Prove that for every root of unity $z$ of size $n$, the vector ($1, z, z^2,..., z^{n-1}$) is an eigenvector of the circulant matrix $A$

Question

Let $a_1, a_2, \dots, a_n \in \Bbb{C}$. Let

$$ A = \begin{bmatrix} a_1 & a_2 & \dots & a_{n-1} &a_n\\ a_n & a_1 & \dots & a_{n-2} & a_{n-1}\\ \vdots & \vdots & \ddots & \vdots & \vdots \\a_2 & a_3 & \dots & a_n & a_1\end{bmatrix} $$

How can I prove that for every root of unity $z$ of size $n$, the vector ($1, z, z^2,\dots, z^{n-1}$) is an eigenvector of $A$?

I'm studying linear algebra and I encountered some difficulties with this task. I would love to get some hints or an answer for this one.

Answer 1

Let $v=(1,z,\ldots,z^{n-1})$.

Begin by computing the first coordinate of $Av$. It is $\lambda=a_1+a_2z+a_3z^2+\cdots+a_nz^{n-1}$. Since the first coordinate of $v$ is 1, $\lambda$ must be the eigenvalue of $A$ corresponding to $v$ if $v$ is an eigenvector of $A$.

Now compute the $i$th coordinate of $Av$ and verify that it is $\lambda z^{i-1}$ (as you want it to be). This shows that $v$ is an eigenvector of $A$ with eigenvalue $\lambda$.