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.

Motivation
What this result tells us is two things:
 * 1) That the relative matrix of a linear transformation can be considered to be the same thing as the transformation itself
 * 2) To determine the relative matrix for the composite of two linear transformations, what you do is multiply the relative matrices of those linear transformations.

Thus one has a means of direct arithmetical manipulation of linear transformations, thereby transforming geometry into algebra.

In fact, matrix multiplication was purposely defined (some would say designed) so as to produce exactly this result.