= Solution
Suppose $v=(v_1,\ldots,v_r)^t$ belongs to the <kernel> of the displayed <matrix> $\Lambda$. Define the <polynomial> $q(z)=\sum_{k=1}^r v_kz^k$. Then $q(\lambda_j)=0$ for all $j$, and $q(0)=0$. These are $r+1$ distinct roots of a <polynomial> of degree at most $r$. Repeated application of the <factor theorem> therefore makes $q$ identically zero, so every $v_k=0$. Hence $\Lambda$ is injective and, by finite-dimensional <rank-nullity theorem>, invertible. Its transpose is invertible with inverse $(\Lambda^{-1})^t$, proving \b[both kernels are trivial].
For the second <matrix>, membership of the all-ones vector in the <kernel> means
$$
\sum_{j=1}^n\lambda_j^k=0\qquad(1\le k\le n).
$$
Group the distinct nonzero values as $\mu_1,\ldots,\mu_r$, with multiplicities $m_1,\ldots,m_r>0$. Zero values contribute nothing to these positive powers. The first $r$ equations are
$$
\sum_{j=1}^r m_j\mu_j^k=0\qquad(1\le k\le r),
$$
or $\Lambda(\mu)^t m=0$. The proved transpose-kernel statement implies $m=0$, contradicting its positive entries if $r>0$. Therefore $r=0$ and
$$
\boxed{\lambda_1=\cdots=\lambda_n=0.}
$$
Conversely, all zero values plainly give a zero <matrix>. This proves the <zero power sums force a finite complex multiset to vanish> criterion, including repetitions and zero values rather than assuming all <eigenvalues> distinct.
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