Trace back orginal matrix from power outcome

Question

I have a $8 \times 8$ unknown transition matrix $T$. I do know the last column and row. The eight state is an absorbing state and the last column is given. Furthermore I know the last columns of $T^i \space for \space i=1...30$ (and of course its last row, since the eight state is an absorbing state.) Is it possible to find the original matrix $T$?

Answer 1

Assume that $T=[t_{i,j}]$ is a random stochastic matrix where the $(t_{i,j})$ follow a continuous law. With probability $1$, $T$ is a cyclic matrix such that the sequence $S=\{e_8,Te_8,\cdots T^7(e_8)\}$ (the last columns of the $T^i$) is a basis of $\mathbb{R}^8$. Let $P$ be the matrix of change of basis (from $(e_i)$ to $S$); then $P^{-1}AP=F$, a Frobenius matrix (for $i<8$, the $i^{th}$ column is $e_{i+1}$); note that $F$ is a known matrix because the last column of $F$ is $B^{-1}T^8(e_8)$.

Finally $A=PFP^{-1}$ and it suffices to know the last columns of $T,\cdots,T^8$.