Timeline for Determinants: periodic entries $0,1,2,3$
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| Mar 11, 2019 at 6:26 | vote | accept | T. Amdeberhan | ||
| Sep 29, 2017 at 6:36 | answer | added | darij grinberg | timeline score: 5 | |
| Sep 24, 2017 at 2:24 | history | edited | darij grinberg |
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| Sep 24, 2017 at 2:13 | history | edited | darij grinberg |
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| Sep 22, 2017 at 16:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 23, 2017 at 19:32 | comment | added | darij grinberg | Equivalently, $\left(M_n\right)_{i,j} \equiv j-i+1 + 2q\left(n-i-j+1\right) \mod 4$, where $q \left(m\right) = \left[m < 0\right] m$ for each integer $m$ (where we are using the Iverson bracket notation). | |
| Aug 23, 2017 at 18:07 | comment | added | darij grinberg | Here is a sort-of-explicit formula for each entry of $M_n$: We have $\left(M_n\right)_{i,j} \equiv 1+i+j + 2 \begin{cases} i-1, & \text{if } i+j-1 \leq n; \\ n-j, & \text{if } i+j-1 \geq n \end{cases} \mod 4$. | |
| Aug 23, 2017 at 17:54 | comment | added | darij grinberg | An observation about the matrix $M_n$ (easy to prove by induction): It has the property that $\left(M_n\right)_{i,j} \equiv \left(M_n\right)_{i+1,j-1} + 2 \mod 4$ for all $i \leq n-1$ and $j \geq 2$. Thus, its values can be easily computed without completing the spiral. | |
| Aug 23, 2017 at 15:31 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 24, 2017 at 15:12 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jun 23, 2017 at 20:37 | comment | added | Zach Teitler | The observation that $M_n$ is the central submatrix of $M_{n+2}$ makes me wonder if Dodgson condensation could be relevant. But these matrices have lots of zeros. | |
| Jun 23, 2017 at 20:16 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jun 9, 2017 at 9:45 | comment | added | user44191 | Extending my above observation: if you do both that column manipulation and the corresponding row manipulation, you are left with an almost 4-antidiagonal matrix (where the 4 antidiagonals are 1 below, 3 below, 5 below, and 7 below the "true" antidiagonal). Those antidiagonals are filled with 2 and -2, in a predictable way. The only exception is the 4 by 4 matrix in the top left corner, which is $M_{2n}$. | |
| May 30, 2017 at 21:12 | comment | added | François Brunault | Adding the first and last rows (resp. columns) in $M_{2n}$ if $n \equiv 0$ resp. $1 \pmod{2}$ shows that $\det M_{2n}$ is divisible by 3. | |
| May 30, 2017 at 16:32 | comment | added | T. Amdeberhan | This is an interest start. | |
| May 30, 2017 at 15:22 | comment | added | François Brunault | If $C_1,\ldots,C_{2n}$ denote the columns of $M_{2n}$ then $C_i \equiv C_{i+2} \pmod{2}$. Replacing $C_{2i+1}$ by $C_{2i+1}-C_1$ and $C_{2i+2}$ by $C_{2i+2}-C_2$ for any $1 \leq i \leq n-1$ yields a matrix where $2n-2$ columns consist of even entries, hence the determinant is divisible by $4^{n-1}$. Probably a similar argument can show that the determinant is divisible by 3. | |
| May 26, 2017 at 10:58 | comment | added | François Brunault | Seems like $M_n$ is the central submatrix of $M_{n+2}$. I don't know how to use this though. | |
| May 25, 2017 at 2:06 | comment | added | T. Amdeberhan | I would be happy to see your claim verified. | |
| May 24, 2017 at 19:43 | comment | added | Shahrooz | By quotient matrix, we can compute the determinants of the above examples simply. Maybe, by some good partitioning of your matrices, we can prove your claim. | |
| May 24, 2017 at 5:07 | comment | added | user44191 | Something I noticed when doing some larger versions: each column is the same as the column 4 before, with 2 exceptions, one of which is 2, the other of which is -2. Other than the first four columns, and switching the signs of some columns, you get 2s and -2s in two diagonals. | |
| May 24, 2017 at 5:01 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| May 24, 2017 at 4:19 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| May 24, 2017 at 3:50 | history | asked | T. Amdeberhan | CC BY-SA 3.0 |