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Paley Construction

Hadamard Matrices ${\hbox{\sf H}}_n$ can be constructed using Galois Field GF($p^m$) when $p=4l-1$ and $m$ is Odd. Pick a representation $r$ Relatively Prime to $p$. Then by coloring white $\left\lfloor{(p-1)/2}\right\rfloor $ (where $\left\lfloor{x}\right\rfloor $ is the Floor Function) distinct equally spaced Residues mod $p$ ($r^0$, $r$, $r^2$, ...; $r^0$, $r^2$, $r^4$, ...; etc.) in addition to 0, a Hadamard Matrix is obtained if the Powers of $r$ (mod $p$) run through $<\lfloor (p-1)/2\rfloor$. For example,

n=12=11^1+1= 2(5+1)=2^2(2+1)

is of this form with $p=11=4\times 3-1$ and $m=1$. Since $m=1$, we are dealing with GF(11), so pick $p=2$ and compute its Residues (mod 11), which are
$\displaystyle p^0$ $\textstyle \equiv$ $\displaystyle 1$  
$\displaystyle p^1$ $\textstyle \equiv$ $\displaystyle 2$  
$\displaystyle p^2$ $\textstyle \equiv$ $\displaystyle 4$  
$\displaystyle p^3$ $\textstyle \equiv$ $\displaystyle 8$  
$\displaystyle p^4$ $\textstyle \equiv$ $\displaystyle 16\equiv 5$  
$\displaystyle p^5$ $\textstyle \equiv$ $\displaystyle 10$  
$\displaystyle p^6$ $\textstyle \equiv$ $\displaystyle 20\equiv 9$  
$\displaystyle p^7$ $\textstyle \equiv$ $\displaystyle 18\equiv 7$  
$\displaystyle p^8$ $\textstyle \equiv$ $\displaystyle 14\equiv 3$  
$\displaystyle p^9$ $\textstyle \equiv$ $\displaystyle 6$  
$\displaystyle p^{10}$ $\textstyle \equiv$ $\displaystyle 12\equiv 1.$  

Picking the first $\left\lfloor{11/2}\right\rfloor =5$ Residues and adding 0 gives: 0, 1, 2, 4, 5, 8, which should then be colored in the Matrix obtained by writing out the Residues increasing to the left and up along the border (0 through $p-1$, followed by $\infty$), then adding horizontal and vertical coordinates to get the residue to place in each square.

\left[{\matrix{\infty & \infty & \infty & \infty & \infty & ...
... & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \infty\cr}}\right]

${\hbox{\sf H}}_{16}$ can be trivially constructed from ${\hbox{\sf H}}_4\otimes{\hbox{\sf H}}_4$. ${\hbox{\sf H}}_{20}$ cannot be built up from smaller Matrices, so use $n=20=19+1=2(3^2+1)=2^2(2^2+1)$. Only the first form can be used, with $p=19=4\times 5-1$ and $m=1$. We therefore use GF(19), and color 9 Residues plus 0 white. ${\hbox{\sf H}}_{24}$ can be constructed from ${\hbox{\sf H}}_2\otimes{\hbox{\sf H}}_{12}$.

Now consider a more complicated case. For $n=28=3^3+1=2(13+1)$, the only form having $p=4l-1$ is the first, so use the GF($3^3$) field. Take as the modulus the Irreducible Polynomial $x^3+2x+1$, written 1021. A four-digit number can always be written using only three digits, since $1000-1021 \equiv 0012$ and $2000-2012\equiv 0021$. Now look at the moduli starting with 10, where each digit is considered separately. Then

x^0\equiv 1\hfill & x^1\equiv 10\hfill & x^2\equiv ...
...uiv 2210\equiv 201\hfill & x^{26}\equiv 2010\equiv 1\hfill\cr}

Taking the alternate terms gives white squares as 000, 001, 020, 021, 022, 100, 102, 110, 111, 120, 121, 202, 211, and 221.


Ball, W. W. R. and Coxeter, H. S. M. Mathematical Recreations and Essays, 13th ed. New York: Dover, pp. 107-109 and 274, 1987.

Beth, T.; Jungnickel, D.; and Lenz, H. Design Theory, 2nd ed. rev. Cambridge, England: Cambridge University Press, 1998.

Geramita, A. V. Orthogonal Designs: Quadratic Forms and Hadamard Matrices. New York: Marcel Dekker, 1979.

mathematica.gif Kitis, L. ``Paley's Construction of Hadamard Matrices.''

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© 1996-9 Eric W. Weisstein