Using factorisation, $A=PJP^{-1}$ to compute $A^k$ Using factorisation, $A=PJP^{-1}$ to compute $A^k$, where $k$ represents an arbitrary positive integer.
$$ \begin{bmatrix} \mathbf{0} & \mathbf{1} \\ \mathbf{-1} & \mathbf{2} \end{bmatrix} = \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{1} & \mathbf{1} \end{bmatrix} \begin{bmatrix} \mathbf{1} & \mathbf{1} \\ \mathbf{0} & \mathbf{1} \end{bmatrix} \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{-1} & \mathbf{1} \end{bmatrix}. $$
Not sure how to finish this problem as the matrix $J$ is not diagonal, but $J^k$ can still be found somehow. How exactly is $J^k$ obtained in this case?
 A: You have basically put your matrix $A$ in Jordan normal form.
This means that you can split your matrix $J$ as follows:
$$J = D + N$$
where $D$ is diagonal and $N$ is nilpotent, i.e. there exists an a positive integer $n$ such that $N^n = 0$.
In your case
$$J = \begin{pmatrix}
1 & 1 \\ 0 & 1
\end{pmatrix}
=
\underbrace{
\begin{pmatrix}
1 & 0 \\ 0 & 1 
\end{pmatrix}}_{=D}
+
\underbrace{
\begin{pmatrix}
0 & 1 \\ 0 & 0
\end{pmatrix}}_{=N}
$$
and in your case you have $N^2 = 0$.
The basic idea is that when you expand your power $J^k = (D+N)^k$ using the binomial theorem, terms with $N^i$ with $i \geq n$ will disappear, since $N^n = 0$. In your case:
$$(D+N)^k = {k \choose 0} D^k + {k \choose 1} D^{k-1}N + \dots$$
where the remaining terms will be zero, e.g. the next term will be
$${k \choose 2}D^{k-2} \underbrace{N^2}_{=0} = 0.$$
More information can be found in this math.SE post.
A: You can easily prove (for example with induction) that 
$${\begin{pmatrix} 1 & 1  \\ 0 & 1  \\  \end{pmatrix}}^k=\begin{pmatrix} 1 & k  \\ 0 & 1  \\  \end{pmatrix}$$
Now $A^k$ can be computed as follows:
$A^k=(S^{-1}BS)^{k}=\underbrace{S^{-1}BS\cdot S^{-1}BS\dots   S^{-1}BS}_{k \text{ times }}=S^{-1}B^{k}S=S^{-1}\begin{pmatrix} 1 & k  \\ 0 & 1  \\  \end{pmatrix}S$
A: Hint: It is related to the Fibonacci numbers. 
