Let $n$ be a natural number and $n\geq2$. Denote $\sigma=\exp\left(\dfrac{2\pi i}{n+1}\right)$ and $\omega=\exp\left(\dfrac{2\pi i}{n}\right)$ and $\{\phi_1,\cdots,\phi_n, \psi_1,\cdots,\psi_n\}$ be $2n$ angles. Now consider the following complex number:\begin{align*} C&=n+1+2(e^{-i\phi_1},e^{-i\phi_2},\cdots,e^{-i\phi_n}) · \begin{pmatrix} e^{i\psi_1}\\ e^{i\psi_2}\\ \vdots\\ e^{i\psi_n} \end{pmatrix}\\ &\quad+(\sigma^{-1},\sigma^{-2},\cdots,\sigma^{-n})\begin{pmatrix} 1&1&1&\cdots&1\\ 1&\sigma^{-1}&\sigma^{-2}&\cdots&\sigma^{-(n-1)}\\ 1&\sigma^{-2}&\sigma^{-4}&\cdots&\sigma^{-2(n-1)}\\ \vdots&\vdots&\vdots&&\vdots\\ 1&\sigma^{-(n-1)}&\sigma^{-2(n-1)}&\cdots&\sigma^{-(n-1)^2} \end{pmatrix}^{-1} ·\\ &\quad\begin{pmatrix} 1&1&1&\cdots&1\\ 1&\omega^{-1}&\omega^{-2}&\cdots&\omega^{-(n-1)}\\ 1&\omega^{-2}&\omega^{-4}&\cdots&\omega^{-2(n-1)}\\ \vdots&\vdots&\vdots&&\vdots\\ 1&\omega^{-(n-1)}&\omega^{-2(n-1)}&\cdots&\omega^{-(n-1)^2} \end{pmatrix}\begin{pmatrix} e^{i\psi_1}\\ e^{i\psi_2}\\ \vdots\\ e^{i\psi_n} \end{pmatrix}\\ &\quad+(e^{-i\phi_1},e^{-i\phi_2},\cdots,e^{-i\phi_n})\begin{pmatrix} 1&1&1&\cdots&1\\ 1&\omega^{1}&\omega^{2}&\cdots&\omega^{(n-1)}\\ 1&\omega^{2}&\omega^{4}&\cdots&\omega^{2(n-1)}\\ \vdots&\vdots&\vdots&&\vdots\\ 1&\omega^{(n-1)}&\omega^{2(n-1)}&\cdots&\omega^{(n-1)^2} \end{pmatrix} ·\\ &\quad\begin{pmatrix} 1&1&1&\cdots&1\\ 1&\sigma^{1}&\sigma^{2}&\cdots&\sigma^{(n-1)}\\ 1&\sigma^{2}&\sigma^{4}&\cdots&\sigma^{2(n-1)}\\ \vdots&\vdots&\vdots&&\vdots\\ 1&\sigma^{(n-1)}&\sigma^{2(n-1)}&\cdots&\sigma^{(n-1)^2} \end{pmatrix}^{-1}\begin{pmatrix} \sigma\\ \sigma^{2}\\ \vdots\\ \sigma^{_n} \end{pmatrix}. \end{align*} I guess this complex number $C$ cannot be equal to zero. But I do not know how to prove that $C\neq0$. Do you have any idea about the complicated number?

Thank you.

  • $\begingroup$ I think maybe the matrix norm can do something about this question. Unfortunately, I am not farmilar with the norm of matrices. $\endgroup$ – liwolf May 8 '18 at 9:00

At the first glance, I could not see any reason why $C=0$ would be impossible. Let's see if we can actually show it to be possible.

We are not really dealing with the angles $\phi_1,\ldots,\phi_n$ and $\psi_1,\ldots,\psi_n$ but rather with the corresponding complex numbers of modulus $1$, so it will be useful to name these explicitly: $u_k:=e^{i\phi_k}$ and $v_k:=e^{i\psi_k}$.

Most of the matrices only depend on $n$ and it turns out their products can be simplified considerably: $$C=(n+1)+2\sum_{k=1}^n v_ku_k^{-1}+\alpha\sum_{k=1}^n \omega^{-k}v_k+\alpha^{-1}\sum_{k=1}^n \omega^ku_k^{-1}$$ where $\alpha:=\exp\left(\dfrac{2i\pi}{n(n+1)}\right)$ is the $n(n+1)$-th root of unity.

We can simplify it even further: If we let $x_k:=\alpha^{-1}\omega^ku_k^{-1}$ and $y_k:=\alpha\omega^{-k}v_i$ (note that $x_i$ and $y_i$ are also complex numbers of modulus $1$), the whole expression simplifies as: $$C=(n+1)+\sum_{k=1}^n \left(x_k+y_k+2x_ky_k\right)$$ or $$C=(n+1)+\frac{1}{2}\sum_{k=1}^n \left((2x_k+1)(2y_k+1) - 1\right)$$

Setting this equal to zero yields: $$\sum_{k=1}^n (2x_k+1)(2y_k+1)=-(n+2)$$

We have more than enough freedom to find some solutions; we can even set $y_k=(-1)$ for all $k$ and still find some. Doing so reduces the equation to $$\sum_{k=1}^n x_k=1$$ and one of its many solutions is $x_k=\rho^{k-1}$, where $\rho=\exp\left(\dfrac{2i\pi}{n-1}\right)$.

Substituting all the way back gives us $$ \begin{align*} \phi_k & = & 2\pi\left(\frac{1}{n+1}-\frac{k-1}{n(n-1)}\right)\\ \psi_k & = & 2\pi\left(\frac{1}{2} + \frac{k}{n}-\frac{1}{n(n+1)}\right) \end{align*}$$

which indeed show $C=0$ is a possibility for any choice of $n\geq 2$.

  • $\begingroup$ Thankyou@Peter Košinár. Yesterday, I had got a similar formular about $C$. With a little different that the $"\alpha"$ in my formular is $\sigma^{-1}$. But I only compute this. Thanks for your following answer. $\endgroup$ – liwolf May 13 '18 at 12:42
  • $\begingroup$ @liwolf Indeed, $\alpha\times\omega^{-1}=\sigma^{-1}$, so that's another way to rewrite the sum (by having $k$ go from $0$ to $(n-1)$ instead of $1$ to $n$, or by adjusting the exponent of $\omega$ by one). Nevertheless, it was an interesting expression you were looking at! Would you mind sharing some context about it? (i.e. what problem were you looking into when you encountered it?) $\endgroup$ – Peter Košinár May 13 '18 at 22:34
  • $\begingroup$ I'm so sorry to reply you several days later @Peter Košinár. I am happy to share the context to you. Well, since $C$ might be equal to zero, my original question becomes difficult again. If you are interested in the question, I'd like to send you an email. Thankyou. $\endgroup$ – liwolf May 19 '18 at 10:16

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