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Fedor Petrov's answer of my preceding question shows that my question reduces to the famous Hadamard conjecture about Hadamard matrices of order $4k$. So I decided to study this conjecture and I got the following raw idea of constructing Hadamard matrices:

If $V_i$ is a row of $H_n$ then add $V_i$ to the $i$th column.
For example: Start with $$H_4=\begin{pmatrix}1&1&1&1\\ 1&1&-1&-1\\ 1&-1&1&-1\\ 1&-1&-1&1 \end{pmatrix} $$ Let $H_{4n-4}$ be a symmetric (if it exists?) Hadamard matrix. Then by the above idea we extend this matrix to $H_{4n}$ as follows: $$H_{4n }=\begin{pmatrix}H_{4n-4}&A_{(4n-4)\times 4}\\ A^T_{4\times (4n-4)}& B_{4\times 4} \\ \end{pmatrix} $$ where $B$ is a symmetric matrix and the first row of $A$ is $(1,1,1,\dots,1)$. Thus $$AA^T=4I_{4n-4},\quad H_{4n-4}A+AB=0,\quad A^T A+B^2=4n I_4.$$

Question: Does this idea give an algorithm for constructing Hadamard matrices?

Any suggestion for improving this idea would be greatly appreciated.

Wolfgang
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C.F.G
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    There is no symmetric Hadamard matrix of order $12$. – Wojowu Sep 30 '17 at 15:57
  • Are you sure? why? – C.F.G Sep 30 '17 at 16:00
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    http://neilsloane.com/hadamard/ – Wojowu Sep 30 '17 at 16:15
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    The question of how large M can be such that 1) M is a Hadamard matrix of order k, and 2) M is a minor of matrix H, and 3) H is a Hadamard matrix of order n leads to 2k being at most n. So your augmentation idea does not produce a Hadamard matrix when M has order of 8 or greater. I think a book of Wallis may have this result. Of course, Hadamard product allows n to be 2k. Gerhard "Augmentation Helps In Determinant Maximizing" Paseman, 2017.09.30. – Gerhard Paseman Sep 30 '17 at 17:18
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    Also, it seems your system of matrix equations gets problematic for n larger than 2, as AA^t should have rank at most 4, and can't be the multiple of a large identity matrix. You might try researching augmentation as an approach to one of Fedor's questions, see https://mathoverflow.net/a/261531 . Gerhard "Can Provide Some Technical Assistance" Paseman, 2017.09.30. – Gerhard Paseman Sep 30 '17 at 17:28
  • Finally, while @Wojowu is correct about their being no order 12 symmetric Hadamard matrix, the link provided is not enough, as there may be a symmetric version of the unique one presented. Trying to make one by hand will convince one otherwise. Gerhard "Experience Is A Good Teacher" Paseman, 2017.09.30. – Gerhard Paseman Sep 30 '17 at 17:35
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    There exist symmetric Hadamard matrices of order 12, though they're not easy to find by hand. See, for example, figure 1 of https://arxiv.org/pdf/1512.01732.pdf – Padraig Ó Catháin Oct 10 '17 at 02:26
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    If $H$ is Hadamard of order $n$, and $M$ is a minor of $H$ which is Hadamard, then $M$ has order at most $n/2$. For this, see https://arxiv.org/pdf/1208.3819.pdf but note that what the authors refer to as Szollosi's theorem is better known as the Jacobi determinant identity. – Padraig Ó Catháin Oct 10 '17 at 02:33
  • How can you have $AA^T=4I_{4n-4}$ if $A$ has rank 4? Don't you mean $AA^T=4I_{4}$ and later $A^T A+B^2=4n I_{4n-4}$? – Wolfgang Jan 27 '18 at 13:03

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