TITLE: Eigenvalues of anti-circulant matrices using 1-circulant matrices
QUESTION [7 upvotes]: 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 ?

REPLY [5 votes]: 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.