $|\mathrm{orb}(x)|=|\im A|=|F|^{\dim\im A}$, $|\mathrm{stab}(x)|=|F|^{\dim\ker A}$ and $|G|=|V|=|F|^{\dim V}$ and the orbit-stabiliser equation is then: $$|F|^{\dim\im A}\cdot|F|^{\dim\ker A}=|F|^{\dim V}$$Taking logarithms (say) produces: $$(\dim\im A+\dim\ker A)\cdot\log|F|=\dim V\cdot\log|F|$$And we conclude: $$\dim\im A+\dim\ker A=\dim V$$