I have a doubt about a proof I've found in my notes. Suppose $M$ is the transition matrix of an absorbing Markov chain, that is $m_{ij}$ is the probability of reaching state $i$ from state $j$ in one step. Let's resort the elements in a way that the first m elements are the absorbing states: $$ M= \left[ {\begin{array}{cc} I_m & B\\ O_h & A\\ \end{array} } \right] $$ where $I_m$ is the identity matrix relative to the absorbing states, $O_h $ is the null matrix. $B$ can't be the null matrix since the Markov chain is absorbing
$\bf{Thm.}$ The submatrix $A$ is substochastic and $ |\lambda| < 1, \forall \lambda $ eigenvalue of $A$.
In the proof, $ 0 \le \lambda \le 1$ is first proved, and I'm okay wiht that part. Then it says: let's assume there exists $\lambda_A = 1$. Then there exists an invariant probability distribution $\bf{p}$ for $A$ ( normalized eigenvector), and a corresponding invariant distribution for $M$, $\bf{\hat{p}} = \left[ {\begin{array}{cc} \bf{0} & \bf{p} \end{array} } \right] \\ $, such that: $$ M \bf{\hat{p}} = \left[ {\begin{array}{cc} I_m\bf{0} & A \bf{p} \end{array}} \right] = \bf{\hat{p}}$$ and this should contradict the fact that M is absorbing. My first doubt is why $\bf{\hat{p}}$ is invariant for $M$ if $\bf{p}$ is invariant for $A$. The second is why in the equation above it is $I_m \bf 0$ and not $I_m {\bf 0} + B \bf{p}$. The third is why this contradicts the fact that $M$ is absorbing.