# Centrosymmetric Group

Consider the set of $$N\times N$$ matrices that satisfy the property

$\mathcal{H}=\{H\,|\,H_{ij}=H_{N+1-i,N+1-j},\det H\neq0\}$

or in matrix forms

$\begin{split}\begin{pmatrix}a_{1} & a_{2} & \cdots & a_{N-1} & a_{N}\\ b_{1} & b_{2} & \cdots & b_{N-1} & b_{N}\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ b_{N} & b_{N-1} & \cdots & b_{2} & b_{1}\\ a_{N} & a_{N-1} & \cdots & a_{2} & a_{1} \end{pmatrix}\end{split}$

Let

$\begin{split}J=\begin{bmatrix}. & . & \dots & . & 1\\ . & . & \dots & 1 & .\\ \\ . & 1 & \dots & . & .\\ 1 & . & \dots & . & . \end{bmatrix}\qquad J_{ij}=\begin{cases} 1 & \text{if }i=N-j+1\\ 0 & \text{otherwise} \end{cases}\end{split}$

The matrix $$J$$ is sometimes referred to as the exchange matrix.

Using the exchange matrix the above set of matrices can also be conveniently defined as

$\mathcal{H}=\left\{ A\,|\,AJ=JA\right\}$

Proposition. The set of matrices $$\mathcal{H}$$ forms a group.

Proof. It can be easily shown that the identity is in the group, i.e. $$I\in\mathcal{H}$$. Also the set is closed under multiplication, i.e. if $$A\in\mathcal{H}$$ and $$B\in\mathcal{H}$$ then $$AB\in\mathcal{H}$$. To see why, consider

$ABJ=AJB=JAB$

If $$A\in\mathcal{H}$$, then $$AJ=JA$$

$(AJ)^{-1}=(JA)^{-1}$
$J^{-1}A^{-1}=A^{-1}J^{-1}$
$JA^{-1}=A^{-1}J$

Therefore, $$A^{-1}\in\mathcal{H}$$. This completes the proof that $$\mathcal{H}$$ forms a group.