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Fedor Petrov
Fedor Petrov
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It looks to be enough to require this for $\lambda=1$.

Indeed, assume that the polynomial $Q(Z)=\sum_{k=0}^n \beta_k Z^k$, $\beta_0=\beta_n=1$, has all roots on the unit circle. Denote the roots by $-\theta_i^{-1}$, $i=1,\ldots,n$. Then $Q(Z)=\prod (1+\theta_i Z)$ and $\prod_i \theta_i=1$.

There exists an Hermitian matrix $(a_{ij})$ with $|a_{ij}|=1$ such that $\theta_i=\prod_{j\ne i} a_{ij}$. Indeed, we may put $a_{ij}=1$ if $\max(i,j)<n$, $a_{in}=\overline{a_{ni}}=\theta_i$ for $i=1,\ldots,n$$i=1,\ldots,n-1$, $a_{nn}=1$. Then $\theta_i=\prod_{j\ne i} a_{ij}$ for all $i=1,\ldots,n-1$ by construction and also for $i=n$ due to $\theta_n=\prod_{i=1}^{n-1} \overline{\theta_i}=\prod_{i=1}^{n-1} a_{ni}$.

Now $$\beta_k=\sum_{|S|=k}\prod_{i\in S} \theta_i=\sum_{|S|=k}\prod_{i\in S} \prod_{j\ne i} a_{ij}=\sum_{|S|=k}\prod_{i\in S} \prod_{j\notin S} a_{ij}$$ since for $j\in S$ the multiples $a_{ij}$ and $a_{ji}$ cancel out.

It looks to be enough to require this for $\lambda=1$.

Indeed, assume that the polynomial $Q(Z)=\sum_{k=0}^n \beta_k Z^k$, $\beta_0=\beta_n=1$, has all roots on the unit circle. Denote the roots by $-\theta_i^{-1}$, $i=1,\ldots,n$. Then $Q(Z)=\prod (1+\theta_i Z)$ and $\prod_i \theta_i=1$.

There exists an Hermitian matrix $(a_{ij})$ with $|a_{ij}|=1$ such that $\theta_i=\prod_{j\ne i} a_{ij}$. Indeed, we may put $a_{ij}=1$ if $\max(i,j)<n$, $a_{in}=\overline{a_{ni}}=\theta_i$ for $i=1,\ldots,n$. Then $\theta_i=\prod_{j\ne i} a_{ij}$ for all $i=1,\ldots,n-1$ by construction and also for $i=n$ due to $\theta_n=\prod_{i=1}^{n-1} \overline{\theta_i}=\prod_{i=1}^{n-1} a_{ni}$.

Now $$\beta_k=\sum_{|S|=k}\prod_{i\in S} \theta_i=\sum_{|S|=k}\prod_{i\in S} \prod_{j\ne i} a_{ij}=\sum_{|S|=k}\prod_{i\in S} \prod_{j\notin S} a_{ij}$$ since for $j\in S$ the multiples $a_{ij}$ and $a_{ji}$ cancel out.

It looks to be enough to require this for $\lambda=1$.

Indeed, assume that the polynomial $Q(Z)=\sum_{k=0}^n \beta_k Z^k$, $\beta_0=\beta_n=1$, has all roots on the unit circle. Denote the roots by $-\theta_i^{-1}$, $i=1,\ldots,n$. Then $Q(Z)=\prod (1+\theta_i Z)$ and $\prod_i \theta_i=1$.

There exists an Hermitian matrix $(a_{ij})$ with $|a_{ij}|=1$ such that $\theta_i=\prod_{j\ne i} a_{ij}$. Indeed, we may put $a_{ij}=1$ if $\max(i,j)<n$, $a_{in}=\overline{a_{ni}}=\theta_i$ for $i=1,\ldots,n-1$, $a_{nn}=1$. Then $\theta_i=\prod_{j\ne i} a_{ij}$ for all $i=1,\ldots,n-1$ by construction and also for $i=n$ due to $\theta_n=\prod_{i=1}^{n-1} \overline{\theta_i}=\prod_{i=1}^{n-1} a_{ni}$.

