Using the standard basis, the image of a linear transformation is the span of the columns of its matrix

Let $F$ be a field and let $\varphi \in \mathsf{Hom}_F(F^n, F^m)$ be a linear transformation. Fix the standard bases $B \subseteq F^n$ and $E \subseteq F^m$. Prove that the image of $\varphi$ is precisely the span of the columns of $M^E_B(\varphi)$, considered as elements of $F^m$. Deduce that the rank of $\varphi$ as a linear transformation is the column rank of $M^E_B(\varphi)$.

Suppose $\varphi(x) \in \mathsf{im}\ \varphi$, and say $M^E_B(\varphi) = [a_{i,j}]_{i=1,j=1}^{m,n}$. Now $\varphi(x) = M^E_B(\varphi) \times x = [a_{i,j}][x_j]$ $= [\sum_{k=1}^n a_{i,k}x_k]_{i=1}^m$ $= \sum_{k=1}^n [a_{i,k}x_k]_{i=1}^m$ $= \sum_{k=1}^n [a_{i,k}]_{i=1}^m x_k$ $\in \mathsf{col\ span}\ M^E_B(\varphi)$. (Note that the $[a_{i,k}]_{i=1}^m$ are precisely the columns of $M^E_B(\varphi)$.) Conversely, we can see that every linear combination of the columns of $M^E_B(\varphi)$ is in $\mathsf{im}\ \varphi$.

Recall that $\mathsf{rank}\ \varphi = \mathsf{dim}\ \mathsf{im}\ \varphi$, and that $\mathsf{col\ rank}\ M^E_B(\varphi)$ is the maximal number of linearly independent columns of $M^E_B(\varphi)$. Because the columns of $M^E_B(\varphi)$ are a generating set for $\mathsf{im}\ \varphi$, they contain a basis for $\mathsf{im}\ \varphi$. The size of this basis is precisely $\mathsf{dim\ im}\ \varphi$, so that in particular, all such subsets of the columns of $M^E_B(\varphi)$ have the same size and are maximal linearly independent subsets. Thus $\mathsf{col\ rank}\ M^E_B(\varphi) = \mathsf{rank}\ \varphi$.