I understood that for a $m*n$ matrix $A$, if its rank is $r$, $(n-r)$ pivot free columns correspond to $(n-r)$ free variables. But if thats the case, how do I go about proving that the dimension of the null space is also $(n-r)$?
I did: Let the solution of the matrix equation $A\vec x=\vec0$ be $\vec x=\sum_{n-r} \vec c_ix_i$ where $\vec c_i$ is just any coefficient vector with the property that $A(\sum_{n-r} \vec c_ix_i)=\vec 0$ for all $x_i$. So $A\vec c_i=\vec 0$ for all given $i$. But what now?
Because of the $n-r$ free variables, the equation $Ax=b$ has an $n-r$-dimensional solution space, if it has any solutions at all. But $Ax=0$ is always consistent, because $x=0$ is a solution. $\therefore\operatorname{null}A=n-r$.