Timeline for pseudo-Hadamard matrix
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Feb 26, 2020 at 2:49 | comment | added | Padraig Ó Catháin | Let $G$ be the additive group of the field and $\phi: G \rightarrow \{-1, 1\}$ be any function such that $\sum_{g \in G} \phi(g) = 0$. Then the matrix $[\phi(g+h)]_{g, h \in G}$ satisfies your conditions. Every matrix satisfying your conditions arises in this way. One can build Hadamard matrices (necessarily with regular row sums) by imposing stronger conditions on the function $\phi$: it should be the support of a Menon-Hadamard difference set. | |
| Feb 6, 2020 at 21:02 | comment | added | Zach Teitler | @BrendanMcKay Oh, that makes sense. Now it seems to me condition (3) is automatically satisfied: $M_{i,j} = M_{i-j,0}$, so the $i$-th row is a permutation of the $0$-th (first) row; and so is the $j$-th column. So if the first row sums to zero, so do automatically every row and column. | |
| Feb 6, 2020 at 6:17 | comment | added | Brendan McKay | @ZachTeitler OP will clarify, but I think you need to use the same action on the rows and columns simultaneously. | |
| Feb 6, 2020 at 6:04 | comment | added | Zach Teitler | Doesn't condition (4) make all entries of the matrix equal? To get $M_{i,j} = M_{k,\ell}$, act by $k-i$ on the rows and by $\ell-j$ on the columns. | |
| Feb 6, 2020 at 4:45 | review | Close votes | |||
| Feb 21, 2020 at 3:05 | |||||
| Feb 6, 2020 at 4:15 | comment | added | Brendan McKay | Maybe I misunderstand the question, but: Write the first row any way you like that sums to 0. Use condition (4) to fill in the rest of the matrix. Then condition (3) is satisfied already. Or not. | |
| Feb 6, 2020 at 3:45 | review | First posts | |||
| Feb 6, 2020 at 4:26 | |||||
| Feb 6, 2020 at 3:44 | history | asked | N.S.N.Sastry | CC BY-SA 4.0 |