Let $\lambda_1,\ldots,\lambda_n$ be complex numbers such that for each positive integer $k\geq 0$, $$\sum_{i=1}^n \lambda_i^k=0.$$ Here I am supposed to show that $\lambda_i=0$ for each $i\in 1,\ldots,n$. One of my friends said it sounds that I should use Vandermonde determinant. I tried to find an appropriate version of Vandermonde determinant that I may apply here, but I could not. I appreciate if you let me know a suitable version.

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    $\begingroup$ $$\pmatrix{\lambda_1 && \ldots && \lambda_n \\ && \ddots && \\ \lambda_1^n && \ldots && \lambda_n^n } \pmatrix{1 \\ \vdots \\ 1} = \pmatrix{0 \\ \vdots \\ 0}.$$ $\endgroup$ – user2468 Feb 13 '12 at 17:57
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    $\begingroup$ An application of the Vandermonde determinant would be: If for a triangular matrix $A$, the trace of $A^k$ is zero for every $k$, then $A$ is nilpotent. The result follows from the OP's question when one takes $\lambda_i$'s to be its eigenvalues. $\endgroup$ – Abhishek Parab Feb 13 '12 at 19:33

Denote $\mu_1,...,\mu_k$ the non-zero distinct values of $\lambda_i$ (supposing they exist), with (non-zero) multiplicities $m_1,...,m_k$.

Then you have $$\sum_{i=1}^km_i\mu_i^p=0, p=1,2,...k $$

Consider this as a system with unknowns $m_i$. Then the determinant of this system is

$$ V=\left| \begin{matrix} \mu_1 & \mu_2 & \cdots & \mu_k \\ \mu_1^2 & \mu_2^2 & \cdots & \mu_k^2 \\ \vdots & \vdots & \ddots &\vdots \\ \mu_1^k & \mu_2^k & \cdots & \mu_k^k \end{matrix} \right| $$

But since this is a Vandermonde determinant, this determinant is zero if and only if one of $\mu_i$ is zero, or there exists $i \neq j$ such that $\mu_i=\mu_j$. This is a contradiction to the way we have chosen $\mu_i$. Therefore, the assumption that not all $\lambda_i$ are zero gives us a contradiction.

This proves that $\lambda_i=0,\ i=1..n$.

This determinant is zero since the system has non-trivial solution.

  • $\begingroup$ Nice proof. I was just about to post it too :) $\endgroup$ – Isomorphism Feb 13 '12 at 19:24
  • $\begingroup$ I guess we both posted a solution almost simultaneously. $\endgroup$ – Abhishek Parab Feb 13 '12 at 19:26
  • $\begingroup$ Yes, it happens. :) $\endgroup$ – Beni Bogosel Feb 13 '12 at 19:52
  • $\begingroup$ Thanks, that was the thing that I was looking for. $\endgroup$ – Mahmood Al Feb 14 '12 at 7:22

An elegant way to solve this question would be to use this lemma

If $P \in \mathbb{C} [X_1,...X_n]$ is symetric there exists $Q \in \mathbb{C}[X_1,...,X_n]$ such that $P = Q(\Sigma^1,...,\Sigma^n)$ where $\Sigma^i(X_1,...,X_n) = \sum_{k=1}^n{(X_k)^i}$.

Let $P = X_1...X_n$. By previous lemma $P(\lambda_1,...,\lambda_n) = 0$ because the hypothesis tels us that $\forall i$ $\Sigma^i(\lambda_1,...,\lambda_n) = 0$. So at least one of the $\lambda_i$ is $0$.

And the result follows by induction.


It suffices to assume that each $\lambda_i$ is nonzero. (Just ignore the zeros). Suppose there are $r_i$ copies of $\lambda_i$. Assume that $r_i>0$ for all $i$ and $\sum_{i=1}^s r_i = n$.

For $k = 1, \cdots, s$, write the equations $$ \sum_{i=1}^s \lambda^k = 0 $$ gives a matrix similar to the one J.D. mentioned in the comments except with $r_i$ 's instead of 1's. Use the property of the Vandermonde matrix to get a contradiction to the assumption that there is some non-zero $\lambda$.


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