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 | |
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| Mar 8, 2013 at 10:07 | vote | accept | Yemon Choi | ||
| Mar 8, 2013 at 3:20 | comment | added | Benjamin Steinberg | The high level description is that the transformation A sends a subset to the sum of all susets minus the sum of all subsets of the complement of the subset and so this is essentially triangular with respect to the order. My base change removes the "essentially" part. | |
| Mar 8, 2013 at 2:00 | comment | added | Yemon Choi | Thanks - I have been too busy to digest all the answers, but with hindsight this kind of M&oouml;bius inversion comes closest to what I was hoping to hear. | |
| Mar 8, 2013 at 0:53 | comment | added | Benjamin Steinberg | From this proof it is fairly easy to compute the inverse. The inverse of the incidence matrix of the poset is the Mobius function which for the subset poset is very simple. The change of basis matrix for the basis change above is also very simple. Then one just has to multiply it out. | |
| Mar 6, 2013 at 4:28 | comment | added | Benjamin Steinberg | The matrix for $A$ in these bases is in fact the incidence matrix of the poset $(S,\subseteq)$. | |
| Mar 6, 2013 at 4:26 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
added 323 characters in body
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| Mar 6, 2013 at 3:34 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
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| Mar 6, 2013 at 3:28 | history | answered | Benjamin Steinberg | CC BY-SA 3.0 |