# Does this idea give an algorithm for constructing Hadamard matrices?

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.

• There is no symmetric Hadamard matrix of order $12$. – Wojowu Sep 30 '17 at 15:57
• neilsloane.com/hadamard – Wojowu Sep 30 '17 at 16:15
• 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
• 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. – Gerhard Paseman Sep 30 '17 at 17:28
• 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 – Padraig Ó Catháin Oct 10 '17 at 2:26