<s>[INST] 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?
 [/INST] Hint: It is related to the Fibonacci numbers. 
</s>