How to prove that the matrix P has eigenvalues $1,-\frac{1}{2},\cdots,(-1)^{n-1}\frac{1}{n}$?

Question

I came up with this when trying to solve a problem of a markov chain with the transition matrix $$P=\begin{bmatrix} 0,0,0,\cdots,1\\ 0,0,\cdots,\frac{1}{2},\frac{1}{2}\\ 0,\cdots,\frac{1}{3},\frac{1}{3},\frac{1}{3}\\ \cdots\cdots\\ \frac{1}{n},\cdots,\frac{1}{n},\frac{1}{n} \end{bmatrix}$$ and it asked me to find $$\lim\limits_{k \rightarrow +\infty}{P^k}\alpha$$ where $\alpha=(0,1,\cdots,n-1)^\top$.

So, I tried to diagonalize $P$ and surprisedly found that it has eigenvalues $1,-\frac{1}{2},\cdots,(-1)^{n-1}\frac{1}{n}$ when $n\leq 7$.So I wonder if this is true for all $n$ from $N^+$ and then how to calculate $$\lim\limits_{k \rightarrow +\infty}{P^k}\alpha$$

Thanks!

Answer 1

$P$ is an irreducible stochastic matrix. Therefore, by Perron-Frobenius theorem, $\lim_{k\to\infty}P^k=\frac{vu^T}{u^Tv}$, where $u$ and $v$ are respectively a left eigenvector and a right eigenvector of $P$ corresponding to the eigenvalue $1$. By inspection, we see that up to scaling, $v=(1,1,\ldots,1)^T$ and $u=(1,2,\ldots,n)^T$. Thus $$ \lim_{k\to\infty}P^k=\frac{2}{n(n+1)}\pmatrix{1&2&\cdots&n\\ 1&2&\cdots&n\\ \vdots&\vdots&&\vdots\\ 1&2&\cdots&n}\tag{1} $$ and $\lim_{k\to\infty}P^k\alpha=\frac{2(n-1)}{3}(1,1,\ldots,1)^T$.

Without using Perron-Frobenius theorem, one may consider the following matrix first: $$ M(t,n)=\pmatrix{ &&&&-(n-1)&t+n\\ &&&-(n-2)&t+(n-1)\\ &&\cdots&\cdots\\ &-2&t+3\\ -1&t+2\\ t+1}\in\mathbb R^{n\times n}. $$ Let $V\in\mathbb R^{n\times n}$ be the upper triangular matrix of ones. Then $V^{-1}$ is the bidiagonal matrix whose main diagonal entries are ones and whose superdiagonal entries are minus ones. By direct calculation, we get \begin{aligned} M(t,n)V &=\pmatrix{ &&&&-(n-1)&t+1\\ &&&-(n-2)&t+1&t+1\\ &&\cdots&\cdots&\cdots&\cdots\\ &-2&t+1&\cdots&\cdots&t+1\\ -1&t+1&\cdots&\cdots&\cdots&t+1\\ t+1&\cdots&\cdots&\cdots&\cdots&t+1},\\ V^{-1}M(t,n)V &=\left(\begin{array}{ccccc|c} &&&n-2&-(t+n)&0\\ &&n-3&-(t+n-1)&0&0\\ &\cdots&\cdots&\cdots&\cdots&\cdots\\ 1&-(t+3)&0&\cdots&\cdots&0\\ -(t+2)&0&\cdots&\cdots&\cdots&0\\ \hline t+1&\cdots&\cdots&\cdots&\cdots&t+1\end{array}\right)\\ &=\pmatrix{-M(t+1,n-1)&0\\ \ast&t+1}. \end{aligned} So, recursively, we have \begin{aligned} \operatorname{spectrum}\left(M(t,n)\right) &=\{t+1\}\cup\operatorname{spectrum}\left(-M(t+1,n-1)\right)\\ &=\{t+1,-(t+2)\}\cup\operatorname{spectrum}\left(M(t+2,n-2)\right)\\ &=\{t+1,-(t+2),t+3\}\cup\operatorname{spectrum}\left(-M(t+3,n-3)\right)\\ &\vdots\\ &=\{t+1,\,-(t+2),\,t+3,\,\ldots,\,(-1)^{n-1}(t+n)\}.\\ \end{aligned} It follows that \begin{aligned} \operatorname{spectrum}(P)=\operatorname{spectrum}\left(M(0,n)^{-1}\right) =\left\{1,\,-\frac12,\,\frac13,\,\ldots,\,\frac{(-1)^{n-1}}{n}\right\}. \end{aligned} Since $1$ is a simple eigenvalue and the moduli of all other eigenvalues are strictly smaller than $1$, we again conclude that $\lim_{k\to\infty}P^k=\frac{vu^T}{u^Tv}$, where $u$ and $v$ are respectively any left and right eigenvectors of $P$ corresponding to the eigenvalue $1$.

Answer 2

About calculating $\lim_{k \to + \infty} P^k \alpha$:

Notice that your state space is finite (which I'll assume is $1, \dots, n$) and your corresponding Markov chain irreducible (every state can transition to $n$ and from $n$ to any state). So, there exists a unique stationary distribution $\pi$ with $\pi P = \pi$.

By solving explicitly for smaller $n$, one can guess that $\pi = (\frac{1}{m}, \frac{2}{m}, \dots, \frac{n}{m})$ where $m = \sum_{k = 1}^n k = \frac{n (n+1)}{2}$.

Now, also notice that $P$ is aperiodic as for example $p_{n,n} > 0$.

So, by the convergence theorem, $\lim_{k \to + \infty} P^k = \Pi $ where $\Pi$ is the matrix with $\pi$ as columns.

Hence, $\lim_{k \to + \infty} P^k \alpha = \Pi \alpha = c \cdot (1, \dots, 1 )^\top$, where $$c = \pi^\top \alpha = \frac{1}{m} \sum_{k = 1}^n k \cdot (k-1) = \frac{1}{m} \left( \sum_{k = 1}^n k^2 - \sum_{k = 1}^n k \right) = \frac{1}{m} \left(\frac{n (n+1)(2n+1)}{6} - m \right) = \frac{2(n-1)}{3}.$$