The basic principal is that elementary row options do not change the rowspace of a matrix. To prove that, you could demonstrate that the processes of scaling a basis element, interchanging two basis elements, and adding multiples of one basis element to another, all result in a new basis.
So, if you perform elementary row operations and change it entirely to row-echelon form, you are left with a matrix that has some zero rows and some nonzero rows. The row-echelon shape makes it obvious that the nonzero rows are linearly independent. Can you see why?
Since this is the case, the remaining rows are a basis for the rowspace of the matrix.
Basic row operations do not change the (right) nullspace either, so you can use the new matrix to work with the nullspace. Using that matrix instead of the original matrix, it will be particularly easy to solve the equation $A'x=0$ with back substitution to discover what is in the nullspace.