So, I'm trying to prove the following result:
The map $f:M(m \times n,F) \to Hom(F^n,F^m)$, which associates a linear map from $F^n$ to $F^m$ to a matrix from the set of $m \times n$ matrices with entries from a field F is an isomorphism of vector spaces.
Proof Attempt:
Now, we just have to prove that it is linear and that it is bijective. Let $A_1,A_2 \in M(m \times n,F)$. Then:
$f(A_1+A_2) = L_0:F^n \to F^m$
We have used the fact that every matrix has a linear map associated with it. In fact, the above means that $f$ is additively linear. So:
$f(A_1+A_2) = f(A_1)+f(A_2)$
Similarly, let $A \in M(m \times n, F)$ and $\lambda \in F$. Then:
$f(\lambda A) = L_1 : F^n \to F^m$
Since $L_1$ is a linear map, it follows that:
$f(\lambda A) = \lambda \cdot f(A)$
We, now, have to prove bijectivity. First, we prove that it is injective. Let $A_1,A_2 \in M(m \times n, F)$ such that:
$f(A_1) = f(A_2)$.
Now, this is going to be a linear map between $F^n$ and $F^m$. From a previously proven result, it is clear that every linear map has a unique matrix associated with it such that $x \to Ax$ for every $x \in F^n$. Hence:
$A_1 = A_2$
To prove surjectivity, we need to show that $f(M(m \times n, F)) = Hom(F^n,F^m)$. Let $L \in Hom(F^n,F^m)$. Since every linear map has a unique matrix associated with it, it suffices to say that there exists an $A \in M(m \times n, F)$ such that $f(A) = L$. However, that means that $L \in f(M(m \times n, F))$. This proves that $Hom(F^n,F^m) \subset f(M(m \times n,F))$ and, thus, establishes equality of the two sets.
That proves the desired result.
^Does my argument above work or no? What could I fix?
You haven't defined $f$. Hints:
$(1).\ $ Let $A=(a_{ij})_{1\le i\le n;\ 1\le j\le m}$ and choose bases $\{e_i\}_{1\le i\le n}$ and $\{e'_i\}_{1\le i\le m}$ for $F^n$ and $F^m$ respectively.
$(2).\ f$ will send $A$ to a $\textit{function}\ \phi: F^n\to F^m.$ So, to define $f$, you need to say how $\phi$ acts.
$(3).\ $But the action of $\phi$ is specified by defining it on the basis $\{e_i\}_{1\le i\le n}$ and extending by linearity (this way, you force $\phi$ to be linear), which means that $\phi$ is determined by $(\alpha_{ij})_{1\le i\le n;\ 1\le j\le m}$ such that $\phi(e_i)=\sum^m_{j=1} \alpha_{ij}e_j'.$
$(4).\ $ Make the obvious choice for the $\alpha_{ij}$, and $\textit{now}$ check that $f$ is linear.