Matrices with rotational symmetry
I've seen a claim without proof that the characteristic polynomials of matrices with rotational symmetry (i.e. $n \times n$ matrices $A$ with $A_{i,j} = A_{n+1-i,n+1-j}$) always factor into the product of the characteristic polynomials of smaller matrices which can be derived from blocks of the original matrix. Is there an elementary proof, and can the result be generalised?
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Notation: $A^{\leftarrow}$ denotes $A$ with the columns reversed; $A^{\uparrow}$ denotes $A$ with the rows reversed; $A^{\leftarrow \uparrow} = A^{\uparrow \leftarrow}$ is denoted $A^{\circ}$ and is the rotation of $A$ by $180^{\circ}$.
Consider first a $(2n+1)\times(2n+1)$ block matrix $\begin{pmatrix} A & v & B^{\circ} \\ h & c & h^{\leftarrow} \\ B & v^{\uparrow} & A^{\circ} \end{pmatrix}$ where $A, B$ are $n \times n$, $v$ is $n \times 1$, $h$ is $1 \times n$, and $c$ is $1 \times 1$. We have \begin{eqnarray*}\det \begin{pmatrix} A & v & B^{\circ} \\ w & c & w^{\leftarrow} \\ B & v^{\uparrow} & A^\circ \end{pmatrix} &=& (-1)^{n(n+1)/2} \det \begin{pmatrix} A & v & B^{\circ} \\ B^{\uparrow} & v & A^{\leftarrow} \\ w & c & w^{\leftarrow} \end{pmatrix} \tag{permuting rows} \\ &=& \det \begin{pmatrix} A & B^{\uparrow} & v \\ B^{\uparrow} & A & v \\ w & w & c \end{pmatrix} \tag{permuting cols} \\ &=& \det \begin{pmatrix} A - B^{\uparrow} & B^{\uparrow}-A & 0 \\ B^{\uparrow} & A & v \\ w & w & c \end{pmatrix} \tag{subtracting rows} \\ &=& \det \begin{pmatrix} A - B^{\uparrow} & 0 & 0 \\ B^{\uparrow} & A + B^{\uparrow} & v \\ w & 2w & c \end{pmatrix} \tag{adding cols} \\ &=& \det (A - B^{\uparrow}) \det \begin{pmatrix} A + B^{\uparrow} & v \\ 2w & c \end{pmatrix} \tag{by Leibniz formula} \end{eqnarray*}
Note that the characteristic polynomial is a special case, since $\chi_M(x) = \det (M - xI)$.
The $(2n)\times(2n)$ block matrix $\begin{pmatrix} A & B^{\circ} \\ B & A^{\circ} \end{pmatrix}$ has an almost identical derivation yielding $$\det \begin{pmatrix} A & B^{\circ} \\ B & A^{\circ} \end{pmatrix} = \det (A - B^{\uparrow}) \det (A + B^{\uparrow})$$
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