TITLE: Eigenvalues of anti-circulant matrix
QUESTION [8 upvotes]: Is there any theorem to find the eigenvalues of any anti-circulant matrix with real entries? 
By anti-circulant matrix, I mean any $n\times n$ matrix  of the form :
$$ \begin{pmatrix}a & b & c & d & e & f \\ b & c & d & e & f & a \\ c & d & e & f &  a & b \\ d & e & f & a & b & c \\ e & f & a & b & c & d  \\ f & a & b & c & d & e  \end{pmatrix}$$

REPLY [8 votes]: Let $v_k=\pmatrix{1&\zeta^k&\zeta^{2k}&\cdots&\zeta^{-k}}^\top$
where $\zeta=\exp(2\pi i/n)$. Of course the $v_k$ are the eigenvectors
of the $n$-by-$n$ circulant matrix, but for an anticirculant $A$ one
has $Av_k=\lambda_k v_{-k}$ instead. Here $\lambda_k=\sum_j \zeta^{jk}a_j$
where $\pmatrix{a_0&a_1&\cdots&a_{n-1}}$ is the top row of $A$. So
$\lambda_0$ is an eigenvalue and so is $\lambda_{n/2}$ if $n$ is even.
The rest of the space of column vectors split up into subspaces with
bases $\{v_k,v_{-k}\}$, where $0<k<n/2$. On this space $A$ acts
like $\pmatrix{0&\lambda_k\\\lambda_{-k}&0}$ which gives eigenvalues
$\pm\sqrt{\lambda_k\lambda_{-k}}$.