Timeline for The proof of the invertibility of $\Big( \sin\frac{8kl\pi}{2n+1} \Big)_{k,l=1}^\frac{n}{2}$
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Dec 28, 2022 at 13:44 | comment | added | ABB | After all, you can imagine the anti-diagonal matrix to check that S-M formula does not hold! | |
| Dec 28, 2022 at 13:42 | comment | added | ABB | But I think the formula $\det M= der A\cdot det(D-CA^{-1}B)$ holds if both $M$ and $A$ are invertible. When $M$ is considered as DST-I, one may check (some details are needed) $A$ is invertible as well. So one may apply S-M formula for the first step. But when we block-wise $A$ into four blocks, we are no longer sure concerning the invertibility of the first block. | |
| Dec 28, 2022 at 13:33 | comment | added | Steve Huntsman | The answer by Denis Serre points out that by S-M, $\det M = \det A \cdot \det(D-CA^{-1}B)$ where $M = \left ( \begin{smallmatrix} A & B \\ C & D \end{smallmatrix} \right )$. We are extracting an $A$ block two times, with the first $M$ as DST-I with rows and columns permuted, so that $\det M \ne 0$ both times, so $\det A \ne 0$ both times, hence each $A$ is invertible. The second $A$ is the matrix you want. | |
| Dec 28, 2022 at 7:02 | comment | added | ABB | I am really following to get the result but could not yet. Do you address "Equality of the determinants of certain submatrices of an orthogonal matrix" for the version of S-M theorem that you are applying? If yes, I can not see the required assumption to get the desired result! Please give some more clear explanation or let us know what version of S-M you are using. | |
| Dec 27, 2022 at 21:13 | comment | added | Steve Huntsman | 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. | |
| Dec 27, 2022 at 19:59 | comment | added | ABB | You are applying twice of the Sherman-Morrison. In the first time, it is okay, but why do you allow to apply it for the second time? What is ambient matrix in the second time? | |
| Dec 27, 2022 at 19:49 | comment | added | Steve Huntsman | DST-I is the ambient matrix. | |
| Dec 27, 2022 at 19:12 | comment | added | ABB | What is the ambient matrix here? | |
| Dec 27, 2022 at 18:37 | comment | added | Steve Huntsman | The Sherman-Morrison part implies nonzero determinant because the ambient matrix has nonzero determinant. | |
| Dec 27, 2022 at 18:35 | comment | added | ABB | If diagonal entries of the matrix $W$ are all non-zero, it seems that det$X\neq0$, does not it?! | |
| Dec 27, 2022 at 17:32 | comment | added | ABB | 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$X\neq0$. | |
| Dec 27, 2022 at 17:32 | comment | added | ABB | 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. | |
| Dec 27, 2022 at 15:23 | comment | added | Steve Huntsman | Note that I am also using the Sherman-Morrison result in the MO link | |
| Dec 27, 2022 at 14:35 | history | answered | Steve Huntsman | CC BY-SA 4.0 |