Invariant Probability Vector I'm reading through my textbook, Introduction to Stochastic Processes (Lawler), before the semester begins in hopes of getting ahead, and I've run into something I just plain cannot figure out: How to compute the invariant probability vector for a transition matrix.  I was hoping that one (or many) of you would be able to walk me through how you would do this for just a simple matrix:
$$\begin{bmatrix} .4&.2&.4 \\\\ .6&0&.4 \\\\ .2&.5&.3 \end{bmatrix}$$
I know that you can compute it by raising the matrix to a large power, but this practice problem says to "compute the invariant probability vector as a left eigenvector."  How would one go about doing this?  
Thanks for your help!
 A: If the transition matrix is $A$ and the probability vector is $\mu$, "invariant" means that $\mu A = \mu$.  Another way of saying this is that $\mu$ is a left eigenvector of $A$ with eigenvalue 1.
$\mu A = \mu$ is really just a system of linear equations.  If we write $\mu = [\mu_1, \mu_2, \mu_3]$ then we have
$$[\mu_1, \mu_2, \mu_3] \begin{bmatrix} .4&.2&.4 \\\\ .6&0&.4 \\\\ .2&.5&.3 \end{bmatrix}= [\mu_1, \mu_2, \mu_3]$$
or in other words
$$\begin{align*}
.4 \mu_1 + .6 \mu_2 + .2 \mu_3 &= \mu_1 \\
.2 \mu_1 + 0 \mu_2 + .5 \mu_3 &= \mu_2 \\
.4 \mu_1 + .4 \mu_2 + .3 \mu_3 &= \mu_3.
\end{align*}
$$
Since $\mu$ is to be a probability vector we also have to have
$$\mu_1 + \mu_2 + \mu_3 = 1.$$
So you have a system of 4 linear equations in 3 unknowns.  Now you just have to solve this system.
A: It means, find a vector $v$ with the following properties: all entries between 0 and 1, entries add up to 1, and $vA=v$ (where $A$ is your transition matrix). 
If you know how to find (left) eigenvectors, you just find one for the eigenvalue 1, and then normalize it so the sum of the entries is 1. 
A: If T is the transistion probability matrix the stationary (or invariant) distribution satisfies the matrix equation
P(X)= T P(X)
