Originally Posted by

**Drexel28** Think about it like this. Let $\displaystyle (x_1,\cdots,x_m)$ be the selected basis for $\displaystyle \ker N$ and $\displaystyle \left\{x_1,\cdots,x_n\}$ the full basis constructed. Then, as the author said one has that for each element $\displaystyle x_j$ of basis vectors one has that $\displaystyle N(x_j)\in \ker N$ so that there exists $\displaystyle \alpha_1,\cdots,\alpha_m\in F$ such that $\displaystyle \displaystyle N(x_j)=\sum_{r=}^{m}\alpha_r x_r\quad\mathbf{(1)}$ and so in particular if $\displaystyle \beta_1,\cdots,\beta_n$ we form the $\displaystyle x_j^{\text{th}}$ column by listing them as the corresponding coefficients of $\displaystyle \displaystyle \sum_{r=1}^{n}\beta_r x_r$...but since the representation with respect to basis elements is unique we may conclude by $\displaystyle \mathbf{(1)}$ that $\displaystyle \beta_r=0$ for $\displaystyle r>m$. But, since $\displaystyle j>m$ this implies that the first element which is necessarily zero occurs at or above the basis. Make sense?