show that
where zeta is a primitive n-th root of unity.
hint:first compute,for example
My attempt:i think that we will use induction to prove,but I have done any way I think related,just don't effective….And the root of unity?how can I deal the property with the matrix?
Thank you for advance!
Put $$ C= \begin{bmatrix} 0 & 1 & 0 & \dots & 0\\ 0 & 0 & 1 & \dots & 0\\ \vdots & \vdots & \vdots &\ddots &0\\ 0 & 0 & 0 & \dots & 1\\ 1 & 0 & 0 & \dots & 0\\ \end{bmatrix}.$$
Then the matrix you are discussing is $$ A= x_0 I +x_1 C + x_2 C^2+\dots +x_{n-1} C^{n-1}. $$
Now let $\zeta$ be a primitive $n$-th root of unity. It's clear that $C^n=I$ and for all smaller $k$, $C^k\not=I$. So the characteristic polynomial of $C$ is $X^n-1$, whose roots are $1,\zeta,\zeta^2,\dots,\zeta^{n-1}$.
Now in the usual way, $Cv=\zeta^{i}v$ implies $C^{j}v=\zeta^{ij}v$. Hence the eigenvalues of $A$ are $\sum_{j=0}^{n-1} x_j \zeta^{ij}$ for $i=0,\dots, n-1$. Taking the product of these gives you what you want.