I have seen several proofs of the fact that column and row rank of a matrix over a commutative field are equal and seem to understand them — except the one provided by Emil Artin in his book on Galois theory. He takes an $m \times n$ matrix, equates its $A \cdot x$ product with a zero vector [system (1)] and then writes:
... we may assume that the first $r$ rows are independent row vectors. ... Thus, each row after the $r^{th}$ is linearly expressible in terms of the first $r$ rows. Consequently, any solution of the first $r$ equations in (1) will be a solution of the entire system since any of the last $n-r$ equations is obtainable as a linear combination of the first $r$ [I am afraid there is a typo there and it should be $m-r$ since the matrix is $m \times n$]. Conversely, any solution of (1) will also be a solution of the first $r$ equations. This means that the matrix \begin{pmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{r,1} & a_{r,2} & \cdots & a_{r,n} \end{pmatrix} consisting of the first $r$ rows of the original matrix has the same right column rank as the original.
This should be absolutely obvious, but I cannot see the connection between the sets of solutions for these two systems and their respective column ranks.