Finding $n$-th power of transition matrix Is there any shortcut to find $P^{n}=\begin{pmatrix} 1-p & p \\ q & 1-q \end{pmatrix}^n$ quickly and elegantly? 
This type of matrix often comes up while dealing with Markov chains. Diagonalization takes way too much time. I need to find eigenvalues, eigenvectors and all that...

The eigenvalues are $1, -(p+q-1)$
$P = \frac {1}{p+q}\begin{bmatrix} 1&-p\\1&q \end{bmatrix}\begin{bmatrix} 1\\&-(p+q-1) \end{bmatrix}\begin{bmatrix} q&p\\-1&1 \end{bmatrix}\\
P^n= \frac {1}{p+q}\begin{bmatrix} 1&-p\\1&q \end{bmatrix}\begin{bmatrix} 1\\&\lambda^n \end{bmatrix}\begin{bmatrix} q&p\\-1&1 \end{bmatrix}\\
P^n =\frac {1}{p+q}\begin{bmatrix} q+p\lambda^n&p(1-\lambda^n)\\q(1-\lambda)^n&p+q\lambda^n \end{bmatrix} $
