The Question:

It is possible to rearrange the matrix equation $\pi^TP= \pi^T$ into a linear system $Ax = b$ where $x = \pi$ is the unique solution to the system. Such a system could be solved by, among other methods, Gaussian Elimination. Deduce the expressions for the coefficient matrix A and the right hand side vector b.

My Thoughts:

I have considered that the second equation can be rearranged to $ x = A^{-1} b $ and therefore $ \pi = A^{-1} b$ but I am unsure if this is even relevant to the question. Overall, I am extremely stuck. Any hints or help would be greatly appreciated!!

  • 1
    $\begingroup$ If $P$ is the transition matrix for a Markov chain and $\pi$ is supposed to be an invariant probability vector, you have an additional equation $\pi^T {\bf 1} = 1$, where $\bf 1$ is a vector of all $1$'s. $\endgroup$ – Robert Israel May 23 '16 at 4:54

$$\pi^TP=\pi^T$$ $$P^T\pi=\pi$$ $$(P^T-I)\pi=0$$


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