Matrix Decomposition $B = B^* + \sum_{i>1}\lambda_i B_i$ - Cross Validated most recent 30 from stats.stackexchange.com 2019-07-21T11:03:57Z https://stats.stackexchange.com/feeds/question/315786 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://stats.stackexchange.com/q/315786 2 Matrix Decomposition $B = B^* + \sum_{i>1}\lambda_i B_i$ CobbDG https://stats.stackexchange.com/users/186255 2017-11-26T20:40:54Z 2017-11-27T01:47:14Z <p>I recently encountered the following decomposition in a paper, may I know whether anyone on CV happened to know the name of the decomposition and a proof of it?</p> <p>$B$ is a regular stochastic matrix and can be decomposed as:</p> <p>$$B = B^* + \sum_{i&gt;1}\lambda_i B_i$$ where $\lambda_i$ is the eigenvalue, $B_i = f_i \pi_i$ ($f_i$ and $\pi_i$ are the left and right eigenvectors respectively).</p> <p>Further details:</p> <p>B is a $mn \times mn$ matrix and has strictly positive entries and $lim_{k \to \infty} B^k = B^*$ where $B^*$ is the unique stationary distribution.</p> <p>The address to the paper is: <a href="https://papers.ssrn.com/sol3/papers.cfm?abstract_id=2970694" rel="nofollow noreferrer">https://papers.ssrn.com/sol3/papers.cfm?abstract_id=2970694</a></p> <p>and i'm referring to Theorem 2 on page 9. The author didn't specify how the above equation is derived.</p>