A more fundamental (and clear) way to prove that two vector spaces are isomorphic is to show that there exists an invertible linear transformation between the vector spaces (this is actually the definition of isomorphism). So, in general, for two arbitrary finite dimensional vector spaces we can state the following
Theorem
Let $V$ and $W$ be finite dimensional vector spaces over a field $F$, with $\mathrm{dim}(V)=n$ and $\mathrm{dim}(W)=m$, and let $\beta$ be an ordered basis (o.b.) of $V$ and $\gamma$ be an ordered basis of $W$, then the transformation $\Phi:\mathcal{L}(V,W)\to \mathcal{M}_{m\times n}$ given by
\begin{equation} \Phi(T)=\left[ T \right]_{\beta}^{\gamma} \end{equation}
is an isomorphism.
Note: $\left[ T \right]_{\beta}^{\gamma}$ is the matrix representation of the linear transformation $T$. Let $\beta=\{v_1,v_2,\dots,v_n\}$ and $\gamma=\{w_1,w_2,\dots,w_m\}$, if $T(v_j)=\sum_{k=1}^m B_{kj} w_k$ then the $j$-th column of $B:=\left[ T \right]_{\beta}^{\gamma}$ (denoted as $B_{\dot \ ,\ j}$) is $(B_{1j},B_{2j},\dots,B_{mj})^T$ (for $j\in \{1,2,\dots,n\}$).
Proof
The path is to show that $\Phi$ is linear and invertible (then it's an isomorphism).
(I) $\Phi$ is linear. Let $U$, $T$ $\in \mathcal{L}(V,W)$ and $c\in F$. Then
\begin{align} \Phi(U+cT) & = \left[ U+cT \right]_{\beta}^{\gamma} \end{align}
But, if $U(v_j)=\sum_{k=1}^m A_{kj} w_k$ then the $j$-th column of $A:=\left[ U \right]_{\beta}^{\gamma}$ (denoted as $A_{\dot \ ,\ j}$) is $(A_{1j},A_{2j},\dots,A_{mj})^T$. So
\begin{align} (U+cT)(v_j) & = U(v_j)+cT(v_j) \\ & = \sum_{k=1}^m A_{kj} w_k+c\sum_{k=1}^m B_{kj} w_k \\ & = \sum_{k=1}^m (A_{kj} +c B_{kj}) w_k \end{align}
Which implies that the $j$-th column of $C:=\left[ U+cT \right]_{\beta}^{\gamma}$ (denoted as $C_{\dot \ ,\ j}$) is
\begin{align} C_{\dot \ ,\ j}&=(A_{1j}+cB_{1j},A_{2j}+cB_{1j},\dots,A_{mj}+cB_{1j})^T \\ &=(A_{1j},A_{2j},\dots,A_{mj})^T+c(B_{1j},B_{2j},\dots,B_{mj})^T \\ &=A_{\dot \ ,\ j}+cB_{\dot \ ,\ j} \end{align}
Since this is valid for $j\in \{1,2,\dots,n\}$, we conclude that $C=A+cB$, or $\left[ U+cT \right]_{\beta}^{\gamma}=\left[ U \right]_{\beta}^{\gamma}+c\left[ T \right]_{\beta}^{\gamma}$, i.e. $\Phi(U+cT)=\Phi(U)+c\Phi(T)$.
(II) $\Phi$ is invertible. To show that this is true, it's enough to show that $\Phi$ is both injective and surjective.
A standard theorem (you can check it in any linear algebra book) says that if two linear transformations have the same matrix representation, then they are the same linear transformation. This implies that $\Phi$ is injective.
On the other hand $\Phi$ is surjective because given $M \in \mathcal{M}_{m\times n}$ (with entries $M_{ij}$), there exists a unique linear transformation $L\in \mathcal{L}(V,W)$ such that $\Phi(L)=\left[ L \right]_{\beta}^{\gamma}=M$. This is the transformation that does the following: $L(v_j)=\sum_{k=1}^m M_{ki}w_k$ (which is the same as the $j$-th column of $M$ and is defined by the entries of $M$).
We conclude that $\Phi$ is an isomorphism. $\blacksquare$