# 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.

### Corollary

Let $R$ be a commutative ring with unity.

Let $M: \struct {\map {\LL_R} G, +, \circ} \to \struct {\map {\MM_R} n, +, \times}$ be defined as:

- $\forall u \in \map {\LL_R} G: \map M u = \sqbrk {u; \sequence {a_n} }$

Then $M$ is an 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.

This needs considerable tedious hard slog to complete it.In particular: Surjectivity needs proving, so does homomorphismTo discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## 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 $29.1$