Let $V$ be a vector space over a field $\mathbb F$ containing all the $m \times n$ matrices ,show that $\text{dim}(V)=mn$.
Define a family of matrices by $M_{i,j}=(m_{rq})=(\delta_{ri}\delta_{jq})\in \mathbb F^{m \times n}$
Take an arbitrary element $A \in V$ ,it's true to claim that $$A=\sum_{i=1}^{m}\sum_{j=1}^{n}a_{i,j}M_{i,j}$$
Besides the set $\left \{M_{i,j} \mid1\le i\le m,1\le j\le n\right\}$ is a linearly independent set since if for some $\lambda_{i,j} \in\mathbb F$ $$\sum_{i=1}^{m}\sum_{j=1}^{n}\lambda_{i,j}M_{i,j}=0$$
Then the LHS is indeed a $m \times n$ matrix with the entry placed in the $i$th row and the $j$ th column equal to $\lambda _{i,j}$ which follows that all such scalars are zero,and hence the set is a lineary independent set.
Based on the definition of a basis it implies that $\left \{M_{i,j} \mid1\le i\le m,1\le j\le n\right\}$ is a linearly independent set sppaning $V$ and so form a basis of $V$,so the dimension of $V$ is determined by its basis which contains $mn$ elements.
The proof is the one that I've learned before and changed it a bit to make the proof easier to follow,but I'm not sure if the representation $M_{i,j}=(m_{rq})=(\delta_{ri}\delta_{jq})$ is true (this is what I've added to the original proof with some extra explanations).
It would be appreciated if someone checks that and if it's possible make the proof more precise (if it's not as much as it can be ).