Ok so whilst I was reading my notes on Markov chains, this is a particular part in the notes I don't completely understand. I don't get how its clear from (1.17) that rows of U are the normalized left eigenvectors for P, simiarly how the columns of U inverse are the normalized right eigenvectors of P. Lastly, where P here is a transition matrix, could someone explain why a transition matrix must have an eigenvalue of 1 always?
Thanks
You've diagonalized the matrix $P$ as $P=U^{-1}DU$. Multiply on the left by $U$ to get $UP=DU$. Now take the $i$th row of both sides to get $r_i P= d_i r_i$ where $r_i$ is the $i$th row of $U$. This shows that $r_i$ is a left eigenvector of $P$ with eigenvalue $d_i$ (the $i$th entry in the diagonal matrix $D$.)
The argument for the columns of $U$ is similar, starting with $PU^{-1}=U^{-1}D$.
The value 1 is an eigenvalue for $P$ since $P$ is stochastic, i.e., its row sums are all equal to 1. Therefore if you let $v$ be the vector of ones, we have $P v= 1 v.$