$\forall k\in\mathbb{Z},\sum_{i=1}^n\lambda_i^k=(\sum_{i=1}^n\lambda_i)^k$ implies at most one number of $\{\lambda_i\}$ doesn't equal $0$?

Question

Given $\{\lambda_i:\lambda_i\in\mathbb{R},1\leq i\leq n\}$. If $$\forall k\in\mathbb{Z^+},\sum_{i=1}^n\lambda_i^k=\left(\sum_{i=1}^n\lambda_i\right)^k,$$ does this imply at most one number of $\{\lambda_i\}$ doesn't equal $0$?

I attempted to use the Newton's identity and proved the case for $n=2$ and $n=3$, however for $n=4$ it becomes complicated. When $n=4$, it is (use $a,b,c,d$ for simplicity): $$4abcd=(abc+acd+bcd+abd)(a+b+c+d)-(ab+ac+ad+...)(a^2+b^2+c^2+d^2).$$

Thanks for the answer. I have one more question: can this be extended to $\{\lambda_i\}\in\mathbb{C}$?

Answer 1

The assumptions imply $$\left (\sum \lambda_i^{2k}\right)^{1/2k} =\left |\sum \lambda_i\right |$$ Therefore $$\left (\sum \lambda_i^{2}\right)^{1/2} =\left (\sum \lambda_i^{4}\right )^{1/4}$$ i.e. $$\left (\sum \lambda_i^{2}\right)^2=\sum \lambda_i^{4}$$ Thus at most one summand can be nonzero, since the last equation implies that the pairwise products $\lambda_i\lambda_j$ are all zero.

Answer 2

The answer is “yes” even if $\mathbb R$ is replaced by an arbitrary field $\mathbb F$ such that either $\operatorname{char}(\mathbb F)=0$ or $\operatorname{char}(\mathbb F)>n$.

Let $\lambda_{n+1}=0$. The condition in your question now becomes $\sum_{i=1}^{n+1}\lambda_i^k=\left(\sum_{i=1}^{n+1}\lambda_i\right)^k$ for every positive integer $k$. If we put $A=\operatorname{diag}(\lambda_1,\ldots,\lambda_{n+1})$, we obtain $\operatorname{tr}(A^k)=\operatorname{tr}(A)^k$ for every positive integer $k$. In turn, we have $\operatorname{tr}\big(\sum_{k=1}^{n+1}c_kA^k\big)=\sum_{k=1}^nc_k\operatorname{tr}(A^k)=\sum_{k=1}^{n+1}c_k\operatorname{tr}(A)^k$ for any scalars $c_1,c_2,\ldots,c_{n+1}$. That is, $\operatorname{tr}(p(A))=p(\operatorname{tr}(A))$ for every polynomial $p$ of degree $n+1$ with a zero constant term.

In particular, if we take $p$ as the characteristic polynomial of $A$ (which does have a zero constant term because $A$ is singular), we obtain $p(\operatorname{tr}(A))=\operatorname{tr}(p(A))= 0$. Hence $\operatorname{tr}(A)$ is an eigenvalue of $A$, i.e., $\operatorname{tr}(A)=\lambda_j$ for some $j$. It follows that $\sum_{i=1}^{n+1}\lambda_i^k=\left(\sum_{i=1}^n\lambda_i\right)^k=\lambda_j^k$ for every $k\ge1$, i.e., $\sum_{i\ne j}\lambda_i^k=0$ for every $k\ge1$. So, if we put $\Lambda=\operatorname{diag}(\lambda_1,\ldots,\lambda_{j-1},\lambda_{j+1},\ldots,\lambda_n)$, we obtain $\operatorname{tr}(\Lambda^k)=0$ for all $k\ge1$. Since $\Lambda$ is $n\times n$ and $\operatorname{char}(\mathbb F)$ is either zero or greater than $n$, the previous trace condition implies that $\Lambda$ is nilpotent. Thus $\lambda_j$ is the only possible nonzero value among all $\lambda_i$s.

