**Definition 1**: A Hadamard matrix is an $n\times n$ matrix $H$ whose entries are either $1$ or $-1$ and whose rows are mutually orthogonal.


**Definition 2**: A matrix $A$ is half-skew-centrosymmetric if there exist two square matrices $B$ and $C$ of order $n$ such that


\begin{equation}
 A = \begin{bmatrix}
 B & R C R \\
 C & -R B R
 \end{bmatrix}.
 \end{equation}


where $R$ is the reverse identity matrix.


One day I found that these two definitions can be considered simultaneously. For example, \begin{bmatrix}
 1 & 1 \\
 1 & -1
 \end{bmatrix} is the simplest half-skew-centrosymmetric Hadamard matrix.


Here comes my questions:



> 
> First, can you give some references about half-skew-centrosymmetric Hadamard matrices? Are there other names for these matrices?
> 
> 
> 



> 
> Second one is about construction methods. It is easy to figure out how to construct a half-skew-centrosymmetric Hadamard matrix of order $2^k\cdot n$ based on a Hadamard matrix of order $n$ by using a variant of Sylvester's construction.
> Can you propose more methods to construct half-skew-centrosymmetric Hadamard matrices?
> 
> 
> 



> 
> Third, can you prove the following conjecture or give a counter-example?
> 
> 
> **Conjecture**: A half-skew-centrosymmetric Hadamard matrix exists for $n=2$ or $n$ is a multiple of $4$.
> 
> 
> 



On the half-skew-centrosymmetric Hadamard matrices
Let $H\_n$ be an $n×n$ Hadamard matrix and $R\_n$ the $n×n$ reverse identity matrix.


The matrix $X= \begin{pmatrix}
H\_n & R\_nH\_n \\
H\_n & -R\_nH\_n
\end{pmatrix}$ has entries of length $1$ and $$XX^\* = 2nI\_{2n} + ((nI\_n - R\_nH\_nH\_n^\*R\_n) \otimes R\_2)$$ which is simply $2nI\_{2n}$ so it is a Hadamard matrix. Permute the last $n$ columns with $R\_n$ and you have it in the form you give. This gives half-skew-centrosymmetric Hadamard matrices of twice the size of a Hadamard matrix. (It also works for complex Hadamard matrices or even Hadamard matrices over \*-rings)


Another construction (edit): suppose we apply the [Paley construction](https://en.m.wikipedia.org/wiki/Paley_construction) II to a finite field with $q=4k+1$ elements giving an $2n×2n$ Hadamard matrix $H= \begin{pmatrix}
H\_1 & H\_2 \\
H\_3 & H\_4
\end{pmatrix} $ after permuting the odd columns and rows to the first $n$ rows/columns, the even as last $n$ AND put the first even row/column at the last position. Define $X := X\_{i,j} = (i-j)^{0.5(q-1)}$ (in the finite field this is $\pm 1$ with zero diagonal and symmetric because $-1$ is a square). Note that $X\_{i,j+k}=X\_{j+k,i}=X\_{n-i+k,n-j}$. Then for Paley construction II, using $j$ as all-ones vector $$H\_2^T = H\_3 = \begin{pmatrix}
j & X-I \\
-1 & j^T
\end{pmatrix}$$ $$H\_1 = \begin{pmatrix}
1 & j^T \\
j & X+I
\end{pmatrix}$$ $$H\_4 = \begin{pmatrix}
-X-I & -j \\
-j^T & -1
\end{pmatrix}$$The $H\_i$ now have the properties $R\_nH\_2R\_n = H\_3$ and $-R\_nH\_1R\_n = H\_4$ due to our note above. Thus $H$ is a half-skew-centrosymmetric Hadamard matrix!


