Theorem 11, Section 3.4 of Hoffman’s Linear Algebra 
Let $V$ be an $n$-dimensional vector space over the field $F$ and $W$ be an $m$-dimensional vector space over the field $F$. Let $B=\{\alpha_1,…\alpha_n\}$ and $B’=\{\beta_1,…,\beta_m\}$ be ordered basis of $V$ and $W$, respectively. $\forall T\in L(V,W)$, $\exists A\in M_{m\times n}(F)$ such that $[T(\alpha)]_{B’}=A\cdot [\alpha]_B$, for all $\alpha \in V$. Furthermore, $f:L(V,W)\to M_{m\times n}(F)$ defined by $f(T)=A=[T]_{B}^{B’}$ is bijective map.

My attempt: First we need to check $f$ is a well-defined function, i.e. $\forall T\in L(V,W)$, $\exists !$ $A\in M_{m\times n}(F)$ such that $f(T)=A$. Let $C=(C_{ij})\in M_{m\times n}(F)$ such that $[T(\alpha)]_{B’}=C\cdot [\alpha]_B$. By uniqueness of coordinates matrix of $T(\alpha)$ relative to ordered basis $B’$, $\sum_{j=1}^n A_{ij}x_j$ $=\sum_{j=1}^n C_{ij}x_j$, for all $i\in J_m$, where $(x_1,…,x_n)\in F^n$ is coordinate of $\alpha$ relative to ordered basis $B$. How to show $A_{ij}=C_{ij}$, $\forall (i,j)\in J_m\times J_n$?
There are two ways to show $f$ is bijective, $(1)$ $f$ is injective and surjective, $(2)$ $f$ is invertible, i.e. $\exists g:M_{m\times n}(F) \to L(V,W)$ such that $g\circ f=\text{id}_{L(V,W)}$ and $f\circ g=\text{id}_{M_{m\times n}(F)}$.
Proof using $(1)$. Approach(1): If $f(T)=f(U)$, for some $T,U\in L(V,W)$. So $f(T)=f(U)$ $=[T]_{B}^{B’}=[U]_{B}^{B’}$. It’s easy to check $T=U$, here is proof. Hence $f$ is injective. Let $A=(A_{ij})\in M_{m\times n}$. Define $\gamma_j =\sum_{i=1}^m A_{ij}\cdot_W \beta_i$, for all $j\in J_n$. By theorem 1 section 3.1, $\exists !$ $T\in L(V,W)$ such that $T(\alpha_j)=\gamma_j= \sum_{i=1}^m A_{ij}\cdot_W \beta_i $, for all $j\in J_n$. By uniqueness of coordinates matrix of $T(\alpha_j)$ relative to ordered basis $B’$, $A_j=[T(\alpha_j)]_{B’}$, $\forall j\in J_n$. Thus $f(T)=[T]_{B}^{B’}$ $=([T(\alpha_1)]_{B’}\dotsb [T(\alpha_n)]_{B’})$ $=(A_1\dotsb A_n)=A$. Hence $f$ is surjective. So $f$ is bijective. Actually in proof of surjectivity, we showed $\forall A\in M_{m\times n}(F)$, $\exists !$ $T\in L(V,W)$ such that $f(T)=A$. Which is precisely the definition of bijective map. Approach(2): $V, W$ are $m,n$-dimensional vector space over field $F$, respectively. By theorem 5 section 3.2, $\mathrm{dim}(L(V,W))=mn$. By exercise 12 section 2.3, $\mathrm{dim}(M_{m\times n}(F))=mn$. So $\mathrm{dim}(V,W)= \mathrm{dim}(M_{m\times n}(F))$. If $f(T)=f(U)$, for some $T,U\in L(V,W)$. So $f(T)=f(U)$ $=[T]_{B}^{B’}=[U]_{B}^{B’}$. It’s easy to check $T=U$, here is proof. Hence $f$ is injective. By theorem 9 section 3.2, $f$ is bijective.
Proof using $(2)$. In proof of $f$ is surjective, we showed $\forall A\in M_{m\times n}(F)$, $\exists !$ $T\in L(V,W)$ such that $f(T)=[T]_{B}^{B’}=A$. Define $g:M_{m\times n}(F)\to L(V,W)$ such that $g(A)=T$. So $g$ is a well-defined function. Let $T\in L(V,W)$. Then $g\circ f(T)$ $=g(f(T))$ $=g(A)=T$. Thus $g\circ f(T)=T$, $\forall T\in L(V,W)$. Hence $g\circ f=\text{id}_{L(V,W)}$. Let $A\in M_{m\times n}(F)$. Then $f\circ g(A)$ $=f(g(A))$ $=f(T)=A$. Thus $f\circ g(A)=A$, $\forall A\in M_{m\times m}(F)$. Hence $f\circ g=\text{id}_{M_{m\times n}(F)}$. Is my proof correct? IMO, this proof is extremely “artificial” since I use $f$ is bijective to prove $f$ is bijective.
Hoffman’s prove existence of inverse map $g$ in following way:
Let $A=(A_{ij})\in M_{m\times n}(F)$. Define $T:V\to W$ such that $T(\alpha)$ $=T(\sum_{j=1}^nx_j \cdot_V \alpha_j)$ $=\sum_{i=1}^m (\sum_{j=1}^n A_{ij}x_j) \cdot \beta_i$, for all $\alpha \in V$. It’s easy to check, $T$ is a linear map and matrix of $T$ relative to ordered bases $B,B’$ is $A$. So $T\in L(V,W)$ and $f(T)=[T]_{B}^{B’}=A$. By theorem 1 section 3.1, $T$ is unique.
Edit:

