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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 :

$$ \begin{pmatrix}a & b & c & d & e & f \\ b & c & d & e & f & a \\ c & d & e & f & a & b \\ d & e & f & a & b & c \\ e & f & a & b & c & d \\ f & a & b & c & d & e \end{pmatrix}$$

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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 $\pm\sqrt{\lambda_k\lambda_{-k}}$.

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  • $\begingroup$ Yes, just wow that was cool. $\endgroup$ Feb 2, 2018 at 6:13

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