In a paper, i saw a statement that the smallest eigenvalue of $P$($P$ is reversible Markov chain with stationary distribution $\pi$) is greater than $2 \beta - 1$ with the condition, $P \geq \beta I$. I can't understand this statement, because there isn't enough explanation for that. Please help me for this problem, thanks.
Sorry for missing some definitions. $\beta$ is an arbitary positive constant satisfying $P \geq \beta I$. And, for two square matrices $A$ and $B$($A = [a_{ij}],\ B = [b_{ij}], \ for\ i,\ j=1,\ldots,n$), $A \geq B$ means $a_{ij} \geq b_{ij},\ for\ i,\ j=1,\ldots,n$.
First we must justify that comparing the eigenvalues of $P$ with a real number makes sense. Ideally, we would like the spectrum of $P$ to be real.
A Markov chain is represented by a (row) stochastic matrix $P$, that is, $P\geq 0$ and $Pe=e$, where $e=[1,\ldots,1]^T$. Digging in Wikipedia showed that the Markov chain is reversible if there is a vector $\pi>0$, $\pi^Te=1$ such that $$ \pi_i p_{ij}=\pi_j p_{ji}\quad\text{for $i,j=1,\ldots,n$.} $$ If $\Pi$ is a diagonal matrix $\Pi=\mathrm{diag}(\pi)$, this can be written in the compact matrix form $$\tag{1} \Pi P=P^*\Pi. $$ Multiplying (1) by $\Pi^{-1/2}$ from both sides ($\pi>0$ so $\Pi$ is invertible), we get $$ \Pi^{1/2}P\Pi^{-1/2}=\Pi^{-1/2}P^*\Pi^{1/2}=(\Pi^{1/2}P\Pi^{-1/2})^* $$ and hence $P$ is similar to a symmetric matrix via diagonal scaling. Consequently, all eigenvalues of $P$ are real.
By Gershgorin theorem (and using the fact that $P\geq 0$ has real spectrum), the eigenvalues of $P$ are contained in the union $$ \bigcup_i\left[p_{ii}-\sum_{j\neq i}p_{ij},p_{ii}+\sum_{j\neq i}p_{ij}\right] $$ so the minimal eigenvalue of $P$ is bounded from below by $$ \min_i \left(p_{ii}-\sum_{j\neq i}p_{ij}\right). $$ Since $P$ is stochastic, $\sum_jp_{ij}=1$, so $\sum_{j\neq i}p_{ij}=1-p_{ii}$. In addition, $p_{ii}\geq\beta$, and hence $$ \min_i \left(p_{ii}-\sum_{j\neq i}p_{ij}\right)=\min_i\left(2p_{ii}-1\right)\geq 2\beta -1. $$