Now $$\beta_k=\sum_{|S|=k}\prod_{i\in S} \theta_i=\sum_{|S|=k}\prod_{i\in S} \prod_{j\ne i} a_{ij}=\sum_{|S|=k}\prod_{i\in S} \prod_{j\notin S} a_{ij}$$ since for $j\in S$ the multiples $a_{ij}$ and $a_{ji}$ cancel out.

Bounty Ended with 500 reputation awarded by Ofir Gorodetsky
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Fedor Petrov
Fedor Petrov
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It looks to be enough to require this for $\lambda=1$. 

Indeed, assume that the polynomial $Q(Z)=\sum_{k=0}^n \beta_k Z^k$, $\beta_0=\beta_n=1$, has all roots on the unit circle, denote. Denote the roots by $-\theta_i^{-1}$, $i=1,\ldots,n$. Then $Q(Z)=\prod (1+\theta_i Z)$ and $\prod_i \theta_i=1$. 

There exists an Hermitian matrix $(a_{ij})$ with $|a_{ij}|=1$ such that $\theta_i=\prod_{j\ne i} a_{ij}$. Indeed, we may put $a_{ij}=1$ if $\max(i,j)<n$, $a_{in}=\overline{a_{ni}}=\theta_i$ for $i=1,\ldots,n$. Then $\theta_i=\prod_{j\ne i} a_{ij}$ for all $i=1,\ldots,n-1$ by construction and also for $i=n$ due to $\theta_n=\prod_{i=1}^{n-1} \overline{\theta_i}=\prod_{i=1}^{n-1} a_{ni}$.

Now $$\beta_k=\sum_{|S|=k}\prod_{i\in S} \theta_i=\sum_{|S|=k}\prod_{i\in S} \prod_{j\ne i} a_{ij}=\sum_{|S|=k}\prod_{i\in S} \prod_{j\notin S} a_{ij}$$ since for $j\in S$ the multiples $a_{ij}$ and $a_{ji}$ cancel out.

It looks to be enough to require this for $\lambda=1$. Indeed, assume that $Q(Z)=\sum_{k=0}^n \beta_k Z^k$, $\beta_0=\beta_n=1$ has all roots on the unit circle, denote the roots $-\theta_i^{-1}$, $i=1,\ldots,n$. Then $Q(Z)=\prod (1+\theta_i Z)$ and $\prod_i \theta_i=1$. There exists an Hermitian matrix $(a_{ij})$ with $|a_{ij}|=1$ such that $\theta_i=\prod_{j\ne i} a_{ij}$. Indeed, we may put $a_{ij}=1$ if $\max(i,j)<n$, $a_{in}=\overline{a_{ni}}=\theta_i$ for $i=1,\ldots,n$. Then $\theta_i=\prod_{j\ne i} a_{ij}$ for all $i=1,\ldots,n-1$ by construction and for $i=n$ due to $\theta_n=\prod_{i=1}^{n-1} \overline{\theta_i}=\prod_{i=1}^{n-1} a_{ni}$.

Now $$\beta_k=\sum_{|S|=k}\prod_{i\in S} \theta_i=\sum_{|S|=k}\prod_{i\in S} \prod_{j\ne i} a_{ij}=\sum_{|S|=k}\prod_{i\in S} \prod_{j\notin S} a_{ij}$$ since for $j\in S$ the multiples $a_{ij}$ and $a_{ji}$ cancel out.

It looks to be enough to require this for $\lambda=1$. 

Indeed, assume that the polynomial $Q(Z)=\sum_{k=0}^n \beta_k Z^k$, $\beta_0=\beta_n=1$, has all roots on the unit circle. Denote the roots by $-\theta_i^{-1}$, $i=1,\ldots,n$. Then $Q(Z)=\prod (1+\theta_i Z)$ and $\prod_i \theta_i=1$. 

There exists an Hermitian matrix $(a_{ij})$ with $|a_{ij}|=1$ such that $\theta_i=\prod_{j\ne i} a_{ij}$. Indeed, we may put $a_{ij}=1$ if $\max(i,j)<n$, $a_{in}=\overline{a_{ni}}=\theta_i$ for $i=1,\ldots,n$. Then $\theta_i=\prod_{j\ne i} a_{ij}$ for all $i=1,\ldots,n-1$ by construction and also for $i=n$ due to $\theta_n=\prod_{i=1}^{n-1} \overline{\theta_i}=\prod_{i=1}^{n-1} a_{ni}$.

