Does $\lambda_1^n+ \lambda_2^n+ \dots +\lambda_k^n =0 $ for all $n$ imply that $\lambda_1= \lambda_2= \dots= \lambda_k = 0 $? Suppose $\lambda_1, \lambda_2, \dots, \lambda_k $ are complex numbers that $\forall n \in \mathbb{N}$ satisfy
$$\lambda_1^n+ \lambda_2^n+ \dots +\lambda_k^n =0.$$
Can we deduce that $\lambda_1= \lambda_2= \dots= \lambda_k = 0 $?
 A: For $k = 1$ the result is trivial, and for $k = 2$ note that
$$0 = (\lambda_1 + \lambda_2)^2 - (\lambda_1^2 + \lambda_2^2) = 2 \lambda_1 \lambda_2,$$
so $\lambda_1 = 0$ or $\lambda_2 = 0$. By relabeling we may assume the latter, in which case
$$0 = \lambda_1 + \lambda_2 = \lambda_1 + 0 = \lambda_1,$$ and so, $\lambda_1 = \lambda_2 = 0$.
The general case follows from an induction argument, with induction step similar to the above argument, using Newton's Identities; in fact, we need only the weaker hypothesis that $\sum_{a = 1}^k \lambda_a^n = 0$ for $1 \leq n \leq k$ (rather than for all $n \in \mathbb{N}$).
A: Here is a slightly different approach:
Thanks to @MartinR for catching a bug in my earlier approach.
Let $p$ be a polynomial such that $p(\lambda_i) = |\lambda_i|$ for $i=1,...,k$,
and $p(0) = 0$.
The $p$ has the form $p(x) = \sum_j p_j x^j$. Note that $p_0 = 0$.
Then $\sum_j p_j \sum_i\lambda_i^j = 0 =  \sum_i \sum_j p_j  \lambda_i^j = \sum_i p(\lambda_i) = \sum_i |\lambda_i|$,
hence $\lambda_i = 0$ for all $i$, which is contradiction.
(Note that since $p_0 = 0$, the $p_0(\lambda_1^0+\cdots+ \lambda_n^0)$ term is still
zero, even though $\sum_i \lambda_i^0 = n$).
A: Use Newton Identities to conclude that the elementary symmetric polynomials are all 0.
Use Vieta's Formula to conclude that
$$ \prod ( x - \lambda_i) = x^k $$
Hence, conclude that $\lambda_i = 0$.
A: As shown in this answer,
$$
\det\begin{bmatrix}
\lambda_1&\lambda_1^2&\cdots&\lambda_1^n\\
\lambda_2&\lambda_2^2&\cdots&\lambda_2^n\\
\vdots&\vdots&\ddots&\vdots\\
\lambda_n&\lambda_n^2&\cdots&\lambda_n^n
\end{bmatrix}
=\prod_{j=1}^n\lambda_j\prod_{1\le j\lt k\le n}(\lambda_k-\lambda_j)\tag1
$$
If any $\lambda_j=0$, then simply delete that row. This ensures that $\lambda_j\ne0$. If there are repeated $\lambda_j$, collect them into a single row and let $m_j\ge1$ be the multiplicity. This ensures that $\lambda_j\ne\lambda_k$ for $j\ne k$.
Formula $(1)$ then says that the determinant is not $0$.
The equations in the question can then be written as
$$
\begin{bmatrix}
m_1&m_2&\cdots&m_n
\end{bmatrix}
\begin{bmatrix}
\lambda_1&\lambda_1^2&\cdots&\lambda_1^n\\
\lambda_2&\lambda_2^2&\cdots&\lambda_2^n\\
\vdots&\vdots&\ddots&\vdots\\
\lambda_n&\lambda_n^2&\cdots&\lambda_n^n
\end{bmatrix}
=0\tag2
$$
Since the determinant is not $0$, the vector of multiplicities must be $0$. However, $m_j\ge1$. 
Thus, we cannot have distinct, non-zero $\lambda_j$ so that $(2)$ holds.
Thus, to satisfy the equations in the question, all the $\lambda_j=0$.
