Finding eigenvalues of circulant matrices

Question

$$ A=\begin{pmatrix} \alpha^R&\beta^R &\gamma^R&-\alpha^I&-\beta^I &-\gamma^I\\ \gamma^R&\alpha^R&\beta^R &-\gamma^I&-\alpha^I&-\beta^I \\ \beta^R &\gamma^R&\alpha^R&-\beta^I&-\gamma^I&-\alpha^I \\ \alpha^I&\beta^I &\gamma^I&\alpha^R&\beta^R &\gamma^R\\ \gamma^I&\alpha^I&\beta^I &\gamma^R&\alpha^R&\beta^R \\ \beta^I &\gamma^I&\alpha^I&\beta^R&\gamma^R&\alpha^R \\ \end{pmatrix} $$ The above matrix is a block circulant matrix. How to find the eigenvalues of $A$?

Answer 1

According to Maple, two of the eigenvalues are $$ \alpha^R + \beta^R + \gamma^R \pm i (\alpha^I + \beta^I + \gamma^I)$$ The other four are roots of a rather unpleasant irreducible quartic.

Answer 2

I assume that all matrix-entries are real numbers.

Your matrix can be written in the form $$ A = \pmatrix{B & -C\\C & B}. $$ Via the standard identification of $\Bbb C$ with $\Bbb R^2$, we can see that the eigenvalues of $A$ have the form $\lambda_1,\lambda_2,\lambda_3,\bar\lambda_1, \bar \lambda_2, \bar \lambda_3$, where $\bar z$ denotes the complex conjugate of $z$ and $\lambda_1,\lambda_2,\lambda_3$ are the eigenvalues of the $3 \times 3$ complex matrix $M = B + iC$.

Because $B + i C$ is itself a circulant matrix, we can find its eigenvalues systematically. In particular, we will have $$ \lambda_j = (\alpha^R + i\alpha^I) + (\beta^R + i\beta^I)\omega^j + (\gamma^R + i \gamma^I)\omega^{2j} \quad j = 1,2,3, $$ where $\omega = e^{2 \pi i/3}$. The eigenvalues in Robert's answer are $\lambda_3,\bar \lambda_3$.