Set of Linear Transformations is Isomorphic to Matrix Space

Theorem
Let $$R$$ be a commutative ring with unity.

Let $$F$$, $$G$$ and $$H$$ be finite-dimensional $R$-modules with ordered bases $$\left \langle {a_p} \right \rangle$$, $$\left \langle {b_n} \right \rangle$$ and $$\left \langle {c_m} \right \rangle$$ respectively.

Let $$\mathcal L_R \left({G, H}\right)$$ be the set of all linear transformations from $$G$$ to $$H$$.

Let $$\mathcal M_{R} \left({m, n}\right)$$ be the $m \times n$ matrix space over $$R$$.

Let $$\left[{u; \left \langle {c_m} \right \rangle, \left \langle {b_n} \right \rangle}\right]$$ be the matrix of $u$ relative to $\left \langle {b_n} \right \rangle$ and $\left \langle {c_m} \right \rangle$.

Let $$M: \mathcal L_R \left({G, H}\right) \to \mathcal M_{R} \left({m, n}\right)$$ be defined as:


 * $$\forall u \in \mathcal L_R \left({G, H}\right): M \left({u}\right) = \left[{u; \left \langle {c_m} \right \rangle, \left \langle {b_n} \right \rangle}\right]$$

Then $$M$$ is an isomorphism, and:


 * $$\forall u \in \mathcal L_R \left({F, G}\right), v \in \mathcal L_R \left({G, H}\right): \left[{v \circ u; \left \langle {c_m} \right \rangle, \left \langle {a_p} \right \rangle}\right] = \left[{v; \left \langle {c_m} \right \rangle, \left \langle {b_n} \right \rangle}\right] \left[{u; \left \langle {b_n} \right \rangle, \left \langle {a_p} \right \rangle}\right]$$.

Corollary
Let $$R$$ be a commutative ring with unity.

Let $$M: \left({\mathcal L_R \left({G}\right), +, \circ}\right) \to \left({\mathcal M_{R} \left({n}\right), +, \times}\right)$$ be defined as:


 * $$\forall u \in \mathcal L_R \left({G}\right): M \left({u}\right) = \left[{u; \left \langle {a_n} \right \rangle}\right]$$

Then $$M$$ is an isomorphism.

Proof

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


 * Let $$\left[{\alpha}\right]_{p n} = \left[{u; \left \langle {b_n} \right \rangle, \left \langle {a_p} \right \rangle}\right]$$ and $$\left[{\beta}\right]_{n m} = \left[{v; \left \langle {c_m} \right \rangle, \left \langle {b_n} \right \rangle}\right]$$.

Then:

$$ $$ $$ $$ $$ $$ $$

So $$\left[{v \circ u; \left \langle {c_m} \right \rangle, \left \langle {a_p} \right \rangle}\right] = \left[{v; \left \langle {c_m} \right \rangle, \left \langle {b_n} \right \rangle}\right] \left[{u; \left \langle {b_n} \right \rangle, \left \langle {a_p} \right \rangle}\right]$$.

Proof of Corollary

 * Follows directly.

Comment
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.