Now $$\beta_k=\sum_{|S|=k}\prod_{i\in S} \theta_i=\sum_{|S|=k}\prod_{i\in S} \prod_{j\ne i} a_{ij}=\sum_{|S|=k}\prod_{i\in S} \prod_{j\notin S} a_{ij}$$ since for $j\in S$ the multiples $a_{ij}$ and $a_{ji}$ cancel out.

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Fedor Petrov
Fedor Petrov
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It looks to be enough to require this for $\lambda=1$. Indeed, assume that $Q(Z)=\sum_{k=0}^n \beta_k Z^k$, $\beta_0=\beta_n=1$ has all roots on the unit circle, denote the roots $-\theta_i$$-\theta_i^{-1}$, $i=1,\ldots,n$. Then $Q(Z)=\prod (Z+\theta_i)$$Q(Z)=\prod (1+\theta_i Z)$ and $\prod_i \theta_i=1$. There exists an Hermitian matrix $(a_{ij})$ with $|a_{ij}|=1$ such that $\theta_i=\prod_{j\ne i} a_{ij}$. It worksIndeed, we may put $a_{ij}=1$ if $\max(i,j)<n$, $a_{in}=\overline{a_{ni}}=\theta_i$ for $i=1,\ldots,n$. Then $\theta_i=\prod_{j\ne i} a_{ij}$ for all $i=1,\ldots,n-1$ by construction and for $i=n$ due to $\theta_n=\prod_{i=1}^{n-1} \overline{\theta_i}=\prod_{i=1}^{n-1} a_{ni}$.

Now $$\beta_k=\sum_{|S|=k}\prod_{i\in S} \theta_i=\sum_{|S|=k}\prod_{i\in S} \prod_{j\ne i} a_{ij}=\sum_{|S|=k}\prod_{i\in S} \prod_{j\notin S} a_{ij}$$ since for $j\in S$ the multiples $a_{ij}$ and $a_{ji}$ cancel out.

It looks to be enough to require this for $\lambda=1$. Indeed, assume that $Q(Z)=\sum_{k=0}^n \beta_k Z^k$, $\beta_0=\beta_n=1$ has all roots on the unit circle, denote the roots $-\theta_i$, $i=1,\ldots,n$. Then $Q(Z)=\prod (Z+\theta_i)$ and $\prod_i \theta_i=1$. There exists an Hermitian matrix $(a_{ij})$ with $|a_{ij}|=1$ such that $\theta_i=\prod_{j\ne i} a_{ij}$. It works.

It looks to be enough to require this for $\lambda=1$. Indeed, assume that $Q(Z)=\sum_{k=0}^n \beta_k Z^k$, $\beta_0=\beta_n=1$ has all roots on the unit circle, denote the roots $-\theta_i^{-1}$, $i=1,\ldots,n$. Then $Q(Z)=\prod (1+\theta_i Z)$ and $\prod_i \theta_i=1$. There exists an Hermitian matrix $(a_{ij})$ with $|a_{ij}|=1$ such that $\theta_i=\prod_{j\ne i} a_{ij}$. Indeed, we may put $a_{ij}=1$ if $\max(i,j)<n$, $a_{in}=\overline{a_{ni}}=\theta_i$ for $i=1,\ldots,n$. Then $\theta_i=\prod_{j\ne i} a_{ij}$ for all $i=1,\ldots,n-1$ by construction and for $i=n$ due to $\theta_n=\prod_{i=1}^{n-1} \overline{\theta_i}=\prod_{i=1}^{n-1} a_{ni}$.

Now $$\beta_k=\sum_{|S|=k}\prod_{i\in S} \theta_i=\sum_{|S|=k}\prod_{i\in S} \prod_{j\ne i} a_{ij}=\sum_{|S|=k}\prod_{i\in S} \prod_{j\notin S} a_{ij}$$ since for $j\in S$ the multiples $a_{ij}$ and $a_{ji}$ cancel out.

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Fedor Petrov
Fedor Petrov
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