We have $n$ people standing on a circle, each one has written on his head a real number, and the sum of the numbers on the people's heads is $M$. We are also given that $n$ is odd. Each second, each of the people deletes the number on his head and replaces it with the average of the numbers of his neighbors. Define the number on the head of the i'th guy after $k$ seconds as $a_i(k)$. Show that $lim_{k \to \infty} a_i(k) = M/n$.
Hint: Look at $T:\mathbb{C}^n \rightarrow \mathbb{C}^n$ defined by $Te_i = e_{i+1}, Te_n=e_1$. Let $\lambda_n$ be a primitive $n$'th root of unity.
Then $(w_1,...,w_n)$ is eigenbasis where $(w_i)_j = \lambda_n^{-(i-1)(j-1)}$ with eigenvalues $1,\lambda_n,\lambda_n^2,...,\lambda_n^{n-1}$. It is also an eigenbasis for $T^{-1}$ with eigenvalues $1,\lambda_n^{-1},,,,\lambda_n^{-(n-1)}$
Use that to find eigenbasis for $R = \frac{1}{2} (T+T^{-1})$. Use that to solve the problem.
I did everthing in the hint and I got that $w_1,...,w_n$ is eigenbasis for $R$ with eigenvalues $1,\cos(\theta),\cos(2\theta),...,\cos((n-1)\theta)$ where $\theta = arg(\lambda_n)$, I also realized how $R$ is related to the problem: $R^k$ just gives us the vector with the $a_i(k)$'s.
Now, take the vector $v$ of the $a_i(0)$'s, and write it as $v = \sum_{j=1}^{n}a_iw_i$.
Then we have $R^k(v)= \sum_{j=1}^{n}a_i\cos^k((j-1)\theta)w_i$. Now I don't know how to continue.
The next step is to realize that every eigenvalue of $R$ is less than $1$ in absolute value, except for the eigenvalue $1$, which corresponds to the eigenvector $(1,...,1)$. This means that the limit $R^\infty=\lim_{k\to\infty}R^k$ exists and has the same eigenbasis, but all eigenvalues are zero, except that the eigenvector $(1,...,1)$ still has eigenvalue $1$. This uniquely determines $R^\infty$ as the transformation $(z_1,...,z_n)\mapsto(z,...,z)$ with $z=(z_1+...+z_n)/n$, since this transformation has the same eigenbasis and eigenvalues. Because $a_i(k)=(R^ka)_i$, the result follows.