An alternative, basis free, proof which perhaps makes use of quotient spaces? Consider the following statement:

Let $V$ be a finite-dimensional vector space over a field $\Bbb{F}$, and let $W$ be a subspace of $V$. Let $E = \{T \in \mathcal{L}(V)| \,\, W \subseteq \ker(T)\}$. Then, $E$ is a subspace of $\mathcal{L}(V)$ and $\dim E = \dim V \left( \dim V - \dim W\right)$.

I already have a proof of this fact: Proving $E$ is a subspace is easy. Next, we simply pick a basis for $W$ and then extend it to a basis for $V$. This basis for $V$ provides us with an isomorphism of $\mathcal{L}(V)$ with the set of $(\dim V \times \dim V)$ matrices. The image of $E$ under this isomorphism is the set of block matrices of the form $\begin{pmatrix} 0\, |\, A\end{pmatrix}$ (with $0, A$ being block matrices of appropriate sizes). From this, $\dim E$ is simply the number of matrix elements in $A$, which is a simple counting exercise.
My question is whether there is a basis-free proof of this fact, perhaps using quotient spaces or some other argument. I think that quotient spaces might be helpful, because my intuitive understanding is that quotient spaces "collapse/ignore" part of a vector space to get a smaller one; and in the proof above, we're pretty much ignoring the "first block" of zero matrices to figure out the dimension. 
Obviously, in this finite-dimensional case, $E$ and $V \times (V/W)$, have the same dimension, and hence are isomorphic, but I haven't been able to construct a canonical isomorphism. Any insight is appreciated.
 A: I would argue as follows: such a map $T$ necessarily induces a unique well-defined map $\widetilde{T}: V/W \longrightarrow V$ such that $\widetilde{T} \circ \pi = T$. Here $\pi: V \longrightarrow V/W$ is the natural projection. 
Indeed, if you denote by $[v]$ the elements of $V/W$, then you put
$$
\widetilde{T}([v]) = T(v) \ .
$$
This is well-defined, because if $[v] = [v']$, then $v - v' = w$, for some $w \in W$. Thus
$$
T(v) - T(v') = T(w) = 0  \qquad \Longleftrightarrow \qquad T(v) = T(v´) \ .
$$
So, no matter which representative you choose for $[v]$, you get the same value $T(v)$.
Hence you have a bijection $T \mapsto \widetilde{T}$ and $S \mapsto S \circ \pi$ between maps in $T \in E$ and maps $S\in\mathrm{Hom}(V/W , V)$. 
Indeed, if you start with some $T \in E$, produce your $\widetilde{T}$ and then go back $\widetilde{T} \circ \pi = T$, by definition of $\widetilde{T}$.
The other way around: start with some $S: V/W \longrightarrow V$, compose with $\pi$ and get $\widetilde{S\circ \pi}$. You necessarily have $\widetilde{S\circ \pi} = S$ (exercise  :-) ).
The dimension of the latter is $\mathrm{dim}(V/W) \times \mathrm{dim}(V)$.
EDIT. (Answering a question in the comments.) Let $T: V \longrightarrow U$ be a linear map. It doesn't matter if $V = U$, or not. Let $W \subset V$ be a subspace. You can always build a basis of $V$ as follows: take a basis $w_1, \dots, w_p$ of $W$ and then add the necessary vectors to get a basis of the whole space $V$: $w_1, \dots, w_p, v_{p+1}, \dots, v_n$. Take any basis $u_1, \dots, u_m$ of $U$. Then the matrix associated with $T$ in these bases has the form $(A | B)$, where the first $p$ columns $A$ are the coordinates of $Tw_1, \dots, Tw_p$ in the basis $u_1, \dots, u_m$ and the remaining $n-p$ columns $B$ are the coordinates of $Tv_{p+1}, \dots, Tv_n$. If $W \subset \mathrm{ker}T$, then the first $p$ columns are obviously zero, and your matrix has always the form $(0 | B)$. Submatrix $B$ is also the matrix of $\widetilde{T}: V/W \longrightarrow U$ in the bases $[v_{p+1}], \dots, [v_n]$ and $u_1, \dots, u_m$ of $V/W$ and $U$, respectively.
