Set of Linear Transformations is Isomorphic to Matrix Space

Theorem
Let $R$ be a ring with unity.

Let $F$, $G$ and $H$ be free $R$-modules of finite dimension $p, n, m > 0$ respectively.

Let $\sequence {a_p}$, $\sequence {b_n}$ and $\sequence {c_m}$ be ordered bases

Let $\map {\LL_R} {G, H}$ denote the set of all linear transformations from $G$ to $H$.

Let $\map {\MM_R} {m, n}$ be the $m \times n$ matrix space over $R$.

Let $\sqbrk {u; \sequence {c_m}, \sequence {b_n} }$ be the matrix of $u$ relative to $\sequence {b_n}$ and $\sequence {c_m}$.

Let $M: \map {\LL_R} {G, H} \to \map {\MM_R} {m, n}$ be defined as:


 * $\forall u \in \map {\LL_R} {G, H}: \map M u = \sqbrk {u; \sequence {c_m}, \sequence {b_n} }$

Then $M$ is a module isomorphism.

Proof
Let $u, v \in \map {\LL_R} {G, H}$ such that:
 * $\map M u = \map M v$

We have that the matrix of $u$ relative to $\sequence {b_n}$ and $\sequence {c_m}$ is defined as the $m \times n$ matrix $\sqbrk \alpha_{m n}$ where:


 * $\ds \forall \tuple {i, j} \in \closedint 1 m \times \closedint 1 n: \map u {b_j} = \sum_{i \mathop = 1}^m \alpha_{i j} \circ c_i$

and it is seen that $\map M u$ and $\map M v$ are the same object.

That is:
 * $\map M u = \map M v \implies u = v$

and $M$ is seen to be injective.