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What can one say about the matrices M with the following properties: 1) the rows and the columns are indexed by the elements of a finite field with an even number q of elements; 2) All the matrix entries are either 1 or -1; 3) the sum of the entries in each row and sum of the entries in each column is zero. 4) the additive group of the field in (1) acts regularly on the rows and the columns of the matrix M and the action preserves M. that is, the (i,j)- entry of M is equal (g(i),g(j))-th entry of M for each element of the field.

As comprehensive an answer or a reference is very much appreciated. -N.S.N.Sastry, IIT Dharwad (nnsastry@gmail.com)

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  • $\begingroup$ 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. $\endgroup$ – Brendan McKay Feb 6 at 4:15
  • $\begingroup$ 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. $\endgroup$ – Zach Teitler Feb 6 at 6:04
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    $\begingroup$ @ZachTeitler OP will clarify, but I think you need to use the same action on the rows and columns simultaneously. $\endgroup$ – Brendan McKay Feb 6 at 6:17
  • $\begingroup$ @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. $\endgroup$ – Zach Teitler Feb 6 at 21:02

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