The row rank of an $m\times n$ matrix $A$ is at most $\min\{m,n\}$. Why? Ok, so let $A$ be an $m\times n$ matrix. I understand by intuition that the row rank has to be $\le m$, but why also $n$? Is this because there can be no more leading ones than $m$ or $n$?
 A: Elementary row operations don't change the row space, that is, the set of linear combinations of the rows.
The number of leading ones in the elimination is at most equal to the number of columns, because they fall in distinct columns. So the number of nonzero rows in the row echelon form of $A$ (which is the row rank) is at most equal to the number of columns.
A: It is so because the column rank of a matrix equals the row rank of the matrix.
A: A matrix $A\in\mathcal M_{nm}(\Bbb R)$ is a representation of a linear transformation
$$f\colon\Bbb R^m\rightarrow \Bbb R^n$$
so the image of $f$ is a subspace of $\Bbb R^n$ hence 
$$\operatorname{rank} f=\dim (\operatorname{Im} f)\le n$$
but also by the rank-nullity theorem we have
$$\operatorname{rank} f\le m$$
hence we find the desired result
$$\operatorname{rank} f\le\min(m,n)$$
A: The rank of a matrix is the dimension of the vector space generated by the columns, hence it is smaller than the number of columns. This involves vectors with a number of coordinates equal to the number of the rows $m$, hence a linearly independent family of such vectors cannot have greater than $m$. 
