Proof of Theorem CRN

Theorem BCS provides a basis for the column space by choosing columns of $A$ that that have the same indices as the pivot columns of $B$. In the analysis of $B$, each leading 1 provides one nonzero row and one pivot column. So there are $r$ column vectors in a basis for $\csp{A}$.

Theorem BNS provides a basis for the null space by creating basis vectors of the null space of $A$ from entries of $B$, one basis vector for each column that is not a pivot column. So there are $n-r$ column vectors in a basis for $\nullity{A}$.