# Square root of a positive-definite Markov matrix

Let $M$ be a positive definite Markov matrix, meaning that $M_{ij}\geq0$, $\sum_j{M_{ij}}=1$ and all eigenvalues of $M$ are positive. Given the spectral decomposition $M=V\Lambda V^{-1}$, the matrix $R=\sqrt M$ can be defined as $R=V\sqrt\Lambda V^{-1}$. I would like to know when is $R$ a Markov matrix. By the Perron-Frobenius theorem, $M \bf{1}=\bf{1}$, where $\bf{1}$ is the all-ones vector. Using this fact, it is easy to prove that $\sum_j{R_{ij}}=1$ for all $M$ defined as above. However, when is $R_{ij}\geq0$?

• I added a clarification. Apr 3, 2014 at 15:18
• If is markov matrix also the sum of the elements in a row is one or not? Apr 3, 2014 at 15:20
• Yes, the sum in a row is 1. Apr 3, 2014 at 15:39
• This article could help you to answer your question. A sufficient condition for $\sqrt{M}$ to be a stochastic (Markov) matrix is that $M$ is the inverse of an M-matrix. Apr 3, 2014 at 16:54