Rank is Dimension of Subspace

Theorem
Let $$K$$ be a field.

Let $$\mathbf A$$ be an $m \times n$ matrix over $$K$$.

Then the rank of $$\mathbf A$$ is the dimension of the subspace of $$K^n$$ generated by the rows of $$\mathbf A$$.

Proof
Let $$u: K^n \to K^m$$ be the linear transformation such that $$\mathbf A$$ is the matrix of $u$ relative to the standard ordered bases of $$K^n$$ and $$K^m$$.

Let $$\rho \left({\mathbf A}\right)$$ be the rank of $$\mathbf A$$.

Let $$\mathbf A^t$$ be the transpose of $$\mathbf A$$.

Similar notations on $$u$$ denote the rank and transpose of $$u$$.

We have $$\rho \left({\mathbf A}\right) = \rho \left({u}\right)$$ and $$\rho \left({\mathbf A^t}\right) = \rho \left({u^t}\right)$$, but $$\rho \left({u^t}\right) = \rho \left({u}\right)$$ from Rank and Nullity of Transpose.