Canonical decomposition of absorbing chains An absorbing Markov chain on $n$ states for which $t$ states are transient and $n-t$ states are absorbing can be reordered in a canonical decomposition with transition matrix 
$$\boldsymbol{P}=
\left[
\begin{array}{c|c}
\boldsymbol{Q} & \boldsymbol{R} \\
\hline
\boldsymbol{0} & \boldsymbol{I}
\end{array}
\right]$$
where $\boldsymbol{Q}$ is a $t\times t$ matrix and $\boldsymbol{R}$ is a $t\times (n-t)$ matrix.  As usual, $\boldsymbol{0}$ is a matrix of zeros and  $\boldsymbol{I}$ is the identity matrix. 
An example would be the transition matrix
$$\boldsymbol{P'}=\begin{bmatrix}
0 & .6 & 0 &  0& .4 & 0\\ 
 .4&  0&  .6&  0& 0 &0 \\ 
 0&  .4&  0&  .6& 0 & 0\\ 
 0&  0&  .4&  0&  0&.6 \\ 
 0&  0&0  & 0 & 1 &0 \\ 
 0& 0 &0  & 0 &0  & 1
\end{bmatrix}.$$
Now, using the fundamnetal matrix $\boldsymbol{F}=(\boldsymbol{I}-\boldsymbol{Q})^{-1}$, we can calculate absorbtion probabilities and expected times until absorption.  
One example gives the doubly stochastic transition matrix
$$\boldsymbol{T} = \begin{bmatrix}
.3 & .5 & .2\\ 
 .2& .4 & .4\\ 
.5 & .1 & .4
\end{bmatrix}$$
and asks to find $E(S \mid X_0 =1)$ where $S$ is the first time that $X_n=3$. The fundamental matrix is used, and the solution defines
$$\boldsymbol{Q}=\begin{bmatrix}
.3 & .5\\ 
 .2& .4
\end{bmatrix}.$$
How does this make sense if the matrix is clearly not in its canonical form, and what is an efficient method to transform a matrix into its canonical form?  
 A: Because we want to know the number of steps until we first reach state $3$, the transition probabilities out of state $3$ don't actually matter. Therefore we can replace the matrix 
$$\boldsymbol{T} = \begin{bmatrix}
.3 & .5 & .2\\ 
 .2& .4 & .4\\ 
.5 & .1 & .4
\end{bmatrix}$$
by the matrix
$$\boldsymbol{T'} = \begin{bmatrix}
.3 & .5 & .2\\ 
 .2& .4 & .4\\ 
0 & 0 & 1
\end{bmatrix}$$
which describes a similar but absorbing Markov chain: here, once you get to state $3$, you never leave.
Again, this is no longer the same Markov chain, but if all we want to know is the expected number of steps until we reach state $3$, the answer is going to be the same whether we use $\boldsymbol T$ or $\boldsymbol {T'}$, because it doesn't matter what happens after we're in state $3$. Moreover, it makes sense to switch to $\boldsymbol {T'}$, because then "number of steps until we reach state $3$" becomes "number of steps until absorption", which we know how to find by using the fundamental matrix.
But $\boldsymbol {T'}$ actually is in the canonical form 
$\left[
\begin{array}{c|c}
\boldsymbol{Q} & \boldsymbol{R} \\
\hline
\boldsymbol{0} & \boldsymbol{I}
\end{array}
\right]$with $\boldsymbol{Q}=\begin{bmatrix}
.3 & .5\\ 
 .2& .4
\end{bmatrix},$
so now we can proceed by using properties of this canonical form.
