Suppose $\lambda_{1,\cdots,n}\in\Bbb C$ and for all $m\in\Bbb N_+$ we have $$\sum_{i=1}^n \lambda_i^m=0.$$ Prove that all $\lambda_i$ are zero.
Is there any easy approach that doesn't play with algebraic manipulations like symmetric polynomials or Vandermonde determinants? A purely analytical approach would be preferred (e.g. observe the equations as $m$ tends to infinity). Thanks!
I don't know if this is acceptable to you, as it uses polynomials. It is the most natural way to me. Given any polynomial $p$ with $p(0)=0$, i.e, $p(z)=\sum_{k=1}^ra_kz^k$, we have $$\tag1 \sum_{j=1}^np(\lambda_j)=\sum_{j=1}^n\sum_{k=1}^ra_k\lambda_j^k=\sum_{k=1}^ra_k\left(\sum_{j=1}^n\lambda_j^k\right)=0. $$ In particular, if $\lambda_j\ne0$ for some $j$ we can choose $p$ with $p(0)=0$ and $p(\lambda_j)=1$ for all $j$ such that $\lambda_j\ne0$. But then the left-hand-side in $(1)$ would be positive, giving a contradiction. So $\lambda_1=\cdots=\lambda_n=0$.
The above also shows that it is enough to have the equalities up to $m=n$, or even less if there are repetitions in the list.