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:

$$\begin{bmatrix} .4&amp;.2&amp;.4 \\\\ .6&amp;0&amp;.4 \\\\ .2&amp;.5&amp;.3 \end{bmatrix}$$

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!
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
$$[\mu_1, \mu_2, \mu_3] \begin{bmatrix} .4&amp;.2&amp;.4 \\\\ .6&amp;0&amp;.4 \\\\ .2&amp;.5&amp;.3 \end{bmatrix}= [\mu_1, \mu_2, \mu_3]$$
or in other words
$$\begin{align*}
.4 \mu_1 + .6 \mu_2 + .2 \mu_3 &amp;= \mu_1 \\
.2 \mu_1 + 0 \mu_2 + .5 \mu_3 &amp;= \mu_2 \\
.4 \mu_1 + .4 \mu_2 + .3 \mu_3 &amp;= \mu_3.
\end{align*}
$$
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.
If T is the transistion probability matrix the stationary (or invariant) distribution satisfies the matrix equation

P(X)= T P(X)