 Solution

Choose any nonzero vector $b$ in the one-dimensional kernel of $A_k$. It gives ${\sum }_{j=1}^k{a}_{ij}{b}_j=0$ for every row. Because no column is zero, $b$ must have at least two nonzero entries: a relation supported at a single index would make that column zero.

Suppose, contrary to the claim, that $A_k\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_k)b=0$. Its kernel is spanned by $b$, so there is a scalar $c$ with ${\lambda }_j{b}_j=c{b}_j$ for every $j$. At the two nonzero entries this gives two equal weights ${\lambda }_j=c$, contradicting their distinctness. Thus the weighted sum is nonzero in at least one row. This proves that distinct diagonal weights break a first column dependence. Some entries of $b$ may be zero; only two nonzero entries are needed.