Find the general expression for the values of a steady state vector of an $n\times n$ transition matrix

Question

I have a question that is asking to find the values of the elements in the steady state vector for a regular transition matrix P of size $n \times n$. All I'm given is that the the elements in each row of the transition matrix sum to 1.

So I guess I'm supposed to find a generic formula for the value of each element in the steady state vector but I have no idea how to even start.

do I need to use the $\left(I-P\right)\mathbf{q}=0$ formula?

I feel like I understand how to find the steady state vector for a given transition matrix, but without knowing anything but the dimensions of P I'm stuck. Any hints would be much appreciated. Thanks guys.

Answer 1

Since all rows sum to $1$, the vector

$$ \frac1n\pmatrix{1\\\vdots\\1} $$

(with $n$ $1$s) is an eigenvector of the transition matrix with eigenvalue $1$. Since the chain is regular, it has a unique stationary distribution, so this is the stationary distribution.

Answer 2

The term "steady state vector" usually refers to a stationary distribution for $P$; that is, a row vector $\pi$ with entries that add to $1$, such that $\pi P=\pi$. Because $P$ is regular, $\lim_nP^n =\Pi$. ($\Pi$ is the $n\times n$ matrix whose every row is $\pi$.) Consequently, if $q$ is any column vector then $\lim_nP^nq$ is a column vector whose every entry is equal to $\sum_k\pi_kq_k$.