Claim: $\forall T\in L(V,W)$, $\exists!$ $A=[T]_B^{B’}\in M_{m\times n}(F)$ such that $[T(\alpha)]_{B’}=A\cdot [\alpha]_B$, for all $\alpha \in V$.

Proof: Let $T\in L(V,W)$. Then $[T]_{B}^{B’}$ $=([T(\alpha_i)]_{B’}\dotsb [T(\alpha_n)]_{B’})\in M_{m\times n}(F)$ is matrix of $T$ relative to ordered basis $B,B’$. It’s easy to check, $[T(\alpha)]_{B’}=[T]_B^{B’} \cdot [\alpha]_B$, for all $\alpha \in V$. Now we show $A=(A_{ij})=[T]_{B}^{B’}$ is unique. Let $C=(C_{ij})\in M_{m\times n}(F)$ such that $[T(\alpha)]_{B’}=C\cdot [\alpha]_B$, for all $\alpha \in V$. In particular, $[T(\alpha_j)]_{B’}=C\cdot [\alpha_j]_B$, $\forall j \in J_n$. Fix $j\in J_n$. Note $[\alpha_j]_B=e_j$ $=(x_1,….x_n)$, where $x_i=0_F$, $\forall i\in J_n\setminus \{j\}$ and $x_j=1_F$. By uniqueness of coordinates matrix of $T(\alpha_j)$ relative to ordered basis $B’$, we have $\sum_{j=1}^n A_{ij}x_j$ $=\sum_{j=1}^n C_{ij}x_j$ $=A_{ij}$ $=C_{ij}$, for all $i\in J_m$. Since $j$ was arbitrary, $A_{ij}=C_{ij}$, $\forall j\in J_n$. Thus $A_{ij}=C_{ij}$, $\forall (i,j)\in J_m\times J_n$. Hence $A=[T]_B^{B’}=C$.
 A: $V, W$ be two finite dimensional vector space over $F$ of dimension $n, m$ respectively. Let us fix a pair of basis $(\beta, \beta') $.
Given a matrix $A\in M_{m×n}(F) $ , consider the associated linear map $L_A:V\to W$ defined by $$L_A(x) =Ax$$
Check :

*

*$L_A$ indeed a linear map.


*$[L]_{\beta}^{\beta'}=A$


*For a fixed pair of basis $(\beta, \beta') $ , $L_A$ is unique.
(Hint: A. Linear map is completely determined by the action on the Basis)


*$L_{A+B}= L_A +L_B$


*$L_{cA}=cL_A$ for $c\in F$

Claim: Two vector spaces $M_{m×n}(F) $ and $\mathcal{L}(V, W) $ are isomorphic i.e there is an invertible linear map between those two spaces.
Proof: Let us consider a map $L : M_{m×n}(F)\to \mathcal{L}(V, W) $ defined by $$L(A) =L_A$$
• Then this map is well defined by $3$.
• $L$ is a linear map by $4$ and $5$.
$\begin{align}\ker L&=\{A\in M_{m×n}(F): L(A) =0\}\\&=\{A\in M_{m×n}(F): L_A=0\}\\&=\{\textbf{0}\} \end{align}$
$L$ is also surjective as every operator in $\mathcal{L}(V, W) $ has a matrix representation w.r.to $(\beta, \beta') $


The inverse map $L^{-1}:=\varphi:\mathcal{L}(V, W) \to 
 M_{m×n}(F)$
defined by $\varphi(T) =[T]_{\beta}^{\beta'}$ is also an isomorphism.
You can check it directly now.
