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.