Suppose that $n$ is even. Any suggestion/appraoch to prove that $S=\Big( \sin\frac{8kl\pi}{2n+1} \Big)_{k,l=1}^\frac{n}{2}$ is invertible?
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1$\begingroup$ It is a submatrix of the discrete Sine transform of type V. The invertibility of this submatrix will lead me to analyze this type of DST. $\endgroup$– ABBCommented Dec 27, 2022 at 11:34
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$\begingroup$ Yes, and I edited in the question. $\endgroup$– ABBCommented Dec 27, 2022 at 11:49
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1$\begingroup$ You might add to the question your reason for being interested in this, and (if any) your attempts to find whether it is already well-known. $\endgroup$– Gerald EdgarCommented Dec 27, 2022 at 11:51
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The DST-I matrix of size $4N$ is $\left ( \sin \frac{\pi k \ell}{4N+1} \right )_{k,\ell = 1}^{4N}$. Up to scale, it is orthogonal. Slightly changing notation, your matrix is $\left ( \sin \frac{8 \pi k \ell}{4N+1} \right )_{k,\ell = 1}^{N}$. By permuting rows and then columns of the DST-I matrix, orthogonality is preserved; by doing this suitably, your matrix can be regarded as a block of a 2x2 block which is itself a block of a 2x2 block. Applying Equality of the determinants of certain submatrices of an orthogonal matrix twice then yields the desired result.
answered Dec 27, 2022 at 14:35
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$\begingroup$ Note that I am also using the Sherman-Morrison result in the MO link $\endgroup$ Commented Dec 27, 2022 at 15:23
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$\begingroup$ Many thanks for your comments. Actually, I could not get the result by this idea yet. As you mentioned (up to some permutations) we may first represent DST-I=$\Big(\begin{array}{cc} A_{11} & A_{12} \\ A_{21} & A_{22}\end{array}\Big)$ where all submatrices/blocks $A_{ij}$s are $2N\times 2N$ matrices and $A_{11}$ includes all entries $s_{i,j}$ of DST-I such that both $i,j$ are even. Up to here, one may check that $A_{11}$ is still invertible. $\endgroup$– ABBCommented Dec 27, 2022 at 17:32
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$\begingroup$ We may also assume that $A_{11}=\Big(\begin{array}{cc}X & Y \\ Z & W\end{array}\Big)$ where $X$ is the matrix $S$ in the problem. I guess we may only conclude that det$X$=det$W$. So, please give some more details why det$X\neq0$. $\endgroup$– ABBCommented Dec 27, 2022 at 17:32
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$\begingroup$ If diagonal entries of the matrix $W$ are all non-zero, it seems that det$X\neq0$, does not it?! $\endgroup$– ABBCommented Dec 27, 2022 at 18:35
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1$\begingroup$ The first time, the ambient matrix is the $4N \times 4N$ DST-I matrix; the $2N \times 2N$ matrix obtained from this has nonzero determinant by Sherman-Morrison. The second time, the ambient matrix is the $2N \times 2N$ matrix just obtained; the $N \times N$ matrix you are interested in has nonzero determinant by S-M. Then you are done. $\endgroup$ Commented Dec 27, 2022 at 21:13