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...

Let $S$ be the matrix
$$
S=
\begin{bmatrix}
1&p\\1&-q
\end{bmatrix}
\ ,
$$
which has on the columns the eigenvectors to the eigenvalues $1, 1-p-q$. Then
$$
\begin{aligned}
S^{-1}PS &=
\begin{bmatrix}
1&0\\0&1-p-q
\end{bmatrix}
\ ,
\\
P &=
S
\begin{bmatrix}
1&0\\0&1-p-q
\end{bmatrix}
S^{-1}\ ,
\\[2mm]
&\qquad\text{ so }
\\[2mm]
P^n &=
S
\begin{bmatrix}
1^n&0\\0&(1-p-q)^n
\end{bmatrix}
S^{-1}\ .
\end{aligned}
$$
