Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 2 / 15E / a / Solution

By the rank-nullity theorem, $\mathrm{nullity}A=n-\mathrm{rank}A\ge n-m>0$. Thus a first dependent column prefix exists. Minimality of $k$ makes the first $k-1$ columns independent, so $\mathrm{rank}{A}_{k-1}=k-1$. Adding one column cannot reduce this rank, while nonzero nullity of $A_k$ implies rank at most $k-1$. HenceFor $k=1$ the earlier prefix is the zero-dimensional empty matrix, and the same argument applies.