Suppose A is a 3 × 5 matrix, and that $dim\ Nul\ A = 2$. Prove the following;
(a) There exists a basis for $R^3$ made up of some of the columns of $A$.
(b) The rows of $A$ form a linearly independent set.
$5 = Dim\ Nul(A) + Rk(A), Rk(A) = 3$ (rank nullity theorem)
So with rank of 3, $rref(A)$ contains 3 pivot columns, and the corresponding columns of $A$ (also the column space of A) form a basis of $R^3$.
If the rows are linearly independent, then $Ax = 0$ IFF $x = 0$.
However since $Dim\ Nul(A) = 2$, there must be two $x \ne 0$ such that $Ax=0$. (contradiction?)
Am I doing this properly? Both answers seem intuitive, and I can't tell if I've proven them sufficiently (if at all). Is there anything I can or should add to improve these proofs?