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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.

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    $\begingroup$ There is no symmetric Hadamard matrix of order $12$. $\endgroup$ – Wojowu Sep 30 '17 at 15:57
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    $\begingroup$ neilsloane.com/hadamard $\endgroup$ – Wojowu Sep 30 '17 at 16:15
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    $\begingroup$ 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. $\endgroup$ – Gerhard Paseman Sep 30 '17 at 17:18
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    $\begingroup$ 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 mathoverflow.net/a/261531 . Gerhard "Can Provide Some Technical Assistance" Paseman, 2017.09.30. $\endgroup$ – Gerhard Paseman Sep 30 '17 at 17:28
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    $\begingroup$ There exist symmetric Hadamard matrices of order 12, though they're not easy to find by hand. See, for example, figure 1 of arxiv.org/pdf/1512.01732.pdf $\endgroup$ – Padraig Ó Catháin Oct 10 '17 at 2:26

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