I am trying to learn about absorbing chains and finding the limiting matrix for absorbing chains. But there is a concept that I am not understanding.
If I am given the standard matrix: $$ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ .1 & .4 & .5 \\ \end{matrix} $$ I understand how to proceed with finding the fundamental matrix F and limiting matrix.
However, often times I am given a matrix in the standard form: $$ \begin{matrix} 1 & 0 & 0 \\ .2 & .6 & .2 \\ .4 & .2 & .4 \\ \end{matrix} $$
Because the top left section of the matrix is not in the form 1 0 (1st line) and 0 1 (second line), it doesn't seem to fit in with the equations in my text book and therefore when I try to find F=(I-Q)^-1 , I get lost. I've been all over youtube trying to find an example like this but something is not "clicking" for me here. If someone could explain what I am not understanding or refer me to another website with a good definition I would really appreciate it. Thanks
So first of all $I-P$, if $P$ is a transition probability matrix, will not be invertible. I think what you are actually asking about is the left eigenvectors of $P$ with eigenvalue $1$, i.e. the solutions to $\pi P = \pi$ where $\pi$ is a row vector. For that you're just solving a system of linear equations.
In the first case your system reads:
$$\pi_1+0.1\pi_3=\pi_1 \\ \pi_2+0.4\pi_3=\pi_2 \\ \pi_3=0.5\pi_3.$$
By the third equation, $\pi_3=0$; substituting that into the first two equations tells you that both $\pi_1$ and $\pi_2$ are free variables, so that any row vector of the form $\begin{bmatrix} x & y & 0 \end{bmatrix}$ is a left eigenvector with eigenvalue $1$.
In the second case your system reads:
$$\pi_1+0.2\pi_2+0.4\pi_3=\pi_1 \\ 0.6\pi_2+\pi_3=\pi_2 \\ 0.2\pi_2+0.4\pi_3=\pi_3.$$
You can do basically the same thing but with a bit more work to find that the solutions are $\begin{bmatrix} x & 0 & 0 \end{bmatrix}$ where $x$ is arbitrary.
Both of these answers make probabilistic sense: they say that the equilibrium distributions are those which are concentrated on absorbing state(s). This makes sense because the absorbing states are reachable from all other states, so eventually no matter where you start you will get absorbed somewhere.
In cases like the second one where there is just one absorbing state and it is accessible from any state, your limiting transition matrix will just tell you that you eventually go to that absorbing state. In cases like the first one, more work is required, which is perhaps why your textbook treats that case in detail. The problem in the first case is that the limiting matrix needs to tell you not just that you eventually get absorbed but also where you get absorbed, if you start at a non-absorbing state (in this case $3$, but there could be more).
In general you need more eigenvectors to figure this out. Of course in this case it is not so hard: the probability to hit $2$ is $4$ times as high as the probability to hit $1$, and there is nowhere else to go other than $3$ without being absorbed, so the third row of the limiting matrix is just $\begin{bmatrix} 0.2 & 0.8 & 0 \end{bmatrix}$