I have an $\frac{N}{2} \times \frac{N}{2}$ matrix $G$ with entries given by

\begin{equation} G_{ij} = \frac{1}{\sin(\frac{\pi}{N}(i+j-\frac{3}{2}))}, \;\;\;\;\;\;\;\; 1 \le i,j \le \frac{N}{2}, \end{equation}

where $N$ is an even integer. By the Fourier transformation the matrix elements can be expressed as

\begin{equation} \begin{split} G_{ij} &= \mathrm{Im} \left[\sum_{n=0}^{N-1} \exp\left(\mathrm{i}\frac{\pi(2n+1)}{N}\left(i+j-\frac{3}{2}\right)\right) \right]\\ &= \sum_{n=0}^{N-1} \sin\left[\frac{\pi(2n+1)}{N}\left(i+j-\frac{3}{2}\right)\right]. \end{split} \end{equation}

Another way to write the entries is as an integral:

\begin{equation} G_{ij} = \frac{1}{\pi} \int_0^\infty \frac{x^{\frac{i+j-3/2}{N}-1}}{1+x} \mathrm{d}x. \end{equation}

Is it possible to find a closed-form expression for the following quantities:

(i) the eigenvalues of $G$ or even its full eigendecomposition?

(ii) $\mathrm{Tr}(G^k)$, for $k \in \mathbb{N}$ (in order to determine, e.g., $\mathrm{Tr}(\log(G))$)?

  • 2
    $\begingroup$ In other words, $$G_{ij} = \frac1{\sin(\frac{\pi}{N}(i+j-\frac32))}.$$ $\endgroup$ Feb 17, 2018 at 5:16
  • 1
    $\begingroup$ That is correct. The entries can also be written as $$ G_{ij} = \mathrm{Im} \left[ \sum_{n=0}^{N-1} \exp\left(\mathrm{i} \frac{\pi(2n+1)}{N} (i+j-3/2)\right) \right] $$ My idea was to use the Fourier transformation to obtain the eigendecomposition. $\endgroup$
    – Marc
    Feb 17, 2018 at 16:04
  • $\begingroup$ @Marc -- apologies, my answer was mistaken, I have deleted it. $\endgroup$ Nov 1, 2019 at 8:59


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