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Here is a statement that seems prima facie obvious, but when I try to prove it, I am lost.

Let $x_1 , x_2, \dots, x_k$ be complex numbers satisfying:

$$x_1 + x_2+ \dots + x_k = 0$$

$$x_1^2 + x_2^2+ \dots + x_k^2 = 0$$

$$x_1^3 + x_2^3+ \dots + x_k^3 = 0$$

$$\dots$$

Then $x_1 = x_2 = \dots = x_k = 0$.


The statement seems obvious because we have more than $k$ constraints (constraints that are in some sense, "independent") on $k$ variables, so they should determine the variables uniquely. But my attempts so far of formalizing this intuition have failed. So, how do you prove this statement? Is there a generalization of my intuition?

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3 Answers 3

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Apply Newton's identities. This shows the power sums form a basis for the symmetric polynomials, since the elementary symmetric polynomials are a basis and the elementary symmetric polynomials can be represented as linear combinations of power sums. So all symmetric polynomials evaluate to $0$ at the $x_i$. Consider a polynomial with the $x_i$ as roots. The coefficients are the elementary symmetry polynomials in the $x_i$, so they are all zero. The polynomial is then $x^k$, which shows $x_i=0$ for all $i$.

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For a slightly different method than Potato's second answer (but the idea is mainly the same):

Without loss of generality, the system of equations can be written as: $$\left\{ \begin{array}{lcl} \lambda_1x_1 + \lambda_2x_2+ \dots + \lambda_k x_k &= &0 \\ \lambda_1x_1^2 + \lambda_2x_2^2+ \dots + \lambda_k x_k^2 & = & 0 \\ & \vdots & \\ \lambda_1x_1^k + \lambda_2x_2^k+ \dots + \lambda_k x_k^k & = & 0 \end{array} \right.$$

where $\lambda_i>0$, $x_i \neq 0$ and $x_i \neq x_j$ for $i \neq j$. Indeed, if $x_i=x_j$ replace $x_i+x_j$ with $2x_i$ and if $x_i=0$ just remove it. By contradiction, suppose $k \geq 1$.

Now, the family $\{ (\lambda_1 x_1^j, \dots , \lambda_k x_k^j) \mid 1 \leq j \leq k \}$ cannot be linearly independent since the vector space $\{(y_1,\dots, y_k) \mid y_1+ \dots+ y_k=0 \}$ has dimension $k-1$ (it is a hyperplane). Therefore, the matrix

$$A:=\left( \begin{matrix} \lambda_1x_1 & \lambda_2x_2 & \dots & \lambda_k x_k \\ \lambda_1x_1^2 & \lambda_2x_2^2 & \dots & \lambda_k x_k^2 \\ \vdots & \vdots & & \vdots \\ \lambda_1x_1^k & \lambda_2x_2^k & \dots & \lambda_k x_k^k \end{matrix} \right)$$

is not invertible. Using Vandermonde formula, $$0= \det(A)= \prod\limits_{i=1}^k \lambda_i \cdot \prod\limits_{i=1}^k x_i \cdot \prod\limits_{i<j} (x_i-x_j) $$

which is nonzero by assumption.

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Rename the $x_i$ as $\alpha_i$ so that I can borrow the notation of this wikipedia article. Multiply the first row of the matrix in the article by $\alpha_1$, the second by $\alpha_2$, and so on. This gives a matrix with determinant

$$\left( \prod_i \alpha_i \right)\left(\prod_{i< j} (\alpha_i-\alpha_j) \right).$$

Since the homogeneous equation with these coefficients has a solution (the vector $(1,1,\cdots,1)$), the determinant must be zero. So either $\alpha_i=0$ for some $i$ or $\alpha_i=\alpha_j$ for some pair of distinct indices. In the first case, we reduce to the same problem with $k-1$ terms and can induct. In the second case, we can consider a slightly modified problem: identifying the $i$th and $j$th term, we have the same problem with $n-1$ variables, except the coefficient of $\alpha_j$ is $2$. But, since this still gives a nonzero solution to the corresponding homogeneous equation, we can again "induct" using the method described above. The number of variables must eventually decrease to one, from which we see $\alpha_1=0$, and hence $x_i=0$ for all $i$ in the original problem.

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  • $\begingroup$ Therefore is enough to know $x_1^i+x_2^i+\ldots+x_k^i=0$ for $i=1,2,\ldots,k$. $\endgroup$
    – P..
    May 23, 2013 at 7:56
  • $\begingroup$ @P.. Yes. This is also the case in the other solution I posted. $\endgroup$
    – Potato
    May 23, 2013 at 7:57
  • $\begingroup$ Indeed. With this solution however it can be generalized to: If $c_1x_1^i+c_2x_2^i+\ldots+c_kx_k^i=0$ for all $i=1,2,\ldots,k$ and $c_1,c_2,\ldots,c_k\in\mathbb C$ such that ... $\endgroup$
    – P..
    May 23, 2013 at 8:06
  • $\begingroup$ @P.. Interesting! $\endgroup$
    – Potato
    May 23, 2013 at 8:08

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