Talk:Paley construction

[Untitled]
"Paley's theorem" also appears to refer to a result in functional analysis. Richard Pinch (talk) 15:11, 20 June 2008 (UTC)


 * Ah, it seems to be the Paley–Wiener theorem. Richard Pinch (talk) 15:29, 20 June 2008 (UTC)

Page moved
I have moved the page from “Paley's theorem″ to “Paley construction″ since the former does not appear to be used in the Hadamard matrix literature. I also reworked the text to make it more compatible with the new title, and also because the previous version too closely resembled the article on MathWorld. I hope to add more detail over the next couple of days. Will Orrick (talk) 14:55, 11 July 2008 (UTC)

Skew Hadamard matrix
H+HT = 2I is correct. A skew symmetric matrix is a matrix whose transpose is the same as its negation, or, in other words, a matrix such that M+MT = 0. Hadamard matrices cannot be skew symmetric since they cannot have 0 elements on the diagonal as skew symmetric matrices must. The definition of skew Hadamard matrix takes this into account by requiring skew symmetry only for off-diagonal elements: hij = −hij for i ≠ j. Diagonal elements are required to equal 1. Together these requirements imply to the stated condition. Will Orrick (talk) 18:24, 4 February 2011 (UTC)

Wrong for q = 2 !?
Unless I err, the statement Q QT = q I - J (right after the definition of the Jacobsthal matrix) is wrong for q = 2, so I added "for q > 2". Can anyone confirm? (I don't know whether it's true for q = 2^k, k > 1: did not yet implement this case in my computations. I was confused until I realized that one cannot take the indices of the matrix to be 0, ..., q-1 when q is not a prime.) &mdash; MFH:Talk 22:07, 31 May 2022 (UTC)


 * In the introduction q is defined to be a power of an odd prime, so q = 2 isn't considered. Perhaps that needs to be repeated where the construction is given.
 * Note that q even is quite different from q odd: for q odd half-of the non-zero field elements are squares and half are non-squares; for q even, every non-zero element is a square. To see this, recall that in any finite field the non-zero elements form a cyclic group under multiplication, that is, there is an element a such that every non-zero element of the field can be written as a power of a. If q = 2n, then a2 n&minus;1 = 1. The important thing here is that a has odd order. Obviously the elements a2, a4, a6, and so on, are squares. But if you think about it, so are a1, a3, a5, and so on, since ar = ar+2 n&minus;1 and the latter has even exponent for odd r. So not only can every non-zero element be expressed as a power of a, every non-zero element can be expressed as an even power of a (and also as an odd power). Will Orrick (talk) 00:42, 1 June 2022 (UTC)
 * A consequence of my previous message: if one uses the same definition of Jacobsthal matrix for q even as for q odd, one gets that the Jacobsthal matrix for q even is the rather uninteresting matrix J &minus; I and hence that QQT = (q&minus;2)J + I. Will Orrick (talk) 02:23, 1 June 2022 (UTC)


 * Thank you for the confirmation. Indeed, now that I implemented the case of q = 2^k, I can also confirm that for q = 2^k, Q = J - I, and so the given relation isn't true. However, if makes of course perfectly sense to consider Haramard matrices of size 2^k (the only case implemented in ...) and even though the Jacobsthal matrix of that size might be interesting, it's still well-defined and could therefore be considered... &mdash; MFH:Talk 03:45, 1 June 2022 (UTC)
 * The Paley matrix construction doesn't give Hadamard matrices when q = 2k. (It does, of course, sometimes give Hadamard matrices of size 2k. For that to happen, 2k &minus; 1 must be a prime power. These Hadamard matrices are most always not equivalent to the ones obtained by forming tensor products of 2&times;2 Hadamard matrices, which I'm guessing is what scipi gives you.)
 * If one wanted to discuss the Jacobsthal matrix for even q in the article, one would have to track down a literature reference where it is defined, assuming there is such a reference. The matrix J &minus; I is constant apart from the diagonal, so it contains almost no information about the field elements. For this reason, I'm guessing one won't find any discussion of it. Will Orrick (talk) 05:32, 1 June 2022 (UTC)