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.

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