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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

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    @maximus: Edited2011-10-12

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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?

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    doob, I think you're making this more complicated than it needs to be. You wanted to know (I think) why the entries of $P^n$ have a certain form. I gave you a formula for $P^n$. If you let Q=\pmatrix{a&b\cr c&d\cr} and then calculate $Q^{-1}$ and then multiply out $QD^nQ^{-1}$ (using the formula I gave for $D^n$) then I think you'll see your formula for the entries of $P^n$. Yes, what's *behind* it is a change of coordinates, but you don't need to worry about that to understand and use what I've written.2011-10-12