A $2k\times 2k$ circulant matrix $\ C$ takes the form 


\begin{align}
 C= \begin{bmatrix} c\_0 & c\_{2k-1} & \dots & c\_{2} & c\_{1} \\ 
c\_{1} & c\_0 & c\_{2k-1} & & c\_{2} \\ 
\vdots & c\_{1}& c\_0 & \ddots & \vdots \\ 
c\_{2k-2} & & \ddots & \ddots & c\_{2k-1} \\ 
c\_{2k-1} & c\_{2k-2} & \dots & c\_{1} & c\_0 \\ \end{bmatrix}. 
\end{align}


Now we consider a $2k\times 2k$ generalized circulant matrix $\ C\_{1}$


\begin{align}
 C\_{1}= \begin{bmatrix} c\_0 & -c\_{2k-1} & \dots & c\_{2} & -c\_{1} \\ 
c\_{1} & -c\_0 & c\_{2k-1} & & -c\_{2} \\ 
\vdots & -c\_{1}& c\_0 & \ddots & \vdots \\ 
c\_{2k-2} & & \ddots & \ddots & -c\_{2k-1} \\ 
c\_{2k-1} & -c\_{2k-2} & \dots & c\_{1} & -c\_0 \\ \end{bmatrix}. 
\end{align}


That is the first column shifts with a multiplication of $-1$.


Then what are the eigenvalues of $\ C\_{1}$? Thank you!



the eigenvalues of a generalized circulant matrix
$C\_1$ isn't quite a "generalized $2k \times 2k$ circulant",
because it does not include $C$ as a special case. It is, however,
block-circulant: let $A\_0,\ldots,A\_{k-1}$ be the $2\times 2$ matrices
$$
A\_j = \left[\begin{array}{cc}
 c\_{2j} & -c\_{2j-1} \\ c\_{2j+1} & -c\_{2j}
\end{array}\right]
\quad(j=0,1,\ldots,k-1)
$$
(with $c\_{-1}$ interpreted as $c\_{2k-1}$); then $C\_1$ is a $k\times k$
block-circulant with first column $A\_0,\ldots,A\_{k-1}$. As with
ordinary circulant matrices, a block-circulant matrix is conjugate
over $\bf C$ to a block-diagonal matrix with $n$-th block
$$
\hat A\_n := \sum\_{j=0}^{k-1} e^{2\pi i nj/k} A\_j.
$$
So the eigenvalues of $C\_1$ are the eigenvalues of the $\hat A\_n$,
which in this $2 \times 2$ case are given by the formula
$\frac12(t \pm \sqrt{t^2-4\Delta})$ where $t$ is the trace and
$\Delta$ is the determinant.


In the present case, each $A\_j$ has trace zero, so the same is true of
all the $\hat A\_n$, and the eigenvalues are simply
$\pm \sqrt{-\det \hat A\_n}$.
This suggests an alternative description. Form the matrix $C\_1^2$,
and note that its entries are zero when the row and column index
have opposite parity. Separating odd and even indices we find that
$C\_1^2$ is conjugate with a block-diagonal matrix
$\left[{C \; 0 \atop 0 \; C}\right]$ for some $k \times k$ matrix $C$
that is circulant in the usual sense. Compute the eigenvalues of this
circulant matrix $C$ as usual, and then extract their square roots
to recover the eigenvalues of $C\_1$.


