Timeline for Invertibility of a certain matrix indexed by the Hamming cube
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 9, 2013 at 2:06 | comment | added | Brendan McKay | And this leads to an explicit expression, see my answer (with nod to i707107). | |
| Mar 8, 2013 at 21:20 | comment | added | Sungjin Kim | Yes. I checked it. Together with a property of this inverse matrix: Any column sum is 0 except for the last column, which is 1. | |
| Mar 8, 2013 at 9:20 | comment | added | Denis Serre | Your guess is correct. It is not too difficult to check. | |
| Mar 8, 2013 at 4:26 | history | edited | Benjamin Young | CC BY-SA 3.0 |
Added some sage code.
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| Mar 8, 2013 at 4:16 | comment | added | Benjamin Young | No problem - you're welcome. | |
| Mar 8, 2013 at 2:02 | comment | added | Yemon Choi | My genuine thanks, by the way: I admit to being too technologically incompetent/lazy to stick the matrix into a CAS (more fool me), and as you observe the formula for the inverse is very suggestive. What it suggests to me is that this was indeed Mobius inversion ;-) | |
| Mar 6, 2013 at 6:10 | comment | added | Yemon Choi | All informative answers are welcome. Note that I wasn't the person who used the word "enlightening" ;-) | |
| Mar 6, 2013 at 5:21 | history | answered | Benjamin Young | CC BY-SA 3.0 |