Consider sampling the uniform measure $\mathbb{Z}_L$ by the Markov chain $X^{(k)}$ with $$P_i[X^{(i)}=(i+1)\mod L]=1$$ with initial condition $X^{(0)}=L-1$. Here $P_i$ means conditioning on $X_0=i$. $$P_i[X^{(k)}=j]=\begin{cases}1, &\text{if }j=(i+k)\mod L\\0, &\text{otherwise}\end{cases}$$ First, I compute the transition matrix. The state space is $\{0,1,\ldots,L-1\}$, and the Markov process permutes one element to the right each time, i.e. $$[0,1,2,\ldots,L-1]\rightarrow[1,2,\ldots,L-1,0]\rightarrow[2,3,\ldots,L-1,0,1]\rightarrow\ldots$$ so $$T=\begin{pmatrix}0&1&0&0&\ldots&0&0\\0&0&1&0&\ldots&0&0\\\vdots&\vdots&\vdots&\vdots&\cdots&\vdots&\vdots\\0&0&0&0&\ldots&0&1\\1&0&0&0&\ldots&0&0\end{pmatrix}_{L\times L}$$ I am interested in finding the eigenvalues of $T$, $T^k$ for any $k$, and $F=\frac{1}{L}\sum_{k=1}^{L}T^k$.
For $T$, the eigenvalues are solution of $\lambda^L-1=0$, which are $L$-roots of unity $\lambda=e^{2\pi i\ell/L}$, where $\ell=0,1,\ldots, L-1$.
For $T^k$, the eigenvalues are the same as the first case, since the effect is just permuting the columns of $T$ cyclically, so the determinant of characteristic polynomial remains unchanged.
For $F$, first consider $J=\sum_{k=1}^{L}T^k$. Each column of $J$ consists of sum of all standard unit vectors $e_i$ and all the entries are $1$, so $J$ is $L\times L$ matrix whose entries are all $1$. So the rank of $J$ is just $1$, with eigenvalue $\lambda=0$ with multiplicity $L-1$ and $\lambda=L$, which is the result of $\pi T=\pi$, where $\pi$ is the stationary distribution, and we are viewing it as an eigenvector. Hence, the eigenvalues of $F=J/L$ is just either $0$ (with multiplicity $L-1$) or $1$.
I'd like to know whether my solutions are correct. If there are any flaws, can anyone point that out? Thank you.
Edit:
The second part of my original solution is wrong. Use the fact that if $\lambda_{\ell}$ are eigenvalues of $T$, then $\lambda_{\ell}^k$ are eigenvalues of $T^k$, we have the result.
