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 an isomorphism of modules, and:


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

Proof
The proof that $M$ is an isomorphism is straightforward.

The relation:
 * $\sqbrk {v \circ u; \sequence {c_m}, \sequence {a_p} } = \sqbrk {v; \sequence {c_m}, \sequence {b_n} } \sqbrk {u; \sequence {b_n}, \sequence {a_p} }$

follows from Relative Matrix of Composition of Linear Mappings.