Given a matrix, $P$, why does finding its eigenvalues, say they are $\{\lambda_1, \lambda_2\}$ then the general form of $p_{ij}^{(n)}=A_{ij}\lambda_1^n+B_{ij}\lambda_2^n$? Thanks.
Added: Context: $P$ is a transition matrix
Given a matrix, $P$, why does finding its eigenvalues, say they are $\{\lambda_1, \lambda_2\}$ then the general form of $p_{ij}^{(n)}=A_{ij}\lambda_1^n+B_{ij}\lambda_2^n$? Thanks.
Added: Context: $P$ is a transition matrix
If $P$ is a $2\times2$ matrix with eigenvalues $\lambda_1\ne\lambda_2$ then $P=QDQ^{-1}$ where $Q$ is the matrix whose columns are the eigenvectors of $P$ and $D=\pmatrix{\lambda_1&0\cr0&\lambda_2\cr}$. So $P^n=QD^nQ^{-1}$, and $D^n=\pmatrix{\lambda_1^n&0\cr0&\lambda_2^n\cr}$. Can you take it from there?