Exercise 5.6.1

It’s clear that a matrix of 2 by 2 won’t work here, so we found a 3 by 3 matrix $P=\left[\begin{array}{ccc}\hfill 0.5\hfill & \hfill 0.6\hfill & \hfill 0.5\hfill \\ \hfill 0\hfill & \hfill 0.2\hfill & \hfill 0.25\hfill \\ \hfill 0.5\hfill & \hfill 0.2\hfill & \hfill 0.25\hfill \end{array}\right]$, it’s clear that $P\ge 0$, and its columns add up to 1. It’s easy to verify that $P^2\ge 0$ is strictly positive. By Perron-Frobenius theorem, $P^2$ is a Markov matrix, it has its largest eigenvalue of ${\lambda }_1=1$ and all other eigenvalues $\left|{\lambda }_i\right|<1$ for $i=2,\dots ,n$.

However, $P^2=X{\Lambda }^2{X}^{-1}$, if all entries on ${\Lambda }^2$ are less than 1 except the first one, then their squar roots are also less than 1, and these square roots are the other eigenvalues of the original matrix $P$.