Jordan Canonical Form exponentiation How do I solve this? I am aware that the solution involves using JCF.

 A: Inductively you can prove 
$\bigg(\array{2  & 1\\
-1  & 0}\bigg)^n=\bigg(\array{n+1  & n\\
-n  & -(n-1)}\bigg)$ so the limit diverges.
A: What is needed is a Dunford decomposition: $A=D+N$, where $D$ is a diagonalisable matrix and $N$ a nilpotent matrix and $D$ and $N$ commute. If $D$ is indeed diagonal, this means the diagonal coefficients are equal. A Jordan block has such properties.
Set $A=\begin{bmatrix}2&1\\-1&0\end{bmatrix}$. It has only one eigenvalue, $1$, and the eigenspace has dimension $1$, generated by $e_1=\begin{bmatrix}1\\-1\end{bmatrix}$. To complete it in a Jordan basis, we have to solve $(A-\lambda I)e_1=(A-I)e_2=e_1$. In this basis, as $Ae_1=e_1$, $Ae_2=e_1+e_2$, the Jordan canonical form will be:
$$J=\begin{bmatrix}1&1\\0&1\end{bmatrix}=I+\begin{bmatrix}0&1\\0&0\end{bmatrix}=I+N$$
Now $N^2=0$, hence $(I+N)^n=\begin{bmatrix}1&n\\0&1\end{bmatrix}$.
Let $P=[e_1 \ e_2]$ the change of basis matrix (from the canonical basis to the Jordan basis). Then $A=P(I+N)P^{-1}$, so:
$$A^n=P(I+N)^nP^{-1}$$
You should find, if I'm not mistaken:
$$A^n=\begin{bmatrix}1+n&n\\-n&1-n\end{bmatrix}.$$
