Timeline for Find the inverse of a matrix that is very similar to the Hilbert matrix
Current License: CC BY-SA 3.0
17 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Feb 27, 2017 at 2:05 | vote | accept | Dings | ||
| Feb 27, 2017 at 2:01 | history | edited | Dings | CC BY-SA 3.0 |
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| Feb 27, 2017 at 1:15 | comment | added | T. Amdeberhan | Dings: perhaps you should write a new MO question with your extended matrix. | |
| Feb 26, 2017 at 20:55 | answer | added | Johann Cigler | timeline score: 3 | |
| Feb 26, 2017 at 15:44 | history | edited | Dings | CC BY-SA 3.0 |
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| Feb 23, 2017 at 18:00 | answer | added | T. Amdeberhan | timeline score: 6 | |
| Feb 22, 2017 at 22:24 | answer | added | Pietro Majer | timeline score: 15 | |
| Feb 22, 2017 at 22:03 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Feb 22, 2017 at 21:48 | comment | added | darij grinberg | @RodrigodeAzevedo: It is a direct sum if you properly re-order the coordinate vectors: You need to take the basis $\left(e_1, e_3, e_5, \ldots, e_2, e_4, e_6, \ldots\right)$. Or, to say it in terms of matrices, you move all the columns with odd indices to the front, and all the columns with even indices to the back; then you do the same to the rows. | |
| Feb 22, 2017 at 20:41 | comment | added | Pietro Majer | Note that $M$ splits into a direct sum of two square matrices: $A_{ij}:={1\over i+j-1/2}$ and $B_{ij}:={1\over i+j-3/2}$. | |
| S Feb 22, 2017 at 20:17 | history | suggested | Rodrigo de Azevedo | CC BY-SA 3.0 |
Minor edits
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| Feb 22, 2017 at 19:56 | review | Suggested edits | |||
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| S Feb 22, 2017 at 15:59 | history | suggested | Alex M. |
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| Feb 22, 2017 at 15:35 | review | Suggested edits | |||
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| Feb 22, 2017 at 15:22 | review | First posts | |||
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| Feb 22, 2017 at 15:19 | history | asked | Dings | CC BY-SA 3.0 |