Eigenvalues of anti-circulant matrices using 1-circulant matrices Is there any theorem to find the eigenvalues of any anti-circulant matrix using the equivalent (with the same first row) circulant matrix. I found out that,
for any anti-circulant matrix, the eigenvalues (taken as $\mu$) of the anti-circulant matrix can be written as,
\begin{equation}
\mu = \pm \mid{\lambda_j}\mid
\label{mu_alpha}
\end{equation}
where $\lambda_j$ is an eigenvalue of 1-circulant matrix with the same first row. This seems valid since any anti-circulant matrix should be symmetric resulting in real eigenvalues.
Can anyone send me a link to any reference which has this proof..? or can you please comment if you think that this should not be correct ?
 A: Edit. Presumably the matrix is real, otherwise the claim is false and it is easy to generate a counterexample by computer.
Let $A$ be a circulant matrix and $B$ be the anticirculant matrix such that $A$ and $B$ have identical first rows. Then $B=PA$ for some permutation matrix $P$ (more specifically, $P=1\oplus J_{n-1}$, where $J_{n-1}$ is the reversal matrix obtained by flipping $I_{n-1}$ from left to right). Hence $A$ and $B$ have identical singular values. However, circulant matrices and real anticirculant matrices (which are also real symmetric) are normal matrices, so they can be unitarily diagonalized and their singular values are the moduli of their eigenvalues. Therefore, the eigenvalues $A$ and $B$ have identical moduli. Finally, as real anticirculant matrices are real symmetric, $B$ has real eigenvalues. Hence the assertion.
A: Let $C$ be the circulant matrix whose first row is $(c_0,c_1,,\ldots,c_{n-1})$ and let $P(x)=\sum_{k=0}^{n-1}c_kx^k$. For each $n$-th root of unity $\omega$, let $e_{\omega}=(1, \omega, \ldots, \omega^{n-1})^\top$. Then $Ce_{\omega}=P(\omega)e_{\omega}$ and hence the eigenvalues of $C$ are $P(\omega)$.
For the anti-circulant matrix $A$ with the same first row, we have $Ae_{\omega}=P(\omega)e_{\bar\omega}$. Therefore, for each $\omega\ne\pm1$, the restriction of $A$ to the invariant subspace with ordered basis $\{e_{\omega}, e_{\bar\omega}\}$ has the matrix representation
$$A\begin{bmatrix}e_\omega&e_\bar\omega\end{bmatrix}=
\begin{bmatrix}e_\omega&e_\bar\omega\end{bmatrix}\begin{bmatrix}0&P(\omega)\\P(\bar\omega)&0\end{bmatrix},
$$
then
$$A^2\begin{bmatrix}e_\omega&e_\bar\omega\end{bmatrix}=
\begin{bmatrix}e_\omega&e_\bar\omega\end{bmatrix}\begin{bmatrix}P(\omega)P(\bar\omega)&0\\0&P(\omega)P(\bar\omega)\end{bmatrix}.
$$
Thus the eigenvalues of $A$ are $P(1),\,\pm \sqrt{P(\omega)P(\bar\omega)}$ for each $\omega\neq\pm1$, and when $n$ is even, also $P(-1)$.
When $A$ is real, so are the coefficients of $P$. Therefore $\pm\sqrt{P(\omega)P(\bar\omega)}=\pm|P(\omega)|$ for each $\omega\ne\pm1$ and $P(1),P(-1)$ are also real.
