Is the inverse of an invertible circulant matrix also circulant? The circulant matrices in $M_n(F)$ ($F$ field) form a subspace $\mathcal C_n$ spanned by $I,J,J^2,\cdots,J^{n-1}$ where
$$J=\begin{bmatrix} O & I_{n-1}\\1 &  O\end{bmatrix}$$
This subspace $\mathcal C_n$ is also closed under multiplication, which makes it a subring of $M_n(F)$.
Now my question is, if $A\in \mathcal C_n$ and $A$ is non-singular, can we assert that $A^{-1}\in\mathcal C_n$ too?
I'm in particular encouraged to make this guess by observing  these examples: 
$$\begin{bmatrix}0 & 1 & 1 & \cdots & 1\\ 1 & 0 & 1 & \cdots & 1 \\ 1 & 1 & 0 & \cdots & 1\\ \vdots & \vdots & \vdots & & \vdots \\ 1 & 1 & 1 & \cdots & 0 \end{bmatrix}^{-1}=\frac{1}{n-1}\begin{bmatrix} 2-n & 1 & 1 & \cdots & 1\\ 1 & 2-n & 1 & \cdots & 1 \\ 1 & 1 & 2-n & \cdots & 1\\ \vdots & \vdots & \vdots & & \vdots \\ 1 & 1 & 1 & \cdots & 2-n \end{bmatrix}$$
and
$$
\begin{bmatrix}
   1 & 2 & 3 & \cdots & n-1 & n \\
   n & 1 & 2 & \cdots & n-2 & n-1\\
   n-1 & n & 1 & \cdots & n-3 & n-2 \\
   \vdots & \vdots & \vdots & & \vdots & \vdots \\
   2 & 3 & 4 & \cdots & n & 1
\end{bmatrix}^{-1}=
\frac1{ns}
\begin{bmatrix}
  1-s & 1+s & 1 & \cdots & 1 & 1\\
  1 & 1-s & 1+s & \cdots & 1 & 1 \\
  1 & 1 & 1-s & \cdots & 1 & 1\\
  \vdots & \vdots & \vdots & & \vdots & \vdots \\
  1+s & 1 & 1 & \cdots & 1 & 1-s
\end{bmatrix}
$$
in which $s:=1+2+\cdots+n$.

By Cayley-Hamilton theorem, the inverse of every nonsingular matrix $A$ can be expressed as a polynomial in $A$ of degree $\le n-1$. Therefore, if $A$ is in the linear span of $I,J,\ldots,J^{n-1}$, then $A^{-1}$ is a polynomial in $J$ of degree $\le (n-1)^2$. Since $J^n=I$, it follows that $A^{-1}$ lies inside the linear span of $I,J,\ldots,J^{n-1}$. Hence it is circulant.
