Timeline for dyadically recursive matrices: Part I
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 16, 2017 at 15:42 | vote | accept | T. Amdeberhan | ||
| Mar 16, 2017 at 14:50 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
edited title
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| Mar 16, 2017 at 8:09 | answer | added | Denis Serre | timeline score: 4 | |
| Mar 16, 2017 at 7:58 | answer | added | Wolfgang | timeline score: 5 | |
| Mar 16, 2017 at 3:59 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
added 2 characters in body
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| Mar 16, 2017 at 3:58 | comment | added | T. Amdeberhan | @GerryMyerson: Yes, the matrix is $2^{n-1}\times 2^{n-1}$. | |
| Mar 16, 2017 at 3:52 | comment | added | Gerhard Paseman | Can't you use Schur complement for this? The result should be an iteration of B^2 - C^2. I get $(A^2 - B^2)^2 - (C^2)^2$ for the next iterate. Gerhard "Capitalizes On Change Of Notation " Paseman, 2017.03.15. | |
| Mar 16, 2017 at 3:43 | comment | added | Gerry Myerson | By "dimension" of a (square) matrix, you mean the number of rows? | |
| Mar 16, 2017 at 2:57 | comment | added | T. Amdeberhan | Running in powers of $2$. | |
| Mar 16, 2017 at 2:56 | comment | added | Turbo | why is it 'dyadic'? | |
| Mar 16, 2017 at 2:17 | history | asked | T. Amdeberhan | CC BY-SA 3.0 |