Roots to a Special Homogeneous Polynomial Equations

Question

Consider the following system of polynomial equations: $$\left\{ \begin{array}{rcl} x_1+x_2+\cdots +x_n&=&0 \\ x_1^2+x_2^2+\cdots +x_n^2&=&0 \\ &\vdots & \\ x_1^n+x_2^n+\cdots+x_n^n&=&0 \end{array}\right. $$ Obviously $X=(x_1,\cdots,x_n)=(0,\ldots,0)$ is a root. But is this the only one in $\mathbb{C}^n$, where $\mathbb{C}$ is the complex plane. In other words, does there exist a non-zero root to the system?

Answer 1

Due to Newton identities all elementary symmetric polynomials on $x_1,x_2,...,x_n$ are also zero. Therefore, by Vieta's formulas, $x_1,x_2,..., x_n$ are the roots of the polynomial $x^n=0$.

Hence, all of them are zero.

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Define $$e_k=\sum_{1\leq i_1<i_2<...<i_k\leq n}x_{i_1}x_{i_2}...x_{i_k}$$

These are called the elementary symmetric functions on the variables $x_1,...,x_n$.

They appear in the Vieta's formulas as the coefficients of a monic polynomial in terms of its roots.

$$P(x)=(x-x_1)(x-x_2)...(x-x_n)=x^n-e_1x^{n-1}+e_2x^{n-2}+....+(-1)^ne_n$$

The polynomials that you have in your equations are also symmetric. Often denoted as $$p_k=\sum_{i=1}^{n}x_i^k$$

Newton formulas allow you to relate the values of $e_1,...,e_n$ to the values of $p_1,...,p_n$. Since your system of equations is saying that all the $p_k$ are zero, for $k=1,...,n$, then the Newton identities tell you that also $e_1=e_2=...=e_n=0$. But then you see that the polynomial $P(x)$, of which $x_1,x_2,...,x_n$ are roots, is equal to $P(x)=x^n$.

Since this polynomial only has $x=0$ as a root, it follows that $x_1=x_2=...=x_n=0$.

Answer 2

Alt.hint: consider the linear system in $\,u_1x_1, u_2x_2, \ldots,u_nx_n\,$:

$$\left\{ \begin{array}{rcl} 1 \cdot u_1x_1+1 \cdot u_2x_2+\ldots +1 \cdot u_nx_n &=&0 \\ x_{1}\cdot u_1x_1+x_{2}\cdot u_2x_2+\ldots +x_{n}\cdot u_nx_n&=&0 \\ &\vdots & \\ x_{1}^{n-1}\cdot u_1x_1+x_{2}^{n-1}\cdot u_2x_2+\ldots+x_{n}^{n-1}\cdot u_nx_n&=&0 \end{array}\right. $$

Assuming not all $\,x_i\,$ are $\,0\,$, the system has the non-trivial solution $\,u_1=u_2=\ldots=u_n=1\,$, so its determinant must be $\,0\,$. But the respective determinant is a Vandermonde determinant which evaluates to $\,\prod_{1 \le i \lt j \le n} (x_j-x_i)\,$, so it can only be zero if $\,x_i=x_j\,$ for some pair $\,i \ne j\,$.

Assume WLOG that $\,x_{n-1}=x_{n}\,$, then consider:

$$\left\{ \begin{array}{rcl} 1\cdot u_1x_1+1 \cdot u_2x_{2}+\ldots +1 \cdot u_{n-1}x_{n-1} &=&0 \\ x_{1}\cdot u_1x_1+x_{2}\cdot u_2x_2+\ldots +x_{n-1}\cdot u_{n-1}x_{n-1}&=&0 \\ &\vdots & \\ x_{1}^{n-2}\cdot u_1x_1+x_{2}^{n-2}\cdot u_2x_2+\ldots+x_{n-1}^{n-2}\cdot u_{n-1}x_{n-1}&=&0 \end{array}\right. $$

Again, if not all $\,x_i\,$ are $\,0\,$, then $\,u_1=u_2=\ldots=u_{n-2}=1, u_{n-1}=2\,$ is a non-trivial solution, which implies that another pair of $\,x_i=x_j\,$.

It follows by induction that all $\,x_i\,$ must be equal, and therefore all must be $\,0\,$.

Answer 3

Assume there is a nonzero solution. If some variables are zero, forget them. If two variables are equal say $x_1=x_2$ replace $x_1^m+x_2^m+\cdots$ by $2x_1^m+\cdots$, or more generally, $n_1x_1^m+\cdots$ if there are $n_1$ many variables equal to $x_1$. This leads to having the system $x_1,\ldots , x_k$ of distinct and nonzero numbers such that $$n_1x_1+\cdots +n_kx_k=0$$ $$n_1x_1^2+\cdots +n_kx_k^2=0$$ $$\vdots$$ $$n_1x_1^k+\cdots +n_kx_k^k=0$$ for some multiplicities $n_1,\ldots ,n_k$ then

$$\begin{vmatrix}x_1&\cdots &x_k\\ &\vdots&\\ x_1^k&\cdots&x_k^k \end{vmatrix}=0$$ contradicting the assumption.