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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 $?

Travis Willse
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4 Answers4

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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 $\lambda_i = 0$.

Calvin Lin
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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 $n \in \mathbb{N}$).

Travis Willse
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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 $\sum_i \lambda_i^0 = n$).

copper.hat
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  • Unless I am mistaken, $\sum_i\lambda_i^j=k$ for $j=0$, so you get $kp_0 = \sum_i |\lambda_i|$. – I came across this Q&A because similar one was just posted here: https://math.stackexchange.com/q/2659274/42969) – Martin R Feb 20 '18 at 21:44
  • @MartinR: Thanks for catching that. I do not have a fix at present. – copper.hat Feb 20 '18 at 21:59
  • @MartinR: I have a fix, a bit ugly (implicitly involves recursion), but it works I believe. – copper.hat Feb 20 '18 at 22:06
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    You can always add the constraint $p(0)=0$, even if some $\lambda_i=0$ (in that case it is already implied by $p(\lambda_i) = |\lambda_i|$). – Martin R Feb 20 '18 at 22:11
  • @MartinR: Good point. I will remove the extraneous cruft from my answer. Thanks again. – copper.hat Feb 20 '18 at 22:15
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    I have taken the liberty to apply your approach to https://math.stackexchange.com/q/2659274/42969 :) – Martin R Feb 20 '18 at 22:39
  • (+1) I like this approach. You added the constraint $p(0)=0$ so that $p_0=0$, but it might help the transparency of your answer if you mentioned this explicitly. – robjohn Feb 21 '18 at 09:30
  • @robjohn: Thanks, good suggestion. – copper.hat Feb 21 '18 at 16:20
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    It's $p_0\left(\lambda_1^0+\cdots+\lambda_n^0\right)$ that needs to be $0$ eventhough $\lambda_1^0+\cdots+\lambda_n^0=n$. $\lambda_1^1+\cdots+\lambda_n^1$ is supposed to be $0$ since $1\in\mathbb{N}$. – robjohn Feb 21 '18 at 16:34
  • @robjohn: Sorry, I was looking at Martin's other answer. Will fix. – copper.hat Feb 21 '18 at 16:41
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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 $\lambda_j=0$.

robjohn
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