<s>[INST] Find general form of eigenvalues of a circulant matrix I have an $n \times n$ matrix in a general form:
$$ A = \begin{pmatrix}
  \alpha^{n-1} & 1 & \alpha & \alpha^2 & \dots & \alpha^{n-2} \\
  \alpha^{n-2} & \alpha^{n-1} & 1 & \alpha & \dots & \alpha^{n-3} \\
  \alpha^{n-3} & \alpha^{n-2} & \alpha^{n-1} & 1 & \dots & \alpha^{n-4} \\
  \alpha^{n-4} & \alpha^{n-3} & \alpha^{n-2} & \alpha^{n-1} & \dots & \alpha^{n-5} \\
  \vdots  & \vdots & \vdots & \vdots & \ddots & \vdots \\
  1 & \alpha & \alpha^2 & \alpha^3 & \dots & \alpha^{n-1}
\end{pmatrix},$$
where $\alpha \in (0,1)$ is some parameter. Is there a chance to have eigenvalues of this matrix in some general form (as an expression in $n$ and $\alpha$)? Or, at least, is that possible to say something about them?
I tried to numerically generate some of them but I don't really see any pattern.

 [/INST] Notably, $A$ is a circulant matrix. As such, its eigenvalues will be of the form
$$
\lambda_j = \alpha^{n-2}\omega^{(n-1)j} + \cdots + \alpha \omega^{2j} + \omega^j +\alpha^{n-1},
$$
where $\omega = e^{2 \pi i/n}$ and $j = 0,1,\dots,n-1$. We can simplify $\lambda_j$ to
$$
\lambda_j = \alpha^{-1}((\alpha \omega^j)^{n-1} + \cdots + (\alpha \omega^j)^2 + (\alpha \omega^j) + 1) + \alpha^{n-1} - 1
\\ = \alpha^{-1}\frac{1-(\alpha \omega^j)^n}{1-\alpha\omega^j} + \alpha^{n-1} - 1\\
= \frac{1-\alpha^n \omega^{jn}}{\alpha - \alpha^2 \omega^j} + \alpha^{n-1} - 1.
$$
For $j = 0$, this yields the real eigenvalue
$$
\lambda_0 = 1 + \alpha + \cdots + \alpha^{n-1} = \frac{1 - \alpha^n}{1 - \alpha}.
$$
</s>