 Distinct diagonal weights break a first column dependence

Let the first $k$ columns of a matrix be the first dependent prefix, with no zero column. Their relation space has dimension one, and every nonzero relation $b$ has at least two nonzero coordinates. If both $Ab=0$ and $A\mathrm{diag}({\lambda }_j)b=0$, one-dimensionality gives ${\lambda }_j{b}_j=c{b}_j$ for all $j$. Two nonzero coordinates would force two of the distinct weights ${\lambda }_j$ to agree, a contradiction.