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Definitions: An $n\times n$ Hadamard matrix (HM for short) is a matrix whose entries are either $1$ or $−1$ and whose rows are mutually orthogonal. If $A$ is a symmetric matrix, then $A = A^T$ and if $A= (a_{ij})$ is a skew-symmetric matrix and $i \neq j$ then $a_{ij}=-a_{ji}$. Here the diagonal elements of skew-symmetric matrices can be zeros or nonzeros.

Background: If $H$ is an $n\times n$ skew-symmetric HM, then $n$ is the only choice for $\lvert trace(H) \rvert$.

Questions: Now we focus on symmetric matrices: if $H$ is an $n\times n$ symmetric HM, how many possible choices for $\lvert trace(H) \rvert$ and what are they? Given a pair of numbers $t$ and $n$, in one's opinion, $t$ is an impossible choice for $\lvert trace(H) \rvert$, how can we check it is correct or incorrect?

The books or papers about these questions are welcomed. Thanks a lot.

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    $\begingroup$ How can it be skew-symmetric, should not skew-symmetric matrix have zeros on diagonal? $\endgroup$
    – Fedor Petrov
    Commented Jun 8 at 12:09
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    $\begingroup$ The skew-symmetric definition here is slightly different than $A^T=-A$. @FedorPetrov $\endgroup$
    – user369335
    Commented Jun 8 at 12:22
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    $\begingroup$ Orthogonality means $A A^T=nI$ (where $I$ is identity matrix), if $A^T=A$, we get $A^2=nI$, thus eigenvalues of $A$ are only $\pm \sqrt{n}$. Therefore, if $n$ is not a perfect square, the sum of eigenvalues (i.e., trace) can be integer only if it is 0. $\endgroup$
    – Fedor Petrov
    Commented Jun 8 at 12:26

1 Answer 1

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Currently, this is an open problem. So I only got a partial answer.

We just need to consider the following two different situations:

Case 1: $n$ is not a perfect square. There is no solution if $\lvert trace(H) \rvert \neq 0$; and $0$ is a possible choice for $\lvert trace(H) \rvert$ if and only if symmetric HM conjecture is correct.

Case 2: $n$ is a perfect square. There is no solution if $\lvert trace(H) \rvert \notin \{0,2\sqrt{n},4\sqrt{n},...,n\}$. So we only need to check $0,2\sqrt{n},4\sqrt{n},...,n$ one by one and make sure at least one instance exists. It is completely settled for $n<196$, here is my result:

$n$ All possible choices for $\lvert trace(H) \rvert$
$1$ $ 1 $
$4$ $ 0,4 $
$16$ $ 0,8,16 $
$36$ $ 0,12,24,36 $
$64$ $ 0,16,32,48,64 $
$100$ $ 0,20,40,60,80,100 $
$144$ $ 0,24,48,72,96,120,144 $
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    $\begingroup$ SageMath constructs a symmetric Hadamard matrix for $n=66^2$ with trace $n$. $\endgroup$
    – Max Alekseyev
    Commented Jun 13 at 15:19
  • $\begingroup$ Wow, SageMath is so powerful! Thanks for your helpful comment! @MaxAlekseyev $\endgroup$
    – user369335
    Commented Jun 13 at 22:33
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    $\begingroup$ It got such a matrix by chance. For other square $n$, constructed matrices are either not symmetric or have zero trace. $\endgroup$
    – Max Alekseyev
    Commented Jun 13 at 23:27
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    $\begingroup$ The Paley II construction gives a symmetric Hadamard matrix of order $2q+2$ for any prime power $q\equiv 1 \mod 4$, with trace $0$. A symmetric Bush-type Hadamard matrix can be constructed with order $16n^2$ whenever there is a symmetric Hadamard matrix of order $4n$, these always have trace $16n^2$. It's conjectured that a symmetric Hadamard matrix exists at all orders $4n$ and that a Bush-type matrix exists for all $4n^2$. $\endgroup$
    – Padraig Ó Catháin
    Commented Jun 14 at 15:05

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