Matrix representation of $\varphi$: Hom(V,W) $\to M_{m x n}$, where $\varphi (T) = [T]_{B}^{C}$ Suppose B = ($v_1, ..., v_n$) and C = ($w_1, ... ,w_m$) be ordered bases for V and W respectively. (Finite-dimensional)
I have come to understand that $\varphi$: Hom(V,W) $\to M_{m x n}$, where $\varphi(T) = [T]_{B}^{C}$ is an isomorphism.
I am wondering what a matrix representation of such an isomorphism might look like?
 A: If $T$ is an element of $\hom(V,W)$ one can prescribe
$$Tv_k=\sum_s T_{sk}w_s$$
and then assign $\varphi(T)=[T_{ij}]$.
This map will have
i) $\varphi(T+S)= \varphi(T)+\varphi(S)$,
ii) $\varphi(rT)=r\varphi(T)$ for each scalar $r$.
iii) trivial $\ker(\varphi)$ and
iv) for each matrix in $M_{m\times n}$ a corresponding linear transformation $V\to W$.
But for the matrix of $\varphi$ one constructs the basic linear transformations $V\to W$ (defined at the basis) as:
$$L_{rt}v_k=\delta_{rk}w_t$$
which generates $\hom(V,W)$ and then assign
$$\varphi(L_{rt})=[\delta^i_r\delta^j_t].$$
Here the matrices $[\delta^i_r\delta^j_t]=:E_{rt}$ have and $1$ at the entry $rt$ and zeroes elsewhere, these generate any element in $M_{m\times n}$.
So the $\varphi$'s matrix is constructed by observing something like that
$$L_{11}\mapsto E_{11},$$
$$L_{12}\mapsto E_{12},$$
$$...$$
$$L_{mn}\mapsto E_{mn}.$$
A: To do this, first we need two basis, one for the domain and one for the codomain.
Let $T_{ij}: V \to W$ be defined as $T_{ij}(v_i) = w_j$ and $T_{ij}(v_k) = 0$ if $k \neq i$ and let $\mathcal{B} = (T_{11}, ..., T_{mn})$. If $T$ is a linear transformation from $V$ to $W$, we know $T(v_i) = \alpha_{i1} w_1 + ... + \alpha_{im} w_m$, which is equal to $\alpha_{i1}T_{i1}(v_i) + ... + \alpha_{im}T_{im}(v_i) = \alpha_{11}T_{11}(v_i) + ... + \alpha_{nm}T_{nm}(v_i)$, since every term with $k \neq i$ is equal to $0$. Therefore, $\mathcal{B}$ generates the homomorphism space.
They are also Linearly independent: you can check that by applying an arbitrary combination on a $v_i$, which will result in a linear combination of the $w_j$ yielding 0.
The matrix representation of each $T_{ij}$ is a matrix containing $1$ in the $i,j$ position, and zeroes elsewhere.
For the base of the matrices, we consider $\mathcal{C} = (E_{11}, ..., E_{mn})$, where each $E_{ij}$ has a $1$ in the $i,j$ position, and zeroes elsewhere. But wait! Those are the matrix representations of $T_{ij}$!
So, $\varphi(T_{ij}) = E_{ij}$. Using the ordering as I have, the matrix representation of $\varphi$ is precisely the identity matrix, as it takes the $k$-th element of base $\mathcal{B}$ onto the $k$-th element of base $\mathcal{C}$
A: Here is a small example of representing the above transformation as a matrix:

