I have a transition matrix $P$ (hence it's rows add up to 1) and I would like to compute it's Nth power. I am finding a lot of answers using the 'Diagonalization' method however I want to prove that actually the Nth power can be written as
$P^N=\sum\limits_{j=1}^kC_j \psi_j^N$ where $C_j$ are constant matrices computed from $P$ and its eigenvalues and $\psi_1,\ldots,\psi_k$ are distinct eigenvalues.
If $P$ has distinct eigenvalues, then it is diagonalizable. Any diagonalizable matrix can be expressed in that form:
Let $$ A = M D M^{-1} $$ a diagonalizable matrix with $D$ diagonal. Then you can write $$ D = \sum_i \psi_i e_ie_i^T $$ where $\psi_i$ are the eigenvalues. Then $$ A = \sum_i \psi_i Me_ie_i^TM^{-1} $$ and $$ A^N = MD^NM^{-1}= \sum_i \psi_i^N Me_ie_i^TM^{-1} $$