Constructing an Isomorphism from $\operatorname{Hom}(V,W)$ to $M_{m\times n}(F)$. Prove that $\operatorname{Hom}(V,W)$ and $M_{m \times n}(F)$ are isomorphic by constructing an isomorphism $\alpha:\operatorname{Hom}(V,W) \rightarrow M_{m \times n}(F)$. (Also given that $\dim(V)=n$ and $\dim(W)=m$)
Question, when prompt says construct, does that mean give an explicit formula for $\alpha$? I'm not exactly sure how to proceed, by a theorem in my textbook, I know they are isomorphic. but how to create such a formula? 
 A: Choose basis $\,A\,,\,B\,$ in $\,V\,,\,W\,$ , resp., and let $\,[T]_A^B\,$ denote the matrix representation of a linear transformation $\,T:V\to W\,$ wrt the basis $\,A\,,\,B\,$ , then you can define:
$$\alpha(T):=[T]_A^B$$
It is one of the nicest, most important exercises in basic linear algebra to prove the above is a vector spaces isomorphism. For this, of course, you must know well some basic properties of matrix representations of linear transformations.
BTW, the above isomorphism is not only of vector spaces but in fact also of algebras = vector spaces with a structure of rings, too (The formal definition of algebra is more involved but to begin with the above suffices)
A: You have the dimensions of the two spaces.  That tells you that bases exist: $\{v_1,\ldots,v_n\}$, $\{w_1,\ldots,w_m\}$.  Suppose $T\in\operatorname{Hom}(V,W)$.  Then $T(v_j) = \text{some linear combination of }w_1,\ldots,w_m$.  The coefficients in the linear combination become entries in a matrix that belongs to $M_{m\times n}(F)$.
Later note inspired by "Don Antonio"'s comments: Those coefficients are a column of that matrix, not a row.
A: Let $\{e_i\}$ be a basis for $V$ and $\{f_j\}$ be a basis for $W$. Then define the map $T_{ij}\in \operatorname{Hom}(V,W)$ to be the map that takes $\sum a_ke_k$ to $a_if_j$. Then you can define $\alpha: \operatorname{Hom}(V,W) \to M_{m\times n}$ by setting $\alpha(T_{ij})$ equal to the matrix with a $1$ in the $ji$th entry and zero everywhere else.
Note: There are several details to check...
