It sounds like you're trying to show that, given a finite dimensional vector space $V$ over a field $F$, the space $S$ of linear transformations $T: V \to V$ is finite dimensional.
Here's a suggestion for how to proceed: First, show that $S$ is a vector space (if you haven't already). You can do this straight from the definition. (Show that if $T, U \in S$ and $\alpha, \beta \in F$ then $\alpha T + \beta U \in S$, which is to say that $(\alpha T + \beta U)(v) = \alpha T(v) + \beta U(v)$ is a linear transformation.)
Let $n = \dim V \in \mathbb{N}$. Fixing a basis allows us to identify $V$ with $F^n$ and $S$ with the set of all $n\times n$ matrices with entries in $F$. If you think about it, there should be a clear choice for a basis of this space $S$ over $F$ now.
If you're not seeing it, imagine taking the rows of an $n\times n$ matrix and laying them out in a line. Note you can do this for any $n\times n$ matrix, and so this allows you to think of $S$ as being essentially the same vector space as $F^{n^2}$. You now have an element of $F^{n^2}$. What's a basis for $F^{n^2}$? Relate this back to a basis for $S$. What's the order of the basis you found? It should be a function of $n$.
Ah. I just saw your edit. Note you're trying to find the dimension of $\dim S$, and so it's perfectly fine to set $\dim V = n$. In fact, you should expect that $\dim S$ is going to be a function of $\dim V$, and so this is the natural thing to do.
The rank formula also holds in infinite dimensions, whether you use cardinal arithmetic for the dimensions, or just say $\infty + n = \infty$, and $\infty + \infty = \infty$ (but one should use cardinal arithmetic). The proof is basically the same as in the finite-dimensional case, you choose a basis $\mathcal{B}_1$ of $\ker T$, a basis $\mathcal{B}_2$ of $\operatorname{im} T$, let $\mathcal{B}_3$ consist of preimages of the elements of $\mathcal{B}_2$ (choose one preimage per element), then $\mathcal{B}_1 \cup \mathcal{B}_3$ is a basis of $V$. In the infinite-dimensional case, some form of the axiom of choice is required, while the finite-dimensional case can be proved without that.
Best Answer
The kernel of a non-zero linear form on an $n$-dimensional vector space is of dimension $n-1$ by the rank theorem. Hence both your kernels have the same dimension. As one kernel is included in the other they are equal. So the quotient zero.