# Composition of Linear Transformations is Isomorphic to Matrix Product

## 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:

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

## Proof

Follows directly from Relative Matrix of Composition of Linear Transformations.

$\blacksquare$

## Motivation

What Set of Linear Transformations is Isomorphic to Matrix Space tells us is two things:

- That the relative matrix of a linear transformation can be considered to be
*the same thing as*the transformation itself - 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.

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices: Theorem $29.1$