0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
+50
$\begingroup$

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 $n\geq 2$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.