Linear Transformations 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:

	\(\displaystyle \)	\(\displaystyle \left({v \circ u}\right) \left({a_j}\right)\)	\(=\)	\(\displaystyle v \left({u \left({a_j}\right)}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle v \left({\sum_{k=1}^n \alpha_{k j} b_k}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k=1}^n \alpha_{k j} v \left({b_k}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k=1}^n \alpha_{k j} \left({\sum_{i=1}^m \beta_{i k} c_i}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k=1}^n \left({\sum_{i=1}^m \alpha_{k j} \beta_{i k} c_i}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i=1}^m \left({\sum_{k=1}^n \alpha_{k j} \beta_{i k} c_i}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i=1}^m \left({\sum_{k=1}^n \alpha_{k j} \beta_{i k} }\right) c_i\)	\(\displaystyle \)

So:

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

$\blacksquare$

Proof of Corollary

• Follows directly.

$\blacksquare$

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.

Sources

• Seth Warner: Modern Algebra (1965): $\S 29$: Theorems $29.1, \ 29.2$