Dimension of Set of Linear Transformations

Theorem
Let $R$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $\left({G, +_G, \circ}\right)_R$ be a unitary$R$-module such that $\dim \left({G}\right) = n$.

Let $\left({H, +_H, \circ}\right)_R$ be a unitary$R$-module such that $\dim \left({H}\right) = m$.

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

Then:
 * $\dim \left({\mathcal L_R \left({G, H}\right)}\right) = n m$

Let $\left \langle {a_n} \right \rangle$ be an ordered basis for $G$.

Let $\left \langle {b_m} \right \rangle$ be an ordered basis for $H$.

Let $\phi_{i j}: G \to H$ be the unique linear transformation defined for each $i \in \left[{1 \,. \, . \, n}\right], j \in \left[{1 \,. \, . \, m}\right]$ which satisfies:
 * $\forall k \in \left[{1 \, . \, . \, n}\right]: \phi_{i j} \left({a_k}\right) = \delta_{i k} b_j$

where $\delta$ is the Kronecker delta.

Then:
 * $\left\{{\phi_{i j}: i \in \left[{1 \, . \, . \, n}\right], j \in \left[{1 \, . \, . \, m}\right]}\right\}$

is a basis for $\dim \left({\mathcal L_R \left({G, H}\right)}\right)$.

Proof

 * Let $B = \left\{{\phi_{i j}: i \in \left[{1 \, . \, . \, n}\right], j \in \left[{1 \, . \, . \, m}\right]}\right\}$.

Let $\displaystyle \sum_{j=1}^m \sum_{i=1}^n \lambda_{i j} \phi_{i j} = 0$.

Then $\displaystyle \forall k \in \left[{1 \,. \, . \, n}\right]: 0 = \sum_{j=1}^m \sum_{i=1}^n \lambda_{i j} \phi_{i j} \left({a_k}\right) = \sum_{j=1}^m \lambda_{k j} b_j$.

So $\forall j \in \left[{1 \,. \, . \, n}\right]: \lambda_{k j} = 0$.

Hence $B$ is linearly independent.


 * Now let $\phi \in \mathcal L_R \left({G, H}\right)$.

Let $\left({\alpha_{i 1}, \alpha_{i 2}, \ldots, \alpha_{i m}}\right)$ be the sequence of scalars that satisfies:


 * $\displaystyle \forall i \in \left[{1 \, . \, . \, n}\right]: \phi \left({a_i}\right) = \sum_{j=1}^m \alpha_{i j} b_j$

Then:
 * $\displaystyle \forall k \in \left[{1 \, . \, . \, n}\right]: \phi \left({a_k}\right) = \left({\sum_{j=1}^m \sum_{i=1}^n \alpha_{i j} u_{i j}}\right) \left({a_k}\right)$

by a calculation similar to the preceding.

So, by Linear Transformation of Generated Module, $\displaystyle u = \sum_{j=1}^m \sum_{i=1}^n \alpha_{i j} u_{i j}$.

Thus $B$ is a generator for $\phi \in \mathcal L_R \left({G, H}\right)$.