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

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

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:

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

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

by a calculation similar to the preceding.

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

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