= 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 $A\operatorname{diag}(\lambda_j)b=0$, one-dimensionality gives $\lambda_jb_j=cb_j$ for all $j$. Two nonzero coordinates would force two of the distinct weights $\lambda_j$ to agree, a contradiction.
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