Answer 3

Define the following Hankel Matrix
$H_A=\begin{pmatrix} \text{trace}\big(A^1\big) & \text{trace}\big(A^2\big) & \text{trace}\big(A^3\big) & \cdots & \text{trace}\big(A^{n}\big) \\ \text{trace}\big(A^2\big) & \text{trace}\big(A^3\big)& \text{trace}\big(A^4\big) & \cdots &\text{trace}\big(A^{n+1}\big) \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \text{trace}\big(A^{n-1}\big) & \text{trace}\big(A^{n}\big) &\text{trace}\big(A^{n+1}\big) & \cdots & \text{trace}\big(A^{2n-2}\big)\\ \text{trace}\big(A^{n}\big) & \text{trace}\big(A^{n+1}\big) &\text{trace}\big(A^{n+2}\big) & \cdots & \text{trace}\big(A^{2n-1}\big) \end{pmatrix}$
What OP's condition says is that $\text{rank}(H_A)\leq 1$ since row $i$ is a scalar multiple of the row above for $2\leq i\leq n$ [and said scalar multiple is $\text{trace}\big(A\big)$ in all cases] and the rank of $H_A$ tells us the number of unique non-zero eigenvalues of $A$.

proof:
Let $r$ equal the number of distinct non-zero eigenvalues of $A$ (over a splitting field) and $V:=\begin{bmatrix} 1 & \lambda_0 & \lambda_0^2 & \dots & \lambda_0^{n-1}\\ 1 & \lambda_1 & \lambda_1^2 & \dots & \lambda_1^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \\ 1 & \lambda_{r-1} & \lambda_{r-1}^{2} & \dots & \lambda_{r-1}^{n-1} \end{bmatrix} =\bigg[\begin{array}{c|c|c|c} \mathbf v_0 & \mathbf v_1 &\mathbf v_2 &\cdots & \mathbf v_{n-1} \end{array}\bigg] $
where $V$ is (short and fat) Vandermonde with distinct rows hence surjective. And let $D$ be an $r\times r$ diagonal matrix with $d_{k,k}=m_k\cdot \lambda_k$ where $m_k$ is the respective algebraic multiplicty for eigenvalue $\lambda_k$. Then

Then $H_A=V^TD V $ because for $i, j\in \big\{0,1,\dots, n-1\big\}$ we have $\mathbf e_i^T H_A \mathbf e_j = \text{trace}\big(A^{i+j+1}\big) =\sum_{k=0}^{r-1} m_k\cdot\lambda_k^{i+j+1} =\sum_{k=0}^{r-1} (\lambda_k^{i+1}\cdot m_k)\cdot \lambda_k^{j}=\big(\mathbf v_i^TD\big) \mathbf v_j = \mathbf e_i^T V^T D V \mathbf e_j $

Thus by surjectivity of $V$
$\text{rank}\big(H_A\big) = \text{rank}\big(V^TDV\big)=\text{rank}\big(D\big)= r$

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Technically to finish, we have to flush out the case of $\text{rank}\big(H_A\big)=1$
i.e. for single distinct non-zero eigenvalue $\lambda_0$ with $k:=2$ we have
$m_0\cdot \lambda_0^2 +0 +0 +\cdots +0=\big(m_0\cdot \lambda_0+0 +0 +\cdots +0\big)^2$
$\implies m_0\in \big\{0,1\big\}\implies m_0= 1$
i.e. the sole non-zero eigenvalue has algebraic multiplicity of 1 as claimed.

remark on other fields:
The result in the proof holds over any field except we only know $\text{rank}\big(D\big)\leq r$ when working in prime characteristic $p$ because $m_k$ may be a positive integer that is zero modulo $p$, reducing the rank of $D$ to something $\lt r$. Thus for interpreting $\text{rank}\big(H_A\big)$ in positive characteristic we may insist that $p\gt n$. (This also ensures the 'technically to finish' works as intended since integer $m_0$ modulo $p$ is $\in \big\{0,1\big\}$ but $m_0\in \big\{1,2,\dots, n\big\}\subseteq \big\{1,2,\dots, p-1\big\}$ so $m_0=1$.)

Answer 4

Suppose by way of contradiction that $n\geq 2,\ \lambda_i\in\mathbb{R}\ \forall\ 1\leq i\leq n\ $ and $$\forall\ k\in\mathbb{Z^+},\quad\sum_{i=1}^n{\lambda_i}^k=\left(\sum_{i=1}^n\lambda_i\right)^k,$$ and that (WLOG) $\lambda_1 \neq 0$ and $\lambda_2 \neq 0.$ Then,

$$ \left(\sum_{i=1}^n\lambda_i\right)^4 = \left(\left(\sum_{i=1}^n\lambda_i\right)^2\right)^2, $$

which by supposition, implies,

$$ \sum_{i=1}^n{\lambda_i}^4 = \left( \sum_{i=1}^n{\lambda_i}^2 \right)^2 = \sum_{i=1}^n{\lambda_i}^4 + \underbrace{2{\lambda_1}^2 {\lambda_2}^2}_{>0} + \text{ non-negative terms }, $$

a contradiction.