© Nickolay Balonin, 27.10.2014

Iteration algorithm calculate a chain E-M-E-M-E-M-E... of quasi-orthogonal matrices, starting with two-circulant or two A,B-blocks E=CM(4tЦ2), arasing order by a border of regular M=CM(4tЦ1), preparing the next two blocks A=M, B=M for E=CM(4tЦ2), and so on. Generator gives matrix portraits of several CM(4tЦ2) with time-interval 3 sec (3000 msec), look below these fractals.


(In honour of my wife)


Two strands matrix CM(14) and plait matrix CM(14)

Let us determine two circulant CM(4tЦ2) as a two strands matrix. Other kind of this matrix, constructed by a chain (fractal) ..E-M-E-M-E-M-E.., we will call plait matrix. The plait matrix has a lenght (fractal has memory, it can be started from any even order 4tЦ2 of a two strands matrix E=CM(4tЦ2) and a twin border, giving M=CM(4tЦ1). Then we make the next matrix E=CM(4tЦ2) combining two M=CM(4tЦ1), example is below. Let us measure the lenght by number of E-matrices in the structure of plait matrix.

Two strands matrix E=CM(6) and regular matrix M=CM(7)

Plait matrix E=CM(14) and regular matrix M=CM(15)

Plait matrix E=CM(30) of a chain of length 3 and two strands matrix E=CM(30)

The difference between two strands and plait matrices is the plait matrix has information about all matrices of its chain. Scythes can be constructed (for orders 4tЦ2) going down, to the point, where it was two strands E, or upstairs, there is no any bound for E-M-E-M-E-M-E... chain above.

We will say, that plait matrix CM(4tЦ2) of lenght>1 and regular CM(4tЦ1), based on it, have a rich structure. Other versions of these matrices, based on two strands matrix, have a poor structure.

On every order we have two strands matrix and all plait matrices, constructed up from all previous two strands matrices. For order 22 we have: two strands matrix E=CM(22) of two-circulant structure (just, it looks like "two strands", see catalogue); a plait 1 is E=CM(22), constructed by CM(11) made by two-circulant E=CM(10); a plait 2 is E=CM(22) constructed by CM(11) made by E=CM(10), constructed by CM(5).

Final orders of all plaits, reconstructed down, are odd orders: 4tЦ1 or 4tЦ3 of regular Cretan matrices, or even orders 4tЦ2 of a poor structure (if the plait starts from a two strands matrix).


With it, any two strands and plait matrices lead to CM(4tЦ1) with one and the same SBIBD. Due the same SBIBD for matrices with memory and not, we have a theorem: if there is some two strands matrix, there are and all plaits of all possible lengths, and, the other verse, if there is a plait matrix, there is two strands matrix.

If it is so, then from theorem (ab. exponential exstence of Hadamard matrices, based on SBIBD for matrices CM(4tЦ1), follows: all chains (plaits) exist.

If there exist one Cretan matrix of a chain, all matrices of the chain can be found all along this chain. The theorem of Jennifer Seberry*) says: there are all Hadamard matrices of enough big orders, that means every chain has a regular Cretan matrix based on SBIBD(4tЦ1,2t,t) for enough big t.

*) There are two well illustrated papers La conjecture de Hadamard I, La conjecture de Hadamard II with with examples of Hadamard matrices H12, H20 from the Hadamard's paper and this theorem.

Thus all matrices of Chains exist.

Every E=CM(4tЦ2) with A,B-blocks has two character solutions, one "poor" solution is based on A=B=M, M is regular Cretan matrix order 4tЦ1: complicated structure, it is given by generator; second "rich" solution is based on A≠B of two circulant SBIBDs: simple structure, matrices are searched and placed in catalogue.


For example, let us take Hadamard matrix of order 668. It is based on CM(4tЦ1) of order 667, and, going down along the chain, on CM(4tЦ2) of order 666. For order 666 we can try to find CM(4tЦ1) of order 333, but 333=111×3=37×3×3 (CM(333), with diagonal entry, exists), but it is a problem order, the same, as 45 (rich structure, symilar to Mathon's type). So better to find two strands type matrix CM(333) based on two consecuences.


Hadamard's maximal determinant problem, named after Jacques Hadamard, asks for the largest determinant of a matrix with elements equal to 1 or Ц1. Any two rows of an Hadamard matrix are orthogonal, which is impossible for a {1, Ц1} matrix when order n is an odd number. When n ≡ 2 (mod 4), two rows that are both orthogonal to a third row cannot be orthogonal to each other. Together, these statements imply that an Hadamard matrix can exist only if n = 1, 2, or a multiple of 4.

Hadamard observed that a well known construction of Sylvester produces examples of matrices that attain the boun when n is a power of 2, and produced examples of his own of sizes 12 and 20. He also showed that the bound is only attainable when n is equal to 1, 2, or a multiple of 4. Additional examples were later constructed by Scarpis and Paley and subsequently by many other authors. Such matrices are now known as Hadamard matrices. They have received intensive study.

Matrix sizes n for which n ≡ 1, 2, or 3 (mod 4) have received less attention.

Hadamard estimated bound by ineaquality Цnn/2≤det(A)≤nn/2 for matrices A with real entries taken from the unite disk, and observed a dilemme: touch (or not) matrix with 1, Ц1 integer entries bounds. This dilemme, solved in hard conditions Ц fixed integer levels and fixed weight n Ц has no resonable solution today.

Cretan matrices, for which n ≡ 1, 2, or 3 (mod 4), have soft bound Цωn/2≤det(A(ab))≤ωn/2, where a=1, 0≤b≤1 is a varying level and 0≤ω≤n is a varying weight, found as solution of optimisation task on local maximum of determinat **). So dilemme looks like: touch (or not) matrix (with adapted level Цb(n)) an adapted bound ω(n)n/2. This bound becoming be concrete during the solution of task: non fixed value ω is appeared as a statement, it is important only columns (and rows) orthogonality.

We have a task with free (non fixed) parameters resolved as monotonous functions giving weights and levels for CM(4tЦ1), CM(4tЦ2), these matrices, taken together, organise, as we can show, some chains of non fixed lenghts. All such chains being constructed upstairs touch Jennifer Seberry border order, and theoretically they can be reconstructed down to the two-circulant CM(4tЦ2) (just if we have to go milliard stadiums) and one border two-circulant CM(4tЦ1) giving SBIBD(4tЦ1,2t,t) to construct any Hadamard matrix.

This way conjecture of Hadamard can be decomposed on several looking quite independent tasks: CM(4tЦ2) existence, CM(4tЦ1) existence, CM(4t) existence, solved all together or as independent tasks. Non mentioned here Fermat matrices of orders CM(4t+1) gives own look on existence of regular Hadamard matrices.

**) See interpretation of det(A) as an optimised volume in the paper: Hadamard et Bordeaux.


LOOK CM(1025)

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