Proof of the dimension formula for the space of linear maps

Proof

We write down an explicit basis of $\Hom_F(V, W)$ with $\dim V \cdot \dim W$ elements.

1. Let $n = \dim V$ and $m = \dim W$. Let $(b_1, \dots, b_n)$ be a basis of $V$ and $(c_1, \dots, c_m)$ a basis of $W$.
2. Write $\delta_{jk} = 1$ if $j = k$ and $\delta_{jk} = 0$ if $j \ne k$.
3. For $1 \le i \le m$ and $1 \le j \le n$, the extension of values on a basis gives a linear map $E_{ij} \colon V \to W$ with $E_{ij}(b_k) = \delta_{jk} c_i$ for $1 \le k \le n$.
4. These $mn$ maps lie in the space of linear maps $\Hom_F(V, W)$. Its operations are pointwise.
5. The maps $E_{ij}$ span $\Hom_F(V, W)$. Let $T \in \Hom_F(V, W)$. For each $k$, write $T(b_k) = \sum_{i=1}^m a_{ik} c_i$ with $a_{ik} \in F$.
6. Let $S = \sum_{i=1}^m \sum_{j=1}^n a_{ij} E_{ij}$. By the pointwise operations, $S(b_k) = \sum_{i,j} a_{ij} \delta_{jk} c_i = \sum_{i=1}^m a_{ik} c_i = T(b_k)$ for each $k$.
7. So $S$ and $T$ are linear maps that agree on a basis. By the uniqueness part of the same result, $T = S$.
8. The maps $E_{ij}$ are linearly independent. Suppose $\sum_{i,j} a_{ij} E_{ij} = 0$, the zero map, with $a_{ij} \in F$.
9. Evaluating at $b_k$, as in step 6, gives $\sum_{i=1}^m a_{ik} c_i = 0$. Since $(c_i)$ is linearly independent, $a_{ik} = 0$ for all $i$. This holds for each $k$.
10. By steps 5 to 9, $(E_{ij})_{1 \le i \le m,\, 1 \le j \le n}$ is a basis of $\Hom_F(V, W)$ with $mn$ elements.
11. So $\Hom_F(V, W)$ is finite-dimensional, and its dimension is $mn = \dim V \cdot \dim W$.